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AVOGADRO'S LAW AND ITS APPLICATIONS

Until 1811 Dalton's Atomic theory was held supreme. It was believed that gases aremonoatomic (H,N, O etc) and atoms combine in 1:1 ratio to give compound atom (what we callnow the molecule)

H + Cl --------------> HCl (compound atom)BERZELIUS HYPOTHESIS:Berzelius on the basis of Dalton's theory proposed:

" Equal volumes of all gases under identical conditions of temperature and pressure containequal number of atoms"Failure of Berzelius Hypothesis:According to Berzelius,1 atom of hydrogen(H) combines with 1 atom of chlorine(Cl) to produce 1 compound atom ofhydrogen cloride(HCl)(Note that there was no idea about molecule at that time and he believed that gases exist inmonoatomic state. One HCl molecule was called a "compound atom" at that time)So 'n' atoms of H should combine with 'n' atoms of Cl to produce 'n' compound atoms of HCl.Applying Berzelius hypothesis, if 1 volume of hydrogen gas contains 'n' atoms, then 1 volume ofother gases also would contain 'n' number of atoms. Hence1 volume of hydrogen should combine with 1 volume of chlorine to produce 1 volume of hydrogenchloride (temperature and pressure remaining constant).However this was found to be wrong by Gay Lussac's experimental study according to which,1 volume of hydrogen gas combines with 1 volume of chlorine gas to produce 2 volumes ofhydrogen chloride gas at the same temperature and pressure. Thus Berzelius hypothesis wasrejected and it was challenged by Avogadro who believed that elementary gases exist mostly asdiatomic molecules, i.e. O

2, N

2, H

2 and so on. The idea of compound atom was discarded and

replaced by the term molecule. The concept of MOLECULE was thus first introduced.Avogadro, in 1811, first gave a hypothesis which subsequently was turned into a law. He presumedthat the elementary gases like nitrogen, hydrogen, oxygen etc. exist as diatomic molecules. Hispresumption was later experimentally confirmed by Cannizzaro in 1950 after which the hypothesisturned into a law.Avogadro's LawEqual volumes of all gases under identical conditions of temperature and pressure containequal number of molecules.

Explanation:According to this law we can take any gas, say, hydrogen, chlorine, oxygen, hydrogen chloride,ammonia, carbon dioxide etc. and if the volumes of these gases are same while measured undersame temperature and pressure, then each of the gas must contain equal number of molecules.

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H2 N2 CO

Pressure= P mmTemp. = t0CVolume = V LNo. of molecules = n No. of molecules = n

Volume = V LTemp. = t0CPressure= P mm

No. of molecules = nVolume = V LTemp. = t0CPressure= P mm

Conversely if the gases contain equal number of molecules at the same temperature and pressurethen they possess equal volumes.

SAQ 1: Suppose you have been given three gas cylinders of equal volume (5 litres) containingthree gases N

2, O

2 and H

2. Each of them has a pressure of 2 atmospheres at temperature of

270C. If nitrogen cylinder contains 20, 00, 000 molecules, then find out the number of moleculesof oxygen and hydrogen present in the other cylinders.SAQ 2.There are 1 lack molecules each of NO and SO

2 in two different cylinders of V litre

capacity each at 300C. The pressure in NO cylinder is 860 mm of mercury. What is the pressurein the SO

2 cylinder?

Experimental validity:Avogadro's law explained satisfactorily the Gay Lussac's Law of combining gaseous volumes.In fact the success of the Avogadro's law is based on the Gay Lussac's experimental observations.Gay Lussac experimentally found that:Under identical conditions of temperature and pressure, if gaseous reactants completely reactto give gaseous products then there exists a whole number ratio always between the volumes ofthe reactants and products.(i) 1 volume of hydrogen gas combines with 1 volume of chlorine gas to give

2 volumes of hydrogen chloride gas(ii) 2 volumes of hydrogen gas combine with 1 volume of oxygen gas to give

2 volumes of water vapourAvogadro's law (the then hypothesis) could corroborate the Gay Lussac's findings stated above.Two assumptions made by Avogadro were that elementary gases like hydrogen, chlorineetc. are diatomic and one molecule of hydrogen chloride contains one H atom and oneCl atom.

H2 + Cl

2 → 2 HCl

1 molecule of hydrogen + 1 molecule of chlorine → 2 molecules of hydrogen chlorideSo x molecules of hydrogen + x molecules of chlorine → 2x molecules of hydrogen chlorideApplying Avogadro's hypothesis if 1 volume of hydrogen contains x molecules, 1 volume of

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other gases also would contain x molecules at the same temperature and pressure.1 volume of hydrogen + 1 volume of chlorine → 2 volumes of hydrogen chlorideThis is exactly what Gay Lussac observed experimentally.

SAQ 3: Indicate what volume of gaseous products would be formed from the following gaseousreactions. Assume complete reaction under constant temperature and pressure conditions.

(i)1 litre of nitrogen gas + 3 litres of hydrogen gas → ? litres of ammonia gas(ii)3 l of carbon monoxide gas + 1.5 l of oxygen gas → ? l of carbon dioxide gas(iii)10 l of hydrogen gas + 10 l of iodine vapours → ? l of hydrogen iodide vapours(iv)5 l of nitrogen monoxide gas + 2.5 l of oxygen gas →? l of nitrogen dioxide gas(v)150ml of methane gas + 300ml of oxygen gas →? ml of carbon dioxide +? ml of water vapour

Gram molar volume and Avogadro's LawJohann Josef Loschmidt(1865) and Jean Baptiste Jean Perrin(1909) were the scientists whoare remembered for the discovery of the famous number 6.023 X 1023 called Avogadro'sNumber i.e the number of species present in one mole of any substance.Again Cannizzaro experimentally found that one mole of any gas at NTP occupied 22.4 litres(grammolar volume). So from the concepts of Avogadro's number and gram molar volume, we can saythat:if the number of molecules of different gases taken in different vessels is each 6.023 X 1023 atNTP then each gas will occupy 22.4 litres. Conversely if the volume of each gas is 22.4 litres atNTP then each will contain 6.023 X 1023 molecules.

Pressure= 760 mmTemp. = 00CVolume = 22.4 LNo. of molecules = 6.023 X 10 No. of molecules = 6.023 X 10

Volume = 22.4 LTemp. = 00CPressure= 760 mm

No. of molecules = 6.023 X 10Volume = 22.4 LTemp. = 00CPressure= 760 mm

CO2 O2 Cl2

23 23 23

SAQ 4:(i) Supposing half mole of cows are grazing in a certain field. How many cows are therethen?

(ii)Supposing you purchased 12.046 X 1024 number of apples from the market. Howmany moles of apples you bought?

(iii)A 5 litre N2 gas cylinder was found to contain 1 lack molecules of N

2 and another

cylinder of unknown volume of oxygen was also found to contain 1 lack molecules at the sametemperature and pressure. What is the volume of the oxygen cylinder?

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(iv)Two gases X and Y kept in two different vessels at 270C and 1 atm. pressure. GasX occupied 1.5 litres and found contain 30,000,00 molecules. If the volume of gas B is 3 litres,how many molecules gas B might be having?

(v)How long it will take for a person to count Avogardo's number at the rate of fivecounts per second?

Alternative definition of Avogadro's Law:At constant temperature and pressure, if number of molecules(or moles) becomes more,

the volume of the gas becomes more, Avogadro's law can be stated as follows."The volume of any gas is directly proportional to the number of moles at constanttemperature and pressure".

V ∝ n, (at constant P and T) where 'n' is the number of moles

⇒ V = k n (where k is the proportionality constant)

⇒ Vn = k ⇒ V

n = nV1 2

1 2(at constant P and T)

SAQ 5: 1000 molecules were removed from a 200 ml of gas containing 5000 molecules at 270Cand 800mm fixed pressure. Will there be any change in volume? If so what will be the newvolume.

APPLICATION OF AVOGADRO'S LAW(i) To prove thatVapour Density or Relative Density of gas = (Molecular Mass)/2Before we establish the relationship between the vapour density and molecular mass of any gas,let us know what is meant by vapour density. You know that density of a solid or liquid whencompared with density of water, we call the ratio, specific gravity or relative density of thatsubstance. Since water has a density equals to 1gm per cm3, the specific gravity of any solid orliquid is numerically equal to density of that substance in gm per cm3. Say for example, thedensity of mercury is 13.6gm per cm3 i.e the density of mercury is 13.6 times greater thandensity of water(1gm/cm3). In other words mercury is 13.6 times heavier than water.Unlike liquids and solids, gases have very very low densities. Say for example the density ofhydrogen gas at NTP is 0.000089 gm per cm3, density of oxygen gas at NTP is 0.0014gm/cm3

and the like. Therefore the densities of gases are not compared with water, in stead, comparedwith density of the lightest gas hydrogen. HenceRelative density(R.D) or Vapour Density(V.D) of a gas:

=Density of a gas or vapour

Density of hydrogen gas(both measured under same temperature and pressure)

Can you calculate the relative density of oxygen gas from the densities of hydrogen and oxygengases given above. It is 0.0014/0.000089 = 16. Is it not half the molecular mass(32) of O

2 gas?

Now let us try to prove the relationship between vapour density and molecular mass of a gasusing Avogadro's law.

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=

mass of the gas or vapourvolume of the gas or vapour

mass of hydrogen gas

volume of hydrogen gas

V.D

If both the volumes are same, volume terms are cancelled out to give

=Mass of certain volume of gas or vapour

Mass of same volume of hydrogenV.D

(measured under same temperature and pressure)

=V.D Mass of V cc of the gas or vapourMass of V cc of hydrogen gas (let us take volume=V cc)

Let V cc of the gas contains 'n' molecules . Applying Avogadro's law, we have

V.D =Mass of 'n' molecules of the gas or vapou

Mass of 'n' molecules of hydrogen gasDividing by 'n' to both numerator and denominator, we have

=Mass of 1 molecule of the gas or vapour

Mass of 1 molecule of hydrogen gas

Since hydrogen molecule is diatomic(H2),

V.D =Mass of 1 molecule of the gas or vapour

Mass of 2 atoms of hydrogen gas

=Mass of 1 molecule of the gas or vapour

2 X Mass of 1 atom of hydrogen gas

= Molecular Mass2

Because we know that molecular mass = (mass of one molecule)/ (mass of one Hatom).IMPORTANT: Although density of a gas is dependent on temperature and pressure, itsvapour density is independent of any external conditions. Since molecular mass of asubstance is fixed, its V.D is also fixed. This is because V. D is actually relative density. Ifthe density of a gas changes due to change in external conditions, density of H

2 will also

change in the same proprotion and their ratio will always remain constant.SAQ 6: (i)What are the vapour densities of the following gases

(a) Carbon dioxide (b)nitrogen (c)sulphur dioxide(d)hydrogen chloride (e)nitric oxide(ii) How many times the density of ammonia is greater than the density of hydrogen ata certain fixed temperature and pressure.(iii)A certain unknown gas has a density equals to 0.00125gm/cc at NTP. What are itsvapour density and molecular mass?Can you identify the gas?(density of hydrogen gas

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at NTP=0.000089gm/cc)(iv)The vapour densities of a few gases are listed below. Identify the gases.(a)X= 40(it is gas of S and O) (b)Y= 24(a gas of O) (c)Z=15(a gas of N and O)(v)The vapour density of SO

2 gas at NTP is 32. What is its density?(density of hydrgen

is 0.000089gm/cc at NTP)(vi)The V.D of oxygen gas is 16 at 270C, what would be its VD at 1000C?

(ii) Atomicity of Hydrogen gas is 2Let us now prove on the basis of Avogadro's law that gaseous hydrogen molecule is diatomic i.eits atomicity is 2.Gay Lussac's experiment1 vol. of hydrogen gas + 1 vol. of chlorine gas → 2 vols. of hydrogen chloride gas(measured under same pressure and temperature)If 1 vol. of gas contains 'n' molecules, then according to Avogadro's law'n' molecules of hydrogen + 'n' molecules of chlorine → '2n' molecules of hydrogen chloride1 molecule of hydrogen + 1 molecule of chlorine → 2 molecules of hydrogen chlorideDividing by 2, we have½ molecule of hydrogen + ½ molecule of chlorine → 1 molecule of hydrogen chlorideOne molecule of hydrogen chloride contains only one H atom as it forms only one type of salt(e.g NaCl with NaOH)and not two types of salts as produced by sulphuric acid i.e NaHSO

4 and

Na2SO

4 with NaOH).This H atom of hydrogen chloride must have come from ½ molecule of

hydrogen.So, ½ molecule of hydrogen = 1 atom

1 molecule of hydrogen = 2 atomsAtomicity of hydrogen =2 and the hence formula of hydrogen gas is H

2

(iii)Atomicity of oxygen is 2The following Gay Lussac's experiment is taken2 vol. of hydrogen gas + 1 vol. of oxygen gas → 2 vol. of water vapour(under same temperature and pressure)Applying Avogadro's law'2n' molecules of hydrogen + 'n' molecules of oxygen → '2n' molecules of water vapour2 molecules of hydrogen + 1 molecule of oxygen → 2 molecules of water vapour1 molecule of hydrogen + ½ molecule of oxygen → 1 molecule of water vapourFrom electrolysis experiment, it was known that the number of atoms of H and O in watermolecule is 2 and 1 respectively as 18 gms of water gave 2 gms of hydrogen(correspond to 2 Hatoms) at cathode and 16gms of oxygen(correspond to one O atom) at anode.

So ½ molecule of oxygen = 1 atom1 molecule of oxygen = 2 atoms. Hence oxygen gas is diatomic(O

2)

(iv)Gram Molar Volume(GMV) = 22.4 liters at NTPYou know that one mole of any gas at NTP occupies 22.4 litres or 22400ml(cc). Let us see howit can be proved from Avogadro's law.

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Mass of 1 litre of gas or vapour at NTP Mass of 1 litre of hydrogen gas at NTP

2 XMolecular Mass = 2 X V.D=

(here we have taken the fixed volume to be 1 litre)

= 2 X mass of 1 litre of gas or vapour at NTP

0.089

(because the density of hydrogen at NTP=0.000089gm/ml=0.089gm/l) = 22.4 X Mass of 1 litre of gas or vapour at NTP = Mass of 22.4 litres of the gas or vapour at NTP

Hence the molecular mass is the mass in gm of 22.4 litres of the gas at NTP. In other words, onemole of any gas at NTP will occupy 22.4 litres.

SAQ 7: Experiments show that 1 volume of nitrogen gas reacts with 3 volumes of hydrogen gasto produce 2 volumes of ammonia at the same temperature and pressure. From this prove thatnitrogen molecule is diatomic (Use Avogadro's law). (The formula of ammonia is NH

3)

SAQ 8: What will be volumes of the following gases in cm3 at NTP(i) ½ mole of nitrogen gas (ii)5 moles of CO gas (iii)0.01 moles of SO

2

How Avogadro's Number(N0) was determined ?

It is humanly not possible to calculate this incredibly large number. Nobody has determinedon the basis of real counting. The calculation was made by indirect methods by several scientists.1. First Loschimdt in 1865 determined the number of molecules present in 1 cc of a gas atNTP by using the Kinetic Theory of gases(using molecular diameter and mean free path) to be2.6 × 1019. Subsequently Maxwell in 1873 determined this number to be 1.9 × 1019. It wasalready known by that time that 1 mole of a gas occupies 22400 mL at NTP. So the number ofmolecules present in 22400 mL was calculated to be 4.3 × 1023.2. Slightly later Kelvin determined this number of molecules in 22400 mL of a gas at NTPusing light scattering experiment to be 5 × 1023.3. In 1908 J. Perrin determined this number using his Brownian motion study in liquids to liebetween 6.5 × 1023 to 6.9 × 1023.4. Rutherford and Geiger used radioactive method(emission of alpha particles from radiumand uranium) to determine this number to be 6.2 × 1023.5. By 1933, more than 80 separate determinations of Avogadro's number were made whichgave nearly same value of the constant but there was no clear agreement on that constant. Thebest modern value of Avogadro's number which was universally accepted came from x-raydiffraction studies of crystals from the measurements of lattice parameters(details not discussedhere). From density of a metallic crystals and unit cell edge length avogadro's number wasdertermined. This value is 6.02214199 × 1023 which has been universally accepted.

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RESPONSE TO SAQsSAQ 1: O

2 = 20,00, 000 and H

2=20,00,000 (same in each case)

SAQ 2: 860 mm(same)SAQ 3: (i) N

2 + 3H

2 ------>2NH

3

First balance the equation. Look to the coefficient of the reactants and products. The volumes ofreactants and products are proportional to their coefficients. Note that for the Gay Lussac's lawto hold good it is required that all the reactants and produts are gaseous.So 1 litre of N

2 reacts completely with 3 litres of H

2 to give 2 litres of ammonia

(ii) CO+½ O2 ---------> CO

2

3 litres of CO gas reacts with 1.5 litres of O2 gas to form 3 litres of CO

2 gas

(iii) H2 + I

2 ------->2 HI

So 20 litres of hydrogen iodide gas will be produced.(iv) NO + ½ O

2 --------> NO

2

So 5 litres of NO2 will be formed

(v) CH4 + 2O

2 --------> CO

2 + 2H

2O

So 150ml of CO2 and 300ml of water vapour will be formed.

SAQ 4:(i)If one mole of any substance contains Avogadro's number of that substance then thenumber of cows present in half mole is ½ X 6.023X1023 = 3.0115 X 1023

(ii)6.023X1023 number of apples = 1 mole of applesSo 12.046 X1024 number apples = 20 moles of apples

(iii)If the number of molecules are same for two gases at equal temperature and pressure,then they will occupy the same volume. So the volume of oxygen cylinder is 5 litres.

(iv)At the identical temperature and pressure if the the volume of gas Y is two timesmore than that of X then the number of molecules of Y will be two times greater than X. Hencegas Y will contain 60,000,00 molecules.

(v)Time = (6.023 × 1023) /5 secs = 38.19 × 1014 years.SAQ 5: Yes, there will be change in volume as the pressure and temperature are kept constantin the two cases. According to Avogadro's law, the volumes of the gases will be directly proportionalto the number of molecules.

V1/V

2 = n

1/n

2

So 200ml/V2 = 5000/(5000-1000)= 5/4

⇒ V2 =(200X4)/5 = 160ml. So the new volume is 160ml.

SAQ 6: (i) (a)The Molecular Mass of CO = 12+16=28, hence its V.D =28/2=14(b)The molecular Mass of N

2= 28, So V.D=28/2=14

(c)SO2 = 32+32=64, So VD=32 (d)HCl= 1+35.5=36.5, V.D=18.25

(e)NO=14+16=30, VD= 15(ii)VD of NH

3 = 17/2 = 8.5, so ammonia gas has 8.5 times greater density than hydrogen.

(iii)VD= density of the gas/density of hydrogen = 0.00125/0.000089= 14. So its molecularmass will be 28. The gas could be either CO or N

2.

(iv) (a)M.M of gas X = 80, hence the gas is SO3(32+48)

(b)M.M of gas Y= 48, hence the gas is O3(ozone)

(c)M.M of gas Z= 30, hence the gas is NO(14+16)(v)V.D = (density of a gas) / (density of hydrgen) at the same temperature and pressure

So density of the gas = 0.000089X32=0.00284gm/cc

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(vi) V.D of oxygen will remain same(16). Note that VD is independent of temperatureand pressure. Although the densities of gases are different at different temperatures andpressures, but since VD is a ratio of two densities at the fixed temperature and pressure, it is aconstant quantity for a given gas. Since molecular mass of a gas is fixed, its VD is bound to befixed.SAQ 7: 1 vol. of nitrogen + 3 vols. of hydrogen ----------> 2 volumes of ammoniaApplying Avogadro's law we have,

n molecules of nitrogen + 3n molecules of hydrogen-----> 2n molecules ofammonia

1 molecule of nitrogen+ 3 molecules of hydrogen -----> 2 molecules of ammonia

½ molecule of nitrogen + 32 molecule of hydrogen -----> 1 molecule of ammonia

Since one molecule of ammonia contains one N atom(given), so½ molecule of nitrogen contains 1 atom of N1 molecule of nitrogen contains 2 atoms of N. Hence nitrogen gas is diatomic(N

2).

SAQ 8: (i)22.4/2 = 11.2 l= 11200cc (ii)22.4X5=112l=112000cc(iii)22.4X0.01=0 224 litres =224 ml.

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ATOMIC MASS

Atomic mass of an element is a relative mass of an atom which is expressed as compared to astandard element like, H, O or C.Hydrogen standard:

Atomic Mass = Mass of one H atom Mass of one atom of an element

In this scale the mass of one H atom is arbitrarily fixed at 1.000. Atomic mass of an elementmeans how many times the mass of one atom of an element is heavier than mass of one hydrogenatom.On H scale the atomic mass of oxygen was found to be 15.87 and that of carbon to be 12.02 i.eone atom of oxygen is 15.87 times heavier than one atom of hydrogen and one atom of carbon is12.02 times heavier than one atom of hydrogen. The atomic mass of oxygen is 15.87. This doesnot mean that mass of one oxygen atom is 15.87 gm. How can this be possible? The individualatom is so small and tiny that it cannot weigh 15.87 gm. Atomic mass hence is a relative massas compared to hydrogen or any other element.Carbon standard: Atomic Mass Unit(amu) or Carbon Unit(cu) or Unified Atomic MassUnit(u)From 1960 onwards, this standard is now universally accepted and comparison according tohydrogen standard is not being used nowadays.It is called the Atomic Mass Unit(AMU) or theUnified atomic mass unit(u) or carbon unit(cu) or dalton.

Atomic Mass = Mass of one atom of an element1 of the mass of one C atom12

a.m.u or c.u or u

In this definition the mass of one C atom(C-12 isotope) was presumed to be 12.00 a.m.u(i.enearest whole number of atomic mass of C obtained from H scale).If we make 12 equal slices (pieces) to a carbon atom whose atomic mass is 12 amu(each piecei.e 1/12 part of one carbon atom is called 1 amu), then how many times one atom of any otherelement is heavier than 1/12 part(one small slice) of carbon atom is atomic mass of the element.On the basis of carbon scale the atomic mass of H was found to be 1.0079 a.m.u(not 1.000taken in the hydrogen scale for making comparison with other atoms), and that of O is15.99491a.m.u and that of Na is 22.98977 a.m.u and so on.For the purpose of chemical calculations we approximate these atomic masses to their respectivenearest whole numbers, i.e 1 for hydrogen, 23 for sodium and 16 for oxygen and so on. Rememberthat the atomic mass of element is never a whole number excepting the carbon-12 isotopewhose atomic mass has been fixed arbitrarily at 12.0000. All other elements have non-whole number atomic masses (with figures after decimal point) but for simplicity in rememberingtheir values and for easy chemical calculations we use the value of atomic masses rounded offto the nearest whole number.SAQ 1: (i)In CU scale which element has a whole number atomic mass and what is that value?

(ii)In CU scale, the atomic mass of H is not 1, while in H scale, the atomic mass of C isnot 12. Justify.

(iii)Atomic mass of phosphorus is 31 and fluorine is 19. Is the statement correct? Justify

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your answer.Mass Number: The number of protons and neutrons present in the nucleus is called the massnumber. Mass number of oxygen is 16, that of fluorine 19 and phosphorous 31 and so on. Notethat when atomic mass is rounded off to its nearest whole number we get its mass number.

SAQ 2: The atomic mass of oxygen is 15.99491a.m.u. What is its mass number? If its atomicnumber is 8, how many protons, neutrons and electrons it has?SAQ 3: Below given are the atomic masses of few elements. Find their mass numbers. Alsofind the number of neutrons present in each of them. The number of protons are indicated withinbrackets.

(i) 38.97377(Potassium: 19) (ii)126.9(Iodine: 53) (iii)107.89 (Ag: 47)SAQ 4: The atomic mass of oxygen is 16.0 amu.Is the statement correct? If not why? what isthe mass number of oxygen?

Gram Atomic Mass: Atomic mass when expressed in gm is called gram atomic mass. Whatdoes it actually imply? To understand this, I want to ask you a question. Could you tell the massof a sodium atom? Some of you may instantly answer it is 23gm. But it is wrong. You may say 23amu, which is correct as one atom of sodium is 23 times heavier than 1/12 part of a carbon atombut not 23 grams. How can a tiny atom weigh so much? Actually gm atomic mass is the massin gram containing Avogadro's number(N) of atoms. 23gms of Sodium contain 6.023X1023

atoms of sodium. Then you can calculate the mass of one atom of sodium in gram. We now stopfurther discussion on this and shall come back to it in the mole concept chapter. So we definegram atomic mass as the mass in gm. of Avogadro's number of atoms of the element.

SAQ 5: What is the value of one amu in gm.

ISOTOPES: (Average Atomic Mass)The atomic mass about which we studied before should in fact be called as isotopic masswhich is the relative mass of a particular isotope of the same element.Then the question arises what are isotopes? A simple example will make the point clear. Let ustake the case of chlorine. If you analyse chlorine gas obtained from any source, you will findtwo varieties of chlorine atoms, one having atomic mass 34.96885(mass number 35) and theother having atomic mass 36.96590(mass no. 37) in the ratio 3:1. This means that for every threeatoms of Cl35 atoms we shall find one Cl37 atom. These are called isotopes of chlorine. They arecalled Cl35 and Cl37 isotopes expressed in terms of their mass numbers. So isotopes are thesame elements (not different elements) i.e have the same identity but they differ only intheir atomic masses(Mass number). They differ in the number of neutrons present in theirnuclei while the number of protons is same in the isotopes. The atomic masses of individualisotopes are called isotopic masses. The isotopic mass of Cl35 is 34.96885 and that of Cl37

isotope is 36.96590. But in the periodic table and other places we find the atomic mass ofchlorine to be 35.5 not the isotopic masses any one of the two isotopes.This is average of the twoisotopic masses of the element. Do you know how the average atomic mass is calculated? It isnot the number average we often find i.e (35+37)/2=36. It is found in a different way and iscalled weight average. It is found as follows

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Atomic mass of Cl = 3 X 34.96885 + 1 X 36.96590

3 + 1 ≈ 35.5

Or we can simplify the calculation by using their mass numbers instead of isotopicmasses.

Atomic mass of Cl = 3 X 35 + 1 X 37

3 + 1 ≈ 35.5

Since the Cl35 and Cl37 are always available in the ratio 3:1, the isotopic mass of Cl35 contributes3/4 and Cl37 contributes 1/4 to the average. Therefore we multiplied 3 with the isotopic mass ofCl35 and 1 with isotopic mass of Cl37 and took their sum and finally we divided this sum by thesum of 3 and 1(i.e 4). This gives the average atomic mass of 35.5 that we use in all chemicalcalculation involving chlorine.In terms of percentage we can also do the same thing by taking their relative abundance(availability) ratio as 75:25.

Atomic mass of Cl = 75 X 35 + 25 X 37

100 ≈ 35.5

Isotopes vary with respect to number of neutrons present in their nuclei. For example, Cl35

isotope has 18 neutrons(35-17) and Cl37 isotope has 20 neutrons(37-17), since the atomic numberof Cl is 17.The average atomic mass of a polyisotopic element can be determined from the followingrelationship.

Average atomic mass=(%)1 X (I.M)1 + (%)2 X (I.M)2 + and so on

100where (%)

1 is the percent composition of isotope 1 and (I.M)

1 is the isotopic mass of isotope 1

and so on.

Most of the elements remain as more than one naturally occurring isotopes. Isotopes of anelement have the same chemical identity i.e same atomic number but they differ in themass number i.e the number of neutrons. The isotopes vary in their atomic masses.

(i) OXYGEN: O exists as O16, O17 and O18 isotopes with a natural percent composition(abundance) as 99.78, 0.02 and and 0.2 % respectively. O16 is the most abundant (nearly100%) isotope in the mixture. So when we speak of O it is O16 having mass number 16. Althoughthe mass numbers of the isotopes are whole numbers 16,17 and 18 their atomic masses (isotopicmasses) are fractional i.e 15.99491, 16.99913 and 17.99916 respectively. But for simplicity weoften say that their atomic masses are 16, 17 and 18 respectively.

SAQ 6: How many protons and neutrons are available in the three isotopes of Oxygen.SAQ 7: Can the atomic mass of C13 isotope be exactly 13? If not for which isotope the atomicmass can be exactly a whole number?

13

(ii) CARBON:Carbon has mainly two naturally occurring isotopes C12(98.88%), C13(1.12%)and the most abundant among the two is C12. This C12 isotope has been taken as the standard inthe amu scale for comparing the isotopic masses of all other isotopes. Note that carbon hasanother isotope C14 which is extremely scarce in nature and is radioactive and we shall studyabout it in other chapter.(iii) HYDROGEN: Hydrogen has two naturally occurring isotopes, H1(protium or hydrogen),H2 or D2(Deuterium or heavy hydrogen) with the percent composition of 99.9855 and 0.0145%and 10-15 % respectively. Another isotope of H is H3(or T3 called the tritium)also which isextremely scarce in nature and is radioactive. Excepting hydrogen isotopes of an element havethe same name. Deuterium and tritium are not to be mistaken as different elements.(v) BROMINE: Br has two isotopes Br79 and Br81 in the ratio 1:1(50:50)(vi) SULPHUR: S has two isotopes S32 and S34 in the ratio 95.6: 4.4Most of the elements that you find in the periodic table exists in more than one isotopic forms. Soin these cases we make use of the average atomic mass which we commonly say atomic mass.There are few elements which unfortunately remain in one isotopic form. For example F19 hasonly one isotope having atomic mass 18.99840. So in this case the atomic mass is same as itsisotopic mass.

Monoisotopic Elements: There are only 19 elements which do not have more than one isotope.These are called monoisotopic. Be9 , F19, Na23, Al27, P31, Mn55, Co59, I127, Cs133, Au197 are thecommon ones and Sc45, As75, Y89,Nb93, Rh103, Pr141, Tb159, Ho165, Tm169 are the uncommonones. Try to remember the common ones at least.The remaining elements have at least two or more than two isotopes. Sn(tin) has the largestnumber of naturally occurring stable isotopes of 10 and Xenon has the second highest 9).SAQ 8: Among the following elements which elements are monoisotopic and which arepolyisotopic? Br, O, F, Al, Na, Fe, Cl, S, Mg, Ca, K, I, C, N, Au, PSAQ 9: Si has three isotopes having isotopic masses 27.97693, 28.97650 and 29.97377. Whatare their mass numbers? The atomic number of Si is 14, find the number of neutrons in eachcase.SAQ 10: Is it wrong to take mass numbers in stead of isotopic masses for calculating theaverage atomic mass? Explain.SAQ 11: Br has two isotopes Br79 and Br81 in the ratio 1:1. What is its average atomicmass?SAQ 12: Oxygen has three isotopes: O16, O17 and O18 in the ratio 99.78:0.02:0.2. Is theaverage atomic mass of oxygen close to 16 or 18 ? What will be its average atomic mass?Calculate by using the mass numbers. What is this average called? How it is different from thecommon number average. Show by calculation.SAQ 13: The average mass of O is 15.9994 . What value should we use in chemicalcalculations?SAQ 14: Br has two isotopes Br79 and Br81. What atomic mass of Br we use in chemicalcalculations and why?SAQ 15: Hydrogen has two naturally occurring isotopes (H1 and H2) having isotopicmasses 1.008 and 2.014 respectively with a composition ratio, 99.985 : 0.015. Find the average

14

atomic mass of H. Will the atomic mass be almost same as the atomic mass of H1 i.e 1.008(andnot close to the atomic mass deuterium)? If so why?SAQ 16: For every 10, 000 oxygen atom that you will count how many O16, O17 and O18

isotopes you will find?SAQ 17: Indicate which isotope is most abundant in the following element. Cl, O, S, H, C

Determination of Isotopic and Atomic Masses:In earlier times Cannizzaro and other workers like Dulong and Petit had used very crude methodsto find the atomic masses. They determined the atomic mass values on the basis of some physicaland chemical methods and these were merely average isotopic masses. For example, the averageatomic mass of Cl is 35.5 and Br is 80 and so on. The discussion on these methods will be madein the next chapter(equivalent mass). No method was known at that time to separate the differentisotopes of an element and get the individual isotopic masses. For example there are threedifferent varieties of O atoms having mass numbers 16, 17 and 18. These could not be found outexperimentally in earlier times. But after the discovery of an instrument called Massspectrometer by Aston in 1919, it was possible not only to separate the individual isotopesfrom each other but also find out their isotopic masses. The first success came in determiningthe isotopic masses of two isotopes of Ne as 20 and 22. At that time, Neon was known to havetwo isotopes. But later the third isotope Ne(21) was discovered. Today, powerful massspectrometers are available which give the isotopic masses correct upto many places (at least 7places) of decimal. Table below shows a list of the naturally occurring elements indicating theirnumber of stable and radioactive isotopes and their average atomic masses.Preliminary Idea about separation of isotopes and determination of isotopic masses ina mass spectrometer:

Unipositve ions are produced from an element in its gaseous state in a cathode ray tube. This isachieved by the bombardment of cathode rays which is nothing but a powerful electron beamonto the gaseous atoms. The knocking out of one electron from each atom converts them tounipoistive ions. Note that the masses of these ions are almost same as the masses of the neutralatoms as electron carries negligible mass. Since the element exists in more than one isotopicforms, their masses are different although their charges are same(+1). For instance, if we takechlorine gas, there would be two types of positve ions i.e 35Cl+ and 37Cl+ produced from the twoisotopes of chlorine. Similarly for Neon gas there would be three positive ions 20Ne+, 21Ne+ and22Ne+ from the three isotopes and so on. This mixture of positive ions of an element is thenallowed to passed through an electric field first and then through a magnetic field. The postiveions are deflected in an electric field towards the -ve plate of the field. But all the ions are notdeflected to the same extent. Depending on their masses, the deflections are different. Ionsfrom heavier isotopes are deflected to the lesser extent while the ions from lighter isotopes aredeflected to a larger extent. For instance, for chlorine, the two postive ions 35Cl+ and 37Cl+ getseparated into two beams, the 35 isotope being deflected more towards the positve plate ofelectric field. To achieve better separation and get sharper focus of ions, Aston used a magneticfield in succession to the electric field. The separated beam of isotopic ions get further separatedin the magnetic field and take a curved path. The magnetic field is applied perpendicular to thedirection of motion of the ion beam(current). So the mechanical effect is produced as per Fleming's

15

Left hand rule perpendicular to the plane carrying the magnetic field and direction of motion ofthe beam. Thus all the isotopic ions take different curvatures, the lighter isotope turns to thegreatest extent while the heaviest to the minimun extent. These focussed ions coming at differentradii of curvature are allowed to fall on a calibrated photographic plate to produce sphericalspots at the appropriate isotopic masses. From position of the spot the isotopic masses could beknown and from the intensity of the spot, relative natural abundance could be determined. Chlorinefor instance produced two spots, one at mass number 35 and the other at 37 and intensity ratio ofthese spots were found to be 3:1 which is their relative abuandance. Note that as many numberof spots are observed for a particular element, same would be the number of isotopes.

N

S

IONISATION CHAMBER

Electron Beam

Cl(35) +

Cl(37) +

Cl(37) +

35

37

magnetic field

Nowadays, with the disovery of powerful electromagnets using superconducting materials,and the interfacing of computer, isotopic masses correct upto 7 places of decimal are availabe inthe form of lines in the computer monitor what is called the mass spectrum. The number of massspectral lines give the number of isotopes and their heights give their relative abundances.Elements with their stable and radioactive isotopes and average atomic mass:

Element At.No No. of stable No. of natural av. at. massisotopes radioisotopes

Ac 89 - 2 227Al 13 1 - 26.98Ar 18 3 - 39.948As 33 1 - 74.92At 85 - 3 210Ag 47 2 - 107.87Au 79 1 196.97Ba 56 7 - 137.34Be 4 1 - 9.012Bi 83 1 3 208.98B 5 2 - 10.81Br 35 2 - 79.909

3537

1

3

16

Ca 20 6 - 40.08Cd 48 8 - 112.40C 6 2 1 12.011Ce 58 4 - 140.12Cs 55 1 - 132.905Cl 17 2 - 35.453Cr 24 4 - 52.00Co 27 1 - 58.93Cu 29 2 - 63.54Dy 66 7 - 162.54Er 68 6 - 167.26Eu 63 2 - 152.0F 9 1 - 19.0Fr 87 - 1 223Gd 64 7 - 157.2Ga 31 2 - 69.72Ge 32 5 - 72.59Hf 72 6 - 178.49He 2 2 - 4.003Ho 67 1 - 164.93H 1 2 - 1.00797Hg 80 7 - 200.59In 49 1 1 114.82I 53 1 - 126.9Ir 77 2 - 192.2Fe 26 4 - 55.85K 19 2 1 39.102Kr 36 6 - 83.8La 57 1 - 138.91Li 3 2 - 6.939Lu 71 1 1 174.97Mg 12 3 - 24.31Mn 25 1 - 54.94Mo 42 7 - 95.94Na 11 1 - 22.990Nd 60 6 1 144.24Ne 10 3 - 20.183Ni 28 5 - 58.71Nb 41 1 - 92.91N 7 2 - 14.007Os 76 7 7 190.2O 8 3 - 15.9994Pd 46 6 - 106.4P 15 1 - 30.974

17

Po 84 - 7 209.0Pt 78 4 2 195.09Pr 59 1 - 140.91Pm 61 No naturally occuring isotopes 147

as it is a synthetic elementPb 82 4 4 207.19Pa 91 - 2 231.0Ra 88 - 4 226.0Rn 86 - 3 222Re 75 1 1 186.2Rh 45 1 - 102.905Rb 37 1 1 85.47Ru 44 7 - 101.1Sm 62 6 1 150.35Sc 21 1 - 44.96Se 34 6 - 78.96Si 14 3 - 28.09Sb 51 2 - 121.75Sr 38 4 - 87.62S 16 4 - 32.064Sn 50 10 - 118.69Ta 73 2 - 180.95Tc 43 No naturally occuring isotopes 98

as it is a synthetic elementTe 52 8 - 127.6Tb 65 1 - 158.92Tl 81 2 4 204.37Th 90 - 6 232.04Tm 69 1 - 168.93Ti 22 5 - 47.90U 92 - 3 238.03V 23 2 - 50.94W 74 5 - 183.85Xe 54 9 - 131.30Yb 70 7 - 173.04Y 39 1 - 88.904Zn 30 5 - 65.37Zr 40 5 - 91.22

In chemical calculations, the atomic mass that you shall use should be the nearest whole numberof the average atomic mass given in the above table. For K, for example, you should use 39instead of 39.12 given in the table. For elements having average atomic mass far away from thewhole number value(e.g Cl=35.5), you have to use as such. But from theoretical point of view,you should remember that the actual isotopic masses as well as the average atomic masses arenot whole numbers (excepting a C-12 isotope).

18

DULONG'S AND PETIT'S LAW:(Study this law after completing the chapter 'Equivalent Mass')This is one of the oldest laws used to determine approximate average atomic masses of solidelements at room temperature and above.

In 1819 French scientists Pierre Dulong and Alexis Petit proposed this which can thefollowing alternative definitions.

(1) For solid element, the product of its specific heat capacity and atomic mass isapproximately equal to 6 cals K-1mol-1(25 JK-1mol-1).

Specific heat × atomic mass = 6 cals K-1mol-1(25 JK-1mol-1) (approximately)(2) The molar heat capacity is approximately constant for solid elements and is equal to

3R.(R = gas constant).(3) The amount of heat required to raise the temperature of a single atom of a solid is

independent of the type of atom. In other words the heat required to raise the temperature of 1gram atomic mass( one mole) of the solid element by 10C is approximately constant. This iscalled molar heat capacity which is approximately constant for all solid elements.Limitations of Dulong and Petit's Law:

1. This law is applicable to elements which are solids at room temperatures and mostlyfor metals. Large deviations were observed for solid nonmetals like sulphur, phosphorous, iodineetc.

2. Large deviations occured while measuring specific heats of S, Co, Te and Pt.3. The law is applicable only at ordinary temperature to elements having simple

crystal structures. At low temperatures and for complex crystal structures this law was notvalid.

4. The atomic mass obtained was approximate and not exact.Application of Dulong and Petit's Law:

Determinationa of exact atomic mass of elements:With the help of the equivalent mass of element which is always exact, the exact atomic masscan be determined.Step 1: Equivalent mass of the element is accurately determined by any experimental methodsuch as -

(a) Hydrogen displacement method (b) oxide method (c) chloridemethod (d) displacement method (e) double displacementmethod (f) electrolysis methodStep II: With the help of Dulong and Petit's law the approximate atomic mass is determined fromits specific heat data. Approximate atomic mass = 6/specific heatStep III : The approximate valency is obtained as follows.

Approximate valency = Approximate atomic mass/ equivalent massStep IV: Determination exact valency: Since valency is a whole number, the exact valency isobtained by rounding off the approximate valency to the nearest whole number.Step V : Correct atomic mass is obtained as follows : Exact atomic mass = Exact valency Xequivalent mass(note that the accuracy of this determination depends on the accurary of measurement ofequivalent mass by the experimental method)Example : A metallic element has specific heat 0.11 cal/g/K. Its oxide contains 22.27% oxygen.Calculate the correct atomic mass of the element.Solution: m(metal)/m(oxygen) = E(metal)/E(oxygen) 77.73/22.27 = x/8, So x = 27.92(EM)From Dulong and Petit's law : Approximate atomic mass = 6/0.11 = 58.182Approx. valency = 58.182/27.92 = 2.083

19

So exact valency is 2Exact atomic mass = 27.92 X 2 = 55.92.SAQ 18: An element was found to have specific heat capacity 0.0276 cal/g0C. If 114.79 g of achloride of this element contained 79.34g of the metal element, calculate the exact atomic massof the element.

Do you know why the atomic mass standard was changed fromHydrogen to Carbon ?

Dalton experimentally found the atomic masses of some elements and hydrogen wastaken as the standard for comparision because it is the lightest element. H was taken as 1.0000amu and other elements have fractional atomic masses. Subsequently it was found H did notcombine with many elements, in stead oxygen combined with many more elements. ThenBerzelius suggested to adopt O(16) as standard for comparision. From 1850 oxygen standardwas used both by chemists and physicists. O was taken as 16.0000 and other elements havefractional atomic masses. In 1919, two other isotopes of O namely O(17) and O(18) werediscovered. Their relative abundance was O(16) = 99.78%, O(17) = 0.02% and O(18) = 0.2%.Then chemists took the average atomic mass of O which is slightly greater than 16 as standardfor comparision. One sixteenth of the average atomic mass of O was taken as 1 amu. Butphysicists continued with the same O-16 standard. The atomic masses of elements varied slightlyin Physicist and Chemist standards which was very unfortunate for the rest of the world. Chemistsand physicists did not agree for a common standard for long years until 1956. Chemists arguedthat by converting to physicist standard(O16), the atomic masses would change by 275 ppm. Toresolve the long standing dispute between the two groups, two scientists Alfred Nier and A.Ölander mediated and suggested to make C-12 as the standard. C-12 was already used as astandard in mass spectrometry. The physicists agreed to this proposal but chemists resistedbecause by taking C-12 as standard their atomic masses would change by 42 ppm. But this timetheir resistance was not high. E. Wichers lobbied the chemists and ultimately in 1961 both thegroups compromised to use C12 as standard. Then onwards C-12 standard is being usedinterntationally.That the isotopes of O are available in different proportions in different areas is an absurdthinking.

20

RESPONSE to SAQs(Atomic Mass)

SAQ 1:(i)In amu scale carbon has a whole number of atomic mass which is 12.(ii)Since in amu scale the standard is C(12.0000), the atomic mass of H is 1.008. On the

other hand in hydrogen scale the standard is H(1.0000), the atomic mass of carbon is 12.02.(iii)Atomic masses of elements cannot be whole number values except

carbon(12.000).Hence the statement is wrong in the strictest sense. The atomic masses of Pand F are 18.99840 amu and 30.97377 respectively.SAQ 2: Mass Number=16. The number of protons= no. of electrons=8, the no. of neutrons=16- 8=8SAQ 3: (i)K: Mass No.= 39, No. of protons= no. of electrons=19 and no. of neutrons=39-19=20

(ii)I: Mass No.=127, No. of protons=no. of electrons=53, and no. of neutrons=127-53=74

(iii) Ag: Mass No.=108, No. of protons= no. of electrons=47, No of neutrons=108-47=61SAQ 4:No, the statement is incorrect. Because the atomic mass of an element cannot be awhole number in the amu scale except carbon(12). The Mass number of oxygen is 16.

SAQ 5: It is the mass of 1/12 part of one carbon atom in gm.We know that 6.203 X1023 atoms of carbon weigh 12 gm(one gm. atomic mass)

Hence 1 atom of carbon will weigh 12/(6.023X1023)gm = 1.9923 X 10-

23gmSo 1/12 part of this mass = (1/12) X 1.9923 X 10-23 = 1.66 X 10-24 gm. which is nearly equal tomass of one H atom.SAQ 6: O16: protons=8 and neutron=16-8=8, O17: protons=8, neutrons=17-8=9

O18: protons=8 and neutrons = 18-8=10SAQ 7: No, the atomic mass of C13 is a fraction. Only C12 isotope has the whole number atomicmass(12.00000amu).SAQ 8: Monoisotopic: F, Na, Al, I, P and Au; Polyisotopic: The restSAQ 9: The mass numbers are 28, 29 and 30. The number of neutrons in Si28 isotope is 14(28-14), in Si29 isotope 15(29-14) and Si30 isotope 16(30-14).SAQ 10: No, it is not wrong to use mass number in chemical calculations. Rather itsimplifies the calculation.SAQ 11: Average atomic mass of Br = (1X79 +1X81)/2 = 80SAQ 12: Average atomic mass of O = (99.78X16 + 0.2X17+0.02X18)/100= 16.002So the average atomic mass of O is close to mass number of O16 isotope. This happened becausethe other two isotopes O17 and O18 contribute almost insignificantly.This is called the weight average. The number average would have been (16+17+18)/3= 17. It isdifferent from the weight average value of 16 that we use.SAQ 13: We use a value of 16 for all chemical calculations because the average atomicmass is almost equal to 16.SAQ 14: We use the average atomic mass of 80(SAQ 11), not 79 or 81 which are thetwo isotopic masses.SAQ 15: The average atomic mass of H = (99. 985X1.008+0.015X2.041)/100= 1.009This value is almost same as the isotopic mass of protium(H1) of 1.008. It is not close to the

21

isotopic mass of H2(2.041) because the contribution of H2 isotope is insignificant(0.015%). Forchemical calculation we often use an atomic mass of 1 for H instead of 1.008.SAQ 16: Since the percentage composition of O16, O17 and O18 isotopes are 99.78, 0.2and 0.02 respectively, for every 10,000 atoms we analyse, 9978 atoms will be of O16, 20 atomswill be of O17 and only 2 atoms will be O18.SAQ 17: The most abundant isotopes are

Cl: Cl35 (75%), O: O16(99.78%), S: S32(95.6%), H: H1(99.985%), C: C12(98.88%)SAQ 18: 79.34/35.45 = x/35.5 , So x = 79.228(EM), Approx. atomic mass = 6/0.0276 =217.391Approx. valency = 217.391/79.228 = 2.73 ; So exact valency = 3 and hence exact atomic mass= 79.228X3 = 237.684

22

EQUIVALENT MASS

In many chemical reactions the reactants do not react in 1:1 mole ratio.

(i) Na + 1/2 Cl2 ---------> NaCl

(ii) Zn + Cl2 --------------> ZnCl

2

(iii) 2 Al + 3 Cl2 ----------> 2 AlCl

3

In the first example, one gram atomic mass of Na reacts with 1 gm. atomic mass of Cl(1/2 moleof Cl

2=one gm. atomic mass), in the second example one gm. atomic mass of Zn reacts with 2

gm. atomic mass of Cl2(one mole of Cl

2 = 2 gm. atomic masses), while in the third case 2 gm.

atomic mass of Al reacts with 6 gm. atomic mass of chlorine. In other words, one gm. atomicmass of Al reacts with 3 gm atomic mass of chlorine. So on the basis of balanced equation, it isnot always possible for the reactants to react in the 1:1 ratio.

In order to establish a 1:1 quantitative relationship between reactants a new mass scale, calledthe Equivalent Mass was introduced. One gm. equivalent mass of one reactant always reactswith one gm. equivalent mass of the other reactant.Then let us know what is equivalent mass of an element?

Atomic Mass= Change in ON of the element per atom

ValencyAtomic Mass

Equivalent Mass of an element =

Number of electrons gained or lost per a= Atomic Mass

In the example (i) given above,The equivalent mass of Na = Atomic Mass/1 =23/1=23, because here the

valency of Na is 1. We can also understand this in terms of change in ON and number ofelectrons lost.

Na → Na+ + e-

The change in ON when Na changes to Na+ =|0-1|=1. The number of electrons lost by one Naatom is also 1.

The equivalent mass of chlorine = Atomic mass/valency=35.5/1 =35.5Gm. Equivalent Mass: When equivalent mass expressed in gm. it is called gm. equivalent mass.From balanced equation, we know that 23 gms(1 gm. atomic mass) of Na reacts with 35.5gm(1gm. atomic mass) of chlorine. Now we can say that one gm. equivalent mass of Na reacts withone gm. equivalent mass of Cl, since the equivalent mass of Na and Cl are 23 and 35.5respectively.In the example(ii),

The equivalent Mass of Zn = Atomic mass/valency =65.5/2=32.75 as the valencyof Zn is 2(same as change in ON and number of electrons lost) and

The equivalent mass of chlorine is already known in example (i) to be 35.5.

23

Looking to the balanced equation of (ii), we know that 65.5gm(one gm. atomic mass) of Znreacts with 71gms(2gm atomic mass) of Cl. So 32.75(65.5/2) gms of Zn must react with35.5(71)gms of chlorine. So we find here also that one gm. equivalent mass of Zn(32.75)reactswith one gm. equivalent mass of Cl(35.5).In the example(iii),

The equivalent mass of Al = Atomic mass/valency = 27/3=9(since the valencyof Al is 3 or change in the ON or number of elctrons lost is 3). From the balanced equation, wesee that 54gms(2gm atomic mass) of Al reacts with 6 X 35.5=210gms(6gm atomic mass) of Cl.So 9 gm of Al must react with (210/54) X 9=35.5gm of chlorine. Here again we found that onegm equivalent mass of Al(9gm) reacts with one gm. equivalent mass of Cl(35.5).Conclusion: One gm. equivalent mass of any element will react with one gm. equivalentof any other element.SAQ 1: Find the equivalent mass of the underlined elements from the following reactions. Theatomic masses of elements are given. Use symbol E for equivalent mass.

(i) Fe + Cl2 ------> FeCl

3(Atomic mass of Fe=56)

(ii) Cu + ½ O2 -------> CuO (Atomic mass of Cu =63.5 and O =16)

(iii) 2Cu + ½O2------> Cu

2O

(iv) Fe + H2SO

4 --------> FeSO

4 + H

2

(v) Mg + ½ O2 --------> MgO (Atomic mass of Mg =24)

SAQ 2: Does an element show more than one equivalent masses? Justify with example.SAQ 3: Find the equivalent masses for the following elements. Explain why you can correctlydetermine these even without knowing the actual chemical reactions.

Ca, H, Na, Al, Ag (At. Masses are 40, 1, 23, 27 and 108 respectively)

Note that in a reaction not only 1gm. equivalent mass of one reactant reacts with 1 gm. equivalentmass of the other reactant but also the product formed is 1 gm. equivalent mass. In otherwords there is a 1:1 relationship between reactants and products in terms of equivalent mass.Let us take one example.

Al + 3HCl -------> AlCl3 + 3/2 H

2

From the above balanced equation, 27gm(1gm atomic mass) of Al produces 3 gms of hydrogen(3gm atomic masse). So 9 gm of Al(1gm equivalent mass) will produce (3/27) X9 = 1 gm ofH(=one equivalent mass of H). So we found that the one gm. equivalent mass of Al producedone gm. equivalent mass of hydrogen. So it may be remembered here that one gm. equivalentmass of a reactant not only reacts with 1gm equivalent mass of the other reactant but also forms1 gm equivalent of each product.

That is why there is a famous saying in chemical sciences:Every chemical reaction takes place in equivalents.

Let us try to understand the concept of equivalent mass in a different way.Classical Definition of Equivalent Mass:Equivalent mass of an element is a number which shows how many parts by mass of the elementcan combine with or displace 1.008 parts by mass of hydrogen, 8 parts by mass of oxygen, 35.5parts by mass of chlorine or 1 equivalent mass of any other element.

24

ALTERNATIVE DEFINITION OF EQUIVALENT MASS OF ELEMENT:It is the mass of the element which loses or accepts one mole of electrons (Avogadro'snumber of electrons) when oxidised from the neutral state to the desired oxidation state orreduced from the desired state to the neutral state.

O0 + 2e ---------------> O2-

One O atom requires 2 electrons to become O2-

Conversely, 2 electrons are accepted by 1 atom of OSo Avogadro's number(N)of electrons are accepted by N/2 atoms of O.Since the mass of Avogadro's number of atoms is the gm. atomic mass of the element,

mass of N/2 atoms of oxygen = 16/2=8 g = g. quivalent mass of oxygen.

Al --------> Al3+ + 3e,3 electrons are lost by 1 Al atom,N electrons are lost by N/3 atoms whose mass is equal to one third of its g. atomic

mass(27/3). So to be more precise, equivalent mass of an element is the atomic mass of theelement divided by the number of electrons lost/gained per atom or change in ON per atom inrespect of that element.

This is same as what we knew earlier. Equivalent mass is equal to atomic mass divided byvalency or change in ON.SAQ 4: Find the equivalent mass of the elements from the ion electron equations. The atomicmasses of the elements are given in the brackets.

(i) Co3+ + 3e --------> Co (59)(ii) Fe --------> Fe2+ + 2e (56)(iii) Co-------->Co2+ + 2e (59)(iv) Ba --------> Ba2+ 2e (137)(v) K---------> K+ +e (39)

SAQ 5: Using equivalent mass concept, answer the following(i) How many gms of hydrogen will react with 8 gms of oxygen to form water(ii)How many gms of chlorine will react with 18 gm of aluminium to form aluminium

chloride.(iii)How many gms of Mg will react with 24 gm of oxygen.(iv)How many gms of hydrogen gas will be evolved by 32.5gm of Zn when all the Znreacts with dilute acid. (Zn=65)(v)How many gms. of Ag will react with 5.35gm of Cl. (Ag=108)

SAQ 6: Find the mass in gm. in case of the following and calculate how much of chlorine eachwill need to form their chlorides.

(i) 0.5 gm equivalent of Mg (ii)2 gm equivalents of Na ( i i i ) 0 . 2 5 g mequivalent of Al

25

Determination of Equivalent Mass of Unknown Metal:From the above discussion we found that the equivalent mass of an element can be calculated ifwe know the element i.e know its atomic mass and valency(ON) by using the relationE=Atomic Mass/valency. But if the identity of an element is not known, can we find its equivalentmass? Yes, we can do so by experimental methods. Let us see some simple experiments bywhich we can find the equivalent mass of unknown metals. .

(a)Displacement Method(b)Hydrogen Displacement Method(c)Oxide formation or Oxide reduction method(d)Chloride method(e)Double displacement method

(a) Displacement Method:A more active metal(metal lying in the upper position in the metal activity series) displaces a lessactive metal(metal lying at a lower position in the activity series) from the salt of the latter. Alsothe metals lying above hydrogen in the metal activity series can displace hydrogen from diluteacids. We know that one gram equivalent mass of one metal(or element) can combine or displaceone gram equivalent mass of the other metal or element. If the equivalent mass of one elementis known, then that of the other can be found out by measuring the masses of the displacingmetal(or element) and displaced metal or element.

Zinc(s) + copper sulphate(aq.) -----------> Zinc sulphate(aq.) + Copper(s)Zinc(s) + sulphuric acid(aq.) -------> zinc sulphate(aq.) + hydrogen(g)

Let, m1 = mass of Zn; m

2 = mass of Cu displaced(or mass of hydrogen displaced);

E1 = Eq. mass of Zn; E

2 = Eq. mass of Cu( eq. mass of hydrogen)

m1 gm of Zn displaces m

2 gm of Cu or hydrogen(from experiment)

E1 gm of Zn displaces (m

2/m

1)E

1 gm of Cu or hydrogen. According to the definition of

eq. mass,E

2 = (m

2/m

1)E

1

⇒m1m2

=E1

E2

This is called the Law of equivalents which states that masses of substances are directlyproportional to their equivalent masses. Law of equivalents is the basis of finding theequivalent mass by any method.

Example: 0.26gm of Al displaces 0.94gm of Copper from Copper Sulphate solution.If the equivalent mass of Aluminium is 9, calculate the equivalent mass of copper.Solution:We can find this by unitary method done before or from the law of equivalents discussed now.Let us apply the law of equivalents.

mCu

mAl

=ECu

AlE

26

⇒ 0.94/0.26 = ECu

/9 ⇒ ECu

= 32. 53 .(a)Hydrogen Displacement Method:Active metals which lie above hydrogen in the metal activity series can displace hydrogen gasfrom dilute acids. This method is applicable only to those active metals which can displacehydrogen gas from dilute acids.

Equivalent Mass of an Active Metal:Metal + H

2SO

4 ----------------> Metal sulphate + H

2 (balancing cannot be done as we

do not know the valency of the metal).From the masses of the metal and displaced H2, we can

calculate the equivalent mass of the metal.Determination of equivalent mass of zinc by hydrogen displacement method:

Principle: A known mass of zinc is allowed to react with excess of dilute acid(e.g dil. H2SO4) toproduce its equivalent quantity of hydrogen gas. The mass of zinc which displaces 1.008 gm ofhydrogen is calculated which gives the equivalent mass of zinc.Working Procedure: A small piece of zinc metal is accurately weighed in an electronic balance.The weighed metal piece is put inside a porcelain basin containing water. An eudiometer tube( along narrow calibrated glass tube closed at one end) completely filled with dil. H2SO4 is invertedover the metal piece carefully using the thumb. It is fixed in a vertical position with the help of aclamp and stand. The reaction between zinc and acid immediately starts producing bubbles ofhydrogen gas which is collectd by the downward displacement of water. When no more gasevolves(i.e when the reaction is complete), the eudiometer tube is carefully removed with thehelp of thumb and transferred to a tall jar containing water.

Levelling: Keeping the tube in vertical position inside the tall jar and by making upward anddownward movement the tube is brought to a position at which the level of water inside theeudiometer tube is same as the level of water outside the tube i.e inside the tall jar. This is calledlevelling. At this position, the pressure of the gas inside the tube is equal to atmoshperic peressure.At this position the volume of the hydrogen gas is recorded from the calibrations of the tube. Theatmospheric pressure(=gas pressure) is recorded from the barometer. The room temperature isalso noted.

AirGas Jar

Tall Jar

atm

27

Calculation: Let the mass of the metal= W gThe volume of hydrogen gas collected in the eudiometer tube= V

1 mL

Pressure of the moist gas = atmospheric pressure = Pmm of Hg (recorded from barometer)Room temperature = t0C = (273+t) KAqueous tension at t0C = f mm of HgPressure of dry gas = (P-f) mm of Hg(Since hydrogen gas is collected over water, it is mixed with some water vapour which has afixed vapour pressure at t0C. This is called aqueous tension at t0C. This has to be substractedfrom the pressure of the moist gas(P) to find the pressure of dry hydrogen gas).Applying combined gas equation, the volume of hydrogen gas at NTP(V

2) is calculated.

P V1 2 21

T1=

TP V

2⇒

1(P-f) V

(273+t)=

760 V

2732 ⇒ 1(P-f)

(273+t)

273V =

X XV

760 X = y mL(say)

We know that density of hydrogen gas at NTP is 0.000089 g/mL ( 2.016g/224000cc)1 mL of hydrogen gas at NTP weighs 0.000089 gso, y mL of hydrogen at NTP weighs 0.000089 X y gNow 0.000089 X y g of hydrogen is displaced by W g of the metal

So, 1.008 g of hydrogen is diplaced by W

0.000089 X yX 1.008 g = z g (say) of metal

So, equivalent mass of the metal = zAlternative method(Equivalent volume method):2.016 g of hydrogen gas occupies 22.4 L at NTP1.008 g (i.e 1 g. eq. mass) of hydrogen gas occupies 11.2 L at NTPSo in stead of converting the volume of hydrogen gas NTP into mass using the density data, wecan directly use the equivalent volume of hydrogen(11.2L at NTP) to calculate equivalent massof the metal.y mL of hydrogen gas at NTP is displaced by W g of metalSo, 11200 mL(11.2L) of hydrogen gas is displaced by (W/y) X 11200 g = z g (say)of metalHence equivalent mass of the metal = z

Example: 1.47gm of a metal was treated with excess of dil H2SO

4 and 541.8cc of hydrogen

gas was collected over water at 150C and 752.5mm of Hg pressure. Calculate the equivalentmass of the metal.(Aqueous tension at 150C = 12.5mm and density of hydrogen gas at NTP=0.000089 gm/cc)Solution: Let us first convert the experimental volume into NTP conditions by applying the combinedgas equation;

Experimental: P1= (P-f)= 752.5-12.5)mm V

1=541.8cc

T1=(273+15)K

NTP : P2=760mm V

2=?

T2=273K

⇒ (752.55 - 12.5)X 541.8(273 + 15)

=760 X V

2732

⇒ =V2 500.1 cc

28

Mass of 1 cc hydrogen is 0.000089gm at NTP (density of H2 gas =0.00009gm/cc)

So mass 500.1cc of hydrogen = 500.1 X 0.000089 = 0.0445 gm0.0445 gm of hydrogen gas is displaced by 1.47gm of the metal

1.008 gm of hydrogen is displaced by 1.47 X 1.008= g

0.044533.29 of the metal

Equivalent Mass of the metal = 33.29Alternatively, we can use law of equivalents to solve the above problem.

mmH

=M EM

EH(Where m

M is the mass of metal and m

H is the mass of hydrogen while

EM and EH are the equivalent masses of metal and hydrogen respectively.In the above problem, we have

1.470.0445

E

1.008= M ⇒ E

M = 33.29

For simplicity, we shall hencforth be using the law of equivalents to calculate the equivalent massof an element.For calculation you can take equivalent mass of hydrogen to be 1 instead of 1.008.There might a slight difference in the answer but this can be ignored.EQUIVALENT VOLUME METHOD:500.1 cc of hydrogen gas is produced by 1.47 gm of the metalSo, 11,200 cc of hydrogen(1 eq. volume) is produced by (1.47/500.1) X 11,400 = 32.9 gmEquivalent mass of the metal = 32.9Can you guess why there has been a difference between the answers obtained in two differentmethods? The reason, the experimental density of hydrogen gas at NTP has been taken as0.000089g/cc in the first method while the theoretical density is 0.00009g/cc which can be obtainedby using gram molar volume(22400cc) for 2.016g of hydrogen gas at NTP. The second methodhas made use of equivalent volume(half of molar volume) and that is why the answers aredifferent.

SAQ 7: 0.139gm of a metal when dissolved in a current of hydrochloric acid evolved 29.5ml ofhydrogen gas collected over water at 130C and 741mm pressure. Find the equivalent mass of themetal. If the atomic mass of the element is 59, what will be its valency.(Aq. Tension at 130C =11.2mm)(b) Oxide Formation Method:

(i)Direct oxide formation:We know that many metals when heated in presence of oxygen or air produces their correspondingoxides

Metal + O2 --------------> Metallic oxide (We cannot balance this

equation since the valency of the metal is not known)Example: Al + O

2 --------> Al

2O

3 (equation not balanced)

If the mass of the metal is measured before the reaction and the mass of the metal oxide formedis also measured after the reaction, we can calculate the mass of oxygen that has combined withthe given mass of the metal . From this, the mass of the metal that would combine with 8 gm ofoxygen can be calculated. This is the equivalent mass of the metal since the equivalent mass ofoxygen is 8. Look to the following example.

29

Example: 50 mg of a metal is heated strongly in air until constant mass of 83.3 mg isobtained. Find the equivalent mass of the metal.Solution:

Mass of the metal = 50 mg = 0.05 gMass of the metal oxide = 83.3 mg = 0.0833gmSo mass of oxygen = 0.0833-0.05 = 0.0333 gm0.0333 g of oxygen has combined with 0.05gm of metal8g of oxygen has combined with 12 gm of the metalHence Eq. Mass of the metal = 12.

Law of Equivalent method:E M=0.05

0.0333 8 ⇒ EM

= 12

(Can you guess, which metal has equivalent mass equal to 12 ? It is magnesium).(ii)Indirect Oxide Formation:

In many cases metal oxides are not obtained directly from the metal. In such cases oxides areprepared in two steps. This is called indirect oxide formation method. The equivalent mass ofcopper can be obtained by this method.

Step-I : Metal + HNO3 -----------> Metal Nitrate + NO

2 + H

2O

Step-II: Metal Nitrate-----heat--------> metal oxide + NO2 + O

2

We know the mass of metal in the beginning and also know the mass of the metallic oxide at theend. Then we can know the mass of oxygen from which the equivalent mass of the element canbe calculated by the same method discussed before.Example: A sample of copper metal weighing 0.8g was first dissolved in conc. HNO3 to form itsnitrate. Cupric nitrate thus formed is then strongly heated until constant dry mass is obtained.Reddish brown gas is evolved during ignition. The mass of copper oxide formed was 1.0 g.Calculate the equivalent mass of copper.Solution: Mass of copper metal = 0.8 gMass of oxygen = 1.0 - 0.8 =0.2 g, Using law of equivalents, we have,

E M=8

0.80.2 ⇒ E

M = 32

(c) Oxide Reduction Method:In this case the metal oxide is reduced by H

2 or any other reducing agent to the corresponding

metal.CuO + H

2 -----------> Cu + H

2O

This method is also similar to the oxide formation method. When the masses of the metal oxideand metal are known, it is then easy to calculate equivalent mass of the metal in the same wayas done before.Example: On heating 1.127 gm of a metallic oxide in a current of hydrogen, 0.9g ofmetal was formed. Calculate the equivalent mass of the metal.Solution:

Mass of metallic oxide= 1.127gm Mass of the metal=0.9gmSo mass of oxygen= 1.127- 0.9= 0.227gmUsing law of equivalents,

30

E M=8

0.90.227 ⇒ E

M = 31.71

(d) Chloride Method:In this case the element reacts with chlorine to produce the chloride of the element. The mass ofthe element and chlorine that reacted are experimentally known. The mass of element whichreacted with 35.5 gm of chlorine is calculated. This gives the equivalent mass of the element asthe equivalent mass of chlorine is 35.5. Look to this example.Example: Chloride of a metal M contains 47.23% of the metal.Find the equivalentmass of the metal.Solution: Mass of the metal=47.23gm, Mass of chlorine =100-47.23= 52.77gm

52.77gm of chlorine reacted with 47.23gm of the metal, so35.5gm of chlorine must react with (47.23/52.77)X35.5=31.77gmHence the equivalent mass of the metal is 31.77.

Alternatively, using law of equivalents we have,E M=47.23

52.77 35.5⇒ E

M = 31.77

SAQ 8: 0.475 gm of a metal chloride was formed by the reaction of certain mass of the metalwith excess of chlorine. If the equivalent mass of the metal is 12, what is the mass of the metalreacted?

(e)Double Displacement Method:A known mass of a reactant(A) is allowed to react with excess of another reactant(B) in aqueoussolution to produce a precipitate containing the metal present in A. The precipitate is dried andweighed. From the dry masses of the reactant A and the corresponding product, we can find theequivalent mass of the metal present in A.Example: BaCl

2 + Na

2SO

4 -------> BaSO

4 ↓+ NaCl

This is a double displacement reaction under the main category of metathesis reactions. Bariumchloride reacts with sodium sulphate to produce a white precipitate of barium sulphate andaquous solution of sodium chloride.EQUIVALENT MASS OF A COMPOUND:The equivalent mass(EM) of a compound involved in a metathesis reaction(reaction in whichON remains unchanged) is determined from the following relationship.

EM of a compound = EM of the basic radica(cation) + EM of acid radical(anion)

EM of a radical(ion) =radical or ionic masvalency

Example: EM of barium sulphate(BaSO4) = EM of Ba2+ + EM of SO4

2-

= 137/2 + 96/2 = 68.5 + 48 = 116.5EM of barium chlorideBaCl

2) = EM of Ba2+ + EM of Cl- = 137/2 + 35.5/1= 68.5 + 35.5=104

Now let us apply law of equivalents to barium chloride and barium sulphate.

mBaCl2

BaSO4

m =E

BaCl2E EBaSO4

=E

Ba Cl

SO4

+ E

EBa + =

x 35.5/1x 96/2

+

+

31

(where x is the EM of Ba) From this relationship, the equivalent mass of Ba can be calculated.Example 1: 0.925 g of anhydrous barium chloride was dissolved in water and treated with excesssulphuric acid. The weight of the dry barium sulphate obtained was 1.036. Find the equivalentmass of the barium. Also find the percentage of error with respect to the theoretical value.Solution:

mBaCl2

BaSO4

m =E

BaCl2E EBaSO4

=E

Ba Cl

SO4

+ E

EBa + =

x 35.5/1x 96/2

+

+

⇒ =x 35.5/1x 96/2

+

+0.9251.036

⇒ x = 68.12

Theoretical equivalent mass = Atomic mass/valency = 137/2= 68.5Percentage of experimental error = [(68.5-68.12)/68.5] X 100 =0.554%Examle 2: On heating 1.127 gm of a metallic oxide in a current of hydrogen, 0.9g of metalwas formed. Calculate the equivalent mass of the metal.We can solve this problem by comparing mass of metal oxide and mass of metal with equivalentmass of the metal oxide and that of metal. There is no necessity of using the mass of oxygen.

=mass of metal

mass of metal oxideEM of metal

EM of metal oxide = x

x + 8⇒ 0.9/1.127 = x/(x+8) ⇒ x = 31.71In other words we can compare the masses with the equivalent masses of any pair of substances,may it be elements or compounds.

SAQ 9: 3.31 g of pure and dry lead nitrate was dissolved in water and treated with excess ofpotassium chromate solution. A yellow precipiate of lead chromate was obtained which wasseparated and dried and weighed to 3.23 g. Calculate the equivalent mass of lead.d(Cr=52)SAQ 10: 1.53 g of metal hydroxide on strong heating produced 0.995 g of its oxide. Calculatethe equivalent mass of the metal.

PRACTICE QUESTIONS1. 24gm of a metal produced 22.4litres of H

2 gas at NTP from an acid. What is the equivalent

mass of the metal? Guess which is the metal?2. Enough steam was passed over 5gm of red hot iron till all the iron reacted completely toproduce 2.67 litres of H

2 gas at NTP. Calculate the equivalent mass of iron. Justify this by

writing the equation.3. 0.13gm of a metal combines with 56ml of O

2 gas at NTP. Calculate the equivalent mass

of the metal.4. 0.640gm of an unknown metal gave 0.851gm of its chloride. Calculate the equivalentmass of the metal. Can you guess which is the metal?5. 1.314gm of an unknown metal displaced 2.158gm of sliver from AgNO3 solution. Findthe equivalent mass of the unknown metal. Could you guess the metal?

32

RESPONSE TO SAQsSAQ 1:(i) E= 56/3= 18.66(since the valency of Fe here is 3 i.e same as the change in ON of Fe

=|0-3|=3)(ii) E(Cu) =63.5/2 =31.75(since Cu forms Cupric oxide in which the valency of Cu is 2) E(O)=16/2=8 (since the valency of O is 2 i.e the change in ON in O =|0-2|=2(iii)E(Cu) = 63.5/1 =63.5(since Cu forms cuprous oxide in which the valency of Cu is 1)(iv)E= 56/2 =28(since Fe forms ferrous sulphate in which Fe show a valency of 2)(v)E= 24/2=12(since the valency of Mg is 2)

SAQ 2: Yes, some elements which show variable valency(oxidation state) has more than oneequivalent masses. For example Fe shows two equivalent masses, 56/3 and 56/2 for ferric andferrous respectively. Cu shows two equivlent massse, 63.5/2 and 63.5/1 for cupric and cuprousstates respectively.SAQ 3: E(Ca)=40/2=20 E(H) =1/1=1 E(Na)=23/1=23,

E(Al) =27/3=9, E(Ag)=108/1=108This is because all the elements given in this SAQ show one valency i.e one ON(not variablevalency). Therefore, they have one equivalent mass each.SAQ 4: (i)E(Co)= 59/3 (ii)E(Fe)=56/2 (iii)E(Co)=59/2

(iv)E(Ba)=137/2 (v)E(K)=39/1SAQ 5: (i) E(H)=1 and E(O)=8, So 8gms of O will react with 1 gm of H. This is becausewe know that one gm equivalent mass of any element will react with one gm. equivalent mass ofany other element.

(ii) E(Al)=27/3=9 and E(Cl)=35.5.9gms of Al reacts with 35.5 gms of Cl, so18gm of Al reacts with (35.5/9) X 18=71gm Cl.

(iii) E(Mg)=24/2=12, E(O)16/2=8, So 8gms of O reacts with 12gm of Mg.Therefore, 24gm of O must react with (12/8) X 24=36gms of Mg.

(iv) E(Zn)= 65/2=32.5, E(H)=1, So 32.5gms of Zn will displace 1gm of hydrogen.(v) E(Ag)=108/1=108, E(Cl)35.5, So 35.5 gms of Cl will react with 108gm of Ag,

So 5.35gm of Cl will react with (108/35.5) X 5.35 = 10.8gms of Ag.SAQ 6:

(i) E(Mg)=24/2=12, So 1gm. equivalent =12gms0.5 gm. equivalent =12X0.5=6gmWe know that 12gm of Mg will react with 35.5gm of Cl(E).So 6 gm of of Mg will react with (35.5/12) X6 =17.75gms of Cl.

(ii) E(Na)=23/1=23, So 1 gm. equivalent of Na =23gmSo 2 gm. equivalents of Na =23X2=46 gms.We know that 23gms of Na(E) will react with 35.5gms of Cl.So 46gms of Na will react with (35.5/23) X 46=71gms of Cl.

(iii) E(Al)=27/3=9, E(Cl)=35.5, So, 1 gm. equivalent of Al =9gmSo 0.25 gm equivalent = 9X0.25 = 2.25gmWe know that 9gm of Al will react with 35.5gm of Cl.So 2.25gm of Al will react with 35.5/9X2.25= 8.75gms of Cl.

SAQ 7:

33

Let us first convert the volume from the experimental conditions to NTP condition byusing the combined gas equation.P

1 =(P-f)= (741-11.2)mm, V

1=29.5cc T

1=(273+13)K

NTP: P2=760mm V

2=? T

2=273K

⇒ 2

273760 X V

=(273 + 13)

(741 - 11.2)X 29.5 ⇒ V2 = 27.04cc

1cc of hydrogen gas at NTP weighs 0.000089gm27.04cc of hydrogen gas at NTP weights 27.04 X 0.000089= 0.0024gmSo 0.0024gm of hydrogen is displaced by 0.139gm of the metalSo 1gm of hydrogen will be displaced by (0.139/0.0024) X 1 = 57.9gmSo the equivalent mass of the metal is 57.91.Valency = Atomic Mass/equivalent mass = 59/57.91= 1.018Since valency is always a whole number, the true valency cannot be 1.018, it is 1 i.e thenearest whole number of calculated valency.So the valency of the element =1.

SAQ 8: Let the mass of the metal = x gm; Mass of metal chloride = 0.475gm(given)So mass of chlorine = (0.475 - x)gm

(0.475-x)gm of Cl reacted with x gm of the metal

So 35.5gms of Cl must react with x0.475-x

X 35.5 gm (equivalent mass of

metal)

According to the data: x0.475-x

X 35.5 = 12 ⇒ x = 0.119gm.

So the mass of the metal = 0.119gm.

SAQ 9 : m(metal nitrate)

m(metal chromate) =E(metal nitrate)

E(metal chromate) =E(metal E(NO3-)+

+E(metal E(CrO42-)=

x + 62/1

x + 116/2

x = 103.5

SAQ 10:1.530.995

= E + 17E + 8 ⇒ E = 8.73

ANSWER TO PRACTICE QUESTIONS1. 22.4litres of H

2 at NTP = 2gms (1 mole)

2gms of H2 is displaced by 24gm of the metal(given)

So 1 gm of H2 must be displaced by (24/2) X 1 = 12gm.

So the equivalent mass of the metal is 12. The metal must be Mg.2. 22.4 litres of H

2 at NTP = 2gms

So 2.67 litres of H2 at NTP = (2/22.4) X 2.67 = 0.238gm

0.238gm of H2 is displaced by 5 gm of iron

1 gm of H2 must be displaced by (5/0.238) X 1 = 21.0 gm.

So the equivalent mass of iron = 21The reaction is: Fe + H

2O --------> Fe

3O

4(magnetic oxide) + H

2

34

Here ON of Fe goes from 0 to +8/3 = 2.66(Fe3O

4). The atomic mass of Fe= 56,

So equivalent mass = Atomic mass/ change in ON = 56/2.66 = 21.0

3. 22400ml of O2 gas at NTP = 32gms

So 56ml of O2 at NTP = 0.08gm

0.08gm of Oxygen combines with 0.12 gm of the metalSo 8gm of Oxygen must combine with (0.12/0.08) X 8 = 12. The metal is Mg.

4. Mass of metal chloride = 0.851gm, mass of metal = 0.6490 gmSo mass of chlorine = 0.851 - 0.640 = 0.211gm0.211gm of the chlorine combines with 0.649gm of metalSo 35.5gm of chlorine combines with (0.640/0.211) X 35.5 = 107.6gmSo the equivalent mass of the unknown metal is 107.6 and the meal would be Ag(sliver).

5. We can apply law of equivalents here. In all the questions from 1-3 given before, wecould have used the law of equivalents in stead of solving by unitary method.

m1m2

=E1

E2

where m1 = mass of unknown metal, m

2=mass of of Ag,

E1=Eq.Mass of unnown metal and E

2=equivalent mass of Ag.

⇒ (1.314/2.158) = E1/108 (Since Eq. mass of Ag = 108/

1=108)⇒ E

1 = 65. 76(Eq. mass of the unknown metal).

35

MOLECULAR MASS

When you are asked to calculate the molecular mass of CO2, you immediately add the atomic

mass of one carbon atom(12) with 2 times the atomic mass of oxgyen atom(2X16=32) and theresult is 12+32=44. So the molecular mass of CO

2 is 44. What does it mean? It means that one

molecule of carbon dioxde is 44 times heavier than one atom of hydrogen or in the cu scale onemolecule of carbon dioxide is 44 times heavier than 1/12 part of the mass of a carbon-12isotope. So we write,

Molecular Mass = Mass of one molecule

1/12 part of one atom of C(12) amu

Gm. Molecular Mass is the molecular mass expressed in gm. It is often called one gm. mole orsimply one mole. Can you say how many molecules of the substance is present in one gm.molecular mass(one mole) of the substance? The answer is Avogadro's number of molecules.More about mole concept will be discussed in the next chapter, Mole Concept.Determination of Molecular Mass of a substance:Add the atomic masses of all the atoms present in the molecule. You will get its molecular mass.SAQ 1: Find the molecular masses of the following:

K2Cr

2O

7, CaCO

3, MgS

2O

3, K

2SO

4, Al

2(SO

4)

3.18H

2O, FeSO

4.(NH

4)

2SO

4.6H

2O, SO

3,

H2C

2O

4, CH

3COCH

3, Fe

4[Fe(CN)

6]

3, Ca

3(PO

4)

2, C

6H

12O

6

DETERMINATION OF MOLECULAR MASS

1. Gram Molar Volume(GMV) Method:We know that one mole of any gas or vapour at NTP condition will occupy 22.4l(22400ml). Thisis called gram molar volume(GMV). If the volume of a fixed mass of an unknown gas or vapouris known at certain temperature and pressure, then we can first convert the volume to NTPconditions and then find out the molecular mass by finding out the mass of the vapour occupying22.4 litres at NTP. Look to the following example.Example: 80 mg of liquid on vaporization occupied 24.9 ml at 270C and 740mm pressure.Calculate the molecular mass of the subtance.Solution:

First the volume is changed to NTP conditon.

⇒ =760 X V2

273740 X 24.9

273+27 ⇒ V2 = 22.06 ml

22.06ml of the vapour at NTP weighs 0.08gm22400 ml of the vapour at NTP weighs 81.23 gmHenc molecular mass of the substance is 81.23.

SAQ 2: At STP, 5 litres of a gas weighs 14.4gm. What is its molecular Mass? If the gas is madeup of S and O, could you guess what is the gas?SAQ 3: 380ml of an unknown gas at 270C and 800mm of Hg pressure weighed 0.455gm. Calculate

36

the molecular mass of the gas. If the gas is made up of a single element, could you guess whatthe gas is? Supposing the gas consists of two elements C and O, could you guess what it is?

2. Vapour Density Method (for volatile liquids):Victer Meyer's Method:This method is applicable for volatile substances(usually low boiling liquids) which vapouriseseasily without decomposition.Principle: A known mass of the volatile liquid is vapourised in the Victor Meyer's apparatuswhich dispaces equal volume of air. The displaced air is collected by downward displacementof water. The vapour density of the vapour is determined from the following relationship.

=Mass of certain volume of gas or vapour

Mass of same volume of hydrogenV.D

(at fixed temp. and pr.)Then molecular mass of the the substance is determined from the relationship

Molecular Mass = 2 X V.DDescription of the the Apparatus:(a)Victor Meyer's tube: This is the main component of the apparatus. It is a long (nearly60cm)narrow glass tube with an elongated bulb at the end. The tube is connected with a sidedelivery tube at th top which is inserted inside the beehive shelf immersed in a trough of water.Thereis glass wool at the bottom of the tube.(b)Outer Jacket: The Victor Meyer's tube is kept inside a jacket made of copper or glass. Aliquid which boils at least 20-300C higher than the boiling point of the volatile liquid is taken insidethe outer jacket. Water is usually taken inside this jacket. The outer jacket is connected to anoutlet tube at the upper end for the exit of water vapour.(c)Hoffmann's bottle: A very small glass bottle having stopper is used to contain volatile liquid.

Functioning: A small quantity of the volatile liquid whose V.D is tobe determined is taken in the stoppered Hoffmann's bottle and isweighed accurately. The water inside the outer jacket is boiled sothat within a period of time Victer Meyer tube attaines a temperatureof 1000C. Excess water vapour produced in the outer jacket iscontinously removed through the upper outlet. The air inside theVicter Meyer's tube expands while boiling of water in the outerjacket continues and the excess air passes out of the tube over thetrough of water in the form of gas bubbles. When the tube attainsequilibrium at 1000C, there is no more expansion of air and the airbubbles stops appearing in the trough of water.Then a graduated measuring tube which is completely filled withwater is inverted over the beehive shelf. Then the Hoffmann'sbottle(which is loosely stoppered) containing the volatile liquid iscarefully and quickly dropped into the Victor Meyer's tube by

37

opening its mouth and quickly closing it. While falling down, the loosely stopperd Hoffmannbottle opens and the liquid instantly vapourises inside the tube as the temperature is 1000C whichis greater than the boiling point of the liquid Since the vapour is heavier than air, it remains insidethe tube and displaces equal volume of air from the tube which is collected in the gas tube by thedownward displacement of water. The Hoffmann bottle settles down over the glass wool at thebottom of Victor Meyer's bute without breaking.Levelling and volume measurement: Then the gas tube contining air is carefully taken out andimmersed vertically in tall jar containing water. The tube is brought to a position at which thelevel of water inside the tube is same with the level of water outside in the tall jar. At this position,the volume of the air is read from the calibration. At this position the pressure of the moist airinside the tube is equal to one atmosphere. This is called levelling. The pressure is noted from thebarometer. Room temperature(temperature of water in the tall jar) is recorded.

Calculation:Let the mass of the volatile liquid = xgm.(Note that this mass will remain unchanged in both the liquid and vapour states)Let the volume of air dislaced by the vapour of the volatile liquid= VmlTemperature=t0C = (273+t)0CPressure = Pmm of Hg, and Aqueous Tension = f mm at t0C.So pressure of dry air = (P-f)mmFirst the volume of air displaced(Vml) is converted from the experimental conditions to NTPconditions by using the combined gas equation.

273760 X V2=

273+t

(P-f) X V

⇒ V2 =173 X (P-f) X V(273+t) X 760

= y ml (say (volume at NTP)

Note that the volume of displaced air = volume of the vapour of the volatile liquidNow we have to find out the mass of y ml of hydrogen gas at NTP. We know thatthe mass of 1ml of hydrogen gas at NTP =0.000089gm (since density = 0.000089g/cc at NTP)Hence the mass of y ml of hydrogen gas at NTP = 0.000089 X y gm

Vapour Density = Mass of y ml of unknown gas at NTPMass of y ml of hydrogen gas at NTP

=x

0.000089 X y(note that the mass of volatile liqiud will not change when vapourised)

Molecular Mass = 2 X V.D

Example: 2.0 gm of a certain unknown gas occupies 418 ml at 270C and 755mm Hgpressure. Find out the molecular mass of the gas.(aqueous tension at 270C =15mm)

38

Solution:V

1= 418ml, P

1= 740mm, T

1= 273+27

Converting to the NTP condition,

273+27 273760 X V2

=(755-15) X 418

V2 = 370 .37 ml(volume at NTP)

V.D = Mass of 370.37ml of the gas at NTP

Mass of 370.37ml of hydrogen gas at NP

=0.000089 X 370.37

2

(Note that the initial mass of the unknown gas(here it is 2gm) cannot change despite itsvapourisation)

V.D = 60.67, Hence Molecular Mass = 2 X 60.67= 121.34

Note that the molecular mass could also be calculated by Gram Molar Volume Methodwithout determining vapour density. You can adopt any method you like. But experimentalmethod(Victor Meyer) demands that you must first find out the VD and then MM.

SAQ 4: A certain vapour at a certain temperature and pressure was found to be 29 timesheavier than hydrogen gas at the same temperature and pressure. What is the molecular mass ofthe substance.SAQ 5: 0.45 g of a volatile liquid in Victer Meyer's apparatus displaced 98.75 mL of moist air at200C and 718 mm of Hg pressure. Calculate the molecular mass of the liquid. (aqeous tension at200C = 17.4 mm of Hg)SAQ 6: The density of some gases are given below at NTP. Find their molecular masses. Thedensity of hydrogen gas at NTP=0.000089gm/cc. If all the gases are elemental in nature. Identifythem.

(i)0.0014 gm/cc (ii)0.00316gm/cc (iii)0.00089gm/ccSAQ 7:In a Victor Meyer's experiment, 0.168gm of a volatile liquid dislaced 49.4ml of air measuredover water at 200C and 740mm of pressure. Calculate the vapour density and molecular mass ofof the compound. (Aqeous tension at 200C= 18mm)SAQ 8: Find the molecular mass of the volatile liquid given in SAQ 7 by gram molar volumemethod. Do you find any difference?

PRACTICE QUESTIONS1. When 3.2 gm of sulphur is vapourized at 4500C and 723 mm pressure, the vapour occupiesa volume of 780ml. What is the molecular formula of sulphur vapour under these conditions?2. 250ml of ozonised oxygen(ozone + oxygen)at NTP weighed 0.393gm. On passing thesample through turpentine oil there was contraction in volume by 50ml. Find the molecular massof ozone.3. If a gas has a density of 0.5 gm per litre at NTP, then find the mass of one mole of the

39

gas.4. The molecular mass of a compound is 46. What will be the volume of air displaced by0.1665g of the substance at 150C and 773.3 mm pressure in Victor Meyer's apparatus? (AqueousTension at 150C =13.3 mm)5. What is the mass of one mole of a gas, one gm. of which occupies 0.9822 litres at1000C and 740mm pressure?

RESPONSE TO SAQs(Molecular Mass)

SAQ 1: K2Cr

2O

7 = 2X 39 + 2X52 + 7 X 16= 294, CaCO3 = 40 + 12 + 3X16=100

K2SO

4 = 2X39 + 32+4X16 = 174; MgS

2O

3(136), Al

2(SO

4)

3.18H

2O(666),

FeSO4.(NH

4)

2SO

4.6H

2O(392), SO

3(80), H

2C

2O

4(90), CH

3COCH

3(58) Fe

4[Fe(CN)

6]

3(860)

Ca3(PO

4)

2(310), C

6H

12O

6 (180)

SAQ 2:If 5 litres of the gas weighs 14.4gm at STPThen 22.4 litres of the gas will weigh (14.4/5) X 22.4= 64.5 (Molecular mass)If the gas consist of S and O , it has to be SO

2. Its molecular mass is 64 which is close

to the experimental value. Note that there would definitely ramain some error in any experimentalmethod.SAQ 3: First we have to convert the volume into STP condtions.

273+27 273760 X V2

=800 X 380

⇒ V2 = 364ml

365ml of the gas weighs 0.455gm at STP22400 ml of the gas will weigh (0.455/365) X 22400 = 27.9So the molecular mass of the gas is 27.9 and it is undoubtedly N

2(2X14=28) as it consists

of only one element. If it consists of C and O, it has to be CO(12+16=28).SAQ 4: The vapour density of the vapour =29, because the density of the vapour 29 timesgreater than density of hydrogen at the same temperature. So its Molecular Mass

=2X29=58SAQ 5:NTP volume(V2) = 84.8 mL, V.D = 0.45/(0.000089X84.8) = 59.62, So MM=119.24SAQ 6:(i) VD = (density of the gas or vapour)/(density of hydrogen) (at same temperatureand pressure, here it is NTP)

VD= (0.0014/0.000089)= 15.73, So MM= 2X15.73=31.46 ≈ 32. So the gas isO

2

(ii) VD= 0.00316/0.000089= 35.5, MM=71, so the gas is Cl2

(iii) VD=0.00089/0.000089=10, MM=20, so the gas is Ne.

SAQ 7: V1= 49.4ml, T

1=(273+20), P

1 = (740-18)mm

Let us first convert the volume to NTP conditions.

273760 X V2

=(740-18) X 49.4

273+20⇒ V

2= 43.72ml

=mass of 43.72ml of the vapour at NTPmass of 43.72ml of hydrogen at NTP

0.168

43.72X0.000089= 43.175V.D =

40

M.M = 2XV.D = 2X43.175= 86.35SAQ 8: Let us take the NTP volume V

2 from the SAQ 7 = 43.72ml

43.72ml of the vapour weighs 0.168gmSo 22400ml of the vapour will weigh (0.168/43.72)X22400 = 86.075gm (M.M)

We noticed that the M.M found by two methods are little different. This is obvious as thedifference is due to the difference in methods adopted. In VD method we made use of thedensity of hydrogen gas as 0.000089gm/cc which is not very exact. That is why this difference.But you are advised to adopt any method unless otherwise asked for specifically.

ANSWERS TO PRACTICE QUESTIONS1. Convert the volume to NTP condition first. The NTP volume V

2=280.18ml(see for

yourself)280.18ml of the vapour at NTP weighs 3.2gm.22400ml of the vapour at NTP must weigh (3.2/280.18) X 22400 = 255.83gm.So the the molecular formula (S)

n can be known: n=255.83/32 ≈ 8 : S

8

2. Turpentine oil absorbs ozone only. Since the contraction of volume was held by 50ml, thevolume of ozone =50ml. So the volume of oxgyen = 250-50=200mlV.D of the mixture = 0.393/(250X0.000089) = 17.66, So MM = 2X 17.66=35.32

Av. M.M = 35.32 = 200 X 32 + 50 X x250 ⇒ x = 48.6 (where x=M.M of ozone)

Note that the actual molecular mass of ozone is 48 which is close to the result.3. 1litre weighs 0.5gm at NTP, So 22.4litres will weigh 11.2gms. This is the mass of one

mole(MM).4. Let the volume of air displaced = x ml at 150C and (773.3-13.3)mm pressure. Let us

convert this volume to NTP condition. The NTP volume V2 =0.957x ml.

V.D = 0.1665/(0.957x X 0.000089) = 46/2=23 ⇒ x = 85.5mlSo the volume of air displaced in the Victor Meyer's experiment is 85.5ml

5. Similar to No.3. Ans: 32g (Try yourself)

41

Mole Concept and Avogadro lasw and i

MOLE CONCEPT

Mole is a chemical unit( like a dozen, a gross etc. used as family units) which can be defined intwo ways.(i) Number wise: one mole of anything contains Avogadro's Number (6.023 X 1023)of species of that thing.(ii) Mass wise: It is the mass of the substance which contains Avogadro's number ofspecies.Mole can be determined for atoms, molecules as well as for ions.(a) For atoms : e.g C, Na, H, Fe etc. (b)For molecules :e.g N

2, H

2, CO

2, H

2SO

4 etc.

(c) For ions : e.g SO42-, NO

3- etc.

(a) For atoms:The atomic mass expressed in gm(gm atomic mass) is the mass of Avogadro's number ofatoms. So one mole of atoms weigh one gm. atomic mass(e.g : one mole of Na atoms weighs 23gms , one mole of H atoms weigh 1.008 gm and one mole of C weigh 12 gms. and so on)N.B: Some authors write one mole of atoms as one gm. atom and according to them theterm mole should not be used for atoms. However, we have used both gm. atom and molefor atoms.SAQ 1:(i)How many bananas are there in one mole of banana?

(ii)You bought 1/10 mole of cricket balls for your team. How many balls you bought. Ifyou require only 20 balls per year, how long your team can play cricket?(iii)A cigarette smoker smoked 12.046 X1023 cigarettes before his death in lung cancer.How many moles of cigarettes did he smoke during his life time?

SAQ 2: Find the mass of 6.203 X 1023 atoms of (a)Phosphorus (b)Calcium(c)Helium (d)Boron

SAQ 3: Find the number of atoms present in (i) 19 gms of Fluorine (ii)39 gms of Potassium(iii)63.5 gms of Cu

NO. OF ATOMS AND MOLES(GM. ATOMS)PRESENT IN A GIVEN MASS OFAN ELEMENTExample: Calculate the number of atoms present in 2.3 gms of NaSolution:

23 gms(gm. atomic mass) of Na contain 6.203 X 1023 atoms of Na

2.3 gms of Na therefore contain 6.023 X 1023

23X 2.3 = 106.023 X

22atoms of Na

SAQ 4:Calculate the number of atoms present in the following:(i) 0.12 gm of C (ii)40 gms of He (iii)4 gms of Oxygen (iv)0.001 gm of

42

Sulphur (v)40mg of Calcium (vi)6.35 kgs of CopperThe atomic masses are as follows: C=12, He=4, O=16, S=32, Ca=40, Cu=63.5

SAQ 5: Calculate the mass of the following(i) 1 million carbon atoms (iii) 12X 1030 of K atomsAtomic Masses: C =12 K=39

No. of moles(gm. atoms) of atoms of an element present in a given mass of element:

You already know that 1 mole of atoms (also called 1 gm. atom of an element) contains Avogadro'snumber of atoms which weigh one gm atomic mass. For examle, the atomic mass of Na is 23. So23 gms of Na = 1 mole of Na atoms =1 gm atom of Na which contains 6.023 X 1023 atoms ofNa. So if you are asked to find the number of moles of atoms or gm. atoms of Na present in 2.3gm of Na, it is 1/10 of a mole =0.1 mole of atoms(gm. atom). Also if you are asked to find thenumber of atoms present in it, you can find out by the method described earlier. See this example.

Example: Find the number of moles(gm. atoms) of sodium atoms present in 0.023 gm of NaSolution:

23 gms of Na = 1 mole of Na atoms(1gm atom of Na)

0.023gm of Na = 23X =

10.023 0.001 mole of Na atoms or 0.001 gm atom of Na

SAQ 6: Calculate the number of moles(gm. atoms) of atoms present in the following(i)1.27 gms of Iodine (ii)400 mgs of Oxygen (iii)5.04 gms of H (iv)0.28gm

of Silicon (Atomic Masses: I =127, O=16, H=1, Si=28)

(b) For molecule and Ions:

Molecular mass expressed in gms(gm. molecular mass) is the mass of Avogadro's numberof molecules. Similarly ionic mass expressed in gm. is the mass of Avogadro's number ofions. Hence one mole of molecules will weigh one gm. molecular mass and one mole of ions willweigh one gm. ionic mass. One mole of N

2 gas will weigh 14X2= 28gms, one mole of CO

2 will

weigh 44 gms and one mole of SO42- will weigh 32+4X16=96 gms and so on.

SAQ 7:Calculate the mass of 6.023X1023 molecules of the following:(i)CaCO

3(ii)S

8(iii)HCl (iv)H

2SO

4

( Atomic masses: Ca=40 , C=12, S=32, Cl=35.5, S=32, O=16, H=1)SAQ 8: Calculate the mass of 6.023X1023 number of ions in case of the following

(i)NO3

- (ii)Cr2O

7 2- (Atomic masses: N=14, Cr=52)

NUMBER OF MOLECULES PRESENT IN GIVEN MASS OF A SUBSTANCEAND VICE VERSA

If 36.5 gms (gm molecular mass) of HCl contains 6.023X1023 molecules of HCl, then x gm of

43

HCl will contain (6.023X1023/36.5)X x molecules. So if the mass of a substance is given, we cancalculate the number of molecules. We can also get the mass of a substance which will containa fixed number of molecules. See the following examples.Example: Calculate the number of molecules present in 7 gms of N

2 gas.

Solution:The molecular mass of N

2 is 28.

28 gms of N2

contains 6.023X1023 molecules

7 gms of N2 contains 6.023 X 10

23

X =23

1028

7 1.505 X molecules.

SAQ 9:Find the mass of the following.(i)2.0076 X 1020 molecules of Na

2CO

3(ii)3.0115 X 1040 molecules of H

2SO

4

NO. OF MOLES PRESENT IN A GIVEN MASS OF A SUBSTANCESince the gm. molecular mass is the mass of 1 mole of a substance, we can find out the numberof moles present in given mass of a substance. Look to the following example.Example: Find the number of moles of sulphuric acid molecules present in 490 gmsof H

2SO

4 .

Solution:Molecular Mass of H

2SO

4 = 98

98 gms of H2SO

4 = 1 mole

490 gms of H2SO

4= 490/98=5 moles

SAQ 10: Find the number of moles in case of the following:(i)18.6 gms of Phosphorous(P

4) (ii)1.47kg of H

2SO

4

(iii)3.55gm of Cl2 gas

SAQ 11: Calculate the number of moles and molecules present in(i)0.106 gmsNa

2CO

3(ii)48gms of Oxygen (iii)15.5 gms of P

4

(iv)0.49 gms of H2SO

4(v)2.8 kg of Nitrogen gas

SAQ 12: Find the mass of the following(i) 1023 molecules of hydrogen (ii)100 million molecules of K

2SO

4

(iii) 1.2 X 10 69 molecules of NitrogenSAQ 13: Find the mass of the following.

(i)0.001mole of Ca3(PO

4)

2(ii)2.5moles of H

2SO

4

(iii)1/20 mole of MgCO3

MASS OF ONE ATOM OF AN ELEMENTSince the mass of Avogadro's number of atoms is gm. atomic mass. So from this the absolutemass of one atom of an element can be calculated. Note that this is the actual mass of an atomwhich is incredibly minute and small. When you are asked, what is the mass of one oxygen atom,you often say 16gm. But just think how the mass of one tiny oxygen atom could be 16gm?? With16gm of sugar you can prepare a cup of tea and could one atom of oxygen can weigh anincredibly large 16gm?? No, what we say 16gm; is its gm. atomic mass, which is the mass ofAvogadro's number of atoms. So from this the actual mass of one atom in gm can be calculated.See this example.

44

Example: Find the mass of one oxygen atom:Solution:

6.023X1023 atoms of oxygen weigh 16 gms1 atom of oxygen weighs 16/( 6.023X1023)= 2.66 X 10-23 gm.Do you notice how small the mass of one atom is!!!!

SAQ 14: Find out the mass of one atom of each of the following.The atomic masses are givenwithin brackets.

(i)C(12) (ii)Al(27) (iii)H(1) (iv)S(32)(v)Cl(35.5) (vi)Ag(108)

SAQ 15: What is the mass of 1 amu in gm? How this is related with the mass of hydrogen atom?

MASS OF ONE MOLECULE OF A SUBSTANCESince the gm. molecular mass is the mass of Avogadro's number of molecules in the same wayas gm atomic mass is the mass of Avogadro's number of atoms, we can calculate the actualmass of one molecule of a substance. See this example.Example: Calculate mass of one molecule of CO

2.

Solution:Molecular mass of CO

2 = 12 +16X2 = 44

6.023X1023 molecules of CO2 weigh 44 gms

So 1 molecule of CO2 weighs 44/(6.023X1023) = 7.305 X 10-23 gm.

You also found that like the mass of one atom, the mass of one molecule is also very small.

SAQ 16:.Find the mass of one molecule in gm for the following(i)NH

3(ii)CaCO

3(iii)(NH

4)

2SO

4(iv)CH

3COOH (v)NaCl

NUMBER OF MOLECULES AND MOLES OF A GAS PRESENT IN A GIVENVOLUME OF GAS AT A PARTICULAR TEMPERATURE AND PRESSURE

Gram Molar Volume(GMV) at NTP = 22.4 litresone mole of any gas occupies 22.4 litres at NTP.e.g 32 gms of O

2 gas, 2 gms of H

2 gas, 44 gms of CO

2 gas, 28 gms of N

2 gas etc. each occupies

22.4 litres at NTP (00C and 760mm Hg Pressure). So the number of moles and molecules of agas present in a given volume of gas can be easily calculated. Also we can do the opposite thing.We can calculate the volume of gas provided we know the number of molecules or moles ormass. The examples below will make these more clear.

Example : Calculate the number of molecules, moles and mass of oxygen gas present in224cc of oxygen gas at NTP.Solution:

22400 cc of O2 gas at NTP contain 6.023X1023 molecules of O

2

224 cc of O2 gas at NTP contain 6.023X1021 molecules of O

2

Again 22400 ml of O2 gas contains 1 mole

45

224 ml of O2 contains 224/22400 = 0.01mole

1 mole of oxygen weigh 32 gms0.01mole of oxygen weigh 32X0.01 = 0.32 gm

N.B : For gas present at conditions other than NTP, the combined gas equation(P

1V

I/T

1=P

2V

2/T

2 )is to be used to convert the given volume to NTP or volume at NTP to

volume at given conditions depending on the requirement. . If the volume data is given,theconversion is made first and if the volume is to be found out, then the conversion is done atthe end.Example : Calculate the volume in cc of nitrogen gas containing 2.4 X 1020 molecules at270C and 800mm pressure.Solution:Here volume is to be calculated at some other temperature and pressure. So the conversionis to be made at the end.

6.023X1023 molecules of N2 at NTP occupies 22400 cc

2.4 X 1020 molecules of N2 occupies 8.96cc at NTP

But we need to find the volume at the conditions given in the question. Let us apply gaslaw.

=760 X 8.96273

800 X V2(273+23) ⇒ V

2= 9.22cc

Example 3: Calculate the number of molecules and mass in gm. of CO2 present in 200 ml

of the gas at 270C and 800mm pressure.Solution: Since the volume data is given here, the gas equation is to be used first to get thevolume at NTP from the volume at given conditions.

273760 X V2

273+27800 X 200 = ⇒ V

2 = 191.6 ml(Volume at NTP)

22400 ml of CO2 gas at NTP weighs 44 gms

191.6 ml of CO2 weighs (44/22400)191.4 = 0.376gm =376 mg.

Again 22400 ml of the gas at NTP contains 6.023 X1023 molecules at 191.6 ml of the gas contains 5.1 X1021 molecules .

SAQ 17:(i) Find the volume of 4gms of CO

2 gas at NTP. How many molecules are present in it?

(ii) Calculate the mass of N2 gas present in 560ml of it at NTP. Also find out the number of

moles and molecules present in it.(iii) Find the number of molecules present in 680ml of O

2 gas at 270C and 900mm of Hg

pressure. Also calculate the mass in gm and number of moles.(iv) Calculate the volume of H

2 at 270C and 900 mm pressure if it contains 0.5 mole of

hydrogen gas.

46

PRACTICE QUESTIONS(Mole Concept)

SET-I1. Find the number of gm. atoms(mole of atoms) present in the following. Also calculate thenumber of atoms present in it. The atomic masses are given inside brackets.

(i)4Kg of Ca (40) (ii) 32.7gm of Zn(65.4) (iii)7.09gm ofCl(35.45)

(iv)95.4gm of Cu(63.55) (v)8.62gm of Fe(55.85)2. How many H

2 molecules are present in 8.5 gm of H

2? How many H atoms are in it?

3. Find the mass of one S atom in gm.4. Find the mass of the following:

(i)12.046 X 1024 atoms of H(1) (ii)3.0115 X 1030 atoms of Zn(65.5)(iii) 3.6138 X 1022 atoms of Ag(108)

5. How many moles and how many molecules of H2O are present in 48gms of H

2O?

6. What is the mass of 0.004mole of glucose(C6H

12O

6)? What is the mass of one glucose

molecule?7. What is the mass of the following. Also find their volumes at NTP.( N stands for Avogadro'snumber)

(i) 3N molecules of CO2(ii)0.5N molecules of N

2(iii)0.0002 N molecules of O

2

8. How many molecules of water and atoms of H and O are present in 9 gms of H2O?

9. Calculate the number of moles in each of the following. Also find the number of molecules ineach case.

(i)10gms of CaCO3

(ii)3.5kg of H2SO

4(iii)3.16gm of KMnO

4

10. Find the number of NO3- ions present in 62gm of NO

3- . What is the mass of one NO

3- ion?

11. Calculate the number of H atoms, S atoms and O atoms present in 7gms of H2SO

4.

12. Calculate the gm. atoms(mole of atoms) of carbon and oxygen present in 22gms of CO2 gas

at NTP. Also find the number of carbon and oxygen atoms present in it.13.Calculate the number of oxygen atoms present in 25gm of CaCO

3. Also find the number of

gm. atoms(mole of atoms) of oxgyen present.14. How many H atoms are present in 25gm of NH

4Cl? Also find the the number of moles of

NH4Cl present.

SET-II

1. Calculate(i)The number of boron atoms in 594gms of Boron(B=11)(ii)the number of calcium atoms in 10gms of calcium(Ca=40)(iii)the number of iron atoms in 8.37gms of iron(Fe=56)(iv)the number of gm. atoms in 1.0gm of Lithium(Li =7)(v)the number of Uranium atoms in 1 gm of U238 isotope( U = 238)

2. A sample of pure element that had a mass of 1.0g was found to contain 1.5 X 1022

atoms. Find out the atomic mass of the element.3. How many gm. atoms and atoms of S are present in 20gm of sulphur?(S=32)

47

4. How many moles are there in each of the following:(i)0.14gm of nitrogen gas (ii)9.2gm of NO

2 gas (iii)3.5 X1022 molecules of

N2O

5. Calculate the number of moles in 2.2gm of CO2.

6. Calculate the number of molecules in 2.4gm of CO.7. Calculate the mass of

(i)one CO2 molecule (ii)One Au atom (At. mass of Au = 197)

(iii)One H2SO

4 molecule (iv)One hydrogen atom

8. An atom of Hg weighs 33.3 X 10-23 gm. Calculate the atomic mass of Hg.9. One atom of an element(X) weighs 6.6 X 10-23 gm. Calculate the number of gm. atomsin 20kg of it.10. Calculate the number of methane molecules and the number of hydrogen and carbon

atoms present in 20gms of methane?11. How many atoms of each kind are present in 2.56gms of sucrose(C

12H

22O

11)?

12. Which of the following contains the greatest number of molecules and which containsthe least.

(i)8 gm of CO2

(ii)8gm of O2

(iii)8gm of N2

(iv)8gm of H2

13. How many moles and molecules are present in the following.(i)19.2 gm of H

2SO

4(ii)4.4 Kg of CO

2(iii)1 kg of ammonium

dichromate(Cr=52)14. Calculate the number of atoms of each kind in

(i)4.4gm of CO2

(ii)2.7gm of Al (iii)10 gm of CaCO3

15. A piece of Zn weighs 0.65gm. How many atoms of Zn does it contain?(Zn =65)16. Calculate the volume at NTP occupied by 2.4gm of SO

2.

17. The volume of a gas in a tube is 1.15 X 10-7 ml at NTP. Calculate the number of H2

molecules of the gas present in the tube.18. What is the number of atoms in 2gm of He and also find its volume at STP.(He=4)19. How many molecules of water are present in one cc of water?(density of water=1gm/cc)20. Calculate the number of molecules present in

(i)100 ml of CO2 at NTP (ii)1 litre of HCl at NTP

(iii)250 ml of H2 at 270C and 800 mm pressure

21. How many gm atoms are contained in(atomic masses are given within brackets)(a)32.7g of Zn(65.5) (b)7.09g of Chlorine(35.5) (c)95.4g of Cu(63.5)(d)4.31gm of Fe(56) (e)0.328g of S(32)

22. What is the volume of :(i)CO gas containing 3 X 1021 molecules at STP(ii)N

2 gas containing 2.4 X 1025 molecules at 270C and 800mm

pressure.23. Calculate the number of moles of carbon atoms present in

(a)1.0gm of C (ii)12 gm of carbon (iii)5.66 X 1020 carbon atoms24. Calculate the mass in gm for the following

(i)3 g atom of Ag (ii)0.5 mole of Ne (iii)2.5 moles of CO2

( i v ) 0 . 0 1mole of K

2CO

3

48

25. Find the number of g. atoms and mass of an element having 3 X 1024 atoms. Atomicmass of the element is 32.26. How many atoms of Na, S, O are present in 0.25 mole of Na

2SO

4?

27. Find the molecular mass of a gas, 7.525 X 1023 molecules weigh 35 gms. Also find thevolume of the 5gm of the gas at NTP.28. Calculate the number of oxygen atoms present in 17.6 g of CO

2.

29. Calculate the mass in gm of HNO3 present in 0.005moles of it. Also calculate the number

of molecules of HNO3 present in it.

30. Calculate the number of SO4

2- ions present 2.5gm of Na2SO

4.

RESPONSE TO SAQs(Mole Concept)

SAQ 1:(i) One mole of banana contains 6.023 X 1023 number of bananas. What a huge number!!!(ii) one mole of cricket balls = 6.023 X 1023 , so 1/10 mole of it will contain ten times less

i.e 6.023 X 1022 number of balls. If you need 20 balls per annum, then the number ofyears required to exhaust the stock = 6.023 X 1022 /20= 3.1 X 1021 years, which meansthousands of generations will play cricket with that stock of balls!!!!

(iii) 6.023 X 1023 number of cigarettes =1 mole of cigarettesSo 1 cigarette = 1/(6.023 X 1023)mole12.046 X 1025 cigarettes = 12.046 X 1023 X(1/6.023 X 1023)=2 molesSo the man smoked 2 moles of cigarettes during his life time.

SAQ 2: The mass of Avogadro's number of atoms= one gm. atomic mass(a)31gms (b)40gms (c)4gms (d)11gms.All these are the atomic masses of the elements.

SAQ 3: (i)6.023 X 1023 atoms of F (ii)6.023 X 1023 atoms of K(iii)6.023 X 1023 atoms of Cu

Note that in all these cases, mass of the substance given were their respective gm. atomicmasses. So the amount must contain Avogadro's number of atoms.SAQ 4: (i) 12 gms(gm. atomic mass) of C contain 6.023 X 1023 number of C atoms

0.12gms of C must contain (6.023 X 1023/12)X0.12= 6.023 X 1021 atoms(ii) 4 gms(gm atomic mass) of He contains 6.023 X 1023 atoms of He

40gms of He must contain (6.023 X 1023/4)X40= 6.023 X 1024 atoms of He(iii) 16gms(gm. atomic mass) of oxygen contains 6.023 X 1023 atoms of O

4gms of oxgyen must contain (6.023 X 1023/16)X4= 1.5 X 1023 atoms(iv) 32gms(gm. atomic mass)of S contains 6.023 X 1023 number of S atoms

0.001gms of S must contain (6.023 X 1023/32)X0.001= 1.88 X 1019 atoms of S(v) 40gms(gm atomic mass) of Ca contain 6.023 X 1023 atoms of Ca

40mg i.e 0.04gm of Ca must contain (6.023 X 1023/40)X0.04=6.023 X 1020

atoms of S(vi) 63.5gm(gm atomic mass) of Cu contains 6.023 X 1023 atoms of Cu

6.35Kg i.e 6350 gm of Cu must contain (6.023 X 1023/63.5)X6350=6.023 X 1025

atoms

49

SAQ 5:(i) 6.023 X 1023 number of carbon atoms weigh 12gms(gm atomic mass)1 million i.e 106 atoms will weigh [12/ (6.023 X 1023)]X106≈ 2X12-17 gms.

(How much small the mass is!!!!)(ii) 6.023 X 1023 atoms of K weigh 39gms (gm atomic mass)

12 X 1030 atoms of K will weighs [39/( 6.023 X 1023)]X12X1030 =78X107 gms= 7.8 X 108gms

SAQ 6:(i) 127gm(gm atomic mass) of I = 1 mole of I atoms=1 gm atom So 1.27gm of I = (1/127)X1.27= 1/100=0.01 mole of I atoms =0.001gm atom

(ii) 16 gms(gm atomic mass) of O = 1 mole of O atoms= 1gm atom400mg i.e 0.4gm of O = (1/16) X 0.4= 4/160 = 0.025 mole of atoms(gm atom)

(iii) 1 gm(gm atomic mass) of H = 1 mole of H atoms =1gm atom5.04gms of H = (1/1) X 5.04= 5.04 moles of H atoms =5.04 gm atoms

(iv) 28gms(gm atomic mass) of Si = 1 mole of Si atoms(1gm atom)0.28gm of Si = (1/28)X0.28= 1/100 =0.01 mole of Si atoms(0.01gm atom)

SAQ 7: (i) Molecular Mass of CaCO3 = 40 + 12 + 3X16= 100

The mass of 6.023X1023 molecules is gm molecular mass =100gms(ii) Molecular Mass of S

8= 8X32=256, So the mass of 6.023X1023 molecules of S

8

is 256gms.(iii) Molecular Mass of HCl = 1+35.5 = 36.5, So the mass of 6.023X1023 molecules

of HCl = 36.5gms.(iv) Molecular Mass of H

2SO

4 = 2+32 +64 =98, So the mass of 6.023X1023 molecules

of H2SO

4 is 98 gms.

SAQ 8:(i) Ionic Mass of NO3

- = 14 + 48 = 62, So 6.023X1023 number of NO3- ions wil

weigh 62gms.(ii) Ionic Mass of Cr

2O

7 2- = 2X52+ 7X16 = 216, So the mass of 6.023X1023 number

of Cr2O

7 2- is 216gms.

SAQ 9:(i) Molecular Mass of Na2CO

3 = 2X23+12+3X16=106,

So 6.023X1023 molecules of Na2CO

3 weighs 106 gms.

2.0076 X 1020 molecules of Na2CO

3 must weigh (106/6.023 X 1023)X2.0076 X

1020 =35.33 X10-3gm = 0.03533 gm = 35.33 mg(ii) Molecular Mass of H

2SO

4 = 2+32+64=98,

6.023X1023 molecules of H2SO

4 weighs 98 gms.

3.0115 X 1040 molecules of H2SO

4 will weigh (98/6.023 X 1023)X3.0115 X 1040

= 490 X 10 16gm = 490 X 10 13 kg = 490 X 10 10 quintals(!!!!).SAQ 10:(i) Molecular Mass of P

4 = 4X31=124

124 gms of P4 = 1 mole ⇒ 18.6gms of P

4 = (1/124)X18.6= 0.15 mole

(ii) Molecular Mass of H2SO

4 = 98

98 gms of H2SO

4 =1 mole

1.47kg i.e 1470gm of H2SO

4 = (1/98)X1470 = 15 moles.

(iii) Molecular Mass of Cl2 = 2X35.5=71

71gm of Cl2 = 1 mole ⇒ 3.55gm of Cl

2 = (1/71)X3.55= 1/2 =0.05 mole.

Note that in this case we had to find out the number of molecules(not atoms) as chrorine is adiatomic molecule.

50

SAQ 11:(i) Molecular Mass of Na2CO

3 = 46 + 12+48=106,

106gm of Na2CO

3 = 1 mole

0.106 gm of Na2CO

3 = (1/106)X0.106=0.001 mole

1 mole of Na2CO

3 contains 6.023 X 1023 molecules

So 0.001 mole of Na2CO

3 must contain 0.001X6.023 X 1023= 6.023 X 1020 molecules

Alternatively:106gm of Na

2CO

3 contains 6.023 X 1023 molecules

0.106 gm of Na2CO

3 contains (6.023 X 1023/106)X0.106 = 6.023 X 1020

moleculesYou are advised to adopt any method you like.

(ii) Molecular Mass of O2 =32,

32 gms of Oxgyen = 1 mole of O2 .

48gms of oxygen = (1/32)X 48 = 1.5 moles of O2

1 mole of O2 conatins 6.023 X 1023 molecules

1.5 moles of O2 must contain 2 X6.023 X 1023= 9.0345 X 1023 molecules.

(iii) Molecular Mass of P4 = 4X31=124

124gms of P4 = 1 mole

15.5 gms of P4 = 1/124 X 15.5 = 0.125 mole.

1 mole of P4 contains 6.023 X 1023 molecules of P

4

0.125 mole of P4 must contain 0.125 X6.023 X 1023 = 7.528 X 1022 molecules.

(iv) 98gms(gm molecular mass) of H2SO

4 = 1mole

0.49gm of H2SO

4 = 1/98 X 0.49 = 0.005 mole

1 mole of H2SO

4 contains 6.023 X 1023 molecules

0.005 mole of H2SO

4 must contain 0.005 X 6.023 X 1023 = 3.01 X 1022 molecules.

(v) Mol. Mass of N2 = 28,

28 gm of N2 = 1 mole

2.8kg i.e 2800gm of N2 = 1/28 X 2800 = 100 moles.

1 mole of N2 conains 6.023 X 1023 molecules of N

2

100 moles of N2 must cotain 100X6.023 X 1023= 6.023 X 1025 molecules.

SAQ 12:(i) Mol. Mass of H2 = 2,

6.023 X 1023 number of H2 molecules 2gm (gm. molecular mass)

1023 number of molecules will weigh (2/6.023 X 1023) X 1023 = 0.332gm(ii) M.M of K

2SO

4 = 2X39+32+4X16=174

6.023 X 1023 molecules of K2SO

4 weigh 174gm

So 100 million i.e 100 X106 molecules will weigh (174/6.023 X 1023 )X(100X106)=29 X10-15 gm

(iii) M.M of N2 =28

6.023 X 1023 molecules of N2 weigh 28 gm

So 1.2 X 10 69molecules of N2 will weigh

(28/6.023 X1023)X(1.2X1069)=5.578X1046gmSAQ 13:(i)M.M of Ca

3(PO

4)

2 = 3X40+2X31+8X16= 310

1 mole of Ca3(PO

4)

2= 310gms

0.001mole = 310X0.001=0.31gm.

51

(ii) M.M of H2SO

4= 98,

1 mole of H2SO

4 = 98gms ⇒ 2.5moles of H

2SO

4= 2.5X98=245gms.

(iii) M.M of MgCO3 =24+12+48=84

1mole of MgCO3 = 84gms

1/20 mole of MgCO3 = 1/20X84 = 4.2 gms.

SAQ 14: (i) Mass of 6.023X1023 atoms of Carbon = 12gmSo the mass of 1 atom of carbon = 12/(6.023X1023 ) = 2 X10-23gm.

(ii) Similarly mass of one Al atom = 27/(6.023X1023 )=4.5 X10-23 gm(iii) Mass of 1 H atom = 1/(6.023X1023 )= 1.66 X 10-24gm.(iv) Mass of 1 S atom = 32/(6.023X1023 )= 5.31 X10-23gm.(v) Massof 1 Cl atom = 35.5/(6.023X1023 )= 5.9X10-23g.(vi) Mass of 1 Ag atom = 108/(6.023X1023 )= 17.93 X 10-23gm=1.79X10-22gm

SAQ 15: 1 amu is the 1/12 part of one atom of C-12 isotope.First let us find out the mass of 1 C atom which is equal to 12/(6.023X1023 )=1.992X10-23gm.

This the mass of 12 amu.So the mass of 1 amu = (1.992X10-23)/12 = 1.66 X 10-24 gmLook to the response of previous SAQ 14 (iii), the mass of H atom was found to be 1.66X 10-24 gm. Hence mass of 1 amu and 1 H atom are same.

SAQ 16:(i) M.M of NH3 = 14+3=17

6.023X1023 molecules of NH3 weigh 17gm

So, 1 molecule of NH3 will weigh 17/(6.023X1023 )= 2.82 X10-23gm.

(ii) M.M of CaCO3 = 40 + 12 + 48 =100,

So the mass of 1 molecule of CaCO3 = 100/(6.023X1023 )= 1.66X10-22gm

(iii) M.M of (NH4)

2SO

4 = 2(14+4)+32+64=132.

So the mass of 1 molecule of (NH4)

2SO

4 = 132/(6.023X1023)= 2.19X10-22gm

(iv) M.M of CH3COOH = 2X12+4+32=60

So the mass of 1 molecule of CH3COOH= 60/(6.023X1023) = 9.96X10-23gm

(v) Formula Mass of NaCl = 23+35.5 = 58.5So the mass of 1 NaCl = 58.5/(6.023X1023)=9.71X10-23gm

SAQ 17:(i) M. M of CO

2= 44

So 44gms of CO2 occupy 22.4 litres at NTP

So 4 gm of CO2 will occupy (22.4/44)X4 = 2.0363 litres= 2036.3 mls.

44gms. of CO2 contains 6.023X1023 molecules

4 gm will contain (6.023X1023 /44)X4 = 5.47X1022 molecules.(ii) M.M of N

2 =28

22400ml of N2 gas at NTP will weigh 28gm

560ml of N2 gas at NTP will weigh (28/22400)X560 = 0.7gm.

For finding the number of moles, we can either start from gm. molecular massor gm. molar volume.

22400 ml of N2 at NTP = 1mole

560 ml of N2 at NTP = (1/22400)X560 = 1/40 =0.025 mole

Alternatively,28gms of N

2 = 1mole

52

0.7gm of N2 = 0.7/28= 0.025 mole

22400 ml of N2 at NTP contains 6.023X1023 molecules

560 ml of N2 at NTP must contain (6.023X1023 /22400)X560 = 1.5X1022

molecules.You can also calculate the number of molecules from the mass of N

2(0.7gm)

(iii) Since volume data is given, we have to convert it to NTP condition at once.The NTP volume = V

2 = 732.79 ml

22400ml of O2 at NTP contains 6.023X1023 molecules

732.79ml of O2 at NTP must contain (6.023X1023/22400) X 732.79 =

1.97X1022molecules(iv) Since the volume has been asked, the conversion is to be done at the end.

1 mole of H2 occupies 22.4 litres at NTP

So 0.5 mole of H2 must occupy 11.2 litres at NTP.

Now convert this volume to the volume at required conditions:(11.2 l X 760mm)/273K = (V

2 X 900)/300K ⇒ V

2 = 10.393litres.

ANSWERS TO PRACTICE QUESTIONS

1. (i) 40 gms of Ca = 1gm atom(1 mole of atoms)So 4kg i.e 4000gm of Ca = (1/40)X4000= 100 gm atoms.1 gm. atom contains 6.023X1023 of Ca atoms100gm atoms must contain 6.023X1023 X100 = 6.023X1025 atoms of Ca

(Note that you can also find the number of atoms from gm. atomic mass, i.e 40 gms ofCa contains 6.023X1023 of atoms and so 4000 gms will contain how many atoms?)

(ii) 65.4 gms of Zn = 1gm atom(mole of atoms)So 32.7 gms of Zn = (1/65.4)X32.7 = 0.5 gm atom65.4 gms of Zn contains 6.023X1023 of atoms32.7gms of Zn must contain (6.023X1023 /65.4)X32.7= 3.0115 X1023 atoms.

[Note that you can also find the number of atoms from the number of gm. atoms calculatedas in (i)](iii) 35.45gms of Cl= 1 gm atom

7.09 gms of Cl =(1/35.45)X7.09= 0.2 gm atom1 gm atom of Cl contains 6.023X1023 atoms of Cl0.2 gm atom of Cl must contain 6.023X1023 X 0.2= 1.2046 X 1023 atoms

(iv) 63.55 gms of Cu = 1 gm atom(mole of atoms)95.4gms of Cu= (1/63.55)X95.4=1.5 gm atoms.1 gm. atom of Cu contains 6.023X1023 of atomsSo 1.5 gm atoms of Cu must contain 6.023X1023 X1.5= 9.03 X 1023 atoms.

(v) 55.85gms of Fe= 1gm atom(mole of atoms)8.62gms of Fe = (1/55.85)X8.62=0.154 gm atom1 gm atom of Fe contains 6.023X1023 atoms0.154 gm. atom of Fe must contain 6.023X1023 X0.154=0.927 X 1023 atoms.

2. M.M of H2= 2

2gms of H2 contains 6.023X1023 molecules of H

2

53

8.5 gms of H2 must contain (6.023X1023 /2)X8.5=25.597X 1023 molecules

1 H2 molecule contains 2 H atoms(since hydrogen molecule is diatomic)

25.597X 1023 molecules of H2contains 25.597X 1023 X 2 =51.195 X1023 atoms

of H3. Atomic Mass of S = 32

6.023X 1023 atoms of S weighs 32 gms1 atom of S must weigh 32/(6.023X 1023)= 5.313 X 10-23 gm.

4. (i) 6.023X1023 atoms of H weigh 1gm (gm atomic mass)12.046 X 1024 atoms of H must weigh (1/6.023X1023 )X12.046 X 1024 =20gms

(ii) 6.023X1023 atoms of of Zn weighs 65.5gm(gm atomic mass)3.0115 X 1030 atoms of Zn must weigh (65.5/6.023X1023 )X3.0115 X 1030

=32.75X107gm.(iii) 6.023X1023 atoms of of Ag weighs 108gms(gm atomic mass)

3.6138 X 1022 atoms of Ag must contain (108/6.023X1023 )X3.6138 X 1022

=6.48gm.

5. M.M of H2O=18

18gms of H2O = 1 mole ⇒ 48 gms of H

2O = (1/18)X48= 2.67 moles

1 mole of H2O contains 6.023X1023 number of molecules

2.67 moles will contain 6.023X1023 X2.67 =16.08 X 1023 molecules.

6. M.M of glucose(C6H

12O

6) = 6X12 + 12X1+ 6X16=180

1 mole of glucose = 180 gms0.004 mole of glucose = 180X0.004 = 0.72gm6.023X1023 molecules of glucose weigh 180 gms.1 molecule of glucose must weigh 180/6.023X1023 =29.88 X10-23=2.98X10-22gm.

7. (i) N molecules of CO2 weigh 44gms ( N = 6.023X1023 )

3N molecules of CO2 must weigh 44X3=132 gms.

N molecules at NTP occupy 22.4 litres3N molecules at NTP must occuply 22.4X3= 67.2litres.

Alternatively:44 gms. of CO

2 occupies 22.4 litres at NTP

132gms of CO2 occupies (22.4/44)X132 = 67.2 litres.

(ii) N molecules of nitrogen gas weighs 28 gm(N =6.023X1023 )0.5 N molecules of nitrogen gas must weigh 0.5X28=14gms.N molecules of nitrogen gas at NTP occupy 22.4 litres0.5N molecules of nitrogen must occupy 0.5X22.4=11.2 litres at NPT.

(iii) N molecules of O2 weighs 32 gms.

0.0002N molecules of O2 must weigh 32X0.0002 = 0.0064gm

N molecules of O2 gas occupy 22.4litres at NTP

0.0002N molecules of O2 gas must occupy 22.4X 0.0002=0.00448litres=4.48ml

8. M.M of H2O= 18

18gms of water contains 6.023X1023 molecules.

54

9gms of water must contain 3.0115X 1023 molecules1 H

2O molecules contain 2 H atoms and 1 O atom

3.0115X1023 molecules contain 2X3.0115X1023 H atoms and 3.0115X1023 Oatoms

Hence the number of hydrogen atoms= 6.023X1023 and number oxygenatoms=3.0115X1023

9. (i) M.M of CaCO3 =40+12+48=100

100gms of CaCO3 = 1 mole

10gms of CaCO3 = 1/10=0.1 mole.

1 mole of CaCO3 contains 6.023X1023 molecules

0.1 mole of CaCO3 must contain 6.023X1023 X0.1= 6.023X1022 molecules

Alternatively:100gms of CaCO

3 cotanins 6.023X1023 molecules

10gms of CaCO3 must contain (6.023X1023 /100) X 10 = 6.023X1022 molecules.

(ii) M.M of H2SO

4 = 98,

98gms of H2SO

4 = 1 mole

3.5kg i.e 3500gm of H2SO

4 = 3500/98=35.71 moles

1 mole contains 6.023X1023 molecules35.71 moles must contain 6.023X1023 X35.71= 2.15 X1025 molecules.

(iii) M.M of KMnO4= 39+55+64=158

158gms of KMnO4 = 1 mole

3.16gms of KMnO4 = 3.16/158= 0.02 mole

1 mole contains 6.023X1023 molecules0.02 mole must contain 6.023X1023 X 0.02= 1.2046X1022 molecules.

10. Ionic Mass of NO3

- = 14+48=6262gms of NO

3- contains 6.023X1023 ions. This is what the question asks.

6.023X1023 number of NO3

- ions weigh 62gm1 ion of NO

3- must weigh 62/(6.023X1023 ) = 1.029X10-22 gm.

11. M.M of H2SO

4 = 98

98 gms of H2SO

4 contains 6.023X1023 molecules

7gm of H2SO

4 must contain (6.023X1023 /98)X7= 4.3X 1022 molecules

1 molecule of H2SO

4 contains 2 H atoms, 1 S atom and 4 O atoms.

So 4.3X 1022 molecules must contain 2X4.3X 1022 H atoms, 1X4.3X 1022 Satoms and 4X 4.3 X 1022 O atoms.

So the number of H atoms = 8.6X1022, number of S atoms= 4.3X 1022

and the number of O atoms =1.72 X 1023

12. 44gms of CO2 contains 1gm atom of C and 2 gm atoms of O.

22gms of CO2 contains 0.5gm atom of C and 1gm atom of O.

No. of C atoms = 0.5X 6.023X1023 and no. of O atoms = 1X 6.023X1023

13. 100gms of CaCO3 contains 3X 6.023X1023 number of O atoms

25gms of CaCO3 must contain (3/4) X 6.023X1023 number of O atoms.

100gms of CaCO3 contains 3 gm. atoms(moles of atoms) of O.

25gms of CaCO3 must contaian 3/4=0.75 gm atoms of O.

55

14. M. M of NH4Cl = 14 + 4 + 35.5= 53.553.5gms of NH

4Cl contains 4 X 6.023X1023 atoms of H

25gms of NH4Cl must contain 11.26 X1023 atoms of H

53.5gms of NH4Cl contains 1 mole of it.

25gms of NH4Cl contains 25/53.5= 0.467 mole.

SET-II1 (i) 11gms of B contains 6.023X1023 atoms.

So 594gms of B contains (6.023X1023)/11 X 594= 3.252X 1025 atoms.(ii) No of Ca atoms = (40/10)X 6.023X1023 = 2.409 X1024 atoms.(iii) No. of Fe atoms = (8.37/56)X 6.023X1023 = 9.0 X1022 atoms(iv) 7gms of Li = 1gm. atom of Li

1 gm of Li = 1/7 = 0.143 gm atom.(v) No. of U atoms = (1/238)X6.023X1023 = 2.53X1021 atoms.

2. 1.5 X1022 atoms of an element weighs 1gmSo 6.023X1023 atoms of the element must weigh (6.023X1023)/(1.56 X1022)=38.6So the atomic mass of the element= 38.6.

3. 32gms of S contains 6.023X1023 atoms.20gms of S must contain (20/32) X6.023X1023 = 3.76 X1023 atoms.32gms of S = 1 gm atom ⇒ 20gms of S= 20/32=5/8 gm.

atom.4. (i) 0.14/28=0.005 mole (ii)M.M of NO

2=14+32=46,

The number of moles= 9.2/46= 0.2(iii) 6.023X1023 molecules of N

2O weighs 1 mole

3.5 X1022 molecules of N2O weights (3.5X1022)/(6.023X1023)= 0.058 mole.

5. 44gms of CO2 = 1mole, So 2.2 gm= 2.2/44= 0.05 mole.

6. M.M of CO=12+16=28; 28gms contain 6.023X1023 molecules,So 2.4 gms of CO contains (2.4/28)X6.023X1023= 5.16 X1022 molecules.

7. (i) 6.023X1023 number of molecules weigh 44gms So 1 molecule must weigh 44/(6.023X1023) = 7.3 X 10-23gm.

(ii) 197/(6.023X1023)=3.27X10-22gm. (iii) 98/(6.023X1023) = 1.627X10-

22 gm.(iv) 1/(6.023X1023)= 1.66 X10-24gm.

8. 1 atom of Hg weighs 33.3 X 10-23gmsSo 6.023X1023 atoms of Hg must weigh (6.023X1023)X(33.3 X10-23)= 200.56(atomicmass)

9. 1 atom of the element weighs 6.6 X 10-23 gm6.023X1023 atoms of the element must weigh 6.023X1023 X 6.6 X 10-23 =39.7539.75gms = 1 gm atom ⇒ 20kg i.e 20,000gms = 20000/39.75= 503.144 gm. atoms.

10. M.M of CH4=16.

16gms of CH4 contains 6.0.23 X 1023 molecules.

20gms of CH4 must contain (20/16) X6.0.23 X 1023 =7.528 X1023 molecules.

1 CH4 molecule contains 1 C atom

7.528 X1023 molecules must contain 7.528 X1023 C atoms

56

1 CH4 molecule contains 4 H atoms

7.528 X1023 molecules must contain 4X7.528 X1023 =3.011X1024 H atoms.11. M.M of C

12H

22O

11= 12X12+22+11X16=342

342gms of sucrose contains 6.023 X1023 molecules2.56gms of sucrose must contain (2.56/342)X6.023 X1023 =4.5X1021 molecules1 molecule of sucrose contains 12 C atoms.4.5X1021 molecules of sucrose must contain 12 X 4.5X1021 C atoms.1 molecule of sucrose contains 22 H atoms4.5X1021 molecules of sucrose must contain 22 X4.5X1021 H atoms1 molecule of sucrose contains 12 O atoms4.5X1021 molecules of sucrose must contain 12 X4.5X1021 O atoms.

12. (i)8/44=0.18mole (ii)8/32=0.25mole (iii)8/28=0.285mole(iv)8/2=4moles

Since the number of molecules is directly proportional to the number of moles, 8gms ofH

2(iv)contains highest number of molecules and 8gms of CO

2(i) contains the lowest

number of molecules.13. (i) No. of moles = 19.2/98=0.196; No. of molecules =0.196 X6.023 X 1023

(ii) No. of moles = 4400/44=100; Hence no of molecules =100X.6.23 X 1023

(iii) M.M of (NH4)

2Cr

2O

7 = 2(18)+2X52 + 7X16= 252

So the number of moles = 1000/252=3.97; No. of molecules=3.97X6.023 X 1023

14. (i) 44gms contain 6.023 X 1023 molecules of CO

2

4gms of CO2 contains 6.023 X 1022

molecules

1 CO2 molecule contains 1 C atom

6.023 X 1022 molecules must contain 6.023 X 1022

C atoms.

1 CO2 molecules cotains 2 O atoms

So 6.023 X 1022 molecules must contain 2X6.023 X 1022

O atoms.

(ii) 27gms of Al contains 6.023 X 1023 atoms

So 2.7gms of Al must contain 6.023 X 1022 atoms.(iii) M.M of CaCO

3 =100,

100 gms of CaCO3 cotains 6.023 X 1023

molecules

10gms of CaCO3 must contain 6.023 X 1022

molecules

So the number of Ca atoms= 6.023 X 1022

The number of C atoms =6.023 X 1022 and number of O atoms = 3X6.023 X

1022

15. 65gms of Zn contains 6.023 X 1023 atoms of Zn

0.65gm of Zn must contain 6.023 X 1021 atoms of Zn

16. 64gms of SO2 occupy 22.4litres, So 2.4gm must occupy 0.84 litre at NTP.

17. 22400ml of H2 gas at NTP contains 6.023X1023molecules, So 1.15 X 10-7ml will contain

3.1 X 1012 molecules.18. 4gms of He contains N atoms, so 2gms of He must contain N/2 atoms(N=6.023X1023)

4gms occupy 22.4litres, so 2gms must occupy 11.2 litres.(note that He monoatomic)19. density = m/v ⇒ mass of 1cc of water = 1gm.. We know that 18gms of H

2O contains

N molecules, So 1 gm must contain N/18 molecules of H2O. (N=6.023X1023)

20. (i)N/224 molecules (ii)N/22.4 molecules

57

(iii) First the volume is converted to the NTP conditions. V2= 239. 47ml.

22400ml at NTP contains N molecules, so 239.47ml must contain 0.01 N molecules.(N=6.023X1023)

21. (a)32.7/65.5= 0.5 (b)7.09/35.5= 0.2 (c)95.4/63.5=1.5(d)4.31/56=0.0769 (e)0.328/32= 0.01

22. (i) 6.023 X 1023 molecules of CO occupy 22.4 litres3 X 1021 molecules of CO must occupy 0.112 litre= 112ml

(ii) First let us find the volume at NTP.Volume at NTP = 892.5 litres. Then convert to given conditions.(760mm X 892.5 l)/273 = (800mm XV

2)/300 ⇒ V

2 = 931.73litres.

23. (a) 1/12 mole(gm atom) (b)1 mole(1gm atom)(c) 5.66 X 1020/N=9.4X10-4 mole(gm atom) (N=6.023X1023)

24. (a) 3X108gms (ii)0.5 X20gms (iii)2.5X44gms (iv)0.01X 138gms25. N atoms = 1gm atom, So 3X1024 atoms = 5 gm atoms. The mass of 5gm.

atoms=5X32=160 gms.26. 1 mole of Na

2SO

4 contains 2 N atoms of Na and N atoms S and 4N atoms of O.

So 0.25 mole of Na2SO

4 must contain 0.5 N atoms of Na and 0.25atoms of S and N

atoms of O. (N=6.023X1023)27. If 7.525 X 1023 molecules weigh 35 gms, N molecules must weigh 27.9gms(MM).

27.9gms of the gas occupy 22.4 litres, So 5gms will occupy (22.4/29.9)X5= 3.74 litres28. 44gms of CO

2 contains 2X N atoms of O. So 17.6gms will contain 0.8 X N atoms of O.

29. 1 mole of HNO3 weighs 63gms, So 0.005mole must weigh 0.315gm. The no. of moleucles

present is 0.005 X N. (N=6.023X1023)30. MM of Na

2SO

4 = 142; 142gms of Na

2SO

4 contain N SO

42- ions,

So 2.5 gms of Na2SO

4 must contain 0.0176N sulphate ions.(N=6.023X1023).