1.3 Formulae, equations and amounts of substance1.3a particles, substances and formulaeAn atom is the smallest particle of an element that has that elements properties.

An atom contains protons, neutrons and electrons. The protons and neutrons are found in the nucleus of the atom while the electrons exist in the shells, energy levels or quantum shells around the nucleus. The existence of the nucleus was proved by Geiger and Marsden in their experiment when alpha particles were fired at gold foil. Deflection of the alpha particles could only be explained by the existence of a nucleus containing all of the atom's mass.

particle relative mass relative chargeelectron 1/1840 -1proton 1 +1neutron 1 0

Story1.1a

A molecule is a particle with two or more atoms chemically ``joined together.An ion is a particle of one or more atoms with an electric charge.An element is a pure substance which cannot be separated by chemical means.A compound is a substance with two or more different elements chemically joined.An empirical formula is the simplest formula for a substance showing the ratio of atoms that have been joined.A molecular formula shows the number of atoms of each type in a molecule of a substance.Task 1.3a.1 Classify the following: He, O2, Na+, bromine, sodium bromide, C2H4, CH2, H, H2O, CO3

2-, neon, lithium chloride, CH3, C2H6.Task 1.3a.2 Give examples of each defined word above.

1.3b Full and ionic equationsWhen iron is added to copper (II) sulphate, copper and iron (II) sulphate is formed. 

The full equation is Fe(s) + CuSO4(aq) ---> Cu(s) + FeSO4(aq) In terms of ions this is:

Fe(s) + Cu2+(aq) + SO42-(aq) ----> Cu(s) +Fe2+(aq) + SO4

2-(aq) Sulphate ions are spectator ions as they are unaffected by the reaction. Therefore they are left out.  The ionic equation is: Fe(s) + Cu2+ (aq) ----> Fe2+ (aq) + Cu(s) Both the number of atoms and charges balance.Common ionsCationsSodium Na+, potassium K+, lithium Li+, magnesium Mg2+, calcium Ca2+, Aluminium Al3+, ammonium NH4+, hydrogen H+, AnionsOxygen O2-, fluoride F-, chloride Cl-, bromide Br- iodide I-, hydroxide OH-, carbonate CO3-, sulphate SO4-, nitrate NO3-,

Task 1.3b Balancing equations

Balance the following equations:Write ionic equations where appropriate

1) Ca + H2O Ca(OH)2 + H2 2) CO + O2 CO2 3) Ca + O2 CaO 4)* Fe2O3(s) + HCl(aq) FeCl3(aq) + H2O (l) 5)* NH3(aq) + H2SO4(aq) (NH4)2SO4 (aq) 6)* Al + H2SO4 Al2(SO4)3 + H2 7)* CaO + HCl CaCl2 + H2O

8) NH3 + O2 NO + H2O   9) Na2O + H2O NaOH 10)* Na2CO3 + HCl NaCl + CO2 + H2O 11)* Br2 + KI KBr + I2 12)* Ca(OH)2 + HNO3 Ca(NO3)2 + H2O 13) Fe + H2O Fe3O4 + H2

14)* Pb3O4 + HNO3 Pb(NO3)2 + PbO2 + H2O

15)* Cu + HNO3 Cu(NO3)2 + H2O + NO 16) HNO3 NO2 + H2O + O2

Answers1) Ca + 2H2O Ca(OH)2 + H2 2) 2CO + O2 2CO2 3) 2Ca + O2 2CaO (s) 4) Fe2O3(s) + 6HCl(aq) 2FeCl3(aq) + 3H2O(l) Fe2O3(s) + 6H+ + 6Cl- 2Fe2+ + 6Cl- + 3H2O(l)Fe2O3(s) + 6H+ 2Fe2+ + 3H2O(l)

5) 2NH3 + H2SO4 (NH4)2SO4 6) 2Al + 3H2SO4 Al2(SO4)3 + 3H2 7) CaO + HCl CaCl2 + H2O

8) NH3 + O2 NO + H2O   9) Na2O + H2O NaOH 10) Na2CO3 + HCl NaCl + CO2 + H2O 11) Br2 + KI KBr + I2 12) Ca(OH)2 + HNO3 Ca(NO3)2 + H2O 13) Fe + H2O Fe3O4 + H2

14) Pb3O4 + HNO3 Pb(NO3)2 + PbO2 + H2O

1.3c Chemical quantitiesThe unit of mass is 1/12 of the mass of an atom of the isotope Carbon-12 (12C=12.0000 exactly) The Relative Atomic Mass (Ar)of an element is the ratio of the mass of an average atom of that element to 1/12 of the mass of an atom of the nuclide Carbon-12.Experiment 1.3c.1Cut squares, to represent different atoms, from graph paper with areas of 4cm2, 8cm2 and 16cm2. Weigh each square on a two decimal place balance. Calculate the relative masses of your “atoms”.

Atom Mass/g Relative massA 0.12 0.12g/0.12g = 1.00B 0.23 0.07g/0.02g = 3.5C 0.48 0.14/0.02 = 7

(graph paper, scissors, 2dec place balances)

1.3c Amount of substance

The mole is the amount of a substance that contains the same number of particles as there are atoms in 12.00g of carbon-12. This number of atoms is 6.02 x 1023 and is called the Avogadro constant. 

One mole of carbon atoms.

Molar mass can be found. The number of g in one mole is the same number as the relative formula mass of a substance E.g. 23gmol-1 of Na atoms and 28g mol-1 of N2 molecules all contain the same number of particles, they are all one mole of atoms or molecules.

Molar mass = mass of 1 mole

amount of substance = mass of substance/molar mass of substance

Task 1.3c.1 What are molar masses of (a) K, (b) O2, (c) Br, (d) Br2, (e) H2O, (f) H2SO4

Task 1.3c.2 What amount is (a)12g C, (b)14g Li, (c)12g Mg, (d)18g H2O, (e)4.4g CO2?

Task 1.3c.3 What mass is (a)1mol O, (b)2mol O2, (c)0.5mol N2, (d) 0.1mol CO, (e)0.2mol H2SO4?

Task 1.3c.4 What is the molar mass of (a)X if 1mol X has mass of 5g, (b)2mol X has mass of 20g, (c)0.25 mol X has 25g.

Answers:

1(a) 39gmol-1, (b) 32 gmol-1 (c) 80 gmol-1 (d) 160 gmol-1, (e) 18 gmol-1, (f) 98 gmol-1

2(a)1mol, (b)2mol, (c)0.5mol, (d)1mol, (e)0.1mol

3(a)16g, (b)64g, (c)14g, (d)2.8g, (e)19.6g

4(a)5gmol-1 , (b)10gmol-1 , (c)100gmol-1

1.3c CO - Carbon Monoxide USA Air Quality Standards These levels may not be exceeded more than once per year:

← 1-hour average concentration -- 35 ppm ← 8-hour average concentration -- 9 ppm ←

NO2 - Nitrogen Dioxide ← Air Quality Standard ← Annual average concentration -- 0.053 ppm

http://www.airquality.co.uk/

Pollutants in air are measured in parts per million. Air that contains 4 cm3 of carbon dioxide in 1 million cm3 (1000dm3) has a carbon dioxide concentration of 4ppm.

Pollutants in water are also measured as parts per million (ppm). A solution that contains 2 g of lead in 1 million grams of water (1,000 dm3) is a 2 ppm solutionTask 1.3c.5

(a) A car produces CO in the concentration 3ppm from its exhaust. What volume of CO is present in 2000dm3 of exhaust gas?

(b) What is the concentration in ppm of nitrogen dioxide if 32 cm3 is found in 500dm3 of air?

(c) What mass of water is there in 1 tonne (1000kg) of aviation fuel if the concentration is 1ppm?

(d) What is the concentration of nitrate ion in water from a stream in which 3g of nitrate is found in 2000dm3 of water?

6cm3, (b) 64ppm, (c) 1g, (d) 1.5ppm.

Answers: (a) 2000dm3 is 2000000cm33ppm 1000000cm3 exhaust contains 3cm3 of CO so 2000000cm3 exhaust contains 6cm3 CO

1.3d Solution concentration dataThe concentration of a solution can be stated as the mass of solute per cubic decimetre of solution (g/dm3) or the amount in moles of a solute present in 1dm3 of solution (mol/dm3).Concentration = mass of solute/volume

Concentration (Molarity) = amount of solute /volume of solution

1dm3 = 1000cm3

Task 1.3d Concentration and Molarity Problems

1. Calculate the mass of 1 mol of (a) NaOH, (b) NaCl, (c) H2SO4, (d) NH4OH, (e) HCl.

2. Calculate the amount present in (a) 10.6g Na2CO3, (b) 12.6 g of HNO3, (c) 21.3g Cl2, (d) 8.5g NH3, (e) 14.8g Ca(OH)2.

Answers:Q1 40g, 58.5g, 98g, 35g, 36.5g

Answers:Q2 0.100mol, 0.200mol, 0.300mol, 0.500mol, 0.200mol

3. Calculate, to three significant figures, the concentrations (molarity) in mol dm-3  of

(a) 3.65g of hydrogen chloride, HCl, in 2.00dm3 of solution(b) 73.0g of hydrogen chloride, HCl,  in 2.00dm3 of solution.(c) 14.8g Ca(OH)2 in 3.00dm3 of solution.(d) 4.25g NH3 in 1dm3 of solution.(e) 17.4g glucose, C6H12O6 in 0.5dm3 of blood.(f) 5.85g NaCl in 200cm3 aqueous solution

Answers; 0.0500 mol dm-3 , 1.00 mol dm-3 , 0.0667 mol dm-3 , 0.250 mol dm-3, 0.193 moldm-3 

1.3e Using reacting masses and chemical equationsWhat mass of sodium carbonate (molar mass = 106g/mol) can be made by heating 16.8g of sodium hydrogencarbonate (molar mass = 84g/mol)?Method 1 (recommended)

2NaHCO3(s) ----> Na2CO3(s) + CO2(g) + H2O(g)mass NaHCO3 =16.8gamount NaHCO3 = mass NaHCO3/molar mass NaHCO3 =16.8g/84gmol-1   = 0.20molfrom equation: amount Na2CO3/amount NaHCO3 =1/2amount Na2CO3 =1/2*0.2mol = 0.1 molmass Na2CO3 = amount Na2CO3*molar mass Na2CO3 =0.1mol*106gmol =10.6gMethod 2formula mass 2NaHCO3 = 168  formula mass Na2CO3  = 106so 168g of sodium hydrogen carbonate forms 106g of sodium carbonateso     1g of sodium hydrogen carbonate forms 106/168g of sodium carbonateso 16.8g of sodium hydrogen carbonate forms 16.8*106/168g of sodium carbonateso 16.8g of sodium hydrogen carbonate forms 10.6g of sodium carbonate

Task 1.3e1. What amounts of substance are the following masses?(a) 8g of S, (b) 8g of O, (c) 8g of O2, (d) 0.5g of H2, (e)25g of NaOH2. What are the masses of the following amounts?(a) 2mol of Cl2, (b) 0.5 mol of CuO, (c) 0.1 mol of H2SO4, (d) 2.5mol of NaHCO3.3. What mass of glucose can be fermented to give 5.00g of ethanol?C6H12O6(aq) -----> 2C2H5OH(aq) + 2CO2(g)4. What mass of silver chloride can be precipitated from a solution which contains 1.00*10-3 mol of silver ions?Ag+(aq) + Cl-(aq) -----> AgCl(s)5. The pollutant, sulphur dioxide, can be removed from the air by the reaction 2CaCO3(s) + 2SO2(g) +O2(g) -----> 2CaSO4(s) + 2CO2(g)What mass of calcium carbonate is needed to remove 10.0g of SO2?6. What mass of sodium carbonate can be made by heating 100g of sodium hydrogencarbonate?2NaHCO3(s)  -----> Na2CO3(s) + CO2(g) H2O(g)7. What mass of KOH is fromed from 5 g of K when it reacts with water8. What mass of HCl forms 4.4g carbon dioxide in a reaction with calcium carbonate.

Answers:  

1. (a) 8g/32gmol-1 = 0.25mol, (b) 0.5mol (c)0.25mol (d) 0.25mol (e) 0.625mol2. (a) 2mol*71gmol-1 = 142g, (b) 40g, (c)9.8g, (d) 210g.3. 9.81g4. 0.144g

5. 15.6g6. 63.1g7. amount K = 5/39 =0.128molAmount of KOH/amount of K =2/2So amount of KOH = 0.128molMass of KOH = amount KOH * molar mass KOH = 0.128mol*56gmol-1 = 7.18g8. amount CO2 = 4.4/44 = 0.1molAmount HCl/amount CO2 =2/1So amount HCl = 2* amount CO2 =2*0.1 = 0.2molMass HCl = amount * molar mass = 0.2 * 36.5= 7.3gCalculating percentage yieldMass of lead ethanoate = 3.78gMass of lead sulfate precipitated from solution = 2.80g(CH3COO)2Pb.3H2O(aq) + H2SO4(aq) PbSO4(s) + 2CH3COOH + 3H2O(l)amount of (CH3COO)2Pb.3H2O = mass (CH3COO)2Pb.3H2O / molar mass of (CH3COO)2Pb.3H2Oamount of (CH3COO)2Pb.3H2O = 3.78g / 378 g/mol = 0.0100molChemical equation shows that 1 mol (CH3COO)2Pb.3H2O forms 1 mol PbSO4

so 0.0100 mol of (CH3COO)2Pb.3H2O forms 0.0100 mol PbSO4

So theoretical mass of PbSO4 = amount PbSO4 * molar mass PbSO4

= 0.0100 mol * 302 gmol-1 = 3.02g% yield of PbSO4 = actual mass of PbSO4 obtained*100 / theoretical mass of PbSO4

= 2.80g *100 / 3.02g = 92.7%Predicted massMass empty tube = 44.0gMass tube + NaHCO3 = 45.90gMass tube + solid product = 45.40gMass of NaHCO3 = 1.90gMass of solid product = 1.40g2NaHCO3 Na2CO3 +H2O + CO2Amount NaHCO3 = mass NaHCO3/molar mass NaHCO3 = 1.90g/84gmol-1 = 0.226molFrom equation: expected amount of Na2CO3 / Amount NaHCO3 = 1/2expected amount of Na2CO3 = amount NaHCO3 * ½ = 0.226/2mol = 0.113molExpected mass of Na2CO3 = amount NaCO3 * molar mass Na2CO3 = 0.113mol * 106gmol-1 = 1.20gAbove the expected mass is less than the actual mass so perhaps decomposition was not complete.Calculation for Standard solutionMass of sodium carbonate = 1.30g Na2CO3

Volume of solution = 250cm3

Amount of Na2CO3 = mass Na2CO3/ molar mass Na2CO3

= 1.30g/106gmol-1 = 0.0123molConcentration of Na2CO3 = amount Na2CO3/ vol Na2CO3 solution = 0.0123mol/(250/1000)dm3 = 0.0491moldm -3 Titration of HCl v sodium carbonateVol of sodium carbonate = 25.0cm3

Conc of sodium carbonate = 0.0500moldm-3

Volume of HCl used = 9.40cm3Amount of Na2CO3 = conc of Na2CO3 solution * vol Na2CO3 solution = 0.0500 moldm-3* (25.0/1000)dm3

= 0.00125mol2HCl(aq) + Na2CO3 2NaCl(aq) +H2O(l) + CO2(g)From equation amount HCl/amount Na2CO3 = 2/1so amount HCl =(2/1)* amount Na2CO3

= 2*0.00125mol = 0.00250molConc of HCl = amount HCl/vol HCl solution = 0.0025mol/(9.40/1000)dm3 = 0.266moldm -3 Conc of NaOH by titration against HClConc of acid = 0.100moldm-3Vol of acid = 24.90cm3Vol of alkali = 25.0cm3Amount of HCl = conc HCl * vol HCl = 0.100moldm-3 * (24.90/1000) dm3 = 0.00249molHCl + NaOH NaCl +H2OFrom equation amount of alkali/amount of HCl = 1/1So amount of NaOH = amount HCl = 0.00249molConc NaOH = amount NaOH/vol NaOH solution = 0.00249mol/(25.0/1000)dm3

= 0.0996moldm -3

1.3f The molar volume of a gasAvogadro's Law states that equal volumes of all gases under the same conditions (temperature and pressure) contain the same number of particles. One mole of any ideal gas occupies 22.4dm3 at stp (Standard temperature and pressure) and 24dm3 at rtp (Room temperature and pressure). This is the molar volume of a gas. amount of gas = volume of gas/molar volume

H2(g) + Cl2(g) ----> 2HCl(g) (At stp) 22.4dm3 of hydrogen reacts with 22.4dm3 of chlorine forms 44.8dm3 of hydrogen chloride 1 volume of hydrogen reacts with 1 volume of chlorine to form 2 volumes of hydrogen chloride 1 mole of hydrogen molecules reacts with 1 mole of chlorine molecules to form 2 moles of hydrogen chloride molecules.task 1.3f.1

Equations can be used to deduce reacting gas volumes: What volume of nitrogen dioxide can be formed when 16.4g of Calcium nitrate (molar mass 164g/mol) decomposes on heating? 2Ca(NO3)2(s)----> 2CaO(s) + 4NO2(g) + O2(g) amount of calcium nitrate = mass/molar mass = 16.4/164 = 0.100mol from equation amount of nitrogen dioxide/amount of calcium nitrate = 4/2

so amount of nitrogen dioxide = 0.200mol volume of nitrogen dioxide = amount *molar volume volume of nitrogen dioxide = 0.200mol*22.4dm3 mol-1 = 4.48dm3 .Task 1.3f.2 Answers1.(a) O2 20cm3, CO2 10cm3, H2O 20cm3.(b) O2 25cm3, CO2 20cm3, H2O 10cm3.(c) H2 30cm3, NH3 20cm3.

1.3g Percentage yields and atom economiesMany organic reactions produce side products and during purification some of the main product is normally lost.  The result is that the amount of product obtained, the yield, is less than the theoretical maximum yield.The theoretical yield is calculated using the balanced chemical equation.  E.g.C2H5OH + HBr ----> C2H5Br + H2O 1 mol of ethanol forms 1 mol of bromoethane so46.0g of ethanol forms 109g of bromoethane so if we start with 2.30g of ethanol1g of ethanol forms 109/46.0g of bromoethane2.30g of ethanol forms 2.30*109/46.0g = 5.45g of bromoethane sothe theoretical yield of bromoethane is 5.45g if we start with 2.30 g of ethanol.If only 4.00g are actually formed (actual yield = 4.00g) in the preparation thenpercentage yield = 100*actual yield/theoretical yield percentage yield = 100*4.00/5.45 = 73.4%Task 1.3g.1(b) What is the % yield if 20.2g ethanol is oxidised to 15.0g ethanoic acid.C2H5OH +[O] CH3COOH

amount of C2H5OH =mass /molar mass = 20.2/46 =0.439molAmount of CH3COOH = mass/molar mass = 35.0/60 = 0.583molSo only 0.439mol of ethanol reacts

% atom economy =

100% atom economy = all atoms in reactants turn into products. The atom economy can be low even if percentage yield is high. The atom economy idea is popular as high raw material costs and environmental concerns make waste an increasing problem.Task 1.3g.2 Calculate the percentage atom economy for each of the following organic products:(a) In the reaction shown below, 4.60g of ethanol gave 8.50g of bromoethane.  Find the % yield (78.0%) and % atom economy(41.4%) for the organic product. C2H5OH + KBr + H2SO4 ---> C2H5Br + KHSO4 + H2O(b) In a reaction to make ethyl ethanoate by the reactionC2H5OH + CH3COOH ---> CH3COOC2H5 + H2O20.2g of ethanol was heated under reflux with 35.0g of ethanoic acid.  15.0g of ethyl ethanoate was obtained.  Find the theoretical and percentage yield of ethyl ethanoate(83.0%) . Find the atom economy of the organic product.(c) What is the theoretical yield of cyclohexene when dehydrating 10g of cyclohexanol by heating it with phosphoric acid?  What is the percentage yield is the actual yield of cyclohexene is 7.1g? C6H11OH  --> C6H10 + H2O(d) A synthesis of 1-bromobutane produced 6.5g of product from 6.0g of butan-1-ol using excess sodium bromide and conc. sulfuric acid.  Calculate the theoretical and % yields of the product.

1.3h Avogadro Constant

Experiment 1.3c.2Describe the amounts of substance in each of 4 bottles. Consider the masses and the number of atoms. (bottles containing 12g of carbon, 32g sulphur, 20 black model atoms, 20 yellow model atoms)

amount of substance = number of particles/Avogadro constantChemist cannot find numbers of particles easily so mass is used.  The mass of a substance which contains the Avogadro number of particles is called the molar mass. The Avogadro number is about 6*1023mol-1

1. Calculate the number of atoms in (a) 1mol of carbon, C.(b) 0.5mol of copper, Cu.(c) 7.20g of sulfur, S

2. Calculate the number of molecules in(a) 1.00g of ammonia, NH3.(b) 3.28g of sulfur dioxide, SO2.(c) 7.83g of hydrogen chloride, HCl.3. Calculate the number of ions in(a) 0.500mol of sodium chloride, NaCl [Na+,Cl-](b) 14.6g of sodium chloride, NaCl.(c) 18.5g of calcium chloride, CaCl2.4. Calculate the amount of hydrogen molecules, H2 in 3*1023 molecules if the Avogadro number is 6*1023.5. Calculate the Avogadro number if 1.5*1023 mol of helium atoms He has a mass of 1g.

 

Answers

 

 

1(a) 9.03*1022, (b) 1.71*1022, (c) 1.35*1023.2(a) 3.54*1022  , (b) 3.08*1022  (c) 1.29*1023  .3(a)6.02*1023. (b)3.01*1023 ,  (c) 3.01*1023 .4. 0.5mol5. 6*1023

1.3i Formulae and equations by experiment

elements reacting magnesium chlorinesymbols of elements Mg Cl

masses reacting (from experiment)

2.4g 7.1g

molar mass (look up relative atomic    

mass in periodic table)

24gmol-1   35.55gmol-1  

amounts (amount = mass/molar mass

2.4g/24gmol-1   =0.1mol  

7.1g/35.5gmol-1  0.2mol

ratio of atoms (divide by smallest)

1 2

formula MgCl2

Task 1.3i.1mgburnpic.JPG      mgburnvid.3gp1.What is the formula of magnesium oxide given experimental results belowMass an empty crucible = 10.00gMass of crucible + magnesium = 11.20gMass of heated crucible + magnesium oxide = 12.00gMass of magnesium =Mass of magnesium oxide =Mass of oxygen reacting with magnesium =

2.Work out formulae of compounds formed when the following react:(a) 56g of iron and 32g of sulphur (Fe =56, S =32)(b) 2g of hydrogen and 16g of oxygen (H=1, O=16)(c) 14g of lithium and 16g of oxygen (Li=7)(d) 32g of copper and 8g of oxygen (Cu=64)(e) 6.4g of copper and 0.8g of oxygen.

Lithium and water react to form lithium hydroxide and hydrogen. The expected reaction might be:2Li + 2H20 2LiOH + H2

This can be checked by experiment.Draw apparatus to mix lithium and water and measure the volume of hydrogen gas formed.

If the molar mass of Li is 7.0gmol-1 and the 0.35g Li reacts with water to form 600cm3 of hydrogen at 25oC is the equation correct? Assume molar vol =24000cm3 at 25oC.Amount of Li = mass Li/molar mass Li =0.35g/7.0gmol-1 =0.05molAmount of H2 = vol H2/molar vol = 600cm3/24000cm3mol-1 = 0.025molSo by experiment 0.05 mol Li forms 0.025 mol H2

So by experiment 1 mol Li forms 0.5 mol H2

So by experiment 2 mol Li forms 1 mol H2

The equations shows 2 mol Li forms 1 mol H2

So the equation must be correct.

Titanium can be made from titanium chloride using sodium. Experiment shows that 1.92 g of Na is needed to make 1g of Ti.Which of the following equations must be correct?TiCl4 + 4Na Ti + 4NaClTiCl3 + 3Na Ti + 3NaCl

1.3j Making salts and percentage yieldsMethods include:

1. metal + acid salt + hydrogen2. metal oxide +acid salt + water3. metal hydroxide + acid salt + water4. metal carbonate + acid salt +water + carbon dioxide5. soluble salt1 + soluble salt2 insoluble salt + soluble salt3

Exp 1.3j making sodium chloride using method 41dm3 1.00M HCl, anhydrous sodium carbonate, balances, matches

1. Add solid anhydrous sodium carbonate, with stirring, to 25.0cm3 of 1.00M HCl until the effervescence ends.

2. Weigh an evaporating basin.3. Filter the mixture and collect the filtrate in the evaporating basin.4. Evaporate the solution to dryness.5. Reweigh the evaporating basin.

Results:Mass of basin =Mass of basin + salt =Analysis and conclusionActual mass of sodium chloride formed (in basin) =Amount of HCl used in experiment = vol HCl * concentration HCl =Na2CO3 +2HCl 2NaCl + H2O + CO2

Amount NaCl formed theoretically/amount HCl = / Amount of NaCl formed theoretically =Mass of NaCl formed theoretically = amount of NaCl * molar mass NaClPercentage yield NaCl = actual mass of NaCl formed *100/theoretical mass NaCl =Sample results: mass of basin =50.00g, mass of basin + salt = 51.20g.1.3k Simple test tube reactionsMethod: Pour about 5cm3 of dilute sulphuric acid into a test tube and add a small piece of magnesium ribbon. Use a lighted splint to identify the gas given off.

Observations: Effervescence, the gas burns with a squeaky pop.Inferences: The gas evolved is hydrogen.Magnesium + sulphuric acid magnesium sulphate + hydrogenMg(s) + H2SO4(aq) MgSO4(aq) + H2(g)Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)

Method: Pour about 5cm3 of dilute hydrochloric acid into a test tube and add an approximately equal volume of dil. sodium hydroxide.Observations: No observable change. Test tube warms slightly.Inferences: Reaction exothermic. Acid base reaction.Hydrochloric acid + sodium hydroxide sodium chloride + waterHCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)H+(aq) + OH-(aq) H2O(l)

Method: Put half a spatula measure of copper (II) oxide into a boiling tube and add dilute sulphuric acid so that the tube is about 1/3 full. Carefully warm the mixture.Observations: The copper (II) oxide dissolves on heating to form a blue solution.Inferences: Acid base reaction. Neutralisation. Blue solution is aqueous copper sufate.Copper (II) oxide + sulfuric acid copper (II) sulfate + waterCuO(s) + H2SO4(aq) CuSO4(aq) + H2O(l)CuO(s) + 2H+(aq) Cu2+(aq) + H2O(l)

Method: Mix equal volumes of lead (II) ethanoate solution and potassium iodide solution in a boiling tube. Filter the mixture.Observations: A yellow precipitate forms. The filtrate is a colourless solution.Inferences: A precipitation reaction. The precipitate is insoluble lead iodide. The filtrate contains aqueous potassium iodide.Lead(II) ethanoate + potassium iodide lead iodide + potassium ethanoate (CH3COO)2 Pb (aq) + 2KI(aq) PbI2(s) +2CH3COOK(aq)Pb2+(aq) + 2I-(aq) PbI2(s)

Method: To about 5 cm3 of dilute hydrogen peroxide in a boiling tube add approximately half a spatula of manganese (IV) peroxide. This black

solid acts as a catalyst in this reaction. Identify the gas given off using a glowing splint.Observations: Vigorous effervescence. Gas relights glowing splint.Inferences: Gas is oxygen. Hydrogen peroxide water + oxygen2H2O2(aq) 2H2O(l) + O2(g)

Method: Pour about 5cm3 of copper sulphate solution into a boiling tube. Add a spatula of zinc powder, stir thoroughly and then leave to stand. Note any colour changes in both the solution and solid.Observations: Blue colour of solution changes to colourless. Red solid appears on surface of zinc.Inferences: Displacement reaction. Copper is the red solid displaced from solution. Colourless zinc sulfate replaces blue copper sulfate in solution.Zinc + copper sulfate copper + zinc sulfateZn(s) + CuSO4(aq) Cu(s) + ZnSO4(aq)Zn(s) + Cu2+(aq) Cu(s) + Zn2+ (aq)

Revision

Expressions (write as is or rearranged)Amount = mass/molar massAmount = number of particles/Avogadro numberConcentration of solution = amount/volume of solutionAmount = volume of gas /Molar volume of gasAtom economy = 100*molar mass of product/sum of molar masses of reactantsYield = 100*actual mass of product/maximum theoretical mass of product

Empirical formulae

1. A compound is found to contain the following % by mass: 69.58% Ba, 6.090% C, 24.32% O. What is the simplest (i.e. empirical) formula?

2. A compound consists of 29.1%Na, 40.5% S, and30.4%O. Determine the simplest formula.

3. Determine the simplest formula if a compound is composed of 7.20 g carbon, 1.20 g hydrogen, and 9.60 g oxygen.

4. Combustion analysis gives the following:26.7% C, 2.2% hydrogen, 71.1% oxygen. If the molecular mass of the compound is 90 g/mol, determine its molecular formula.

5. A compound’s empirical formula is CH, and it weighs 104g/mol. Give the molecular formula.

6 . A substance is decomposed and found to consist of 53.2% C, 11.2% H, and 35.6% O by mass. Calculate the empirical formula

2HCl(aq) + Na2CO3(s) 2NaCl(aq) + H2O(l) + CO2(g)

Mass of HCl = 3.65gMolar mass HCl =?Amount of HCl =?Vol HCl(aq) = 0.5dm3

Conc HCl(aq) =?

tingMass reacNa2CO3=5.3gMolar mass Na2CO3= ?Amount of Na2CO3= ?Show that the equation is correct by showing thatAmount HCl/amount Na2CO3 =2/1

2HCl(aq) + Na2CO3(s) 2NaCl(aq) + H2O(l) + CO2(g)

Find x and y in the following expressionsAmount NaCl/amount HCl =x/yAmount of Na2CO3/amount H2O =x/yAmount CO2/amount HCl =x/yAmount HCl =0.100molAmount NaCl =?Amount of Na2CO3=?Amount CO2=?Volume of CO2=?Molar volume of gas = 24dm3mol-1

Calculate the theoretical maximum masses of each product.Calculate the atom economy of each product.If 2.0g of CO2 form what is its percentage yield?

A 25.07 mL sample of vinegar is titrated with 37.31 mL of 0.5119 M NaOH. What is the molarity of the acetic acid in vinegar? (0.7978 M)CH3COOH + NaOH CH3COONa + H2O

Volumetric calculationshttp://www.docbrown.info/page06/Mtestsnotes/ExtraVolCalcs1.htm

Q1 A solution of sodium hydroxide contained 0.25 mol dm-3. Using phenolphthalein indicator, titration of 25.0 cm3 of this solution required 22.5 cm3 of a hydrochloric acid solution for complete neutralisation.

(a) write the equation for the titration reaction.NaOH + HCl NaCl + H2O

(b) what apparatus would you use to measure out (i) the sodium hydroxide solution? pipette(ii) the hydrochloric acid solution?burette

(c) what would you rinse your apparatus out with before doing the titration ?pipette NaOH, burette HCl

(d) what is the indicator colour change at the end-point?pink to colourless

(e) calculate the moles of sodium hydroxide neutralised.amount NaOH = conc NaOH * vol NaOH = 0.25*25/1000 = 0.0625 mol

(f) calculate the moles of hydrochloric acid neutralised.amount HCl = amount NaOH = 0.0625 mol

(g) calculate the concentration of the hydrochloric acid in mol/dm3 (molarity).conc HCl = amount HCl/vol HCl = 0.0625mol/22.5/1000 dm3 = 0.278 moldm-3

Q2 A solution made from pure barium hydroxide contained 2.74 g in exactly 100 cm3 of water. Using phenolphthalein indicator, titration of 20.0 cm3 of this solution required 18.7 cm3 of a hydrochloric acid solution for complete neutralisation. [atomic masses: Ba = 137, O = 16, H = 1)

(a) write the equation for the titration reaction.Ba(OH)2 + 2HCl BaCl2 + 2H20

(b) calculate the molarity of the barium hydroxide solution.amount Ba(OH)2 = 2.74g/171gmol-1 = 0.160 molconc Ba(OH)2 = 0.160/100/1000 = 1.6moldm-3

(c) calculate the moles of barium hydroxide neutralised.amount Ba(OH)2 neutralised = conc * vol = 1.6 * 20/1000 =0.032mol

(d) calculate the moles of hydrochloric acid neutralised.amount HCl = amount Ba(OH)2 *2/1 = 0.032*2/1 = 0.064mol

(e) calculate the molarity of the hydrochloric acid.

Conc HCl = amount HCl/vol HCl = 0.064/18.7/1000 = 3.42moldm-3