ps gr 11 session 11 ln

7/28/2019 PS Gr 11 Session 11 LN

1/8

GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME

PHYSICAL SCIENCE Grade 11 SESSION 11 (LEARNER NOTES)

Page 1 of8

MOLE CONCEPT, STOICHIOMETRIC CALCULATIONS

Learner Note: The mole concept is carried forward to calculations in the acid and base

section, as well as in the chemical equilibrium. It is important to know this section well.

Question 1: 5 minutes (Taken from DoE Nov 2007)

(Revise calculations of molecular mass. Remember chemical calculations needformulae, substitution and an answer with the correct unit.)

The compound NaHCO3 is commonly known as baking soda. A recipe requires 1,6 g ofbaking soda, mixed with other ingredients, to bake a cake.

1.1 Calculate the number of moles of NaHCO3 used to bake the cake. (3)1.2 How many atoms of oxygen are there in1,6 g baking soda? (4)

(7)Question 2: 10 minutes (Taken from MED Nov 2009)

(The sum of all the percentages equalsyes, 100. Therefore, in a 100 g of thesubstance the ratio will be the same. Learn the method the steps are alwaysrepeated. The elements are mostly given in the same order as they appear in theformula)

One of the active ingredients in vinegar is Ethanoic acid. Ethanoic acid has a molecular massof 60 g.mol

-1and the following percentage composition

39,9 % carbon6,7 % hydrogen53,4 % oxygen

2.1 Define the concept empirical formula (2)2.2 Determine the empirical formula of Ethanoic acid (5)

2.3 What is the molecular formula of Ethanoic acid? (3)(10)

Question 3: 10 minutes (Adapted from MED Nov 2009)

The contact process is given by the equation below.

SO2 (g) + O2 (g) SO3 (g)

3.1 Balance the chemical equation (2)

In an investigation 256 g SO2 reacts with 80 g O2 in a reaction vessel.3.2 Calculate the number of moles of each reactant present at the start of the reaction(5)3.3 Identify the limiting reagent in the reaction and justify your answer. (2)

(Limiting reagents are frequently asked the one that limits the reaction is the one thatwill be used up first. You must first work out the number of moles represented by thegiven masses of the reactants, then determine the limiting reagent by using the mollratio)

3.4 Calculate the mass of SO3 produced in the reaction (4)(13)


2/8



Page 2 of8

Question 4: 5 minutes

(It is a common mistake to interpret the dot as a multiply function. This is the waterthat is trapped in the crystal during crystallisation and must be ADDED to the mass ofthe ionic compound.)

4.1 Calculate the percentage water of crystallisation in CuSO4 . 5 H2O (4)

4.2 Calculate the concentration of a 250 ml solution of sodium hydroxide if 10 g of thesolute is dissolved. (4)

(8)

Question 1

1.1 M (NaHCO3)= 23 + 1 + 12 + 3(16)= 84 g.mol

-1

n =M

m

n =84

1,6

= 0,02 mol (rounded to 2 decimal places)

1.2 Each atom has three oxygen atoms, there is 0,02 mol of atoms1 mol = 6,023 x 1023 particlestherefore 0,02 mol x 3 atoms x 6,023 x 10

23 particles = 3,44 x 1022 oxygen

atoms

SECTION B: SOLUTIONS AND HINTS

Learner Note: Emphasize the correct use of formulae and layout. Use the correct

unit


3/8



Page 3 of8

Question 2

2.1 Empirical formula the simplest ratio of atoms in a molecule

Elements C H O

In 100g 39,9 g 6,7 g 53,4 g

Convert mass tomol

n =M

m 12

9,39

= 3,325 mol

1

7,6

= 6,7 mol

16

4,53

= 3,3375 mol

Divide bysmallest answer 325.3

325.3

= 1

325.3

7,6

= 2,01

325.3

3375.3

= 1

Ratio ofelements in theempiricalformula

1 2 1

Empirical formula CH2O

2.3 M (CH2O)= 12 + 2(1) + 16= 30 g.mol-1

Molecular mass is double empirical formula mass therefore molecular formula isC2H4O2

Question 3

(It is important to balance chemical equations before doing the calculations. The onlytime where the balancing coefficients are used will be in the ratio step.)

3.1 2 SO2 (g) + O2 (g) 2 SO3 (g)

3.2 n =M

m

n =64

256

n = 4 mol SO2

n =M

m

n =32

80

n = 2,5 mol O2

Learner Note: If the steps are followed, the questions become relatively simple.

Remember we are working out what the ratio of the elements are in the molecule

therefore when you determine the number of mol the element mass is used-


4/8



Page 4 of8

3.3 A ratio of 2 mol SO2 is needed for 1 mol O2 according to reaction. Therefore 2,5 molO2 needs 5 mol SO2 to react completely, the SO2 is therefore the limiting reagent

3.4 4 mol SO2 reacts and 4 mol SO3 is produced

n =M

m

4 =80

m

m = 320 g SO3 made

Question 4

4.1 M (CuSO4 . 5 H2O)= 63,5 + 32 + 4(16) + 5 (1 + 1 + 16)= 159,5 + 90= 249,5 g.mol-1

% water =5,249

90x 100

= 36 % water

4.2 c =MV

m

c =40x0,25

10

c = 1 mol.dm-3


5/8



Page 5 of8

The Mole

The mole is one of the base units of the SI and is the base unit of the amount of matter of

substance.A mole of any substance always has the same number of atoms.Definition: A mole of any substance is that amount of substance which contains as manyelementary particles as there are atoms in 12g of carbon-12.The amount of matter or substance is not the same as the mass of the substance. While onemole of a substance contains the same number of particles, their masses are not the same.The formula used to calculate the number of moles in a substance:

In symbols n =M

m

Avogadros Number

One mole of any substance is approximately equal to 6,02x1023 elementary particles.

This very large number is known as Avogadros number or Avogadros constant and has thesymbol NA or L.Therefore: Number of particles = Avogadros number x number of moles

In symbols Np = NA x n

Molar Volumes of Substances

Molar volume is the volume of one mole of a substance and can be measured in dm 3/mol.One mole of any gas occupies a volume of approximately 22,4dm

3at Standard Temperature

and Pressure (STP). Standard Temperature is 0oC (273K) and Standard Pressure is101,3kPa.Equal volumes of all gases at STP contain the same number of molecules.

Therefore: molar volume = volume of substanceNumber of moles of substance

n =3

dm22,4

V

Empirical formulasmallest ratio of atoms in a molecule

True Formula or molecular formulaactual ratio of atoms in a molecule eg. the Empir ical

Formulaof a substance is COH2 but its molecular formula is C2O2H4

Percentage Compositionshows percentage of each element in a compoundcompared toits molecular mass

SECTION C: ADDITIONAL CONTENT NOTES


6/8



Page 6 of8

Learner Note: Revise steps for each type of calculation and follow the prescribed method.

Attempt all questions and refer to notes when in doubt.

Question 1: 25 minutes

1.1 Calculate the relative formula mass of KCO3 (3)1.2 Calculate how many times a molecule of methanol (CH3OH) is heavier than a molecule

of water (5)1.3 Calculate the empirical formula of the substance with the following composition

45,3 % O; 43 % Na; 11,3 % C (5)1.4 How many potassium atoms are there in 2 g K2SO4 (5)

1.5 Fe + S FeS

Which of the two substances will be used up if 10 g Fe and 10 g S are mixed andheated (7)(25)

SECTION D: HOMEWORK


7/8



Page 7 of8

1.1 When H atoms approach each other, the valence electron of the two atoms attract the

nucleus of the other atom, these attractive forces are stronger than the repulsive

forces between the atoms. The protons and electrons of respective atoms attract theatoms to form the bond. The atoms move closer together and the potential energy

becomes negative. The atoms are most stable at the lowest value of potential energy

when the orbitals overlap and bonding occurs. The two hydrogen atoms each share

an electron during bonding; there is a net electrostatic force of attraction between the

atoms. H2 is formed.

When an H atom approach oxygen atom, the valence electron of the atoms attracts the

nucleus of the other atom, these attractive forces are stronger than the repulsive

forces between the atoms. The protons and electrons of respective atoms attract the

atoms to form the bond. The atoms move closer together and the potential energybecomes negative. The atoms are most stable at the lowest value of potential energy

when the orbitals overlap and bonding occurs. The hydrogen and oxygen atom

share an electron during bonding, there is a net electrostatic force of attraction

between the atoms. H2O is formed.

1.2 Both H atoms require an electron to fill the orbital and obtain the noble structurewhich is of lower energy. Its valence energy level is not filled. The H atoms sharean electron pair, there is a net electrostatic force of attraction, bonding occurs

1.3 The Helium atom is in the noble state, it has a filled last energy level, it is stable andrequires a large amount of energy to remove an electron. No bonding occurs.

1.4 A bond is a net electrostatic force between two atoms. Atoms bond to obtain a filledvalence orbital octet rule 8 electrons in valence orbital increases stability. Except

hydrogen which follows the rule of two its valence orbital can have a maximum of 2

electrons. When two atoms approach each other, the valence electrons of the twoatoms attract the positive nucleus of the other atom; these attractive forces are

stronger than the repulsive forces between the atoms. The protons and electrons of

respective atoms attract the atoms to form the bond when orbitals overlap to get a full

valence orbital. The atoms move closer together and the potential energy becomesnegative. The atoms are most stable at the lowest value of potential energy when the

orbitals overlap and bonding occurs. The two atoms each share electrons during

bonding, there is a net electrostatic force of attraction between the atoms.

1.5a. Different atoms, each with an unpaired valence electron can share these electrons toform a chemical bond

SECTION E: SOLUTIONS TO SESSION 10 HOMEWORK


8/8



The SSIP is supported by

Page 8 of 8

b. Different atoms with paired valence electrons called lone pairs of electrons,cannot share these four electrons and cannot form a chemical bond

c. Different atoms, with unpaired valence electrons can share these electrons andform a chemical bond for each electron pair shared (multiple bond formation)

d. Atoms with an incomplete complement of electrons in their valence shell canshare a lone pair of electrons from another atom to form a co-ordinate or dative

covalent bond Question 2

2.1 a.

b.

2.2 a.

b.

c.

2.3 a.

b. Lewis base

c. Lewis acid

C

He

F F

O H

H

NH H

H

H+

NH H

H

ps gr 11 session 11 ln

Documents