IGCSE Grade (10)

Topic 4
Stoichiometry

• Chemical formula

• Chemical equation

• Chemical calculations

• The mole

• Empirical formula and molecular

formula
IGCSE Grade (10)

Paper 2
1. Which compound has the largest relative molecular mass, Mr? J. 02 (9)
A. CO2
B. NO2 D
C. SiO2
D. SO2

2. What is the formula of copper(II) oxide and of sulfur hexafluoride? J. 02 (10)

3. The relative atomic mass of oxygen is 16 and that of hydrogen is 1. J. 03 (9)

This means that …(i)… of oxygen has the same mass as …(ii)… of hydrogen.

Which words correctly complete the gaps? B

4. Water is formed when 48 g of oxygen combine with 6 g of hydrogen. J. 03 (11)

What mass of oxygen combines with 2 g of hydrogen? B

A. 12 g B. 16 g C. 96 g D. 144 g
IGCSE Grade (10)
5. Two gases react as shown. N. 03 (10)

B
When measured at the same temperature and pressure, what is the value of

A. 1/2
B. 1
C. 2
D. 4

6. Carbon and chlorine form a chloride. N. 03 (11)

N. 08 (11)
What is the formula of this chloride?

A. CCl2 B
B. CCl4
C. CaCl2
D. CaCl4

7. The compound ethyl mercaptan, C2H5SH, has a very unpleasant smell. J. 04 (10)

What is its relative molecular mass?

D
A. 34 B. 50 C. 61 D. 62

8. When propane is burned, carbon dioxide and water are formed, as shown. N. 04 (9)

C3H8 + 5O2 r CO2 + s H2O

C
Which values of r and s balance the equation?
IGCSE Grade (10)
9. For which compound is the formula correct? J 05 (10)

10. The equation shows the reaction that occurs when ethanol burns in air. N. 05 (9)

C2H5OH + x O2 → y CO2 + z H2O

Which values of x, y and z are needed to balance this equation?
D

11. The diagrams show the molecules of three elements. J. 06 (9)

Which of these elements are present in water?

A. 1 and 2 only

B. 1 and 3 only

C. 2 and 3 only

D. 1, 2 and 3
IGCSE Grade (10)
12. Magnesium and sulfur each form a chloride. N 06 (9)

What could be the formulae of these chlorides?

D

13. A gas has the molecular formula NOCl. N. 06 (10)

Which diagram could show molecules of the pure gas NOCl ?

C
IGCSE Grade (10)
14. Boron, B, forms an oxide. J. 07 (10)

Which equation is correctly balanced? D

A. 2B + 3O2 → B2O3
B. 2B + 3O2 → 2B2O3
C. 4B + 2O2 → 2B2O3
D. 4B + 3O2 → 2B2O3

15. For which compound is the formula correct? J. 08 (10)

16. J. 08 (18)
When written as formulae, which compound has the greatest number of oxygen atoms?

A. calcium oxide
C
B. copper(II) oxide

C. iron(III) oxide

D. potassium oxide

17. Nitrogen and hydrogen react together to form ammonia. J. 09 (10)

N2 + 3H2 → 2NH3

When completely converted, 7 tonnes of nitrogen gives 8.5 tonnes of ammonia. C

How much nitrogen will be needed to produce 34 tonnes of ammonia?

A 7 tonnes B 8.5 tonnes C 28 tonnes D 34 tonnes

 IGCSE Grade (10)
18. Which relative molecular mass, Mr, is not correct for the molecule given? J. 09 (11)

19. Hydrogen and chlorine react as shown.

1 molecule of hydrogen + 1 molecule of chlorine → 2 molecules of hydrogen chloride

What is the equation for this reaction? C

A. 2H + 2Cl → 2HCl

B. 2H + 2Cl → H2Cl 2

C. H2 + Cl 2 → 2HCl

D. H2 + Cl 2 → H2Cl 2

J. 2016 p 21 (8-9)
20
D

21

C
IGCSE Grade (10)
J. 2016 p 22 (8-9)
22
B

23
D

J. 2016 p 23 (8-9)
24

25

26. J. 2017 (7)

B
IGCSE Grade (10)
27. J. 2017 (8)

28. J. 2018 (8)

29. J. 2019 (7)

30. J. 2019 (8)

D
IGCSE Grade (10)

31. J. 2021 p 21 (9)

2.56 g of a metal oxide, MO2, is reduced to 1.92 g of the metal, M.
What is the relative atomic mass of M? B

A 48 B 96 C 128 D 192
IGCSE Grade (10)

Paper 4
1. N. 01 (2. c)
(c) Potassium chlorate, which has a formula of the type, KClOn, decomposes to form
oxygen. 2.45 g of the chlorate produced 1.49 g of potassium chloride and 0.72dm3 of
oxygen at r.t.p. Find the value of n.

KClOn KCl + O2

Mass of one mole of KCl = 74.5 g

Number of moles of KCl formed = 1.49 / 74.5 = 0.02 moles

Number of moles of oxygen molecules formed = 0.72 / 24 = 0.03 moles

Number of moles of oxygen atoms = 0.06 moles

Mole ratio KCl : O is .0.02 : 0.06 = 1: 3

n= 3
[4]

2. N. 01 (3. a)
Propane is an alkane. It has the structural formula:

(a) The equation for the complete combustion of propane is given below. Insert the two
missing volumes.

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Volume of gas/cm3 5 25 15
IGCSE Grade (10)
3. J. 02 (5. c, d)
3 3
(c) A 20 cm sample of butyne, C4H6, is burnt in 150 cm of oxygen. This is an excess of
oxygen.

2C4H6(g) + 11O2(g) 8CO2(g) + 6H2O(l)

(i) What volume of oxygen reacts?

0.02 x 11 = 0.11 dm3 ....[1]

2

(ii) What volume of carbon dioxide is produced?

0.02 x 8 = 0.8 dm3 ....[1]

2

(iii) What is the total volume of gases left at the end of the reaction?

[0.15 – 0.11 ] + 0.08 = 0.12 dm3 .....................................[1]

(d) Calculate the mass of water formed when 9.0 g of butyne is burnt. The mass of one
mole of butyne is 54 g.

from the above equation, 1 mole of butyne forms 3 moles of water

number of moles of butyne reacted 9 = 0.16 moles

54

number of moles of water formed 0.016 x 3 = 0.05 moles

mass of water formed . 0.5 x 18 = 9 g [3]

4. N. 02 (1. c)
(c) The results of an investigation into the action of heat on copper(II) sulphate-5-water, a
blue crystalline solid, are given below.

The formula is CuSO4.5H2O and the mass of one mole is 250 g

A 5.0 g sample of the blue crystals is heated to form 3.2 g of a white powder. With
further heating this decomposes into a black powder and sulfur trioxide.

(i) Name the white powder.

Anhydrous copper sulfate .........................................[1]

IGCSE Grade (10)

(ii) What is observed when water is added to the white powder?

Turns blue .......................................................................................[1]

(iii)Name the black powder.

Copper oxide ...................................................................................[1]

(iv) Calculate the mass of the black powder. Show your working.

CuSO4 CuO + SO3

3.2 / 160 = 0.02 0.02

0.02 x 80 = 1.6 g ...................................................[3]

5. N. 02 (3. f)
(f) Sodium reacts with sulfur to form sodium sulfide.

2Na + S Na2S

An 11.5 g sample of sodium is reacted with 10 g of sulfur. All of the sodium reacted but
there was an excess of sulfur.

Calculate the mass of sulfur left unreacted.

(i) Number of moles of sodium atoms reacted = 11.5 / 23 = 0.5 moles

[2 moles of Na react with 1 mole of S]

(ii) Number of moles of sulfur atoms that reacted = 0.5 / 2 = 0.25 moles

(iii) Mass of sulfur reacted = 0.25 x 32 = 8 ...g

(iv) Mass of sulfur left unreacted = .10 – 8 = 2.g [4]

IGCSE Grade (10)
6. J. 03 (2. c)
(c) Each tablet contains the same number of moles of CaCO3 and MgCO3. One tablet
reacted with excess hydrochloric acid to produce 0.24 dm3 of carbon dioxide at r.t.p.

CaCO3 + 2HC l CaCl2 + CO2 + H2O

MgCO3 + 2HCl MgCl2 + CO2 + H2O

(i) Calculate how many moles of CaCO3 there are in one tablet.

number of moles CO2 = .0.24 / 24 = 0.01 moles

number of moles of CaCO3 and MgCO3 = 0.01 moles

number of moles of CaCO3 = 0.01 / 2 = 0.005 moles

[3]
(ii) Calculate the volume of hydrochloric acid, 1.0 mol /dm3, needed to react with one
tablet.

number of moles of CaCO3 and MgCO3 in one tablet = 0.01 moles

Use your answer to (c)(i).

number of moles of HCl needed to react with one tablet = 0.01 x 2 = 0.02 moles

volume of hydrochloric acid, 1.0 mol /dm3, needed to

react with one tablet = .0.02 / 1 = 0.02 dm3

[2]

7. N. 03 (5. d)
(d) Sulfur dioxide reacts with chlorine in an addition reaction to form sulfuryl chloride.

SO2 + Cl2 SO2Cl2

8.0 g of sulfur dioxide was mixed with 14.2 g of chlorine. The mass of one mole of
SO2Cl2 is 135 g.

Calculate the mass of sulfuryl chloride formed by this mixture.

Calculate the number of moles of SO2 in the mixture = 8 / 32 + 2 x 16 = 0.125 moles

Calculate the number of moles of Cl2 in the mixture = 14.2 / 2 x 35.5 = 0.2 moles
IGCSE Grade (10)
Which reagent was not in excess? SO2

How many moles of SO2Cl2 were formed = 0.125 moles

Calculate the mass of sulfuryl chloride formed = 0.125 x 135 = 16.9. g [5]

8. J. 04 (3. a)
An organic compound decomposes to form nitrogen.

C6H5N2Cl(aq) → C6H5Cl(l) + N2(g)

(a) Explain the state symbols.

Aq Aqueous ( dissolved in water )

l Liquid

g Gas [2]

9. J. 04 (7)
Chemists use the concept of the mole to calculate the amounts of chemicals involved in a
reaction.

(a) Define mole.

Mole is the amount of substance that contains Avogadro’s number of particles… [1]

(b) 3.0 g of magnesium was added to 12.0 g of ethanoic acid.

Mg + 2CH3COOH → (CH3COO)2Mg + H2

The mass of one mole of Mg is 24 g.

The mass of one mole of CH3COOH is 60 g.

(i) Which one, magnesium or ethanoic acid, is in excess? You must show your
reasoning.

Mg + 2CH3COOH → (CH3COO)2Mg + H2
3 / 24 = 0.125 12 / 60 = 0.2
.
Mg is in excess as each 0.2 moles of acid needs only 0.1 moles of Mg
.[3]
IGCSE Grade (10)
(ii) How many moles of hydrogen were formed?

0.1
moles……………………………………………………………………………..[1]

(iii) Calculate the volume of hydrogen formed, measured at r.t.p

0. 1 x 24 = 2.4 dm3

…[2]

(c) In an experiment, 25.0cm3 of aqueous sodium hydroxide, 0.4mol /dm3 was neutralised
by 20.0cm3 of aqueous oxalic acid, H2C2O4.

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

Calculate the concentration of the oxalic acid in mol /dm3

(i) Calculate the number of moles of NaOH in 25.0 cm3 of 0.4 mol /dm3 solution.

0.025 x 0.4 = 0.01 moles ……… [1]

(ii) Use your answer to (i) and the mole ratio in the equation to find out the number of
moles of H2C2O4 in 20 cm3 of solution.

0.01 / 2 = 0.005 moles .. [1]

(ii) Calculate the concentration, mol /dm3, of the aqueous oxalic acid.

0.005 / 0.02 = 0.25 mol/dm3

[2]
IGCSE Grade (10)
10. N. 04 (7.
c)
(c) Iron(III) sulfate decomposes when heated. Calculate the mass of iron(III) oxide
formed and the volume of sulfur trioxide produced when 10.0 g of iron(III) sulfate
was heated.

Mass of one mole of Fe2(SO4)3 is 400 g.

Fe2(SO4)3 (s) Fe2O3 (s) + 3SO3 (g)

Number of moles of Fe2(SO4)3 = 10 / 40 = 0.025 moles

Number of moles of Fe2O3 formed = 0.025 moles

Mass of iron(III) oxide formed = 0.025 x [2 x 56 + 3 x 16] = 4 g

Number of moles of SO3 produced = 0.025 x 3 = 0.075 moles

Volume of sulfur trioxide at r.t.p. = 0.075 x 24 = 1.8 dm3 [5]

11. J. 05 (1. c, d)
(c) 0.015 moles of iodine react with 0.045 moles of chlorine to form 0.030 moles of a single
product. Complete the equation.

I2 + ……3….. Cl2 ………2ICl3…………….

[2]

(d) Traces of chlorine can be separated from bromine vapour by diffusion.

Which gas would diffuse the faster and why?

Cl2 as it has lower Mr / density

.[2

12. J. 05 (4. d)
(d) Gypsum is hydrated calcium sulfate, CaSO4.xH2O. It contains 20.9% water by mass.
Calculate x.
Mr: CaSO4, 136; H2O, 18.

79.1 g of CaSO4 = 79.1 / 136 = 0.58 moles

20.9 g of H2O = 20.9 / 18 = 1.16 moles

x = …2 [3]
IGCSE Grade (10)
13. N. 05 (6. a)
(a) The following method is used to make crystals of hydrated nickel sulfate.

An excess of nickel carbonate, 12.0 g, was added to 40 cm3 of sulfuric acid, 2.0
mol/dm3. The unreacted nickel carbonate was filtered off and the filtrate evaporated to
obtain the crystals.

NiCO3 + H2SO4 NiSO4 + CO2 + H2O

NiSO4 + 7H2O NiSO4.7H2O

Mass of one mole of NiSO4.7H2O = 281 g

Mass of one mole of NiCO3 = 119 g

(i) Calculate the mass of unreacted nickel carbonate.

Number of moles of H2SO4 in 40 cm3 of 2.0 mol/dm3 acid = 0.08

Number of moles of NiCO3 reacted = 0.08 moles

Mass of nickel carbonate reacted = 0.08 x 119 = 9.52. g

Mass of unreacted nickel carbonate = 12 – 9.52 = 2. 48 g [3]

(ii) The experiment produced 10.4 g of hydrated nickel sulfate. Calculate the
percentage yield.

The maximum number of moles of NiSO4 .7H2O that could be formed = 0.08 moles

The maximum mass of NiSO4 .7H2O that could be formed =0.08 x 281 = 22.48 .. g

The percentage yield = 10.4 / 22.48 x 100 = 46.3 .% [3]

IGCSE Grade (10)
14. J. 06 (7. d)
(d) Propene reacts with hydrogen iodide to form 2 - iodopropane.

CH3−CH=CH2 + HI CH3−CHI−CH3

1.4 g of propene produced 4.0 g of 2 - iodopropane.

Calculate the percentage yield.

moles of CH3–CH=CH2 reacted = 1.4 / 3 x 12 + 6 = 0.033 moles

maximum moles of CH3–CHI–CH3 that could be formed = 0.033 moles

mass of one mole of CH3–CHI–CH3 = 170 g

maximum mass of 2 - iodopropane that could be formed = 0.033 x 170 = 5.61 g

percentage yield = 4 / 5.61 x 100 = 71.3 % [4]

15. N. 06 (3. b)
(b) When calcium carbonate is heated strongly, it decomposes.

CaCO3 → CaO + CO2

(i) Calculate the relative formula mass of:

CaCO3 = 40 + 12 + 3 x 16 = 100

CaO …= 40 + 16 = 56 … [2]

(ii) 7.00 kg of calcium oxide was formed. What mass of calcium carbonate was heated?

100 → 56
→ 7

100 x 7 / 56 = 12.5 kg …[2]

IGCSE Grade (10)
16. N. 06 (6. a)
An ore of copper is the mineral, chalcopyrite. This is a mixed sulfide of iron and copper.

(a) Analysis of a sample of this ore shows that 13.80 g of the ore contained

4.80g of copper, 4.20 g of iron and the rest sulfur.

Complete the table and calculate the empirical formula of chalcopyrite.

4.80 4.20 13.8 – [4.8 + 4.2]

= 4.80
4.8 / 63.5 = 0.075 4.2 / 56 = 0.075 4.8 / 32 = 0.15
0.075 / 0.075 = 1 0.075 / 0.075 = 1 0.15 / 0.075 = 2

[3]

The empirical formula is……CuFeS2…………………………………… [1]

17. J. 07 (7. d)
(d) A better way of measuring the degree of unsaturation is to find the iodine number
of the unsaturated compound.

This is the mass of iodine that reacts with all the double bonds in 100 g of the fat.
Use the following information to calculate the number of double bonds in one
molecule of the fat.

Mass of one mole of the fat is 884 g.

One mole of I2 reacts with one mole

The iodine number of the fat is 86.2 g.

Complete the following calculation.

100 g of fat reacts with 86.2 g of iodine.

884 g of fat reacts with = 86.2 x 884 / 100 = 762…..….g of iodine.

One mole of fat reacts with……762 / 2 x 127 = 3………….. moles of iodine molecules.

Number of double bonds in one molecule of fat is………………3………………….. [3]

IGCSE Grade (10)
18. N. 07 (7. b)
(ii) One piece of marble, 0.3 g, was added to 5 cm3 of hydrochloric acid,
concentration 1.00 mol / dm3.

Which reagent is in excess? Give a reason for your choice.

mass of one mole of CaCO3 = 100 g

number of moles of CaCO3 = 0.3 / 40 + 12 + 3 x 16 = 0.003 moles

number of moles of HCl = 1.00 x 0.005 = 0.005 moles

reagent in excess is… CaCO3

reason 0.005 moles of acid reacts with only 0.0025 moles of CaCO3 [4]

(iii) Use your answer to (ii) to calculate the maximum volume of carbon dioxide
produced measured at r.t.p.

V = 24 x 0,005 / 2 = 0.06 dm3…. [1]

19. J. 08 (7. b)
(b) Using 25.0cm3 of aqueous sodium hydroxide, 2.24 mol / dm3, 3.95 g of
crystals were obtained. Calculate the percentage yield.

2NaOH + H2SO4 Na2SO4 + 2H2O

Na2SO4 + 10H2O Na2SO4.10H2O

Number of moles of NaOH used = 0.025 x 2.24 = 0.056 moles

Maximum number of moles of Na2SO4.10H2O that could be formed = 0.056 / 2 = 0.028 moles

Mass of one mole of Na2SO4.10H2O = 322 g

Maximum yield of sodium sulphate-10-water = 0.028 x 322 = 9.02 g

Percentage yield =…3.95 / 9.02 x 100 = 43.8 ..…. % [4]

IGCSE Grade (10)
20. N. 08 (3. c)
(c) (i) Calculate the mass of one mole of Fe2O3.2H2O.

5 x 16 + 2 x 56 + = 196 [1]

(ii) Use your answer to (i) to calculate the percentage of iron in rust.

% Fe = 2 x 56 x 100 / 196 = 57.1 %

…..[2]

21. N. 08 (4. b)
(b) Benzene contains 92.3% of carbon and its relative molecular mass is 78.

(i) What is the percentage of hydrogen in benzene?

100 – 92.3 = 7.7 % ….[1]

(ii) Calculate the ratio of moles of C atoms: moles of H atoms in benzene.

C : H = 92.3 / 12 : 7.7 / 1 = 7.7 : 7.7

1 : 1 …….[2]

(iii) Calculate its empirical formula and then its molecular formula.

The empirical formula of benzene is…………… CH …………………………………….

The molecular formula of benzene is…78 / 13 = 6

6 x CH = C6H6 …………………….. [2]

IGCSE Grade (10)
22. N. 08 (7. a)
alkanes are generally unreactive. Their reactions include combustion, substitution and
cracking.

(a) The complete combustion of an alkane gives carbon dioxide and water.

(i) 10 cm3 of butane is mixed with 100 cm3 of oxygen, which is an excess. The mixture
is ignited. What is the volume of unreacted oxygen left and what is the volume of
carbon dioxide formed?

C4H10(g) + 6.5 O2 (g) 4CO2(g) + 5H2O(l)

Volume of oxygen left = 100 – (10 x 6.5 / 1 ) = 35 cm3

Volume of carbon dioxide formed = 4 x 10 / 1 = 40 …..cm3 [2]

IGCSE Grade (10)
23. J. 09 (5. b)
(b) The formulae of insoluble compounds can be found by precipitation reactions.

To 12.0 cm3 of an aqueous solution of the nitrate of metal T was added 2.0 cm3
of aqueous sodium phosphate, Na3PO4. The concentration of both solutions was
1.0 mol / dm3. When the precipitate had settled, its height was measured.

The experiment was repeated using different volumes of the phosphate solution. The
results are shown on the following graph.

What is the formula of the phosphate of metal T ? Give your reasoning.

Metal nitrate : Sodium phosphate

Volume used 12 8
Number of moles 12 / 1 8/1
Mole ratio 3 2
Formula T3 (PO4)2 [3]
IGCSE Grade (10)
24. J. 09 (9)
Quantities of chemicals, expressed in moles, can be used to find the formula of
a compound, to establish an equation and to determine reacting masses.

(a) A compound contains 72% magnesium and 28% nitrogen. What is its empirical
formula?

Mg : N =
72 / 24 = 3 : 28 / 14 = 2
Mg3N2 ……[2]

(b) A compound contains only aluminium and carbon. 0.03 moles of this compound
reacted with excess water to form 0.12 moles of Al(OH)3 and 0.09 moles of CH4

Write a balanced equation for this reaction.

AL4C3 + 12 H2O 4 Al(OH)3 + 3 CH4

0.03 / 0.03 =1 . 0.12 / 0.03 = 4 0.09 / 0.03 = 3 ..[2]

(c) 0.07 moles of silicon reacts with 7.2 g of fluorine.

Si + 2F2 SiF4

(i) Which one is the limiting reagent? Explain your choice.

Si : F2 =

0.07 7.2 / 2 x 19 = 0.187

Silicon is limiting as 0.07 moles of Si reacts completely with 0.14 moles of Fluorine
[3]

(ii) How many moles of SiF4 are formed?

………………0.07 moles……………………………………………..……………[1]
IGCSE Grade (10)
25. N. 2011 P 31[7, C]

3.36/ 84= 0.04

2.21/62 = 0.03

2.21/40 = 0.05

2.21/106 = 0.02

Equation 3
Mole ratio 2:1 agrees with equation
IGCSE Grade (10)
26. J.2013 P 31 [7C]

27. J.2014 P 32 [7b]

0.025 X 2.48 = 0.062

0.062 / 2 = 0.031

0.031 X 128 = 3.97 g

2.2 / 3.97 X 100 = 55.4 %

IGCSE Grade (10)
28. J 2016 P 41 [2, a]

n = 3.4 / ( 23+14+3 x 16) = 3.4 / 85 = 0.04 mol

n = 0.04 / 2 = 0.02 mol

V = 0.02 X 24 = 0.48 dm3

29. N. 2016 P 41 [ 3, e]

C H
88.24/12 = 7.35 11.76 / 1 = 11.76
Divide by smallest number
7.35 / 7.35 = 1 11.76 / 7.35 = 1.6 all times 5
5 8
C5 H8

Relative molecular mass

IGCSE Grade (10)
30. J. 2017 P 41 [3, b]

n = 2.52 / 18 = 0.14

0.14

n = 4.92 - 2.52 / 120 = 0.02

0.02

0.02 0.14
Divide by smallest number
0.02/0.02 0.14/0.02
1 7
1: 7

MgSO4 .7H2O
IGCSE Grade (10)
31. J. 2018 P41 [4, c, d, e]

n = 0.06 / 24 = 0.0025

0.0025

0.095 / 0.0025 = 38

Fluorine

P O
1.68 / 31 = 0.054 (3.87 1.68) / 16 = 0.13
Divide by smallest number
0.054 / 0.045 = 1 0.13 / 0.054 = 2.5 All X 2
2 5

P2O5

P2O3 = 2 x 31 + 3 X 16 = 110
P4O6 = 2 X 110 = 220