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GCSE · Chemistry

C3a Moles, Formulae and Equations

AQA 8464 · Calculator allowed · about 80 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Counting the Uncountable: Atoms, Formulae and the Mole

Original text written for Revision Library.

At the start of the nineteenth century, the chemist John Dalton proposed that every element is made of identical atoms, and that these atoms combine in fixed, whole-number ratios to form compounds. This idea explained something chemists had already noticed: a given compound always contains the same elements in the same proportion by mass, wherever the sample comes from. Dalton went further and assigned each element a relative atomic mass, a number showing how heavy its atoms are compared with one another. A few years later, the Italian scientist Amedeo Avogadro suggested that equal volumes of gases, at the same temperature and pressure, contain equal numbers of particles. His name was later given to a huge number, 6.02 * 10^23, which chemists use to count out a mole of any substance: an amount that always contains this same number of particles, however light or heavy those particles are. Relative formula mass, conservation of mass and the mole are still the basic tools chemists use today, from balancing an equation on paper to working out how many tonnes of limestone a quarry must supply to produce a given mass of quicklime, or how much of each gas is needed to manufacture ammonia in an industrial reactor.

Key words and equations used throughout this worksheet. [Relative atomic masses, Ar: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, Al = 27, S = 32, Cl = 35.5, Ca = 40, Fe = 56]

(a)State what is meant by the term 'relative formula mass' (Mr).(1)

(b)State what is meant by the term 'empirical formula'.(1)

(c)State the law of conservation of mass.(1)

(d)State what is meant by 'one mole' of a substance.(1)

(e)State the equation that links the number of moles (n), the mass in grams (m) and the relative formula mass (Mr) of a substance.(1)

(Total for Question 1 is 5 marks)

For each question, identify the correct answer from the options given.

(a)Which equation below is correctly balanced?(1)

A) Mg(s) + O2(g) -> MgO(s)
B) 2Mg(s) + O2(g) -> 2MgO(s)
C) 2Mg(s) + 2O2(g) -> 2MgO(s)
D) Mg(s) + 2O2(g) -> 2MgO(s)

(b)Which statement correctly describes the law of conservation of mass?(1)

A) The total mass of the products is always greater than the total mass of the reactants, because energy is added during a reaction.
B) Atoms can be created during a chemical reaction, but they can never be destroyed.
C) No atoms are created or destroyed in a chemical reaction, so the total mass of the reactants equals the total mass of the products in a closed system.
D) Mass is only conserved in reactions that produce a precipitate.

(c)Glucose has the molecular formula C6H12O6. Which of the following is its empirical formula?(1)

A) C6H12O6
B) C3H6O3
C) CH2O
D) C2H4O2

(Total for Question 2 is 3 marks)

Complete each word equation.

(a)magnesium + oxygen -> ____________(1)

(b)calcium carbonate -> ____________ + ____________ (on strong heating)(1)

(Total for Question 3 is 2 marks)

Balance each equation by writing the correct number in front of each formula (leave a blank if the number is 1).

(a)__Mg(s) + __O2(g) -> __MgO(s)(1)

(b)__Na(s) + __Cl2(g) -> __NaCl(s)(1)

(c)__Fe(s) + __Cl2(g) -> __FeCl3(s)(1)

(Total for Question 4 is 3 marks)

These equations are harder to balance. Write the correct number in front of each formula (leave a blank if the number is 1).

(a)Complete combustion of propane: __C3H8(g) + __O2(g) -> __CO2(g) + __H2O(g)(2)

(b)Thermite reaction: __Al(s) + __Fe2O3(s) -> __Al2O3(s) + __Fe(s)(2)

(c)Complete combustion of ethane: __C2H6(g) + __O2(g) -> __CO2(g) + __H2O(g)(2)

(Total for Question 5 is 6 marks)

Calculate the relative formula mass (Mr) of each compound.

(a)Calculate the Mr of calcium carbonate, CaCO3.(1)

(b)Calculate the Mr of sodium hydroxide, NaOH.(1)

(Total for Question 6 is 2 marks)

Calculate the relative formula mass (Mr) of ammonium sulfate, (NH4)2SO4. Show your working.

(Total for Question 7 is 2 marks)

A student heats a small amount of copper carbonate in a sealed, closed flask.

Explain, in terms of atoms, why the total mass inside the flask stays the same during the reaction.

(Total for Question 8 is 2 marks)

2.4 g of magnesium ribbon is burned completely in a plentiful supply of oxygen: 2Mg(s) + O2(g) -> 2MgO(s)

Calculate the maximum theoretical mass of magnesium oxide that could be formed.

(Total for Question 9 is 3 marks)

In practice, the magnesium in Question 9 was heated in an open crucible rather than a sealed flask, and the measured mass of the solid product was only 3.7 g.

Suggest and explain why this measured mass is lower than the 4.0 g calculated in Question 9.

(Total for Question 10 is 2 marks)

Required practical: determining the empirical formula of magnesium oxide.
A student, Aaliyah, investigates the empirical formula of magnesium oxide. She:
1. Weighs an empty crucible and lid.
2. Adds a coiled piece of magnesium ribbon and weighs the crucible, lid and magnesium.
3. Heats the crucible strongly on a pipe-clay triangle and tripod, lifting the lid slightly at intervals, until the magnesium has completely reacted and the mass is constant.
4. Allows the crucible to cool, then weighs the crucible, lid and solid product.

Her results:
Mass of empty crucible and lid = 24.62 g
Mass of crucible, lid and magnesium = 24.86 g
Mass of crucible, lid and solid product (after heating) = 25.02 g

(a)Calculate the mass of magnesium used in the experiment.(1)

(b)Calculate the mass of oxygen that combined with the magnesium.(1)

(c)Use your answers to (a) and (b) to determine the empirical formula of magnesium oxide. Show your working.(3)

(d)Give one reason why the lid was lifted slightly at intervals during heating.(1)

(e)Explain why the lid was lifted only slightly, rather than removed completely.(1)

(f)Evaluate this method for determining the empirical formula of magnesium oxide. In your answer, refer to possible sources of error, their effect on the calculated formula, and how the method could be improved.(6)

(Total for Question 11 is 13 marks)

1.2 g of carbon combines completely with 3.2 g of oxygen to form a compound containing only carbon and oxygen.

Determine the empirical formula of this compound. Show your working.

(Total for Question 12 is 3 marks)

A compound contains only carbon, hydrogen and oxygen. A sample of the compound contains 2.4 g of carbon, 0.6 g of hydrogen and 3.2 g of oxygen.

Determine the empirical formula of this compound. Show your working.

(Total for Question 13 is 4 marks)

A compound, X, has the empirical formula CH2. Its relative molecular mass (Mr) is 84.

Determine the molecular formula of X.

(Total for Question 14 is 3 marks)

3.0 g of magnesium ribbon is burned completely in oxygen: 2Mg(s) + O2(g) -> 2MgO(s)

Calculate the mass of magnesium oxide produced.

(Total for Question 15 is 3 marks)

A quarry supplies calcium carbonate (limestone) to a lime kiln, where it is heated until it fully decomposes: CaCO3(s) -> CaO(s) + CO2(g)

Calculate the maximum mass of calcium oxide that could be produced from 500 kg of calcium carbonate.

(Total for Question 16 is 3 marks)

Aluminium reacts with oxygen to form aluminium oxide: 4Al(s) + 3O2(g) -> 2Al2O3(s)

Calculate the mass of aluminium oxide formed when 5.4 g of aluminium reacts completely with oxygen.

(Total for Question 17 is 3 marks)

Calcium oxide reacts with water to form calcium hydroxide: CaO(s) + H2O(l) -> Ca(OH)2(s)

Calculate the mass of calcium hydroxide produced when 5.6 g of calcium oxide reacts completely with water.

(Total for Question 18 is 3 marks)

This question is about the Avogadro constant, which is 6.02 * 10²³ per mole.

(a)State the value of the Avogadro constant, including its units.(1)

(b)Calculate the number of atoms in 0.50 mol of iron.(2)

(Total for Question 19 is 3 marks)

The Avogadro constant is 6.02 * 10²³ per mole. Mr of CO2 = 44.

Calculate the number of molecules in 8.8 g of carbon dioxide, CO2.

(Total for Question 20 is 3 marks)

In an industrial reactor, a chemical engineer, Priya, reacts a mixture of 2.8 g of nitrogen gas and 1.2 g of hydrogen gas: N2(g) + 3H2(g) -> 2NH3(g)

Determine which reactant is in excess, and calculate the maximum mass of ammonia that could be formed. Show all of your working.

(Total for Question 21 is 5 marks)

Mark scheme · C3a Moles, Formulae and Equations

Question 1

(a) B1 the sum of the relative atomic masses of all the atoms shown in the formula of a substance (oe)
(a) Answer: The relative formula mass (Mr) is the sum of the relative atomic masses of all the atoms shown in the formula of a substance.
(b) B1 the simplest whole-number ratio of the atoms of each element present in a compound (oe)
(b) Answer: The empirical formula is the simplest whole-number ratio of the atoms of each element in a compound.
(c) B1 no atoms are created or destroyed in a chemical reaction, so the total mass of the reactants equals the total mass of the products (oe)
(c) Answer: No atoms are created or destroyed in a chemical reaction, so the total mass of the reactants equals the total mass of the products.
(d) B1 the amount of a substance that contains 6.02 * 10²³ particles (atoms, molecules or ions) of that substance (oe); accept reference to the Avogadro constant
(d) Answer: One mole of a substance contains 6.02 * 10²³ particles (the Avogadro constant) of that substance.
(e) B1 moles = mass / Mr (n = m / Mr), oe
(e) Answer: moles = mass (g) / Mr

Question 2

(a) B1 B
(a) Answer: B
(b) B1 C
(b) Answer: C
(c) B1 C
(c) Answer: C

Question 3

(a) B1 magnesium oxide (cao)
(a) Answer: magnesium oxide
(b) B1 calcium oxide and carbon dioxide (both needed)
(b) Answer: calcium oxide + carbon dioxide

Question 4

(a) B1 2Mg(s) + O2(g) -> 2MgO(s), fully correctly balanced
(a) Answer: 2Mg(s) + O2(g) -> 2MgO(s)
(b) B1 2Na(s) + Cl2(g) -> 2NaCl(s), fully correctly balanced
(b) Answer: 2Na(s) + Cl2(g) -> 2NaCl(s)
(c) B1 2Fe(s) + 3Cl2(g) -> 2FeCl3(s), fully correctly balanced
(c) Answer: 2Fe(s) + 3Cl2(g) -> 2FeCl3(s)

Question 5

(a) M1 at least two of the four coefficients correctly balanced relative to each other, e.g. 3 before CO2 and 4 before H2O
(a) A1 C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g), fully correctly balanced
(a) Answer: C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)
(b) M1 correct ratio identified for at least one pair, e.g. 2 Al to 1 Al2O3, or 1 Fe2O3 to 2 Fe
(b) A1 2Al(s) + Fe2O3(s) -> Al2O3(s) + 2Fe(s), fully correctly balanced
(b) Answer: 2Al(s) + Fe2O3(s) -> Al2O3(s) + 2Fe(s)
(c) M1 correct ratio of C2H6 to CO2 (1:2) or to H2O (1:3) shown, even if the whole equation is not yet balanced with whole numbers
(c) A1 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g), fully correctly balanced using whole-number coefficients
(c) Answer: 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g)

Question 6

(a) B1 100 cao (40 + 12 + 3*16)
(a) Answer: Mr(CaCO3) = 100
(b) B1 40 cao (23 + 16 + 1)
(b) Answer: Mr(NaOH) = 40

Question 7

M1 correct working shown, e.g. 2*(14 + 4*1) + 32 + 4*16
A1 132 cao
Answer: Mr((NH4)2SO4) = 132

Question 8

B1 no atoms are created or destroyed in the reaction (oe)
B1 because the flask is sealed/closed, no atoms (mass) can enter or leave the system (oe)
Answer: No atoms are created or destroyed in the reaction, and because the flask is sealed, no atoms can enter or leave, so the total mass stays the same.

Question 9

M1 moles of Mg = 2.4 / 24 = 0.1 mol
M1 correct use of the 1:1 mole ratio between Mg and MgO
A1 4.0 g cao (accept awrt 4.0 g)
Answer: 4.0 g of magnesium oxide

Question 10

B1 some of the white magnesium oxide product (smoke/powder) escaped from the open crucible during heating (oe)
B1 because the crucible was not a closed system, some product mass was lost to the surroundings, so less product remained to be weighed than the theoretical 4.0 g (oe); ft from candidate's Question 9 answer
Answer: Some of the magnesium oxide escaped as white smoke/powder because the crucible was open (not a closed system), so less product was left to weigh than the theoretical mass.

Question 11

(a) B1 0.24 g cao (24.86 - 24.62)
(a) Answer: 0.24 g
(b) B1 0.16 g cao ((25.02 - 24.62) - mass of Mg from (a)); ft from candidate's (a)
(b) Answer: 0.16 g
(c) M1 moles Mg = 0.24 / 24 = 0.01 mol; ft from (a)
(c) M1 moles O = 0.16 / 16 = 0.01 mol; ft from (b)
(c) A1 MgO cao, from a correctly identified 1:1 ratio
(c) Answer: MgO
(d) B1 to let air/oxygen into the crucible so that all of the magnesium can react (oe)
(d) Answer: Lifting the lid lets air (oxygen) into the crucible so that all of the magnesium can react completely.
(e) B1 to minimise the amount of white magnesium oxide smoke/powder that escapes; if too much escaped, the measured mass (and calculated formula) would be inaccurate/too low (oe)
(e) Answer: Removing the lid completely would let more of the white magnesium oxide smoke escape, lowering the measured mass and making the calculated formula less accurate.
(f) L3 (5-6): A detailed and coherent evaluation is given, covering more than one source of error, each clearly linked to its effect on the calculated formula, together with at least one justified, practical improvement. The answer is well structured, using appropriate scientific terminology throughout.
(f) L2 (3-4): Some relevant sources of error and/or improvements are described, with a partial explanation of how they would affect the calculated formula. The answer has some structure.
(f) L1 (1-2): Simple, isolated statement(s) about a source of error or an improvement, with little or no explanation of the effect on the result. The answer may lack a logical structure.
(f) L0 (0): No relevant content.
(f) Answer: See levels/indicative content.

Question 12

M1 moles C = 1.2/12 = 0.1 mol and moles O = 3.2/16 = 0.2 mol
M1 correct simplest whole-number ratio shown, e.g. 0.1 : 0.2 = 1 : 2
A1 CO2 cao
Answer: CO2

Question 13

M1 moles C = 2.4/12 = 0.2 mol
M1 moles H = 0.6/1 = 0.6 mol and moles O = 3.2/16 = 0.2 mol
M1 ratio divided by the smallest value: 0.2 : 0.6 : 0.2 -> 1 : 3 : 1
A1 CH3O cao
Answer: CH3O

Question 14

M1 empirical formula mass of CH2 = 12 + 2*1 = 14
M1 84 / 14 = 6
A1 C6H12 cao
Answer: C6H12

Question 15

M1 moles Mg = 3.0/24 = 0.125 mol
M1 correct use of the 1:1 mole ratio between Mg and MgO
A1 5.0 g cao (accept awrt 5.0 g)
Answer: 5.0 g

Question 16

M1 Mr(CaCO3) = 100 and Mr(CaO) = 56 identified/used
M1 correct use of the 1:1 mole ratio scaled to 500 kg, e.g. 500 * (56/100)
A1 280 kg cao (accept awrt 280 kg)
Answer: 280 kg

Question 17

M1 moles Al = 5.4/27 = 0.2 mol
M1 correct use of the 4:2 (2:1) mole ratio between Al and Al2O3, e.g. moles Al2O3 = 0.2 * (2/4) = 0.1 mol
A1 10.2 g cao (accept awrt 10.2 g)
Answer: 10.2 g

Question 18

M1 moles CaO = 5.6/56 = 0.1 mol
M1 correct use of the 1:1 mole ratio between CaO and Ca(OH)2
A1 7.4 g cao (accept awrt 7.4 g); Mr(Ca(OH)2) = 74
Answer: 7.4 g

Question 19

(a) B1 6.02 * 10²³ per mole (mol^-1) cao
(a) Answer: 6.02 * 10²³ per mole (mol^-1)
(b) M1 correct method: number of atoms = moles * Avogadro constant, e.g. 0.50 * 6.02*10²³
(b) A1 3.01 * 10²³ (atoms) cao (accept awrt 3.01*10²³)
(b) Answer: 3.01 * 10²³ atoms

Question 20

M1 moles CO2 = 8.8/44 = 0.2 mol
M1 correct method: number of molecules = moles * Avogadro constant, e.g. 0.2 * 6.02*10²³
A1 1.20 * 10²³ (molecules) cao (accept awrt 1.20*10²³, e.g. 1.2*10²³)
Answer: 1.20 * 10²³ molecules

Question 21

M1 moles N2 = 2.8/28 = 0.1 mol
M1 moles H2 = 1.2/2 = 0.6 mol
M1 correct comparison shown: moles H2 needed to react with 0.1 mol N2 = 0.1*3 = 0.3 mol; since 0.6 mol H2 is available (more than the 0.3 mol needed), H2 is in excess and N2 is the limiting reactant
M1 correct use of the limiting reactant (N2) with the 1:2 mole ratio to NH3: moles NH3 = 0.1*2 = 0.2 mol; allow ecf if the candidate identifies the wrong limiting reactant but applies the mole-ratio method correctly and consistently from that point
A1 3.4 g cao (accept awrt 3.4 g); Mr(NH3) = 17
Answer: Nitrogen (N2) is the limiting reactant; hydrogen is in excess; maximum mass of ammonia formed = 3.4 g

Moles, Formulae and Equations - Worksheets, Questions and Revision