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IGCSE Chemistry

Chemistry Calculations A

22 RAMs and Moles Relative Atomic Mass

Relative atomic mass is the average mass of the isotopes of an element

compared with one twelfth the mass of a carbon-12 atom. It is also called

RAM or Ar

RAM = average mass of an atom of the element (in Kg)

1/12 x mass of an atom of carbon-12 (in Kg)

Q1 The average mass of an atom of Magnesium is 4.035943 x 10-26 Kg. Given that

the mass of an atom of carbon-12 is 1.9926 x 10-26, find the relative atomic mass

of Magnesium.

Q2 The relative abundances of the isotopes of Tungsten are: 180W 0.12% 182W 26.50% 183W 14.31% 184W 30.64% 185W 28.43%

Find the relative atomic mass of Tungsten

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Relative Formula Mass (RFM of Mr)

Relative formula or relative molecular mass is the mass of a substance in relative

units.

Note: The term relative formula mass is used when calculating the mass of ionic compounds The term relative molecular mass is used when calculating the mass of covalent compounds To find the Mr of a compound, simply add the atomic masses.

e.g.: Mr of NaCl = 23 + 35.5 = 58.5

Q3 Find the Mr of the following compounds

a) Cr2(SO4)3

b) CuSO4.5H2O

c) CuCO3.Cu(OH)2

d) (NH4)2SO4.FeSO4.6H2O

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Moles (short form is mols)

Moles are the units used to measure the amount of a substance.

1 mole =

Amount of that substance that contains as many particles (atoms, ions or

molecules) as there are atoms in exactly 12g of carbon-12 = 6.02 x 1023

1 mole of sodium has 6.02x1023 sodium atoms.

1 mole of Na+ has 6.02x1023 sodium ions.

1 mole of CO2 has 6.02x1023 Carbon dioxide molecules.

1 mole of NaCl has 6.02x1023 Na+ ions and 6.02x1023 Cl- ions (A total of

1.204 x 1024 ions)

Q4 Calculate the no. of particles in:

a) 0.5 mols of CO2

b) 5 mols of FeSO4

c) 0.1 mols of O2

d) 6 mols of H2O

e) 10 mols of Fe2+

1 Mole 6.02 x 1023 Particles

6.02x1023 is called the Avogadros constant and

was named after the Italian scientist Amedeo

Avogadro in honour of his contributions to

Molecular theory

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Relationship between moles, mass and molar mass

Molar mass mass of 1 mole of a substance

Molar mass = Mr but the only difference between them is Mr has no units and molar

mass has units grams per mole (gmol-1)

Moles = Mass (grams)

Molar Mass (gmol-1)

Q5 Calculate the no. of moles (to 3 s.f. where necessary) in:

a) 50 grams of Copper II sulphate crystals, CuSO4.5H2O

b) 1 Kilogram of iron, Fe

c) 0.032 grams of Sulphur Dioxide, SO2

d) 50 milligrams of copper carbonate, CuCO3

e) 1 tonne of Magnesium, Mg

Q6 Calculate the mass in grams of:

a) 1 mol of Lead(II) Nitrate, Pb(NO3)2

b) 4.3 mols of methane, CH4

c) 0.24 mols of Na2CO3.10H2O

d) 5.25 mols of Al2O3

e) 4.1 mols of octacontane, C80H162

Using relative formula mass to find percentage composition

e.g.: Find the percentage by mass of copper in Copper (II) Oxide, CuO

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Q7 Find the percentage composition of Oxygen in each of the following compounds

a) CO(NH2)2

b) (NH4)2SO4

c) FeSO4.7H2O

d) Na2CO3.10H2O

e) CH3CH2COOH

Empirical formula and Molecular Formula

Empirical formula is the formula that shows the simplest ratio of elements in

a compound.

Molecular formula is the formula that shows the actual ratio of elements in a

compound.

Molecular Formula mass = Empirical formula mass x n

e.g.: An organic compound has the composition 38.7% Carbon, 9.70% Hydrogen and

51.6% Oxygen by mass. The relative formula mass of the compound is 62. Calculate

the empirical formula and the molecular formula of the compound.

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Q8 A gaseous compound of C and H contains 80% C by mass. Its molecular

mass is 30.2, Calculate its empirical formula and molecular formula.

Q9 The empirical formula of a compound is C3H6O. It has molar mass 58 gmol-1.

Calculate its molecular formula.

Q10 1.24g of phosphorus was burnt completely in oxygen to give 2.84 g of

phosphorus oxide. The molecular formula of the oxide is 284 gmol-1. Find its

empirical formula and molecular formula.

Concept of molar volume

Molar volume is the volume occupied by 1 mol of a gas

under room temperature and pressure conditions (r.t.p) conditions (25 C and

1atm pressure) , molar volume is

under standard temperature and pressure conditions (s.t.p) conditions (0 C

and 1atm pressure), molar volume is

Q11 Calculate the volumes of each of these in dm3 under r.t.p conditions

a) 5 mols of O2

b) 0.5 mols of CH4

c) 0.25 mols of water vapour, H2O

d) 50 grams of Nitrogen, N2

e) 100 grams of Fluorine, F2

Q12 Calculate the volumes of the same moles of the above gases in dm3 under s.t.p

conditions

Q13 Calculate the mass in grams of the following volumes of gases:

a) 24dm3 of Chlorine gas

b) 200cm3 of Oxygen gas

c) 1.42dm3 of water vapour

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23 Calculations from equations Consider the equation:

2CaCo3 (s) + 2SO2 (s) + O2 (g) 2CaSO4 (s) + 2CO2 (g)

Calculate the mass of CaSO4 (Calcium sulphate) formed from 100 grams of Calcium

Carbonate.

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Percentage Yield

% Yield = Actual mass of product x 100%

Expected mass of product

e.g.:

Heating 12.4 grams of copper (II) carbonate in a crucible produced 7.0 g of copper

(II) Oxide. What is the percentage yield of Copper (II) Oxide?

The equation for the above mentioned reaction is:

CuCO3 CuO + CO2

Questions (1), (2), (4), (6) and (11) from text book