Relative masses of atoms and molecules

Cambridge IGCSE Chemistry 0620 · Core and Extended

Syllabus points

• Describe relative atomic mass, Aᵣ, as the average mass of the isotopes of an element compared with 1/12 of the mass of an atom of carbon-12.
• Define relative molecular mass, Mᵣ, as the sum of the relative atomic masses of the atoms in a molecule.
• State that relative formula mass, also Mᵣ, is the term used for an ionic compound.
• Calculate the relative molecular or formula mass of a compound from its formula.
• Extended only: calculate reacting masses in simple proportions.

Relative atomic mass

Relative atomic mass, Aᵣ, is the average mass of the isotopes of an element compared with 1/12 of the mass of an atom of carbon-12.

Two things follow from that definition and both are examined.

It is a ratio, so it has no units. Writing "Aᵣ = 24 g" is wrong; the answer is simply 24.

It is an average, so it need not be a whole number. Chlorine has Aᵣ 35.5 because a normal sample is about three quarters chlorine-35 and one quarter chlorine-37. Carbon-12 is the standard because it is defined as exactly 12, which is why almost every other value comes out close to a whole number.

The values used in 0620 come from the Periodic Table on the back of the paper, and these are the ones that appear most often:

Relative molecular mass and relative formula mass

Relative molecular mass, Mᵣ, is the sum of the relative atomic masses of all the atoms shown in the formula.

For an ionic compound there are no molecules, so the same number is called the relative formula mass. It is still written Mᵣ and worked out in exactly the same way. Sodium chloride has a relative formula mass, not a relative molecular mass.

Working out Mᵣ

Multiply each Aᵣ by how many of that atom the formula contains, then add. Subscripts multiply the symbol before them; brackets multiply everything inside.

The last two are where marks are lost, so take them slowly.

For (NH₄)₂SO₄, one NH₄ group is 14 + (4 × 1) = 18, and there are two of them, so 2 × 18 = 36. The sulfate group is 32 + (4 × 16) = 96. Adding, 36 + 96 = 132.

For Al₂(SO₄)₃, the aluminium is 2 × 27 = 54. One sulfate group is 32 + 64 = 96, and there are three, so 3 × 96 = 288. Adding, 54 + 288 = 342. The 3 outside the bracket multiplies the sulfur as well as the oxygen, and forgetting the sulfur is the classic slip.

Water of crystallisation is counted too. For hydrated copper(II) sulfate, CuSO₄·5H₂O, the anhydrous salt is 64 + 32 + 64 = 160 and the water is 5 × 18 = 90, giving 160 + 90 = 250.

Reacting masses in simple proportions (Extended)

This is the only Extended requirement in 3.2, and it does not need moles. The idea is that a balanced equation fixes the masses that react, so any other quantity follows by simple proportion.

The method is three steps every time:

1. Work out the Mᵣ of the substance you are given and of the substance you want, and multiply each by its balancing number.
2. Write down the two masses the equation says react.
3. Scale them to the mass in the question.

Worked example 1. What mass of magnesium oxide is made when 12 g of magnesium burns?

2Mg + O₂ → 2MgO

Aᵣ(Mg) = 24 and Mᵣ(MgO) = 24 + 16 = 40. The equation uses 2 Mg and makes 2 MgO, so 2 × 24 = 48 g of magnesium makes 2 × 40 = 80 g of magnesium oxide.

Scaling to 12 g of magnesium:

12 / 48 × 80 = 20 g of magnesium oxide

Worked example 2. What mass of calcium oxide is made by heating 25 g of calcium carbonate, and what mass of carbon dioxide is given off?

CaCO₃ → CaO + CO₂

Mᵣ(CaCO₃) = 40 + 12 + (3 × 16) = 100, Mᵣ(CaO) = 40 + 16 = 56 and Mᵣ(CO₂) = 12 + (2 × 16) = 44. So 100 g of calcium carbonate gives 56 g of calcium oxide and 44 g of carbon dioxide.

25 / 100 × 56 = 14 g of calcium oxide

25 / 100 × 44 = 11 g of carbon dioxide

Check the answer against conservation of mass: 14 + 11 = 25, which is the mass you started with. Every question of this type can be checked this way, and it catches an arithmetic slip in seconds.

Worked example 3, working backwards. What mass of iron(III) oxide is needed to make 112 g of iron?

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Mᵣ(Fe₂O₃) = (2 × 56) + (3 × 16) = 112 + 48 = 160, and the equation makes 2 × 56 = 112 g of iron from it. Since 112 g of iron is exactly what is wanted, the answer is 160 g, and for any other mass of iron you would scale in the same way.

Note that only substances joined by the equation can be compared. Masses do not carry across from one reaction to another.

Common mistakes

• Giving relative atomic mass or relative molecular mass a unit. Both are ratios.
• Calling the Mᵣ of an ionic compound a relative molecular mass.
• Ignoring the number outside a bracket, or applying it to only part of the group.
• Forgetting the water of crystallisation in a hydrated salt.
• Using the Mᵣ of one substance with the balancing number of another.
• Leaving out the balancing number altogether, so 2MgO is treated as 40 rather than 80.
• Adding masses of substances that are not in the same equation.