The relative atomic mass, A of an element is the number of times the average mass of one atom of an element is heavier than one twelfth the mass of one atom of carbon-12. The relative atomic mass is measured accurately using a mass spectrometer.

The relative molecular mass, M of an element or a compound is the number of times the average mass of one molecule of it is heavier than one-twelfth the mass of one atom of carbon-12.
It is the sum of the individual relative atomic masses of the constituent elements that make up the compound.

Example 1 : Calculate the relative molecular mass of Baking soda NaHCO₃(Na = 23, H = 1, C = 12, O = 16)

M = 23 + 1 + 12 + (16 × 3)
= 23 + 1 + 12 + 48
= 84g/mol

Example 2: Calculate the Relative atomic mass of copper tetraoxosulphate(VI) decahydrate CuSO₄.10H₂0
(Cu = 63.5, S = 32, O = 16, H = 1)

M = 63.5 + 32 + (16 × 3) + 10(1 ×2 +16)
= 63.5 + 32 + 48 + 180
= 323.5g/mol

Example 3: Calculate the relative molecular mass of Sodium trioxocarbonate(IV) decahydrate. (Na = 23, O = 16, H =1, C = 12)

Compound formula = Na₂CO₃.10H₂O
M = (23×2) + (12) + (16 × 3) + 10(1×2 + 16)
M = 46 + 12 + 48 + 180
M = 286g/mol

Example 4: Calculate the relative molecular mass of ethanoic acid(C = 12, H = 1, O = 16)

compound formula = CH₃COOH
M = 12 + (1 × 3) + 12 + 16 + 16 + 1
M = 60g/mol

Example 5: Calculate the relative atomic mass of iron(III) sulphate. (Fe = 56, S = 32, O = 16)

Compound formula = Fe₂(SO₄)₃
M = (56 ×2) + (32 ×3) + (16 × 12)
M = 400g/mol

The percentage by mass of an atom or molecule is the percentage composition of the atom or molecule in a given compound.
It can be solved using the formula, $$\frac{\text{mass of atom or molecule}} {\text{molar mass of compound}} × {100} $$

Example 1: Find the percentage by mass of Oxygen in Sodium bicarbonate decahydrate(Na = 23, C = 12, H = 1, O = 16)

Compound formula = Na₂CO₃.10H₂O
Number of atoms of oxygen = 13
mass of oxygen = 16 × 13 = 208g
Molar mass of compound;
M = (23×2) + (12) + (16 × 3) + 10(1×2 + 16)
M = 46 + 12 + 48 + 180
M = 286g/mol $$\text{percentage by mass} = \frac{208}{286} × {100} $$ $$\text{% by mass of oxygen} = \text{72.7%} $$

Example 2: Calculate the percentage by mass of nitrogen in calcium trioxonitrate (V) [Ca = 40, N = 14, O = 16] (JAMB)

compound formula = Ca(NO₃)₂
Atoms of nitrogen = 2
mass of nitrogen = 14 × 2 = 28g
molar mass of compound
M = 40 + 2(14 + 16 × 3)
M = 40 + 124
M = 164g/mol $$\text{% composition} = \frac{28}{164} × {100} $$ $$\text{% composition} = \text{17.07%} $$

Example 3: Calculate the percentage composition of oxygen in calcium trioxocarbonate(IV) [Ca=40, C=12, O=16]

compound formula = CaCO₃
Atoms of oxygen = 3
mass of oxygen = 3 × 16 = 48g
Molar mass of compound;
M = 40 + 12 + 16 × 3
M = 100g/mol $$\text{% composition} = \frac{48}{100} × {100} $$ $$\text{% composition} = \text{48%} $$

Example 4: A solution contains 20g of solute in 180g of solvent. If the solvent is water, what is the concentration of the solution in terms of mass by mass percentage

mass of solute = 20g
mass of solvent = 100g
Solution = solute + solvent $$\text{mass %} = \frac{20}{200} × {100} $$ $$\text{mass %} = \text{10%}$$

Example 5: The percentage of water of crystallization in ZnSO₄.7H₂O is

Water of crystallization is the amount of water present in a salt.
atoms of water = 7
mass of water present = 7(1×2+16)
mass of water of crystallization = 126g
Molar mass of compound;
M = 65 + 32 + (16 × 4) + 7(2 +16)
M = 287g $$\text{% composition}= \frac{126}{287} × {100} $$ $$\text{% composition} = \text{43.9% }$$