Relative Formula Mass and Percentage by Mass

Atoms are far too light to weigh in grams, so chemists compare their masses on a relative scale. The relative atomic mass (Ar) of an element is its average atomic mass compared with one twelfth of a carbon-12 atom. You read Ar values straight from the periodic table.

The relative formula mass (Mr) of a compound is simply the sum of the relative atomic masses of every atom in its formula.

Mr = sum of the Ar of all the atoms in the formula. It has no units.

The Ar values you need for the worked examples below are:

Element	H	C	N	O	Na	Mg	S	Cl	Ca	Fe	Cu
Ar	1	12	14	16	23	24	32	35.5	40	56	63.5

Worked example — find the Mr of water, H₂O:

$M_r=(2\times 1)+16=2+16=18$

Worked example — find the Mr of carbon dioxide, CO₂:

$M_r=12+(2\times 16)=12+32=44$

Getting the Mr right depends on counting atoms correctly. A subscript multiplies only the symbol directly in front of it. A subscript outside a bracket multiplies everything inside that bracket.

A bracket subscript applies to every atom inside the bracket. In Mg(NO₃)₂ the 2 multiplies both the N and the three O atoms.

Worked example — find the Mr of calcium hydroxide, Ca(OH)₂:

The 2 outside the bracket doubles both the O and the H inside it.

$M_r=40+2\times (16+1)=40+2\times 17=40+34=74$

Worked example — find the Mr of magnesium nitrate, Mg(NO₃)₂:

Each nitrate group is N + 3O = 14 + 48 = 62, and there are two of them.

$M_r=24+2\times (14+3\times 16)=24+2\times 62=24+124=148$

Set the calculation out in stages. Working out the mass of one bracket group first, then multiplying, avoids the most common counting slip.

In a balanced equation, the total relative formula mass of the reactants (counting the multipliers) equals the total relative formula mass of the products. This is conservation of mass expressed in numbers, and it is a quick way to check that an equation is balanced correctly.

Worked example — check the totals for the reaction of hydrogen with oxygen:

$2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$

Side	Species	Mr contribution
Reactants	2 × H₂	2 × (2 × 1) = 4
Reactants	1 × O₂	1 × (2 × 16) = 32
Products	2 × H₂O	2 × 18 = 36

Total reactants = 4 + 32 = 36. Total products = 36. The two totals match, confirming the equation balances and mass is conserved.

The percentage by mass of an element tells you what fraction of a compound's mass comes from that element. It matters in real chemistry: a fertiliser is judged partly on its percentage of nitrogen, and an ore on its percentage of metal.

$\%\text{ by mass}=\frac{A_r\times \text{number of atoms of the element}}{M_r\text{ of the compound}}\times 100$

Worked example — find the percentage by mass of calcium in calcium carbonate, CaCO₃.

First the Mr: $40+12+(3\times 16)=40+12+48=100$.

The mass due to calcium is one Ca = 40.

$\%\text{ Ca}=\frac{40}{100}\times 100=40\%$

Worked example — find the percentage by mass of iron in iron(III) oxide, Fe₂O₃.

Mr = $(2\times 56)+(3\times 16)=112+48=160$. Mass due to iron = 2 × 56 = 112.

$\%\text{ Fe}=\frac{112}{160}\times 100=70\%$

An ore containing 70% iron by mass is far richer than one containing 30%, which is exactly the kind of comparison this calculation supports.

Percentage-by-mass questions often compare two compounds, so being methodical pays off. Work out the Mr, identify the total mass contributed by the element you want, then apply the formula.

Worked example — which contains the higher percentage of nitrogen by mass: ammonium nitrate, NH₄NO₃, or ammonia, NH₃?

For NH₄NO₃, Mr = $14+(4\times 1)+14+(3\times 16)=14+4+14+48=80$. Mass of nitrogen = 2 × 14 = 28.

$\%\text{ N}=\frac{28}{80}\times 100=35\%$

For NH₃, Mr = $14+(3\times 1)=17$. Mass of nitrogen = 14.

$\%\text{ N}=\frac{14}{17}\times 100=82.4\%$

Ammonia has a higher percentage of nitrogen by mass (82.4% against 35%). This kind of comparison explains why chemists compare fertilisers on their nitrogen content per kilogram, not just by name.

1. Ignoring a bracket subscript

In Ca(OH)₂ the 2 applies to both the O and the H inside the bracket, giving two O and two H. Forgetting this gives an Mr of 57 instead of the correct 74.

2. Multiplying only the nearest atom in a bracket

The subscript outside a bracket multiplies every atom inside. In Mg(NO₃)₂ it doubles the N and all three O atoms, not just the last O.

3. Using mass number instead of relative atomic mass

Read Ar values from the periodic table. Chlorine's Ar is 35.5, not 35, because it is an average over its isotopes; using 35 gives a wrong Mr.

4. Forgetting to multiply by 100 in a percentage

Percentage by mass is a fraction times 100. Leaving out the ×100 gives a decimal like 0.40 instead of 40%.

5. Using the wrong element mass in the percentage formula

The numerator is the total mass of the chosen element in the whole formula. For Fe₂O₃ that is 2 × 56 = 112, not a single 56.