How Chemists Count Atoms by Weighing Them

A spoonful of water contains more molecules than anyone could ever count one by one. Chemists solve this by counting in bundles called moles — and because each bundle has a known mass, weighing a substance tells you how many particles it holds.

This lesson covers the mole and Avogadro's number, percentage composition, and empirical and molecular formulas.

**A mole is the amount of substance that contains $6.022\times 10^{23}$ particles — Avogadro's number — and the mass of one mole in grams, the molar mass, equals the atomic or molecular mass in u.

Key relations:**

$$n=\frac{m}{M}N=n\times N_A$$

where n is the number of moles, m the mass in grams, M the molar mass in g mol${}^{-1}$, N the number of particles and $N_A=6.022\times 10^{23}$ mol${}^{-1}$.

Worked example 1. How many moles and molecules are in 36 g of water?

$$n=\frac{36\text{ g}}{18{\text{ g mol}}^{-1}}=2\text{ mol},N=2\times 6.022\times 10^{23}=1.204\times 10^{24}\text{ molecules}$$

Worked example 2. What is the mass of $3.011\times 10^{23}$ molecules of carbon dioxide?

$$n=\frac{3.011\times 10^{23}}{6.022\times 10^{23}}=0.5\text{ mol},m=0.5\times 44=22\text{ g}$$

Counting atoms. Each water molecule has 3 atoms, so 2 mol of water contains $3\times 1.204\times 10^{24}=3.612\times 10^{24}$ atoms.

An everyday example. A dozen bananas at a fruit stall always means 12, whether the bananas are big or small — a mole is simply a far larger counting unit.

The substance. Equal masses do not mean equal numbers of particles — 1 g of hydrogen gas and 1 g of oxygen gas hold very different numbers of molecules because their molar masses differ.

Percentage composition gives the mass of each element in 100 g of a compound, found by dividing the mass of that element in one mole by the molar mass and multiplying by 100.

$$\text{Mass per cent of an element}=\frac{\text{mass of the element in 1 mol}}{\text{molar mass of the compound}}\times 100$$

Worked example 1 — water. Molar mass $=2(1.008)+16.00=18.016$ g mol${}^{-1}$.

$$\text{Hydrogen}=\frac{2.016}{18.016}\times 100=11.19\text{Oxygen}=\frac{16.00}{18.016}\times 100=88.81$$

So water is 11.19 per cent hydrogen and 88.81 per cent oxygen by mass.

**Worked example 2 — ethanol, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$.** Molar mass $=2(12.01)+6(1.008)+16.00=46.07$ g mol${}^{-1}$.

$$\text{Carbon}=\frac{24.02}{46.07}\times 100=52.14\text{Hydrogen}=\frac{6.048}{46.07}\times 100=13.13$$

$$\text{Oxygen}=\frac{16.00}{46.07}\times 100=34.73$$

So ethanol is 52.14 per cent carbon, 13.13 per cent hydrogen and 34.73 per cent oxygen.

An everyday example. The nitrogen content printed on a bag of urea fertiliser comes from percentage composition — urea carries a high share of nitrogen by mass.

The substance. The percentages must add up to 100 — a quick check that catches arithmetic slips before you move on.

The empirical formula gives the simplest whole-number ratio of atoms in a compound and is found by converting mass percentages to moles and dividing by the smallest value, while the molecular formula is a whole-number multiple of it, found using the molar mass.

Steps:

- Take 100 g of the compound, so each percentage becomes a mass in grams
- Divide each mass by the element's atomic mass to get moles
- Divide every mole value by the smallest one
- Round to whole numbers, multiplying all values by a small integer if needed
- For the molecular formula, find $n=\frac{\text{molar mass}}{\text{empirical formula mass}}$ and multiply

Worked example. A compound contains 40.0 per cent carbon, 6.7 per cent hydrogen and 53.3 per cent oxygen, and its molar mass is 180 g mol${}^{-1}$.

- Carbon: $\frac{40.0}{12.0}=3.33$ mol
- Hydrogen: $\frac{6.7}{1.0}=6.7$ mol
- Oxygen: $\frac{53.3}{16.0}=3.33$ mol
- Dividing by 3.33 gives C : H : O $=1:2:1$, so the empirical formula is $\mathrm{C}{\mathrm{H}}_2\mathrm{O}$
- Empirical formula mass $=12+2+16=30$
- $n=\frac{180}{30}=6$, so the molecular formula is ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

An everyday example. The glucose in an oral rehydration packet and the acetic acid in vinegar share the empirical formula $\mathrm{C}{\mathrm{H}}_2\mathrm{O}$, even though their molecules are very different.

The substance. Different compounds can share one empirical formula — only the molar mass tells them apart.