Percentage composition and empirical formulae

The law of definite proportions, percentage composition by mass, finding empirical and molecular formulae from percentage or mass data, the magnesium oxide crucible investigation, water of crystallisation in hydrates, and combustion analysis of organic compounds.

When a forensic lab receives an unknown white powder, one of the first things it reports is what percentage of the mass is carbon, hydrogen, nitrogen and oxygen. Those four numbers, plus a molar mass, are enough to pin down the molecular formula. The arithmetic is exactly what you'll learn on this page.

The law of definite proportions

In 1799 Joseph Proust showed that copper carbonate always contains copper, carbon and oxygen in the same proportions by mass, whether it was made in the lab or dug out of the ground. The law of definite proportions (or constant composition) states that a pure compound always contains the same elements in the same proportions by mass, however it is made. Water is always 11.19% hydrogen and 88.81% oxygen by mass.

The particle explanation is the fixed formula: every water molecule is H₂O, so the ratio of atoms, and therefore the ratio of masses, never changes. If a sample's composition doesn't match, it is either a different compound (H₂O₂ is 5.93% H) or not pure.

Percentage composition by mass

$$\%\ \text{element} = \frac{\text{(number of atoms)} \times A_r\ \text{of element}}{M\ \text{of compound}} \times 100$$

Nitrogen in fertilisers

Farmers compare nitrogen fertilisers by their percentage of nitrogen. Calculate %N in ammonium nitrate, NH₄NO₃, and in urea, CO(NH₂)₂.

M(NH₄NO₃) = 2 × 14.01 + 4 × 1.008 + 3 × 16.00 = 80.05 g mol⁻¹; %N = 28.02 ÷ 80.05 × 100 = 35.0%

M(CO(NH₂)₂) = 12.01 + 16.00 + 2 × 14.01 + 4 × 1.008 = 60.06 g mol⁻¹; %N = 28.02 ÷ 60.06 × 100 = 46.7%

Urea delivers more nitrogen per kilogram, which is one reason it is the most widely used nitrogen fertiliser in Australia.

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula is the actual number of each atom in one molecule, and is always a whole-number multiple of the empirical formula. Glucose is C₆H₁₂O₆; its empirical formula is CH₂O. Ionic compounds are written only as empirical formulae (NaCl, MgO, CaCl₂), because they have no discrete molecules.

Working from masses or percentages to a formula. Moles, not masses, give the ratio of atoms.

From percentages to a molecular formula

A compound contains 40.0% C, 6.7% H and 53.3% O by mass. Its molar mass is 180 g mol⁻¹. Find its empirical and molecular formulae.

Assume a 100 g sample, so the percentages become masses.

	C	H	O
mass (g)	40.0	6.7	53.3
n = m ÷ M	40.0 ÷ 12.01 = 3.33	6.7 ÷ 1.008 = 6.6	53.3 ÷ 16.00 = 3.33
÷ smallest	1.00	2.0	1.00

Empirical formula: CH₂O, with an empirical formula mass of 30.03.

k = 180 ÷ 30.03 = 6, so the molecular formula is C₆H₁₂O₆.

If a ratio comes out as about 1.5, 1.33 or 1.25, don't round it: multiply all the ratios by 2, 3 or 4 to make whole numbers. Iron oxide with Fe : O = 1 : 1.5 is Fe₂O₃. Only round values within about 0.1 of a whole number, and only after dividing by the smallest.

Investigation: the formula of magnesium oxide

Heating magnesium in a crucible. Lifting the lid briefly lets oxygen in without letting white smoke escape.

Investigation: empirical formula of magnesium oxide

Aim: to determine the empirical formula of magnesium oxide from the masses of magnesium and oxygen that combine.

Materials: crucible and lid, pipeclay triangle, tripod, Bunsen burner, heatproof mat, crucible tongs, electronic balance (±0.01 g or better), 20 to 30 cm of magnesium ribbon cleaned with steel wool, dropper and distilled water.

Method:

Heat the empty crucible and lid strongly for 2 minutes, cool and weigh.
Coil the cleaned ribbon into the crucible and weigh again.
Heat with the lid on. Every minute, lift the lid slightly with tongs to let air in, then replace it quickly to stop white MgO smoke escaping.
When the magnesium no longer glows, remove the lid and heat for 2 more minutes.
Cool, add a few drops of water (this converts any magnesium nitride, Mg₃N₂, into magnesium hydroxide), heat again to drive off water and ammonia, cool and weigh.
Repeat heating, cooling and weighing until two consecutive masses agree (heating to constant mass).

Typical results:

crucible + lid	25.32 g
crucible + lid + Mg	25.56 g → m(Mg) = 0.24 g
crucible + lid + product (constant mass)	25.72 g → m(MgO) = 0.40 g, m(O) = 0.16 g

Processing: n(Mg) = 0.24 ÷ 24.31 = 0.0099 mol; n(O) = 0.16 ÷ 16.00 = 0.010 mol; ratio 1 : 1.0, so the empirical formula is MgO.

Errors: smoke escaping makes the oxygen mass too small (ratio below 1); unreacted magnesium does the same; magnesium nitride formed from N₂ in air gains less mass than oxide would. With only two decimal places the 0.16 g of oxygen has a percentage uncertainty of about ±6%, so a more sensitive balance or more ribbon improves accuracy.

Safety: do not look directly at burning magnesium; the crucible stays hot long after it stops glowing.

Hydrates and water of crystallisation

Many ionic crystals contain water molecules locked into the lattice in a fixed ratio, shown with a dot: CuSO₄·5H₂O, copper(II) sulfate pentahydrate. Heating drives the water off, and the blue crystals turn into a white powder. Weighing before and after gives the number of water molecules per formula unit.

Finding x in a hydrate

2.50 g of blue CuSO₄·xH₂O is heated to constant mass, leaving 1.60 g of white anhydrous CuSO₄. Find x.

m(H₂O) = 2.50 − 1.60 = 0.90 g, so n(H₂O) = 0.90 ÷ 18.02 = 0.050 mol.

n(CuSO₄) = 1.60 ÷ 159.62 = 0.0100 mol.

x = 0.050 ÷ 0.0100 = 5.0, so the formula is CuSO₄·5H₂O.

Combustion analysis

For organic compounds, chemists burn a weighed sample in a stream of pure oxygen. All the carbon ends up as CO₂ and all the hydrogen as H₂O, which are trapped in separate absorber tubes and weighed. Oxygen in the sample is found by difference.

A combustion train. Each absorber gains the mass of one product.

From combustion data to a molecular formula

0.450 g of a compound containing C, H and O burns to give 0.660 g of CO₂ and 0.270 g of H₂O. Its molar mass is 60.05 g mol⁻¹. Find its molecular formula.

$m(\text{C}) = 0.660 \times \dfrac{12.01}{44.01} = 0.180\ \text{g}$ $m(\text{H}) = 0.270 \times \dfrac{2.016}{18.02} = 0.0302\ \text{g}$

m(O) = 0.450 − 0.180 − 0.0302 = 0.240 g

n: C 0.180 ÷ 12.01 = 0.0150; H 0.0302 ÷ 1.008 = 0.0300; O 0.240 ÷ 16.00 = 0.0150. Ratio 1 : 2 : 1, empirical formula CH₂O (30.03).

k = 60.05 ÷ 30.03 = 2, so the molecular formula is C₂H₄O₂ (ethanoic acid, the acid in vinegar).

InteractiveCombustion analyser

Try this

Burn 0.230 g of Sample A. The tubes should gain 0.4395 g (CO₂) and 0.2698 g (H₂O). Work out the formula yourself, then press Show the working to check: it is C₂H₆O.
Burn 0.200 g of Sample B. It contains no oxygen and its empirical formula is CH. How does the molar mass (78.1 g mol⁻¹) turn that into the molecular formula?
Burn Sample F, another hydrocarbon. The H : C ratio comes out as 2.25. What do you multiply by, and what are the empirical and molecular formulae?
Change the sample mass for Sample C. Do the tube masses change? Does the formula?
Turn on Swap the absorber tubes and burn Sample A again. Explain why the formula goes wrong.

Test yourself

What is the percentage by mass of oxygen in calcium carbonate, CaCO₃?
An oxide of iron contains 2.23 g of iron combined with 0.96 g of oxygen. What is its empirical formula?
A hydrocarbon has the empirical formula CH₂ and a molar mass of 84.2 g mol⁻¹. What is its molecular formula?
In the magnesium oxide experiment, some white smoke escaped each time the lid was lifted. What effect does this have on the calculated ratio of Mg : O?
In combustion analysis, why must the water absorber be placed before the carbon dioxide absorber?

Label the crucible set-up.

Put the steps for finding the formula of a hydrate in order.

The formula shows the simplest whole-number ratio of atoms. To find it, convert masses to and divide by the . A ratio of 1 : 1.33 must be multiplied by . Heating until two weighings agree is called heating to constant .

6 marks

Methyl salicylate (oil of wintergreen) contains carbon, hydrogen and oxygen. Combustion of 0.304 g produced 0.704 g of CO₂ and 0.144 g of H₂O. Its molar mass is 152.1 g mol⁻¹. Determine its empirical and molecular formulae, showing all working.

Your answer