Determining Empirical Formulas from Data

The empirical formula of a compound is the simplest whole-number ratio of its atoms, and it is found from experimental data by converting the mass or percent of each element to moles, dividing by the smallest number of moles and, if needed, multiplying to get whole numbers. In brief: a compound that is 69.94% iron and 30.06% oxygen has 1.2524 mol of Fe and 1.8789 mol of O per 100 g; the ratio is 1 : 1.5, which becomes 2 : 3, so the empirical formula is Fe₂O₃.

When chemists make or discover a new compound, the first thing they determine is its empirical formula from the masses of the elements. This article presents the four-step method, explains how to handle ratios such as 1.5 and 1.33, describes combustion analysis for compounds containing carbon, hydrogen and oxygen, includes a table of ratio multipliers, works through six examples, and gives the history of elemental analysis.

What are the steps?

1. Write down the mass (or percent) of each element. If you have percentages, assume a sample of 100 g, so that the percent of each element becomes grams.
2. Convert the mass of each element to moles by dividing by its atomic mass.
3. Divide all the mole values by the smallest one, to get the ratio of the atoms, with the smallest equal to 1.
4. Make the ratios whole numbers. If a ratio is within about 0.1 of a whole number, round it. Otherwise, multiply all the ratios by a small whole number (2, 3, 4 or 5) that makes them whole.
5. Write the formula with the whole numbers as subscripts.

How do you deal with ratios that are not whole numbers?

Experimental errors cause small deviations, so a ratio of 2.98 is treated as 3, and 1.02 as 1. A decimal that is close to a simple fraction, such as 0.5, 0.33, 0.25 or 0.67, shows that you need to multiply. Never round a ratio such as 1.5 to 2, or 1.33 to 1, since this would give a wrong formula.

Multipliers for common ratios

Worked examples

1. From percent: iron oxide. A compound is 69.94% Fe and 30.06% O. In 100 g: Fe 69.94 g ÷ 55.845 g/mol = 1.2524 mol; O 30.06 g ÷ 15.999 g/mol = 1.8789 mol. Divide by 1.2524: Fe 1.000; O 1.5002. The ratio of 1 : 1.5 is multiplied by 2: Fe 2 and O 3. The empirical formula is Fe₂O₃.
2. From masses. 2.40 g of carbon and 0.60 g of hydrogen form a hydrocarbon. C: 2.40 ÷ 12.011 = 0.1998 mol. H: 0.60 ÷ 1.008 = 0.5952 mol. Divide by 0.1998: C 1.000, H 2.979, which is 3. The empirical formula is CH₃.
3. Three elements. A compound contains 43.4% Na, 11.3% C and 45.3% O. In 100 g: Na 43.4 ÷ 22.990 = 1.888 mol; C 11.3 ÷ 12.011 = 0.9408 mol; O 45.3 ÷ 15.999 = 2.831 mol. Divide by 0.9408: Na 2.007, C 1.000, O 3.009. The formula is Na₂CO₃.
4. Ratio of 1.33. A compound is 72.36% Fe and 27.64% O. Fe: 72.36 ÷ 55.845 = 1.2957 mol. O: 27.64 ÷ 15.999 = 1.7276 mol. Divide by 1.2957: Fe 1.000, O 1.3333. Multiply by 3: Fe 3, O 4. The formula is Fe₃O₄ (magnetite).
5. Percent of oxygen by difference. A compound of phosphorus and oxygen contains 43.64% P. The rest is oxygen: 56.36%. P: 43.64 ÷ 30.974 = 1.409 mol. O: 56.36 ÷ 15.999 = 3.523 mol. Divide by 1.409: P 1.000, O 2.500. Multiply by 2: P₂O₅.
6. Mass changes in an experiment. 1.500 g of copper is heated in air and forms 1.878 g of a copper oxide. The mass of oxygen is 1.878 − 1.500 = 0.378 g. Cu: 1.500 ÷ 63.546 = 0.02360 mol. O: 0.378 ÷ 15.999 = 0.02363 mol. The ratio is 1 : 1, so the oxide is CuO.

How does combustion analysis work?

For compounds of carbon, hydrogen and (maybe) oxygen, a weighed sample is burned in excess oxygen. All the carbon is converted into CO₂ and all the hydrogen into H₂O. The two products are collected, the water in a tube with a drying agent and the carbon dioxide in a tube with a base, and weighed. The masses of C and H in the sample are calculated from the masses of products, and the mass of oxygen is found by difference.

Example: A 0.500 g sample of a compound of C, H and O gives 0.955 g of CO₂ and 0.587 g of H₂O.

1. Carbon: 0.955 g CO₂ × (12.011 ÷ 44.009) = 0.2606 g C, or 0.2606 ÷ 12.011 = 0.02170 mol.
2. Hydrogen: 0.587 g H₂O × (2.016 ÷ 18.015) = 0.06569 g H, or 0.06569 ÷ 1.008 = 0.06517 mol.
3. Oxygen by difference: 0.500 − 0.2606 − 0.0657 = 0.1737 g, or 0.1737 ÷ 15.999 = 0.01086 mol.
4. Divide by the smallest (0.01086): C 2.00, H 6.00, O 1.00.
5. Empirical formula: C₂H₆O (ethanol or dimethyl ether).

What does the empirical formula not tell you?

It gives only the simplest ratio. Different compounds can share the same empirical formula. CH₂O corresponds to formaldehyde (CH₂O), acetic acid (C₂H₄O₂) and glucose (C₆H₁₂O₆). To find the molecular formula, the molar mass is also needed. And to find the structure, spectroscopic methods are used. For ionic compounds, the empirical formula is the usual formula.

How did elemental analysis develop?

The study of the composition of compounds began with Lavoisier, who burned organic substances and measured the carbon dioxide and water produced. Joseph Louis Gay-Lussac and Louis Jacques Thénard improved combustion analysis in 1810. Justus von Liebig invented his five-bulb apparatus in 1831, which allowed accurate and quick determinations. Fritz Pregl developed microanalysis, which needs only a few milligrams of sample, and received the Nobel Prize in Chemistry in 1923. Modern CHN analyzers do the process automatically.

Which mistakes are common?

Mistake 1: Using the percentages as moles

Convert grams to moles first. 40% carbon and 60% oxygen is not a 2 : 3 ratio of atoms.

Mistake 2: Rounding ratios too early

A ratio of 1.5 must be multiplied by 2, not rounded to 2 or 1.

Mistake 3: Forgetting to divide by the smallest number of moles

The ratio is relative to the smallest value.

Mistake 4: Assuming that the empirical formula is the molecular formula

The molecular formula needs the molar mass.