Calculating Percent Composition by Mass

Percent composition by mass is the percentage of the total mass of a compound that comes from each element, calculated as (mass of the element in one mole of compound ÷ molar mass of the compound) × 100%. In brief: in water, H2O (18.015 g/mol), hydrogen makes up 2.016 ÷ 18.015 × 100% = 11.19% of the mass and oxygen 88.81%; in an experiment, percent composition is found from the masses of the elements in a sample, and it is used to find empirical formulas.

H2O H 11.19% O 88.81%

Percent composition tells what a compound is made of by mass. It is used to check purity, to compare fertilizers, to evaluate ores and to determine formulas. This article gives the two ways of calculating it, from a formula and from experimental data, shows how to find the mass of an element in a sample, lists the composition of common compounds in tables, works through examples that include hydrates and fertilizers, and explains how to use the result.

How do you calculate percent composition from a formula?

  1. Find the molar mass of the compound by adding the atomic masses of all atoms in the formula.
  2. For each element, find the mass in one mole of the compound: (number of atoms in the formula) × (atomic mass).
  3. Divide the mass of the element by the molar mass and multiply by 100%.
  4. Check that the percentages add up to 100% (within rounding).

How do you calculate it from experimental data?

If the masses of the elements in a sample of the compound are measured, for example by combustion analysis, then percent of element = (mass of the element ÷ total mass of the sample) × 100%. The percentages from data can be compared with those calculated from a proposed formula to check it.

Percent composition of common compounds

Percent composition by mass
Compound Molar mass (g/mol) Element Percent by mass
Water, H2O18.015H11.19
O88.81
Carbon dioxide, CO244.009C27.29
O72.71
Sodium chloride, NaCl58.44Na39.34
Cl60.66
Iron(III) oxide, Fe2O3159.69Fe69.94
O30.06
Glucose, C6H12O6180.16C40.00
H6.71
O53.28

Worked examples

  1. Water. M = 18.015 g/mol. H: 2 × 1.008 = 2.016, so 2.016 ÷ 18.015 × 100% = 11.19%. O: 15.999 ÷ 18.015 × 100% = 88.81%. The sum is 100.00%.
  2. Ammonium sulfate, (NH4)2SO4. M = 132.13 g/mol. N: 28.014 ÷ 132.13 = 21.20%. H: 8.064 ÷ 132.13 = 6.10%. S: 32.06 ÷ 132.13 = 24.26%. O: 63.996 ÷ 132.13 = 48.43%. The sum is 99.99%, which is within rounding.
  3. From experimental data. A 1.250 g sample of a compound contains 0.500 g of carbon, 0.0840 g of hydrogen and 0.666 g of oxygen. The total is 1.250 g. C: 0.500 ÷ 1.250 = 40.0%. H: 0.0840 ÷ 1.250 = 6.72%. O: 0.666 ÷ 1.250 = 53.3%. These values match glucose (40.00%, 6.71%, 53.28%) and also formaldehyde and acetic acid, which have the same empirical formula CH2O.
  4. Mass of an element in a sample. How many grams of iron are in 250 g of Fe2O3? m(Fe) = 250 × 0.6994 = 175 g.
  5. Mass of a compound for a given amount of an element. How much calcium carbonate contains 10.0 g of calcium? The fraction of Ca in CaCO3 is 40.078 ÷ 100.086 = 0.4004. m(CaCO3) = 10.0 ÷ 0.4004 = 25.0 g.
  6. Which compound has more oxygen by mass? CO: 15.999 ÷ 28.010 = 57.1% O. CO2: 72.71% O. Carbon dioxide has the higher percent of oxygen.
  7. Check a formula with composition. Analysis of a sample gives 43.4% Na, 11.3% C and 45.3% O. For Na2CO3 (105.99 g/mol): Na 45.98 ÷ 105.99 = 43.38%, C 11.33%, O 45.29%. The data agree, so the compound is probably sodium carbonate.

How is percent composition used for fertilizers and ores?

Nitrogen content of common fertilizers and iron content of ores
Material Formula Element Percent by mass
UreaCO(NH2)2N46.65
Ammonium nitrateNH4NO3N35.00
Ammonium sulfate(NH4)2SO4N21.20
MagnetiteFe3O4Fe72.36
HematiteFe2O3Fe69.94
ChalcopyriteCuFeS2Cu34.63

A farmer can compare fertilizers by the nitrogen content: urea has the highest. For 100 kg of urea, the mass of nitrogen is 46.65 kg, while 100 kg of ammonium sulfate provides only 21.20 kg. The amount of urea needed to supply 50.0 kg of nitrogen is 50.0 ÷ 0.4665 = 107 kg. For ores, the percent of the metal helps to decide whether mining is worthwhile.

CO(NH2)2 N 46.65% NH4NO3 N 35.00% (NH4)2SO4 N 21.20%

What about hydrates?

For copper(II) sulfate pentahydrate, CuSO4·5H2O (249.68 g/mol), the composition by element is Cu 25.45%, S 12.84%, O 57.67% (including oxygen from the water) and H 4.04%. By component: CuSO4 is 159.61 ÷ 249.68 = 63.93% and water is 90.075 ÷ 249.68 = 36.08%. Both ways of reporting are useful: percent water shows how much mass is lost on heating.

CuSO4·5H2O CuSO4 63.93% 5H2O 36.08%

How did percent composition lead to the law of definite proportions?

Joseph Proust analyzed many compounds around 1799 to 1806 and found that the percent composition of a pure compound is always the same, whatever its source. For example, copper carbonate prepared in the laboratory or found in nature always contains the same proportions of copper, carbon and oxygen. This became the law of definite proportions, and it was a main support for Dalton’s atomic theory. Combustion analysis, developed by Justus von Liebig in 1831, made it practical to determine the percent composition of organic compounds, and the method is still used in automated analyzers.

Which mistakes are common?

Mistake 1: Forgetting to multiply by the number of atoms

In H2O there are 2 H atoms, so use 2 × 1.008 for hydrogen.

Mistake 2: Using a rounded molar mass inconsistently

Use the same atomic masses throughout so that the percentages add to 100%.

Mistake 3: Mixing up percent by mass and percent by atoms

Water is 11.19% hydrogen by mass but 66.7% hydrogen by number of atoms (2 of 3).

Mistake 4: Dividing by the mass of the element instead of the compound

The denominator is always the molar mass (or total mass) of the compound.