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How to Find the Molecular Formula from Percent Composition

A 100 g basis gives the empirical formula C₄H₅N₂O. Its 97.10 amu formula mass fits twice into 194.2 amu, so the molecular formula is C₈H₁₀N₄O₂.

Question

An unknown compound is 49.47%49.47\% C, 5.201%5.201\% H, 28.84%28.84\% N, and 16.48%16.48\% O by mass. Its molecular mass is 194.2 amu194.2\,\mathrm{amu}. Find the molecular formula.

Answer

The molecular formula is C8H10N4O2\mathrm{C_8H_{10}N_4O_2}.

Use a 100 g100\,\mathrm{g} sample so that each percentage becomes grams:

n(C)=49.4712.01=4.119 mol,n(H)=5.2011.008=5.160 mol,n(N)=28.8414.01=2.059 mol,n(O)=16.4816.00=1.030 mol.\begin{aligned} n(\mathrm C)&=\frac{49.47}{12.01}=4.119\,\mathrm{mol},\\ n(\mathrm H)&=\frac{5.201}{1.008}=5.160\,\mathrm{mol},\\ n(\mathrm N)&=\frac{28.84}{14.01}=2.059\,\mathrm{mol},\\ n(\mathrm O)&=\frac{16.48}{16.00}=1.030\,\mathrm{mol}. \end{aligned}

Divide every amount by the smallest value, 1.030 mol1.030\,\mathrm{mol}. The ratio is approximately 4.00:5.01:2.00:1.004.00:5.01:2.00:1.00, so the empirical formula is C4H5N2O\mathrm{C_4H_5N_2O}.

Its empirical-formula mass is about 97.10 amu97.10\,\mathrm{amu}. The molecular mass is twice that value:

194.2 amu97.10 amu=2.000.\frac{194.2\,\mathrm{amu}}{97.10\,\mathrm{amu}}=2.000.

Multiply every empirical-formula subscript by 2 to obtain C8H10N4O2\mathrm{C_8H_{10}N_4O_2}.

Evidence boundary

This calculation treats the reported percentages as the composition of one pure compound and accepts their small rounding difference from 100%. It uses standard classroom atomic weights and identifies a molecular formula only; it does not determine molecular structure or distinguish isomers.

Sources

These references support the concepts and methods used in the explanation above.

How to Find a Molecular Formula from Percent Composition | Verla