AP Chemistry · Unit 1 · Topic 1.3
Elemental Composition of Pure Substances
A compound's formula is a statement about atom ratios, but a composition experiment usually reports grams. The crucial move is therefore not rounding percentages into subscripts. It is converting each element's mass into an amount before comparing the elements.
What stays fixed in a pure compound
For a specified pure compound with a given isotopic composition, different sample sizes have the same elemental mass fractions. Doubling the sample doubles every constituent mass without changing their ratios. A mixture can vary its component proportions, whereas a chemical formula specifies a particular composition. Your goal is to connect this fixed composition to a defensible whole-number atom ratio.
Percent mass is not percent atoms
Heavy atoms contribute more mass per atom than light atoms. A substance can contain many hydrogen atoms while having a small hydrogen mass percentage. For percent-composition data, choosing a hypothetical 100 g sample makes the reported percentages numerically equal to grams. This is a convenient scale choice, not a claim that the laboratory used exactly 100 g.
Worked example: from percentages to ratios
An invented analysis reports 54.5% C, 9.1% H, and 36.4% O by mass. On a 100 g basis, divide by 12.01, 1.008, and 16.00 g mol−1, respectively. The amounts are approximately 4.54, 9.03, and 2.275 mol. Divide all three by 2.275: the ratio is about 2.00 : 3.97 : 1.00, supporting C2H4O.
Keep the interpretation visible
| Element | Mass in chosen basis / g | Amount / mol | Ratio |
|---|---|---|---|
| C | 54.5 | 4.54 | 2.00 |
| H | 9.1 | 9.03 | 3.97 |
| O | 36.4 | 2.275 | 1.00 |
The empirical formula records the simplest ratio. It does not yet identify a unique molecule or structure. Different substances can share the same empirical formula.
A molecular-mass measurement adds information
The empirical formula mass is approximately 44.05 g mol−1. If an independent experiment gives a molecular molar mass near 88.1 g mol−1, the molecule contains two empirical units.
Multiply every subscript by the same integer. Doubling only carbon would destroy the measured composition.
Use uncertainty instead of wishful rounding
A ratio of 1.50 is not reasonably rounded to 2. Multiply the entire ratio set by two to test for small integers. Conversely, 1.99 may plausibly represent 2 when the measurements have limited precision. In a combustion analysis, incomplete combustion can make collected carbon dioxide underestimate carbon; missing collection products can therefore change the inferred formula systematically.
Check the formula against the measurements
Recalculate predicted mass percentages from the proposed formula and compare them with the data. This checks both arithmetic and interpretation. If the calculated molecular multiplier is far from an integer, revisit the analysis or molar mass instead of inventing fractional atom subscripts. Oxygen by difference is valid only when the sample contains no unmeasured elements besides oxygen.
Practice, then check your reasoning
An empirical ratio is C:H = 1:1.5. What simplest formula follows?
Answer and reasoning: C2H3; multiplying the complete ratio by two removes the half.
Does CH2 mean carbon and hydrogen have equal mass contributions?
Answer and reasoning: No. It means two hydrogen atoms per carbon atom; their molar masses determine the mass contributions.
Scientific references: OpenStax: empirical and molecular formulas. OpenStax Chemistry 2e authors: Paul Flowers, Klaus Theopold, Richard Langley, and William R. Robinson. Consult source pages for their license terms. Accessed September 13, 2026. Explanations, comparison tables, and practice tasks were independently written for this site; numerical teaching datasets are illustrative unless explicitly identified otherwise.
Official curriculum reference: College Board AP Chemistry course page. The explanation and worked example are independently written for this study site.