Chemistry · AP CED Unit 1: Atomic Structure and Properties · 14 min read · Updated 2026-05-11
Elemental composition of pure substances — AP Chemistry
AP Chemistry · AP CED Unit 1: Atomic Structure and Properties · 14 min read
1. Mass Percent Composition★★☆☆☆⏱ 3 min
Mass percent intuitively tells you how many grams of an element you would have in a 100 g sample of the compound, which simplifies further calculations for empirical formulas. The formula is:
\text{Mass percent of X} = \frac{(\text{subscript of X} \times \text{Average atomic mass of X})}{\text{Molar mass of the compound}} \times 100\%
Exam tip: Always sum the mass percents of all elements in a compound at the end of your calculation; a sum that differs from 100% by more than 0.5% indicates an arithmetic error.
2. Empirical Formula from Mass Data★★★☆☆⏱ 4 min
To calculate an empirical formula from experimental data, follow these four core steps:
Convert mass of each element to moles using $n = \frac{m}{M}$
Divide all mole values by the smallest mole value to get a preliminary ratio
Multiply all ratios by a whole number to convert any fractional ratios to whole numbers
Use the whole numbers as subscripts for the empirical formula
Exam tip: If a ratio is within 0.05 of a whole number (e.g., 1.98 instead of 2), round directly to the whole number; only multiply to clear fractions if the ratio is clearly 1.25, 1.33, 1.5, etc.
3. Molecular Formula from Empirical Formula and Molar Mass★★☆☆☆⏱ 3 min
Once you have the empirical formula of a molecular compound, you can find the actual molecular formula if you know the compound's experimental molar mass (usually provided in AP problems from mass spectrometry). Because the molecular formula is a whole-number multiple of the empirical formula, its molar mass is the same multiple of the empirical formula's molar mass. The multiplier $n$ is calculated as:
n = \frac{\text{Molar mass of molecular compound}}{\text{Molar mass of empirical formula}}
$n$ will always be a whole number greater than or equal to 1. If $n=1$, the empirical and molecular formulas are identical. After finding $n$, multiply all subscripts in the empirical formula by $n$ to get the final molecular formula.
Exam tip: Always confirm that the molar mass of your calculated molecular formula matches the given molar mass within rounding error before writing your final answer.
4. Combustion Analysis for Organic Compounds★★★★☆⏱ 4 min
Combustion analysis is an experimental technique used to determine the elemental composition of pure organic compounds (most commonly compounds made of C, H, and O). In the experiment, a known mass of the organic compound is burned completely in excess oxygen, and all $\text{CO}_2$ and $\text{H}_2\text{O}$ produced are absorbed by pre-weighed materials. All carbon in the original compound becomes $\text{CO}_2$, and all hydrogen becomes $\text{H}_2\text{O}$, so we can calculate the mass of C and H in the original compound from product masses. Any remaining mass of the original compound is oxygen, since excess oxygen from the reaction does not contribute to the original sample mass.
Exam tip: Don’t forget to multiply the moles of $\text{H}_2\text{O}$ by 2 to get moles of H; forgetting this step is the most common mistake in combustion analysis problems.
Common Pitfalls
Why: Students implicitly assume a 100 g sample but forget percent values are percentages, not masses, leading to incorrect mole calculations.
Why: Students forget that almost all oxygen in the products comes from the excess oxygen used for combustion, not the original compound.
Why: Students are eager to round to whole numbers and miss common simple fractions that require scaling.
Why: Students mix up the order of division in the multiplier formula.
Why: Students rush and forget to apply the multiplier to all elements.
Why: Students forget the subscript indicates how many atoms of the element are in one formula unit.