Elemental composition of pure substances

AP Chemistry· AP Chemistry CED — Atomic Structure and Properties· 14 min read

1. Mass Percent Composition★★☆☆☆⏱ 3 min

📘 Definition

Mass Percent Composition

The percentage of the total mass of a pure compound that comes from one specific constituent element. All pure substances have fixed mass percent for each element following the law of definite proportions.

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\%

📐 Worked Example

What is the mass percent of oxygen in calcium carbonate, $CaCO_{3}$ , rounded to one decimal place?

1
Look up average atomic masses from the periodic table:
2
$Ca = 40.08 \text{ g/mol}, C = 12.01 \text{ g/mol}, O = 16.00 \text{ g/mol}$
3
Calculate total mass of O in one mole of $CaCO_{3}$ (subscript of O is 3):
4
$3 \times 16.00 = 48.00 \text{ g/mol}$
5
Calculate total molar mass of $CaCO_{3}$ :
6
$40.08 + 12.01 + 48.00 = 100.09 \text{ g/mol}$
7
Plug into the mass percent formula:
8
$\frac{48.00}{100.09} \times 100\% = 47.9\%$
9
The mass percent of oxygen in calcium carbonate is 47.9%.

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

📘 Definition

Empirical Formula

The simplest whole-number ratio of atoms of each element in a compound. Ionic compounds always use the empirical formula as their official chemical formula, while molecular compounds have a molecular formula that is a whole-number multiple of the empirical formula.

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

📐 Worked Example

A 7.50 g pure sample of a nitrogen oxide contains 2.30 g of nitrogen. What is the empirical formula of the compound?

1
Calculate the mass of oxygen by difference:
2
$7.50 \text{ g} - 2.30 \text{ g} = 5.20 \text{ g O}$
3
Convert masses to moles:
4
$n(\text{N}) = \frac{2.30 \text{ g}}{14.01 \text{ g/mol}} \approx 0.164 \text{ mol}; \quad n(\text{O}) = \frac{5.20 \text{ g}}{16.00 \text{ g/mol}} \approx 0.325 \text{ mol}$
5
Divide both values by the smallest mole value (0.164 mol):
6
$\text{N} = \frac{0.164}{0.164} = 1.00; \quad \text{O} = \frac{0.325}{0.164} \approx 1.98 \approx 2$
7
The whole number ratio of N:O is 1:2, so the empirical formula is:
8
$\text{NO}_2$

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.

📐 Worked Example

A molecular compound has an empirical formula of $NO_{2}$ , and an experimental molar mass of 92.01 g/mol. What is its molecular formula?

1
Calculate the molar mass of the empirical formula $NO_{2}$ :
2
$14.01 + (2 \times 16.00) = 46.01 \text{ g/mol}$
3
Calculate the multiplier $n$ :
4
$n = \frac{92.01}{46.01} \approx 2$
5
Multiply each subscript in the empirical formula by 2: N = $1 \times 2 = 2$ , O = $2 \times 2 = 4$ . The molecular formula is:
6
$\text{N}_2\text{O}_4$

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 $CO_{2}$ and $H_{2} O$ produced are absorbed by pre-weighed materials. All carbon in the original compound becomes $CO_{2}$ , and all hydrogen becomes $H_{2} O$ , 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.

📐 Worked Example

A 0.500 g sample of a pure organic compound containing only C, H, and O produces 0.733 g $CO_{2}$ and 0.300 g $H_{2} O$ in combustion analysis. What is the empirical formula?

1
Calculate moles and mass of C (all C from the original sample becomes $CO_{2}$ ):
2
$n(\text{C}) = n(\text{CO}_2) = \frac{0.733 \text{ g}}{44.01 \text{ g/mol}} \approx 0.01666 \text{ mol}; \quad m(\text{C}) = 0.01666 \times 12.01 \approx 0.200 \text{ g}$
3
Calculate moles and mass of H (each mole of $H_{2} O$ has 2 moles of H from the original compound):
4
$n(\text{H}) = 2 \times n(\text{H}_2\text{O}) = 2 \times \frac{0.300 \text{ g}}{18.015 \text{ g/mol}} \approx 0.0333 \text{ mol}; \quad m(\text{H}) = 0.0333 \times 1.008 \approx 0.0336 \text{ g}$
5
Calculate mass and moles of O by difference from the original sample mass:
6
$m(\text{O}) = 0.500 - 0.200 - 0.0336 = 0.2664 \text{ g}; \quad n(\text{O}) = \frac{0.2664}{16.00} \approx 0.01665 \text{ mol}$
7
Divide all mole values by the smallest mole value (0.01665 mol) to get a 1:2:1 ratio, so the empirical formula is:
8
$\text{CH}_2\text{O$

Exam tip:

Don’t forget to multiply the moles of $H_{2} O$ by 2 to get moles of H; forgetting this step is the most common mistake in combustion analysis problems.

5. Common Pitfalls

Wrong move:

When calculating empirical formula from percent composition, you use the percent values directly as moles without converting to mass first.

Why:

Students implicitly assume a 100 g sample but forget percent values are percentages, not masses, leading to incorrect mole calculations.

Correct move:

Always explicitly assume a 100 g sample, so each percent value becomes the mass of the element in grams before converting to moles.

Wrong move:

In combustion analysis, you calculate mass of O by adding oxygen from $CO_{2}$ and $H_{2} O$ instead of using mass difference.

Why:

Students forget that almost all oxygen in the products comes from the excess oxygen used for combustion, not the original compound.

Correct move:

Always calculate mass of O in the original compound by subtracting mass of C and H from the total mass of the original sample.

Wrong move:

You round a 1.33 mole ratio to 1 instead of multiplying all ratios by 3 to get a whole-number ratio.

Why:

Students are eager to round to whole numbers and miss common simple fractions that require scaling.

Correct move:

Recognize common fractions (0.25 = 1/4, 0.33 = 1/3, 0.5 = 1/2, 0.66 = 2/3) and multiply all ratios by the denominator of the fraction before rounding.

Wrong move:

You calculate n as empirical molar mass divided by molecular molar mass, getting a value less than 1, then round incorrectly.

Why:

Students mix up the order of division in the multiplier formula.

Correct move:

Memorize the order: $n = \frac{larger molar mass (molecular)}{smaller molar mass (empirical)}$ , which always gives a whole number ≥ 1.

Wrong move:

You multiply only the first subscript in the empirical formula by n when calculating molecular formula.

Why:

Students rush and forget to apply the multiplier to all elements.

Correct move:

Always multiply every subscript in the empirical formula by n, not just the first one.

Wrong move:

You calculate mass percent of an element using only the atomic mass once, ignoring the subscript.

Why:

Students forget the subscript indicates how many atoms of the element are in one formula unit.

Correct move:

Always multiply the atomic mass of each element by its subscript when calculating total mass of the element.

6. Quick Reference Cheatsheet

Category	Formula / Process	Notes
Mass percent of element X	$Mass % X = \frac{( subscript X \times A _{r} ( X ))}{M ( compound )} \times 100%$	Sum of all mass percents must equal ~100%
Empirical formula from mass	Convert mass to moles; 2. Divide by smallest mole; 3. Scale to whole numbers	Assume 100 g sample if given mass percent
Molecular formula multiplier	$n = \frac{M ( molecular )}{M ( empirical )}$	n is always whole number ≥ 1; multiply all subscripts by n
Moles of C (combustion)	$n (C) = n (CO_{2})$	All C from original compound becomes CO₂
Moles of H (combustion)	$n (H) = 2 \times n (H_{2} O)$	1 mole H₂O has 2 moles H from original compound
Mass of O (C/H/O combustion)	$m (O) = m_{sample} - m (C) - m (H)$	Most O in products comes from excess combustion O₂
Empirical vs Molecular Formula	N/A	Ionic compounds always use empirical formula; only molecular compounds have distinct molecular formulas

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Empirical formula calculation from data
2022 · FRQ
Combustion analysis problem

What's Next

Elemental composition of pure substances is the foundational link between macroscopic mass measurements and microscopic atomic composition, which is required for nearly all quantitative calculations in AP Chemistry. Mastery of these calculation techniques allows you to connect experimental lab data to the chemical identity of unknown pure compounds, a core skill that appears across every unit of the AP Chemistry course. Immediately after mastering this topic, you will move on to composition of mixtures, then apply your empirical formula skills to stoichiometry of chemical reactions, where you will use these elemental ratios to calculate reactant and product yields. Without correctly determining elemental composition and empirical formulas, you will not be able to solve limiting reactant problems, titration calculations, or any other quantitative problem that relies on mole ratios of compounds.