Chemistry 2e · Composition of Substances and Solutions
Determining Empirical and Molecular Formulas
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Given only the elemental composition of a compound, chemists can deduce its formula. The empirical formula The simplest whole-number ratio of atoms in a compound. Full entry → is the simplest whole-number ratio of atoms in a compound; the molecular formula The actual number of each atom in one molecule. Full entry → gives the actual number of each atom in one molecule. Percent-composition data (or the results of a combustion analysis Burning a sample in excess oxygen and measuring the CO₂ and H₂O produced. Full entry →) yield the empirical formula, and one additional piece of information — the molar mass — converts the empirical formula into the molecular formula.
Why this matters
Determining formulas is how chemists identify unknown substances: forensics labs analyze residues, pharmaceutical companies confirm the identity and purity of drug batches, and food-safety laboratories check products against their labels. The empirical/molecular distinction matters because different molecules can share the same ratio: acetylene (C₂H₂) and benzene (C₆H₆) both have the empirical formula CH, yet they are utterly different substances. percent composition The mass percentage of each element in a compound. Full entry → also appears in everyday contexts — nutrition labels report fat, carbohydrate, and protein as percentages of mass, the same mathematics used here.
The college version
Core Concepts
Percent composition
The percent composition of a compound is the mass percentage of each element in it. For a formula unit:
% element = mass of element in one mole of compoundmolar mass of compound × 100%
In water, the hydrogen contribution is 2(1.008) = 2.016 g per mole and the molar mass is 18.016 g/mol, so % H = (2.016 / 18.016) × 100% = 11.19% and % O = 88.81%. The percentages of all elements in a compound sum to 100%.
Empirical formula from percent composition
An empirical formula shows the simplest whole-number ratio of atoms. To find it from percentages: (1) assume a 100 g sample so each percentage becomes a mass in grams; (2) convert each mass to moles by dividing by the element's atomic mass; (3) divide every mole value by the smallest one; (4) if the resulting ratios are not whole numbers, multiply all of them by the smallest integer that makes them whole (×2 for values near 0.5, ×3 for values near 0.33 or 0.67, ×4 for values near 0.25 or 0.75). The whole-number ratio is the empirical formula.
Empirical formula from combustion data
Combustion analysis is the classic laboratory method for C–H (and O) compounds. Burning the sample in excess oxygen converts all carbon to CO₂ and all hydrogen to H₂O. Because each CO₂ molecule contains one carbon atom, moles of C = moles of CO₂; because each H₂O contains two hydrogen atoms, moles of H = 2 × moles of H₂O. Convert the measured masses of CO₂ and H₂O to moles using their molar masses (44.01 and 18.02 g/mol), extract moles of C and H, then apply the same ratio steps. If the compound contains oxygen, its mass is found by difference: mass O = sample mass − mass C − mass H.
Molecular formula from the empirical formula
The molecular formula is a whole-number multiple of the empirical formula:
molecular formula = (empirical formula)n, n = molar massempirical formula mass
For glucose, the empirical formula CH₂O has a formula mass Sum of atomic masses in a formula, in amu. Full entry → of 30.03 amu and the measured molar mass is 180.16 g/mol: n = 180.16 / 30.03 = 6, giving C₆H₁₂O₆. The empirical formula alone cannot distinguish glucose from formaldehyde (CH₂O) or acetic acid (C₂H₄O₂); the molar mass is required.
How It Works / Step-by-Step Process
- From percent composition: assume 100 g; turn each % into grams.
- Convert each element's mass to moles (mass ÷ atomic mass).
- Divide every mole value by the smallest mole value.
- Multiply by the smallest integer that makes all ratios whole numbers.
- Write the empirical formula.
- If the molar mass is known, compute n = molar mass ÷ empirical formula mass and multiply subscripts by n.
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| empirical formula | molecular formula | Empirical is the reduced ratio; molecular is the actual counts (CH₂O vs C₆H₁₂O₆). |
| "percent composition identifies the compound" | it gives only the empirical formula | Different molecules share ratios; molar mass is needed. |
| rounding any decimal | rounding only near integers | 1.33 → multiply by 3; 1.25 → multiply by 4; 1.50 → multiply by 2. Rounding 1.33 to 1 is wrong. |
| "moles of H = moles of H₂O" | mol H = 2 × mol H₂O | Each water molecule contains two hydrogen atoms — forgetting the 2 is the classic error. |
| acetylene and benzene have the same formula | they share only the empirical formula CH | C₂H₂ vs C₆H₆: same ratio, different molecules. |
| combustion analysis tells the whole story | it reports C and H (O by difference) | Other elements require different methods. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
A formula is like a recipe. The empirical formula tells you the ratio of ingredients — "2 parts flour to 1 part sugar" — but not how many batches you made. The molecular formula tells you the real total for one finished molecule. If you know the ratio and you also know how heavy one molecule is, you can figure out exactly how many batches went into it.
Worked example
Example 1: From percent composition to both formulas
A compound analyzes as 40.0% C, 6.7% H, and 53.3% O. Find its empirical formula, then its molecular formula if its molar mass is 180.16 g/mol.
Assume 100 g: 40.0 g C, 6.7 g H, 53.3 g O. Convert to moles:
40.0 g C ÷ 12.01 g/mol = 3.33 mol C
6.7 g H ÷ 1.008 g/mol = 6.65 mol H
53.3 g O ÷ 16.00 g/mol = 3.33 mol O
Divide by the smallest (3.33): C 1.00, H 1.997 ≈ 2, O 1.00 → empirical formula CH₂O. Its formula mass: 12.01 + 2(1.008) + 16.00 = 30.03 amu. Then n = 180.16 / 30.03 = 6.0, so the molecular formula is C₆H₁₂O₆ — glucose.
Example 2: Combustion analysis
A 0.250 g sample of a hydrocarbon (only C and H) burns completely, producing 0.785 g CO₂ and 0.321 g H₂O. Determine the empirical formula; then find the molecular formula if the molar mass is 56.11 g/mol.
Moles of C:
0.785 g CO2 × 1 mol44.01 g = 0.01784 mol CO2 = 0.01784 mol C
Moles of H:
0.321 g H2O × 1 mol18.02 g = 0.01781 mol H2O
mol H = 2 × 0.01781 = 0.03563 mol
Ratio H/C = 0.03563 / 0.01784 = 1.997 ≈ 2, so the empirical formula is CH₂ (formula mass 14.03 amu). n = 56.11 / 14.03 = 4.0, giving C₄H₈. (Both butene and cyclobutane fit this formula — identifying which one requires additional data, a reminder that formulas alone do not always identify a single substance.)
Example 3: Percent composition of a known compound
Calculate the percent composition of NaCl. The molar mass is 22.99 + 35.45 = 58.44 g/mol.
% Na = 22.9958.44 × 100% = 39.34%
% Cl = 35.4558.44 × 100% = 60.66%
The two values sum to 100.00%, as they must.
Key takeaways
- Empirical formula = simplest whole-number ratio; molecular formula = actual atom counts; molecular = (empirical)ₙ with integer n.
- Percent composition → empirical formula: assume 100 g, convert to moles, divide by smallest, multiply to integers.
- Combustion analysis: mol C = mol CO₂; mol H = 2 × mol H₂O; oxygen by difference.
- n = molar mass ÷ empirical formula mass.
- The same empirical formula can describe different molecules (CH → C₂H₂ or C₆H₆; CH₂O → CH₂O, C₂H₄O₂, C₆H₁₂O₆).
- Percentages sum to 100%; check your work with that.
- Safety: combustion analysis involves high temperatures and flammable gases and belongs in a properly equipped lab under supervision. This guide covers the calculation, not the procedure — never attempt lab techniques without instruction and appropriate safety equipment.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
What is the empirical formula of hydrogen peroxide, H₂O₂?
Show answer
HO — the ratio 2:2 reduces to 1:1.
A compound is 27.3% C and 72.7% O. What is its empirical formula?
Show answer
CO₂ — 27.3 ÷ 12.01 = 2.27 mol C; 72.7 ÷ 16.00 = 4.54 mol O; ratio 1:2.
If the empirical formula is CH₂ and the molar mass is 56.11 g/mol, what is the molecular formula?
Show answer
C₄H₈ — n = 56.11 / 14.03 = 4.
What single piece of information converts an empirical formula into a molecular formula?
Show answer
The molar mass (n = molar mass ÷ empirical formula mass).
In combustion analysis, how do you find moles of hydrogen from the mass of water produced?
Show answer
Convert grams of H₂O to moles, then multiply by 2 (two H atoms per water molecule).
What is the percent composition of hydrogen in water?
Show answer
About 11.2% — (2 × 1.008) / 18.016 × 100%.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- percent composition
- The mass percentage of each element in a compound.
- empirical formula
- The simplest whole-number ratio of atoms in a compound.
- molecular formula
- The actual number of each atom in one molecule.
- combustion analysis
- Burning a sample in excess oxygen and measuring the CO₂ and H₂O produced.
- formula mass
- Sum of atomic masses in a formula, in amu.
- simplest ratio
- The reduced whole-number ratio of elements.
Sources & references
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