Difficulty: Introductory to Intermediate | Prerequisites: Mole concept, molar mass calculations, periodic table familiarity
Percent composition tells you how much of each element (by mass) is in a compound. From those percentages you can work backwards to find the empirical formula, which is the simplest whole-number ratio of atoms. If you also know the compound's molar mass, you can scale the empirical formula up to the molecular formula, the one that reflects how many atoms are in an actual molecule.
Percent composition
The percentage by mass of each element in a compound, calculated as (mass of element in one mole of compound / molar mass of compound) × 100.
In simple terms, this means: if you had 100 g of the compound, how many grams of each element would be in it?
Empirical formula
The simplest whole-number ratio of atoms of each element in a compound.
Think of it as the reduced fraction version of the molecular formula. For example, glucose (C₆H₁₂O₆) has the empirical formula CH₂O.
Molecular formula
The formula showing the actual number of atoms of each element in one molecule of a compound.
Think of it as the full picture: the empirical formula tells you the ratio, the molecular formula tells you the real count.
Molar mass (molecular mass)
The mass, in grams, of one mole of a substance, numerically equal to its atomic or molecular mass in amu.
In simple terms, this means: it is the number you look up on the periodic table for an element, or the sum of those numbers for a compound.
Mole (mol)
The SI unit for amount of substance, equal to 6.022 × 10²³ particles (Avogadro's number).
Think of it as a chemist's counting unit, the way a "dozen" means 12.
Step 1: Assume 100 g of the compound. This turns every percentage directly into grams. If a compound is 36.5% Na, you have 36.5 g Na.
Step 2: Convert grams to moles. Divide each element's mass by its molar mass from the periodic table.
Step 3: Divide every mole value by the smallest mole value. This gives you the mole ratio.
Step 4: If the ratios are not whole numbers, multiply all ratios by the smallest integer that makes them whole. Common cases: a ratio ending in .5 means multiply by 2; a ratio ending in .33 means multiply by 3; .25 means multiply by 4.
Step 5: Write the empirical formula using the whole-number subscripts.
When given the actual mass of each element in a sample (rather than percentages), skip Step 1 and begin at Step 2. Convert each mass to moles, then proceed with the divide-and-round process above.
Step 1: Calculate the molar mass of the empirical formula. Add up the atomic masses of the atoms in the empirical formula.
Step 2: Divide the given (actual) molar mass by the empirical formula mass. This gives you a whole-number multiplier, often written as n.
Step 3: Multiply every subscript in the empirical formula by n. The result is the molecular formula.
For example, if the empirical formula is CH₂O (molar mass ≈ 30 g/mol) and the actual molar mass is 180 g/mol, then n = 180 / 30 = 6, so the molecular formula is C₆H₁₂O₆.
Percent composition of element X:
% X = (mass of X in one mole of compound / molar mass of compound) × 100
Moles from mass:
mol = mass (g) / molar mass (g/mol)
Multiplier for molecular formula:
n = molar mass of compound / molar mass of empirical formula
Molecular formula:
Molecular formula = (empirical formula) × n
Pharmaceutical chemists use molecular formulas to confirm the identity and purity of drug compounds; the wrong ratio of atoms means the wrong molecule and potentially a dangerous product.
Environmental scientists use percent composition to identify unknown pollutants in water or soil samples, working backwards from combustion analysis data to pin down what the contaminant is.
Students often forget to assume 100 g when given percentages, and try to convert percentages directly to moles. Percentages are not grams until you set a basis.
A common error is dividing by the largest mole value instead of the smallest. You always divide by the smallest to get whole-number ratios starting near 1.
Students sometimes round 1.5 or 1.33 down to 1 instead of multiplying all ratios by 2 or 3. If the ratio is not within about 0.1 of a whole number, you need to multiply.
Confusing empirical and molecular formulas is frequent. The empirical formula can be the same as the molecular formula (e.g., H₂O), but it does not have to be. They are only the same when the multiplier n equals 1.
⚠️ Expect at least one problem that gives you percentages and asks for an empirical formula, and at least one that then asks you to go further to the molecular formula using a given molar mass.
⚠️ Problems may give you mass data instead of percentages. The method is the same from Step 2 onwards, but students lose marks by trying to "convert to percentages" first, which is unnecessary extra work.
⚠️ Watch for ratios that come out to 1.5, 2.5, 1.33, or 1.67. Exams like to test whether you know to multiply through rather than round.
⚠️ Some questions will give a compound name or context ("a sugar," "an acid used in the automotive industry") and expect you to confirm your answer makes chemical sense.
1. True or false: The empirical formula of a compound is always different from its molecular formula.
False. They are the same when the multiplier n = 1 (e.g., H₂O).
2. Fill in the blank: When given percent composition, the first step is to assume a sample size of ______ g.
100 g.
3. True or false: If you calculate a mole ratio of 1 : 1.5, you should round 1.5 down to 1.
False. Multiply all ratios by 2 to get 2 : 3.
4. Fill in the blank: To find the molecular formula from the empirical formula, divide the ______ molar mass by the ______ formula mass.
Actual (given) molar mass by the empirical formula mass.
5. True or false: You need to convert mass data to percentages before you can find an empirical formula.
False. You can go straight from grams to moles.
Q: A compound is 36.5% Na, 25.4% S, and 38.1% O. What is its empirical formula?
A: Assume 100 g: 36.5 g Na, 25.4 g S, 38.1 g O. Convert to moles: Na = 36.5 / 22.99 = 1.587 mol; S = 25.4 / 32.07 = 0.792 mol; O = 38.1 / 16.00 = 2.381 mol. Divide by the smallest (0.792): Na = 2.00, S = 1.00, O = 3.00. Empirical formula: Na₂SO₃.
Q: A compound is 53.7% Fe and 46.3% S. The molecular mass is 416 g/mol. Find the empirical and molecular formulas.
A: Assume 100 g: 53.7 g Fe, 46.3 g S. Moles: Fe = 53.7 / 55.85 = 0.961; S = 46.3 / 32.07 = 1.443. Divide by 0.961: Fe = 1.00, S = 1.50. Multiply by 2: Fe₂S₃. Empirical formula mass = 2(55.85) + 3(32.07) = 207.91 g/mol. n = 416 / 207.91 ≈ 2. Molecular formula: Fe₄S₆.
Q: A sample contains 1.04 g K, 0.70 g Cr, and 0.86 g O. What is the empirical formula?
A: Moles: K = 1.04 / 39.10 = 0.0266; Cr = 0.70 / 52.00 = 0.01346; O = 0.86 / 16.00 = 0.05375. Divide by 0.01346: K = 1.98 ≈ 2, Cr = 1.00, O = 3.99 ≈ 4. Empirical formula: K₂CrO₄.
Q: 4.04 g of nitrogen combines with 11.46 g of oxygen to produce a compound with a molar mass of 108.0 g. What is the molecular formula?
A: Moles: N = 4.04 / 14.01 = 0.2884; O = 11.46 / 16.00 = 0.7163. Divide by 0.2884: N = 1.00, O = 2.48 ≈ 2.5. Multiply by 2: N₂O₅. Empirical formula mass = 2(14.01) + 5(16.00) = 108.02 g/mol. n = 108.0 / 108.02 ≈ 1. Molecular formula: N₂O₅.
Q: A compound has a molar mass of 92 g and contains 0.606 g N and 1.390 g O. Find the molecular formula.
A: Moles: N = 0.606 / 14.01 = 0.04326; O = 1.390 / 16.00 = 0.08688. Divide by 0.04326: N = 1.00, O = 2.01 ≈ 2. Empirical formula: NO₂. Empirical formula mass = 14.01 + 2(16.00) = 46.01 g/mol. n = 92 / 46.01 ≈ 2. Molecular formula: N₂O₄.
This material connects directly to stoichiometry: once you know a compound's molecular formula, you can write balanced equations and calculate how much reactant you need or product you will get.
It also ties into combustion analysis (a common lab technique where you burn a sample and measure the CO₂ and H₂O produced to work backwards to the empirical formula), which typically appears later in general chemistry courses.
Understanding mole ratios here lays the groundwork for limiting reagent problems, where you compare the mole ratio of reactants to what the balanced equation requires.
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