Why assume a 100 g sample?

With 100 g, each mass percent becomes grams directly (40.0% C = 40.0 g C). The sample size cancels out of the mole ratio, so any mass works; 100 g just saves a step.

What if the mole ratio comes out as 1.5 or 1.33?

Multiply all ratios by the smallest whole number that clears the decimal: 1.5 means multiply by 2, 1.33 or 1.67 means multiply by 3, 1.25 means multiply by 4. Only round values that are within about 0.05 of a whole number.

Chemistry 1: 40.0% C, 6.7% H, 53.3% O, Molar Mass 180 g/mol: Empirical and Molecular Formula

The Problem

A compound contains only carbon, hydrogen, and oxygen. By mass it is 40.0% C, 6.7% H, and 53.3% O. Its molar mass is 180 g/mol. Find (a) the empirical formula and (b) the molecular formula.

This exact set of percentages is a staple of first-semester chemistry homework and midterms, because the answer is a molecule every student knows.

Step 1: Turn Percent Into Grams

Assume a 100.0 g sample. Then:

C: 40.0 g
H: 6.7 g
O: 53.3 g

Step 2: Convert Grams to Moles

Atomic masses: C = 12.011, H = 1.008, O = 15.999 g/mol.

$$ n_{\text{C}} = \frac{40.0}{12.011} = 3.330\ \text{mol} $$

$$ n_{\text{H}} = \frac{6.7}{1.008} = 6.647\ \text{mol} $$

$$ n_{\text{O}} = \frac{53.3}{15.999} = 3.331\ \text{mol} $$

Step 3: Divide by the Smallest Number of Moles

The smallest value is 3.330 mol (carbon):

Element	Moles	÷ 3.330	Whole number
C	3.330	1.000	1
H	6.647	1.996	2
O	3.331	1.000	1

Empirical formula: $\text{CH}_2\text{O}$

Step 4: Empirical Formula Mass

$$ M_{\text{CH}_2\text{O}} = 12.011 + 2(1.008) + 15.999 = 30.03\ \text{g/mol} $$

Step 5: Find the Multiplier

$$ n = \frac{\text{molar mass}}{\text{empirical mass}} = \frac{180}{30.03} = 5.99 \approx 6 $$

Step 6: Molecular Formula

Multiply every subscript in $\text{CH}_2\text{O}$ by 6:

Molecular formula: $\mathbf{C_6H_{12}O_6}$ (glucose). Check: $6(12.011) + 12(1.008) + 6(15.999) = 180.16$ g/mol, which matches the given 180 g/mol.

Traps That Cost Points

Dividing percents by each other. 53.3 / 40.0 is a mass ratio, not a mole ratio. Convert each element to moles first.
Rounding too early. Keep at least three significant figures in the mole values. H comes out as 1.996, which rounds safely to 2. Rounding 6.647 to 6.6 and 3.330 to 3.3 still works here, but on other problems early rounding turns 1.5 into 1 and gives a wrong formula.
Stopping at the empirical formula. If a molar mass is given, the question almost always wants the molecular formula too.
Using the empirical formula as the multiplier. The multiplier is molar mass ÷ empirical mass (180 ÷ 30.03), not the other way around.

Quick Mnemonic

"Percent to mass, mass to mole, divide by small, multiply till whole." Then compare with the molar mass for the molecular formula.

Once you have the formula, the next skill on most midterms is stoichiometry. Try the 2Al + 3Cl₂ limiting reactant and theoretical yield problem.