How to Find Empirical Formula (and Molecular Formula)

• Empirical formula: the simplest whole number ratio. Glucose is $\mathrm{CH_2O}$.
• Molecular formula: the real formula of one molecule. Glucose is $\mathrm{C_6H_{12}O_6}$, which is $(\mathrm{CH_2O})_6$.

The molecular formula is always a whole number multiple of the empirical formula. Sometimes the multiple is 1, so the two are the same. Water, $\mathrm{H_2O}$, is an example.

The method for the empirical formula is often remembered as: percent to mass, mass to mole, divide by small, multiply till whole.

Problem: A compound is 40.0% carbon, 6.71% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.

1. Percent to mass. Assume a 100 g sample, so the percents become grams: 40.0 g C, 6.71 g H, 53.3 g O.
2. Mass to moles. Divide by each atomic mass:
   • C: $40.0 \div 12.01 = 3.331$ mol
   • H: $6.71 \div 1.008 = 6.657$ mol
   • O: $53.3 \div 16.00 = 3.331$ mol
3. Divide by the smallest (3.331):
   • C: $3.331 \div 3.331 = 1.000$
   • H: $6.657 \div 3.331 = 1.998 \approx 2$
   • O: $3.331 \div 3.331 = 1.000$
4. Multiply until whole. The ratio is already 1 : 2 : 1.

Answer: the empirical formula is $\mathrm{CH_2O}$.

Values within about 0.1 of a whole number, like 1.998, can be rounded. Values like 1.5 or 1.33 cannot. See the next section.

If a ratio ends in a familiar decimal, multiply every subscript by the number that clears it:

Decimal ending	Multiply all by
.50	2
.33 or .67	3
.25 or .75	4
.20, .40, .60, .80	5

Example: An iron oxide is 69.9% Fe and 30.1% O. Find the empirical formula.

1. Assume 100 g: 69.9 g Fe and 30.1 g O.
2. Moles: Fe: $69.9 \div 55.85 = 1.252$. O: $30.1 \div 16.00 = 1.881$.
3. Divide by the smallest: Fe: $1.252 \div 1.252 = 1.000$. O: $1.881 \div 1.252 = 1.503$.
4. 1.503 is about 1.5, so multiply both by 2: Fe = 2, O = 3.

Answer: $\mathrm{Fe_2O_3}$. Rounding 1.5 up to 2 would have given FeO, which is a different compound.

If the problem gives masses, skip step 1 and go straight to moles.

Problem: A sample of a nitrogen oxide contains 0.700 g of nitrogen and 1.600 g of oxygen. Find the empirical formula.

1. N: $0.700 \div 14.01 = 0.04996$ mol.
2. O: $1.600 \div 16.00 = 0.1000$ mol.
3. Divide by the smallest (0.04996): N = 1.000, O = 2.002.
4. The ratio is 1 : 2.

Answer: $\mathrm{NO_2}$.

You do not need to convert the masses to percents first. The mole ratio comes out the same either way.

To go from the empirical formula to the molecular formula, you need the compound's molar mass, which the problem will give you.

$$n = \frac{\text{molar mass of compound}}{\text{empirical formula mass}}$$

Then multiply every subscript in the empirical formula by $n$.

Example 1: The empirical formula is $\mathrm{CH_2O}$ and the molar mass is 180.16 g/mol.

1. Empirical formula mass: $12.01 + 2(1.008) + 16.00 = 30.03$ g/mol.
2. $n = 180.16 \div 30.03 = 6.00$.
3. Multiply: $\mathrm{C_{1 \times 6}H_{2 \times 6}O_{1 \times 6}} = \mathrm{C_6H_{12}O_6}$.

Example 2: The nitrogen oxide above has a molar mass of 92.0 g/mol.

1. Empirical formula mass of $\mathrm{NO_2}$: $14.01 + 2(16.00) = 46.01$ g/mol.
2. $n = 92.0 \div 46.01 = 2.00$.
3. Molecular formula: $\mathrm{N_2O_4}$.

$n$ should come out very close to a whole number. If it does not, recheck the empirical formula mass. Paste it into the VeilGPT solver to check your steps.