Empirical Formula
The simplest whole number ratio of atoms of each element in a particle.
e.g. Glucose (\(C_6H_{12}O_6\)) → \(CH_2O\)
Molecular Formula
The actual number of atoms of each element in a molecule.
Formula: \( \text{Molecular Mass} = n \times \text{Empirical Mass} \)
Finding Empirical Formula from Mass
Paper 2 StyleA compound contains 72% Magnesium and 28% Nitrogen by mass. Calculate its empirical formula.
1. Assume 100g Sample:
- Mg: 72g
- N: 28g
2. Calculate Moles (Mass / Ar):
- Mg: \( 72 / 24.31 = 2.96 \text{ mol} \)
- N: \( 28 / 14.01 = 2.00 \text{ mol} \)
3. Divide by Smallest (Ratio):
- Mg: \( 2.96 / 2.00 = 1.48 \approx 1.5 \)
- N: \( 2.00 / 2.00 = 1.00 \)
4. Make Whole Number: (Multiply by 2)
Mg: 3, N: 2 → \( Mg_3N_2 \)
Student Practice Set
1.
What is the empirical formula of Benzene (\(C_6H_6\))?