Frequently Asked Question

Here's how to determine the empirical formulas for the compounds with the given compositions:
General Approach
1. Assume a 100g sample: This simplifies the percentages to grams. For example, 40.1% C in a 100g sample means you have 40.1g of carbon.
2. Convert grams to moles: Divide the mass of each element by its molar mass. Use the provided atomic mass of hydrogen (1.008 g/mol) and the standard values for other elements (e.g., carbon: 12.01 g/mol, oxygen: 16.00 g/mol, etc.).
3. Find the mole ratio: Divide each of the mole values by the smallest mole value calculated. This will give you the simplest whole-number ratio of elements in the compound.
4. Write the empirical formula: The resulting whole-number ratios represent the subscripts in the empirical formula.
Calculations
(a) 40.1% C, 6.6% H, 53.3% O
Moles of C: 40.1 g / (12.01 g/mol) = 3.3401 mol
Moles of H: 6.6 g / (1.008 g/mol) = 6.5536 mol
Moles of O: 53.3 g / (16.00 g/mol) = 3.3313 mol
Divide by smallest mole value (3.3313 mol):

C: 3.3401 / 3.3313 ≈ 1

H: 6.5536 / 3.3313 ≈ 2

O: 3.3313 / 3.3313 ≈ 1
Empirical formula: CH₂O
(b) 18.4% C, 21.5% N, 60.1% K
Moles of C: 18.4 g / (12.01 g/mol) = 1.5322 mol
Moles of N: 21.5 g / (14.01 g/mol) = 1.5347 mol
Moles of K: 60.1 g / (39.10 g/mol) = 1.5371 mol
Divide by smallest mole value (1.5322 mol):

C: 1.5322 / 1.5322 ≈ 1

N: 1.5347 / 1.5322 ≈ 1

K: 1.5371 / 1.5322 ≈ 1
Empirical formula: CKN