Frequently Asked Question

Here's how to determine the empirical formulas for the given compounds:
Understanding Empirical Formulas
An empirical formula represents the simplest whole-number ratio of atoms in a compound. To find it, we follow these steps:
1. Assume a 100g Sample: This makes the percentages directly equivalent to grams.
2. Convert Grams to Moles: Divide the mass of each element by its molar mass.
3. Find the Simplest Mole Ratio: Divide each mole value by the smallest mole value. Round to the nearest whole number if possible. This ratio represents the subscripts in the empirical formula.
Let's Apply This to the Examples:
(a) 2.1% H, 65.3% O, 32.6% S
1. Assume 100g sample: 2.1g H, 65.3g O, 32.6g S
2. Convert to moles:

H: 2.1g / 1.008 g/mol = 2.08 mol

O: 65.3g / 16.00 g/mol = 4.08 mol

S: 32.6g / 32.06 g/mol = 1.02 mol
3. Find simplest ratio:

H: 2.08 mol / 1.02 mol = 2

O: 4.08 mol / 1.02 mol = 4

S: 1.02 mol / 1.02 mol = 1
Therefore, the empirical formula is H₂SO₄.
(b) 20.2% Al, 79.8% Cl
1. Assume 100g sample: 20.2g Al, 79.8g Cl
2. Convert to moles:

Al: 20.2g / 26.98 g/mol = 0.75 mol

Cl: 79.8g / 35.45 g/mol = 2.25 mol
3. Find simplest ratio:

Al: 0.75 mol / 0.75 mol = 1

Cl: 2.25 mol / 0.75 mol = 3
Therefore, the empirical formula is AlCl₃.
(c) 40.1% C, 6.6% H, 53.3% O
1. Assume 100g sample: 40.1g C, 6.6g H, 53.3g O
2. Convert to moles:

C: 40.1g / 12.01 g/mol = 3.34 mol

H: 6.6g / 1.008 g/mol = 6.55 mol

O: 53.3g / 16.00 g/mol = 3.33 mol
3. Find simplest ratio:

C: 3.34 mol / 3.33 mol = 1

H: 6.55 mol / 3.33 mol = 2

O: 3.33 mol / 3.33 mol = 1
Therefore, the empirical formula is CH₂O.
(d) 18.4% 8, 21.5% N, 60.1% K
1. Assume 100g sample: 18.4g 8, 21.5g N, 60.1g K
2. Convert to moles:

8: 18.4g / 79.90 g/mol = 0.23 mol

N: 21.5g / 14.01 g/mol = 1.53 mol

K: 60.1g / 39.10 g/mol = 1.54 mol
3. Find simplest ratio:

8: 0.23 mol / 0.23 mol = 1

N: 1.53 mol / 0.23 mol = 7

K: 1.54 mol / 0.23 mol = 7
Therefore, the empirical formula is K₇N₇O.