Answer to Question #165860 in Chemistry for Ava Brooke

Question

What is the empirical formula if 51.23 % Iron, and 48.77 % Chlorine? 

Answer 1

Solution:

Molar mass of iron (Fe) is 55.845 g mol^-1.

Molar mass of chlorine (Cl) is 35.453 g mol^-1.

Assume a 100g sample, convert the same % values to grams.

Hence,

Mass of Fe = w(Fe) × Mass of sample = 0.5123 × 100 g = 51.23 g

Mass of Cl = w(Cl) × Mass of sample = 0.4877 × 100 g = 48.77 g

Convert to moles.

51.23 g Fe × (1 mol Fe / 55.845 g Fe) = 0.91736 mol Fe

48.77 g Cl × (1 mol Cl / 35.453 g Cl) = 1.3756 mol Cl

Divide both moles by the smallest of the results.

Fe: 0.91736 / 0.91736 = 1.000

Cl: 1.3756 / 0.91736 = 1.499

The "non-whole number" empirical formula of the compound is Fe₁Cl_1.5

Multiply each of the moles by the smallest whole number that will convert each into a whole number.

Fe:O = 2 × (1:1.5) = 2:3

The empirical formula of the compound is Fe₂Cl₃

Answer: The empirical formula of the compound is Fe₂Cl₃.