Answer to Question #165861 in Chemistry for Ava Brooke

Question

What is the empirical formula if 29.16 % Nitrogen, 8.39 % Hydrogen, 12.50 % Carbon, and 49.95 % Oxygen?

Answer 1

Solution:

Molar mass of nitrogen (N) is 14.0067 g mol^-1.

Molar mass of hydrogen (H) is 1.00784 g mol^-1.

Molar mass of carbon (C) is 12.0107 g mol^-1.

Molar mass of oxygen (O) is 15.999 g mol^-1.

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

Hence,

Mass of N = w(N) × Mass of sample = 0.2916 × 100 g = 29.16 g

Mass of H = w(H) × Mass of sample = 0.0839 × 100 g = 8.39 g

Mass of C = w(C) × Mass of sample = 0.1250 × 100 g = 12.50 g

Mass of O = w(O) × Mass of sample = 0.4995 × 100 g = 49.95 g

Convert to moles.

29.16 g N × (1 mol N / 14.0067 g N) = 2.08186 mol N

8.39 g H × (1 mol H / 1.00784 g H) = 8.32473 mol H

12.50 g C × (1 mol C / 12.0107 g C) = 1.04074 mol C

49.95 g O × (1 mol O / 15.999 g O) = 3.12207 mol O

Divide all moles by the smallest of the results.

N: 2.08186 / 1.04074 = 2.000 = 2

H: 8.32473 / 1.04074 = 7.999 = 8

C: 1.04074 / 1.0407 = 1.000 = 1

O: 3.12207 / 1.0407 = 2.999 = 3

The empirical formula of the compound is N₂H₈CO₃

Possible compound formula: (NH₄)₂CO₃ - ammonium carbonate.

Answer: The empirical formula of the compound is N₂H₈CO₃.