Answer to Question #165856 in Chemistry for Ava Brooke

Question

What is the empirical formula if 38.23 % Oxygen, 50.61 % Cu, and 11.16 % Nitrogen?

Answer 1

Solution:

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

Molar mass of copper (Cu) is 63.546 g mol^-1.

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

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

Hence,

Mass of O = w(O) × Mass of sample = 0.3823 × 100 g = 38.23 g

Mass of Cu = w(Cu) × Mass of sample = 0.5061 × 100 g = 50.61 g

Mass of N= w(N) × Mass of sample = 0.1116 × 100 g = 11.16 g

Convert to moles.

38.23 g O × (1 mol O / 15.999 g O) = 2.3895 mol O

50.61 g Cu × (1 mol Cu / 63.546 g Cu) = 0.79643 mol Cu

11.16 g N × (1 mol N / 14.0067 g N) = 0.79676 mol N

Divide both moles by the smallest of the results.

O: 2.3895 / 0.79643 = 3.000

Cu: 0.79643 / 0.79643 = 1.000

N: 0.79676 / 0.79643 = 1.000

The empirical formula of the compound is CuNO₃

Answer: The empirical formula of the compound is CuNO₃.