Question

URGENT !! A substance has 55.80% carbon, 7.04% Hydrogen, and 37.16% Oxygen. What is it's empirical and molecular formula if it has a molar mass of 301.35 grams?

Answer 1

Answer;

Empirical formula = C₂H₃O

Molecular formula = C₁₄H₂₁O₇

Step-by-step explanation;

Empirical formula

Moles of;

Carbon = 55.8 /12 = 4.65 moles

Hydrogen = 7.04/ 1 = 7.04 moles

Oxygen = 37.16/ 16 = 2.3225 moles

We then get the mole ratio;

4.65/2.3225 = 2.0

7.04/2.3225 = 3.0

2.3225/2.3225 = 1.0

Therefore;

The empirical formula = C₂H₃O

Molecular formula;

(C2H3O)n = 301.35 g

(12 ×2 + 3× 1 + 16×1)n = 301.35

43n = 301.35

n = 7

Therefore;

Molecular formula = (C2H3O)7

= C₁₄H₂₁O₇