Question

A compound contains 46.7% N and 53.3% O. If the molecular mass of the compound is 60.0g/mol. What is the molecular formula?

Answer 1

46.7% Nitrogen

53.3% Oxygen

Firstly divide by their respective molar mass

$\dfrac{46.7}{14}= 3.34$ Nitrogen

$\dfrac{53.3}{16}= 3.33$ Oxygen

Then Divide by the lowest result

$\dfrac{3.34}{3.33} \approx 1$ Nitrogen

$\dfrac{3.33}{3.33} \approx 1$ Oxygen

The empirical formula is therefore, $NO\ (N_1O_1)$

The molecular mass is 60.0 g/mol

$\therefore (NO)n = 60\\ (14+16)n = 60\\ (30)n = 60\\ n = 2$

$\therefore$ The molecular formula is $N_2O_2$