Some Basic Concepts of Chemistry - Empirical and Molecular Formula

A compound contains $40.0\%$ Carbon, $6.71\%$ Hydrogen, and $53.29\%$ Oxygen. If the molar mass of the compound is $60\ g/mol$, find its empirical and molecular formulae.

1. Calculate relative moles:

• $C: \frac{40.0}{12.01} = 3.33$
• $H: \frac{6.71}{1.008} = 6.66$
• $O: \frac{53.29}{16.00} = 3.33$

1. Find simple ratio (divide by $3.33$):

• $C: \frac{3.33}{3.33} = 1$
• $H: \frac{6.66}{3.33} = 2$
• $O: \frac{3.33}{3.33} = 1$

Empirical Formula = $CH_2O$

1. Calculate $n$:

• Empirical Formula Mass = $12.01 + (2 \times 1.008) + 16.00 = 30.026\ g/mol$
• $n = \frac{Molecular\ Mass}{Empirical\ Formula\ Mass} = \frac{60}{30.026} \approx 2$

1. Molecular Formula:

• $Molecular\ Formula = 2 \times (CH_2O) = C_2H_4O_2$

First, the percentage of each element is divided by its atomic mass to find the molar ratio. These ratios are simplified by dividing by the smallest value. The empirical formula is determined from this ratio. Finally, the ratio of the given molecular mass to the empirical mass ($n$) is used to find the actual molecular formula.

Determine the molecular formula of a compound with empirical formula $CH_2$ and a vapour density of $42$.

1. Calculate Molecular Mass:

• $Molecular\ Mass = 2 \times Vapour\ Density = 2 \times 42 = 84\ u$

1. Calculate Empirical Formula Mass:

• $Empirical\ Formula\ Mass\ (CH_2) = 12 + (2 \times 1) = 14\ u$

1. Calculate $n$:

• $n = \frac{84}{14} = 6$

1. Find Molecular Formula:

• $Molecular\ Formula = 6 \times (CH_2) = C_6H_{12}$

Vapour density is first converted to molecular mass. Then, the empirical mass is calculated. The ratio of molecular mass to empirical mass gives the multiplier $n$, which is applied to the empirical formula.