Question

Bromoform is 94.85% bromine, 0.40% hydrogen, and 4.75% carbon by mass. Determine its empirical formula.

Short answer

The empirical formula is${\text{CHBr}}_3$

Step-by-step solution

Conversion of mass to number of moles

Mass can be converted to the number of moles by dividing the mass by the molar mass of the elements.

Mass of hydrogen,${\text{m}}_{\text{H}}=0.40\text{g}$

Molar mass of hydrogen,${\text{M}}_{\text{H}}=1{\text{gmol}}^{-1}$

Number of moles of hydrogen,localid="1663422792950" ${\text{n}}_{\text{H}}=\frac{{\text{m}}_{\text{H}}}{{\text{M}}_{\text{H}}}=\frac{0.40}{1}=0.40\text{mol}$

Mass of bromine,localid="1663422338088" ${\text{m}}_{\text{Br}}=94.85\text{g}$

Molar mass of bromine,

${\text{n}}_{\text{Br}}=\frac{{\text{m}}_{\text{Br}}}{{\text{M}}_{\text{Br}}}=\frac{94.85}{79.9}=1.1871\text{mol}$

Step 2: Simplification of ratios

The smallest value is of hydrogen. Now, we divide each number of moles by this smallest value.

$\text{H =}\frac{0.40}{0.40}=1$

$\text{C =}\frac{0.395}{0.40}=1$

$\text{Br =}\frac{1.18}{0.40}=3$

Hence, the empirical formula is${\text{CHBr}}_3$.