# Question #373c7

Mar 27, 2017

Empirical formula is ${C}_{4} {H}_{9}$

Molecular formula is ${C}_{8} {H}_{18}$

#### Explanation:

You've been given the composition in percent which we can write as grams and then divide by the mass of the element.

$C \Rightarrow \frac{84.21 \cancel{g} \cancel{C}}{12 \cancel{g} \cancel{C}} = 7.0175 m o l$

$H \Rightarrow \frac{15.79 \cancel{g} \cancel{C}}{1 \cancel{g} \cancel{C}} = 15.79 m o l$

Now divide each by the smallest mole. i.e: $7.0175$

$C \Rightarrow \frac{7.0175 \cancel{m o l}}{7.0175 \cancel{m o l}} = 1.0$

$H \Rightarrow \frac{15.79 \cancel{m o l}}{7.0175 \cancel{m o l}} = 2.25$

You realize that 2.3 is not a whole number as compared to 1.0. So now we multiply by a number that will give us a whole number for H.

If you multiply by $1 , 2 , \mathmr{and} 3$, you still don't get a whole number until you multiply by 4.

Rememember whatever you do to H, you do same to C.

$C \Rightarrow 1.0 \cdot 4 = 4$

$H \Rightarrow 2.25 \cdot 4 = 9$

$\therefore$ The empirical formula is ${C}_{4} {H}_{9}$

Molecular formula is found by:

$n$(empirical formula)

$n =$ (molecular weight)/(molar mass)

Molar mass of ${C}_{4} {H}_{9} = \left(12 \cdot 4\right) + \left(1 \cdot 9\right) = 57 \frac{g}{m o l}$

$n = \frac{114 \frac{\cancel{g}}{\cancel{m o l}}}{57 \frac{\cancel{g}}{\cancel{m o l}}} = 2$

$n \left({C}_{4} {H}_{9}\right) = 2 \left({C}_{4} {H}_{9}\right)$

$\therefore$ the molecular formula is ${C}_{8} {H}_{18}$