What is the mass percent of hydrogen for C_3H_5OH?

Jan 2, 2017

color(red)(10.41%)

Explanation:

In order to compute the mass percent of hydrogen in ${C}_{3} {H}_{5} O H$ we're going to need two things:

$\textcolor{b r o w n}{\text{1: The formula mass of the entire compound}}$
$\textcolor{b r o w n}{\text{2: The atomic mass of hydrogen}}$

The formula mass of ${C}_{3} {H}_{5} O H$ is $58.08 \frac{g}{m o l}$ because:

$3 \left(12.01 \frac{g}{m o l}\right) + 6 \left(1.008 \frac{g}{\text{mol") + (16.00 g/"mol}}\right)$ $= 58.08 \frac{g}{m o l}$

The atomic mass of the hydrogen atom is $1.008 \frac{g}{m o l}$

Now, we have to use the following equation:

The numerator represents the mass of the desired atom, which is H in our case, and the denominator represents the mass of the entire compound. You just divide the two and multiply by 100 to obtain the percent composition:

"Mass of H"/("Molar Mass of C"_3"H"_5"OH") xx100%

"(6)1.008 g/mol"/"58.08g/mol"xx100% = 10.41%

I multiplied the atomic mass of hydrogen by 6 because you have to account for all of the hydrogen atoms in the formula.