What the mass percent of aluminum in Al(OH)_3?

The molar mass of $A l = 26.98 , H = 1.0079 ,$ and $O = 16.00$.

Jan 2, 2017

$\textcolor{m a \ge n t a}{\text{34.59%}}$

Explanation:

In order to calculate the mass percent of aluminum in that ionic compound, we're going to need two things:

$\textcolor{b l u e}{\text{1: The formula mass of the entire compound}}$
$\textcolor{b l u e}{\text{2: The atomic mass of aluminum}}$

The formula mass of aluminum hydroxide is $78.00 \frac{g}{m o l}$ because:

$26.98 \frac{g}{m o l} + 3 \left(16.00 \frac{g}{\text{mol") + 3(1.0079 g/"mol}}\right)$ $= 78.00 \frac{g}{m o l}$

We already know the atomic mass of aluminum since that's already given.

Now, we have to use the following equation:

The numerator represents the mass of the desired atom, which is Al 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 Al"/("Molar Mass of Al(OH)"_3)" xx100%

"26.98 g/mol"/"78.00g/mol"xx100% = 34.59%