Question 717ba

Feb 10, 2017

${\text{Al(OH)}}_{3}$ is $\text{34.58 % Al}$, $\text{61.54 % O}$, and $\text{3.88 % H}$ by mass.

Explanation:

The formula for percent composition is

color(blue)(bar(ul(|color(white)(a/a)"% by mass" = "mass of component"/"mass of sample" × 100 %color(white)(a/a)|)))" "

We must therefore find the masses of $\text{Al, O}$, and $\text{H}$ in a given mass of the compound.

Let's choose 1 mol of ${\text{Al(OH)}}_{3}$.

We rewrite the formula as ${\text{AlO"_3"H}}_{3}$ for easy calculation.

$\text{Mass of Al"color(white)(l) = color(white)(m) "1 × 26.98 g" color(white)(ll)= "26.98 g}$
$\text{Mass of O" = color(white)(ll)" 3 × 16.00 g" color(white)(ll)= "48.00 g}$
$\text{Mass of H"color(white)(l) = color(white)(ml)3 × color(white)(l)"1.008 g" color(white)(l) = color(white)(l) "3.024 g}$
stackrel(——————————————————)("Mass of Al(OH)"_3color(white)(mmmmml) =color(white)(l) "78.004 g")

"% Al" = "mass of Al"/("mass of Al(OH)"_3) × 100 % = (26.98 color(red)(cancel(color(black)("g"))))/(78.004 color(red)(cancel(color(black)("g")))) × 100 % = 34.58 %

"% O" = "mass of O"/("mass of Al(OH)"_3) × 100 % = (48.00 color(red)(cancel(color(black)("g"))))/(78.004 color(red)(cancel(color(black)("g")))) × 100 % = 61.54 %

"% H" = "mass of H"/("mass of Al(OH)"_3) × 100 % = (3.024 color(red)(cancel(color(black)("g"))))/(78.004 color(red)(cancel(color(black)("g")))) × 100 % = 3.88 %#