What is the percent composition by weight of $A l$ in ${{A l}_{2} ({S O}_{4})}_{3}$ ?

7.9%

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31.6%

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15.8%

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12.7%

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Solution

The correct option is C 15.8%
% by weight $= \frac{M a s s o f c o m p o n e n t}{T o t a l m a s s} \times 100$ %
Mass of $2 A l$ atoms $= 2 \times 26.98 = 53.96 u$
Mass of $3 S$ atoms $= 3 \times 32.06 = 96.18 u$
Mass of $12 O$ atoms $= 12 \times 16 = 192 u$
Mass of $1 A l_{2} (S O_{4})_{2} = (53.96 + 96.18 + 192) u = 342.1 u$
Percentage by weight of $A l = \frac{53.96 u}{342.1 u} \times 100$
$= 15.8$ %

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