Дано:
$Al_{2}O_{3}$
Найти:
ω (Al) − ?
ω (O) − ?
Решение:
$M_{r} (Al_{2}O_{3})$ = 2 * 27 + 3 * 16 = 102
ω (O) = $\frac{3\; * \;16}{102}$ * 100% = 47%
ω (Al) = 100% − 47% = 53%
Ответ: ω (O) = 47%; ω (Al) = 53%.

Дано:
$BaSO_{4}$
Найти:
ω (Ba) − ?
ω (S) − ?
ω (O) − ?
Решение:
$M_{r} (BaSO_{4})$ = 137 + 32 + 4 * 16 = 233
ω (Ba) = $\frac{1\; * \;137}{233}$ * 100% = 59%
ω (O) = $\frac{4\; * \;16}{233}$ * 100% = 27%
ω (S) = 100% − 59% − 27% = 14%
Ответ: ω (Ba) = 59%; ω (O) = 27%; ω (S) = 14%.