Дано:
$A_{r}$ (S) = 32
$A_{r}$ (O) = 16
$SO_{2}$
Найти:
ω (S) − ?
ω (O) − ?
Решение:
$M_{r}$ ($SO_{2}$) = 32 + 16 * 2 = 64
ω (S) = $\frac{n\; * \;A_{r}}{M_{r}}$ = $\frac{1\; * \;32}{64}$ = 50%
ω (O) = $\frac{n\; * \;A_{r}}{M_{r}}$ = $\frac{2\; * \;16}{64}$ = 50%
Ответ: ω (S) = 50%; ω (O) = 50%.

Дано:
$A_{r}$ (K) = 39
$A_{r}$ (Cl) = 35,5
$A_{r}$ (O) = 16
$KClO_{3}$
Найти:
ω (K) − ?
ω (Cl) − ?
ω (O) − ?
Решение:
$M_{r}$ ($KClO_{3}$) = 39 + 35,5 + 16 * 3 = 122,5
ω (K) = $\frac{n\; * \;A_{r}}{M_{r}}$ = $\frac{1\; * \;39}{122,5}$ = 31,8%
ω (Cl) = $\frac{n\; * \;A_{r}}{M_{r}}$ = $\frac{1\; * \;35,5}{122,5}$ = 29%
ω (O) = $\frac{n\; * \;A_{r}}{M_{r}}$ = $\frac{3\; * \;16}{122,5}$ = 39,2%
Ответ: ω (K) = 31,8%; ω (Cl) = 29%; ω (O) = 39,2%.