Дано:
$CH_{4}$
$C_{2}H_{4}$
$C_{2}H_{2}$
Найти:
$ω_{CH_{4}}$ ($H$) − ?
$ω_{C_{2}H_{4}}$ ($H$) − ?
$ω_{C_{2}H_{2}}$ ($H$) − ?
Решение:
$M_{r}$ ($CH_{4}$) = 12 + 4 * 1 = 16
$M_{r}$ ($C_{2}H_{4}$) = 2 * 12 + 4 * 1 = 28
$M_{r}$ ($C_{2}H_{2}$) = 2 * 12 + 2 * 1 = 26
ω = $\frac{n\; * \;A_{r}}{M_{r}}$ * 100%
$ω_{CH_{4}}$ ($H$) = $\frac{4\; * \;1}{16}$ * 100% = 25%
$ω_{C_{2}H_{4}}$ ($H$) = $\frac{4\; * \;1}{28}$ * 100% = 14,3%
$ω_{C_{2}H_{2}}$ ($H$) = $\frac{2\; * \;1}{26}$ * 100% = 7,7%
$ω_{CH_{4}}$ ($H$) > $ω_{C_{2}H_{4}}$ ($H$) > $ω_{C_{2}H_{2}}$ ($H$)
Ответ: $ω_{CH_{4}}$ ($H$) = 25%; $ω_{C_{2}H_{4}}$ ($H$) = 14,3%; $ω_{C_{2}H_{2}}$ ($H$) = 7,7%.