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Вычислите массовые доли элементов в следующих соединениях:
а) CaCO3; б) CuSO4; в) C7H16; г) C5H8O2.
а)
Ответ: `ω(Ca) = 40%`, `ω(C) = 12%`, `ω(O) = 48%`. |
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| Дано: | Решение |
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`CaCO_3` |
`ω(Ca) = (100*k*A_r(Ca))/(M_r(CaCO_3)) = (100*1*40)/100 = 40%` `ω(C) = (100*k*A_r(C))/(M_r(CaCO_3)) = (100*1*12)/100 = 12%` `ω(O) = (100*k*A_r(O))/(M_r(CaCO_3)) = (100*3*16)/100 = 48%` |
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`ω(Ca) = ?` `ω(C) = ?` `ω(O) = ?` |
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б)
Ответ: `ω(Cu) = 40%`, `ω(S) = 20%`, `ω(O) = 40%`. |
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| Дано: | Решение |
|---|---|
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`CuSO_4` |
`ω(Cu) = (100*k*A_r(Cu))/(M_r(CuSO_4)) = (100*1*64)/160 = 40%` `ω(S) = (100*k*A_r(S))/(M_r(CuSO_4)) = (100*1*32)/160 = 20%` `ω(O) = (100*k*A_r(O))/(M_r(CuSO_4)) = (100*4*16)/160 = 40%` |
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`ω(Cu) = ?` `ω(S) = ?` `ω(O) = ?` |
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в)
Ответ: `ω(C) = 84%`, `ω(H) = 16%`. |
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| Дано: | Решение |
|---|---|
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`C_7H_16` |
`ω(C) = (100*k*A_r(C))/(M_r(C_7H_16)) = (100*7*12)/100 = 84%` `ω(H) = (100*k*A_r(H))/(M_r(C_7H_16)) = (100*16*1)/100 = 16%` |
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`ω(C) = ?` `ω(H) = ?` |
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г)
Ответ: `ω(C) = 60%`, `ω(H) = 8%`, `ω(O) = 32%`. |
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| Дано: | Решение |
|---|---|
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`C_5H_8O_2` |
`ω(C) = (100*k*A_r(C))/(M_r(C_5H_8O_2)) = (100*5*12)/100 = 60%` `ω(H) = (100*k*A_r(H))/(M_r(C_5H_8O_2)) = (100*8*1)/100 = 8%` `ω(O) = (100*k*A_r(O))/(M_r(C_5H_8O_2)) = (100*2*16)/100 = 32%` |
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`ω(C) = ?` `ω(H) = ?` `ω(O) = ?` |
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