Формула расчёта: \(w(\text{элемента})=\frac{A_r(\text{элемента})\cdot n}{M_r(\text{вещества})}\cdot100\%\).

NH₃: \(M_r=14+3\cdot1=17\).
\(w(\mathrm N)=\frac{14}{17}\cdot100\%\approx82.35\%\); \(w(\mathrm H)=\frac{3}{17}\cdot100\%\approx17.65\%\).
H₂SiO₃: \(M_r=2\cdot1+28+3\cdot16=78\).
\(w(\mathrm H)=\frac{2}{78}\cdot100\%\approx2.56\%\); \(w(\mathrm{Si})=\frac{28}{78}\cdot100\%\approx35.90\%\); \(w(\mathrm O)=\frac{48}{78}\cdot100\%\approx61.54\%\).
Fe(OH)₃: \(M_r=56+3\cdot(16+1)=107\).
\(w(\mathrm{Fe})=\frac{56}{107}\cdot100\%\approx52.34\%\); \(w(\mathrm O)=\frac{48}{107}\cdot100\%\approx44.86\%\); \(w(\mathrm H)=\frac{3}{107}\cdot100\%\approx2.80\%\).
Mn₂O₇: \(M_r=2\cdot55+7\cdot16=222\).
\(w(\mathrm{Mn})=\frac{110}{222}\cdot100\%\approx49.55\%\); \(w(\mathrm O)=\frac{112}{222}\cdot100\%\approx50.45\%\).

Ответ: NH₃ — N 82.35%, H 17.65%; H₂SiO₃ — H 2.56%, Si 35.90%, O 61.54%; Fe(OH)₃ — Fe 52.34%, O 44.86%, H 2.80%; Mn₂O₇ — Mn 49.55%, O 50.45%.

Поделись с друзьями