Výpočet procentuálního složení propantriolu
Určete procentuální složení následující sloučeniny:
propantriolu
\( \large\begin{array}{l}\mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{3\cdot M_{r}(\mathrm{C})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{36,03}{92,03}=0,3915\Rightarrow\bm{39,15}\%\mathbf{C}\\ \mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{8\cdot M_{r}(\mathrm{H})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{8}{92,03}=0,0869\Rightarrow\bm{8,69}\%\mathbf{H}\\ \mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{3\cdot M_{r}(\mathrm{O})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{48}{92,03}=0,5216\Rightarrow\bm{52,16}\%\mathbf{O}\end{array} \)
\( \large\begin{array}{l}\mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{3\cdot M_{r}(\mathrm{C})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{36,03}{92,03}=0,3915\Rightarrow\bm{39,15}\%\mathbf{C}\\ \mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{9\cdot M_{r}(\mathrm{H})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{8}{92,03}=0,0869\Rightarrow\bm{8,69}\%\mathbf{H}\\ \mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{3\cdot M_{r}(\mathrm{O})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{48}{92,03}=0,5216\Rightarrow\bm{52,16}\%\mathbf{O}\end{array} \)
\( \large\begin{array}{l}\mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{4\cdot M_{r}(\mathrm{C})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{36,03}{92,03}=0,3915\Rightarrow\bm{39,15}\%\mathbf{C}\\ \mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{8\cdot M_{r}(\mathrm{H})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{8}{92,03}=0,0869\Rightarrow\bm{8,69}\%\mathbf{H}\\ \mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{3\cdot M_{r}(\mathrm{O})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{48}{92,03}=0,5216\Rightarrow\bm{52,16}\%\mathbf{O}\end{array} \)
\( \large\begin{array}{l}\mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{3\cdot M_{r}(\mathrm{C})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{36,03}{92,03}=0,3915\Rightarrow\bm{39,15}\%\mathbf{C}\\ \mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{8\cdot M_{r}(\mathrm{H})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{8}{92,03}=0,0869\Rightarrow\bm{8,69}\%\mathbf{H}\\ \mathrm{f}\left(\mathrm{C}_3\mathrm{H}_5(\mathrm{OH})_3\right)=\frac{4\cdot M_{r}(\mathrm{O})}{M_{r}\left(C_3H_5(\mathrm{OH})_3\right)}=\frac{48}{92,03}=0,5216\Rightarrow\bm{52,16}\%\mathbf{O}\end{array} \)