Procentuální složení hydrogensiřičitanu sodného
Určete procentuální složení následující sloučeniny:
hydrogensiřičitanu sodného
\( \large\begin{array}{l}\mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{M_{r}(\mathrm{H})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{1}{104,06}=0,0096\Rightarrow\bm{0,96\%~H~}\\ \mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{M_{r}(\mathrm{~S})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{32,06}{104,06}=0,3081=>\bm{30,81}\%\mathbf{S}\\ \mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{3\cdot M_{r}(\mathrm{O})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{48}{104,06}=0,4613\Rightarrow\bm{46,13}\%\mathrm{O}\end{array} \)
\( \large\begin{array}{l}\mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{M_{r}(\mathrm{H})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{1}{104,06}=0,0096\Rightarrow\bm{0,97\%~H~}\\ \mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{M_{r}(\mathrm{~S})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{32,06}{104,06}=0,3081=>\bm{30,81}\%\mathbf{S}\\ \mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{3\cdot M_{r}(\mathrm{O})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{48}{104,06}=0,4613\Rightarrow\bm{46,13}\%\mathrm{O}\end{array} \)
\( \large\begin{array}{l}\mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{M_{r}(\mathrm{H})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{1}{104,06}=0,0096\Rightarrow\bm{0,96\%~H~}\\ \mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{M_{r}(\mathrm{~S})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{32,06}{104,06}=0,3081=>\bm{30,82}\%\mathbf{S}\\ \mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{3\cdot M_{r}(\mathrm{O})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{48}{104,06}=0,4613\Rightarrow\bm{46,13}\%\mathrm{O}\end{array} \)
\( \large\begin{array}{l}\mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{M_{r}(\mathrm{H})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{1}{104,06}=0,0096\Rightarrow\bm{0,96\%~H~}\\ \mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{M_{r}(\mathrm{~S})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{32,06}{104,06}=0,3081=>\bm{30,81}\%\mathbf{S}\\ \mathrm{f}\left(\mathrm{NaHSO}_3\right)=\frac{3\cdot M_{r}(\mathrm{O})}{M_{r}\left(\mathrm{NaHSO}_3\right)}=\frac{48}{104,06}=0,4613\Rightarrow\bm{46,14}\%\mathrm{O}\end{array} \)