Gọi công thức tổng quát của A là: NaxCyHzO2
Ta có: \(n_A=\frac{14,4}{23x+12y+z+32}\)
\(\Rightarrow n_{Na}=x.\frac{14,4}{23x+12y+z+32}\)
\(\Rightarrow n_C=y.\frac{14,4}{23x+12y+z+32}\)
\(\Rightarrow n_H=z.\frac{14,4}{23x+12y+z+32}\)
Ta lại có:
\(n_{CO_2}=\frac{28,6}{44}=0,65\)
\(n_{H_2O}=\frac{4,5}{18}=0,25\)
\(n_{Na_2CO_3}=\frac{5,3}{106}=0,05\)
\(\Rightarrow n_H=2n_{H_2O}=2.0,25=0,5=z.\frac{14,4}{23x+12y+z+32}\)
\(\Leftrightarrow-23x-12y+27,8z=32\left(1\right)\)
\(\Rightarrow n_{Na}=2n_{Na_2CO_3}=2.0,05=0,1=x.\frac{14,4}{23x+12y+z+32}\)
\(\Leftrightarrow121x-12y-z=32\left(2\right)\)
\(\Rightarrow n_C=n_{CO_2}+n_{Na_2CO_3}=0,65+0,05=0,7=y.\frac{14,4}{23x+12y+z+32}\)
\(\Leftrightarrow-161x+60y-7z=224\left(3\right)\)
Từ (1), (2), (3) ta có hệ: \(\left\{\begin{matrix}-23x-12y+27,8z=32\\121x-12y-z=32\\-161x+60y-7z=224\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=1\\y=7\\z=5\end{matrix}\right.\)
Vậy CT của A là: C7H5O2Na
