Đáp án:
\(\begin{array}{l}
a)\\
hh:{C_3}{H_7}OH,{C_4}{H_9}OH\\
b)\\
\% {m_{{C_3}{H_8}O}} = 54,88\% \\
\% {m_{{C_4}{H_{10}}O}} = 45,12\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{C{O_2}}} = \dfrac{{22,44}}{{44}} = 0,51\,mol\\
{n_{{H_2}O}} = \dfrac{{11,88}}{{18}} = 0,66\,mol\\
{n_{hh}} = 0,66 - 0,51 = 0,15\,mol\\
C = \dfrac{{0,51}}{{0,15}} = 3,4 \Rightarrow hh:{C_3}{H_7}OH,{C_4}{H_9}OH\\
CTCT:\\
{C_3}{H_7}OH:\\
C{H_3} - C{H_2} - C{H_2} - OH:propan - 1 - ol\\
C{H_3} - CH(OH) - C{H_3}:propan - 2 - ol
\end{array}\)
\(\begin{array}{l}
C{H_3} - C{H_2} - C{H_2} - C{H_2} - OH:bu\tan - 1 - ol\\
C{H_3} - CH(C{H_3}) - C{H_2} - OH:2 - metyl\,propan - 1 - ol\\
C{H_3} - C{H_2} - CH(OH) - C{H_3}:bu\tan - 2 - ol\\
C{H_3} - C(OH)(C{H_3}) - C{H_3}:2 - metyl\,propan - 2 - ol
\end{array}\)
\(\begin{array}{l}
b)\\
hh:{C_3}{H_8}O(a\,mol),{C_4}{H_{10}}O(b\,mol)\\
\left\{ \begin{array}{l}
3a + 4b = 0,51\\
4a + 5b = 0,66
\end{array} \right.\\
\Rightarrow a = 0,09;b = 0,06\\
\% {m_{{C_3}{H_8}O}} = \dfrac{{0,09 \times 60}}{{0,09 \times 60 + 0,06 \times 74}} \times 100\% = 54,88\% \\
\% {m_{{C_4}{H_{10}}O}} = 100 - 54,88 = 45,12\%
\end{array}\)
