Đáp án:
\(\begin{array}{l}
a)\\
\% {m_C} = 52,17\% \\
\% {m_H} = 13,04\% \\
\% {m_O} = 34,79\% \\
b)\\
CTPT:{C_2}{H_6}O
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{C{O_2}}} = \dfrac{{0,224}}{{22,4}} = 0,01\,mol\\
{n_{{H_2}O}} = \dfrac{{0,27}}{{18}} = 0,015\,mol\\
{n_C} = {n_{C{O_2}}} = 0,01\,mol\\
{n_H} = 2{n_{{H_2}O}} = 0,03\,mol\\
\% {m_C} = \dfrac{{0,01 \times 12}}{{0,23}} \times 100\% = 52,17\% \\
\% {m_H} = \dfrac{{0,03}}{{0,23}} \times 100\% = 13,04\% \\
\% {m_O} = 100 - 52,17 - 13,04 = 34,79\% \\
b)\\
{m_O} = 0,23 - 0,01 \times 12 - 0,03 = 0,08g\\
{n_O} = \dfrac{{0,08}}{{16}} = 0,005\,mol\\
{n_C}:{n_H}:{n_O} = 0,01:0,03:0,005 = 2:6:1\\
\Rightarrow CTDGN:{C_2}{H_6}O\\
{M_X} = 1,586 \times 29 = 46g/mol\\
\Rightarrow 46n = 46 \Rightarrow n = 1\\
CTPT:{C_2}{H_6}O
\end{array}\)
