Đáp án:

\(CTPT:{C_5}{H_{10}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{m_{{H_2}O}} = m \text{ bình 1 tăng } = 18g\\
{n_{{H_2}O}} = \dfrac{{18}}{{18}} = 1\,mol\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{{100}}{{100}} = 1\,mol\\
{n_{{O_2}}} = \dfrac{{33,6}}{{22,4}} = 1,5\,mol\\
BTNT\,O:{n_{O\,trong\,A}} + 2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}}\\
 \Rightarrow {n_{O\,trong\,A}} = 2 \times 1 + 1 - 1,5 \times 2 = 0\\
 \Rightarrow A:C,H\\
{n_A} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
C = \dfrac{1}{{0,2}} = 5;H = \frac{2}{{0,2}} = 10\\
 \Rightarrow CTPT:{C_5}{H_{10}}\\
b)\\
C{H_2} = CH - C{H_2} - C{H_2} - C{H_3}\\
C{H_2} = C(C{H_3}) - C{H_2} - C{H_3}\\
C{H_3} - CH = CH - C{H_2} - C{H_3}\\
C{H_3} - C(C{H_3}) - CH - C{H_3}
\end{array}\)