\(\begin{array}{l}
7)\\
nC{O_2} = \frac{{2,24}}{{22,4}} = 0,1mol\\
n{H_2}O = \frac{{3,6}}{{18}} = 0,2mol\\
m = 0,1 \times 12 + 0,2 \times 2 = 1,6g\\
\% mC = \frac{{1,2}}{{1,6}} = 75\% \\
\% mH = 25\% \\
b)\\
nC:nH = 0,1:0,4 = 1:4\\
= > CTDGN:C{H_4}\\
CTPT:C{H_4}\\
8)\\
nC{O_2} = \frac{{1,32}}{{44}} = 0,03mol\\
n{H_2}O = \frac{{0,54}}{{18}} = 0,03mol\\
mO = 0,9 - 0,03 \times 12 - 0,06 = 0,48mol\\
nO = \frac{{0,48}}{{16}} = 0,03mol\\
nC:nH:nO = 0,03:0,06:0,03 = 1:2:1\\
= > CTDGN:C{H_2}O\\
MA = 180 = > 30n = 180 = > n = 6\\
= > CTPT:{C_6}{H_{12}}{O_6}
\end{array}\)
