\(\begin{array}{l}
3)\\
nC{O_2} = \frac{{0,44}}{{44}} = 0,01\,mol\\
n{H_2}O = \frac{{0,18}}{{18}} = 0,01\,mol\\
mC + mH = 0,01 \times 12 + 0,02 \times 1 = 0,14g\\
= > mO = 0,3 - 0,14 = 0,16g\\
nO = 0,16/16 = 0,01mol\\
nC:nH:nO = 0,01:0,02:0,01 = 1:2:1\\
= > CTDGN:C{H_2}O\\
MA = \frac{{0,3}}{{0,01}} = 30g/mol\\
30n = 30 = > n = 1\\
= > CTPTA:C{H_2}O\\
4)\\
nC{O_2} = \frac{{4,4}}{{44}} = 0,1\,mol\\
n{H_2}O = \frac{{1,8}}{{18}} = 0,1\,mol\\
mC + mH = 0,1 \times 12 + 0,2 \times 1 = 1,4g\\
= > mO = 2,2 - 1,4 = 0,8g\\
nO = 0,8/16 = 0,05mol\\
nC:nH:nO = 0,1:0,2:0,05 = 2:4:1\\
= > CTDGN:{C_2}{H_4}O\\
MA = \frac{{1,1}}{{0,0125}} = 88g/mol\\
44n = 88 = > n = 2\\
= > CTPTA:{C_4}{H_8}{O_2}
\end{array}\)
