Đặt CTTQ ankan là $C_nH_{2n+2}$

a,

$2n+2=16$

$\Leftrightarrow n=7$

Vậy CTPT ankan là $C_7H_{16}$

b,

$M=12n+2n+2=14n+2$

$\Rightarrow 14n+2=72$

$\Leftrightarrow n=5$

Vậy CTPT ankan là $C_5H_{12}$

c,

$\%m_H=17,24\%$

$\Rightarrow \dfrac{2n+2}{14n+2}=17,24\%$

$\Leftrightarrow 2n+2=17,24\%(14n+2)$

$\Leftrightarrow n=4$

Vậy CTPT ankan là $C_4H_{10}$

d,

Đặt CTPT: $C_{3n}H_{7n}$

Xét $1$ mol ankan. 

Đốt cháy $1$ mol $C_{3n}H_{7n}$ thu được:

$n_{CO_2}=3n (mol)$

$n_{H_2O}=3,5n (mol)$

$\Rightarrow 3,5n-3n=1$

$\Leftrightarrow n=2$

Vậy CTPT ankan là $C_6H_{14}$

Em tham khảo nha :

\(\begin{array}{l}
CTPT:{C_n}{H_{2n + 2}}\\
a)\\
\text{Ankan có 16 nguyên  tử H}\\
2n + 2 = 16\\
 \Rightarrow n = \frac{{16 - 2}}{2} = 7\\
 \Rightarrow CTPT:{C_7}{H_{16}}\\
b)\\
{M_B} = 72dvC\\
 \Rightarrow 12n + 2n + 2 = 72\\
 \Rightarrow n = \dfrac{{72 - 2}}{{14}} = 5\\
 \Rightarrow CTPT:{C_5}{H_{12}}\\
c)\\
\% H = \dfrac{{(2n + 2){M_H}}}{{n{M_C} + (2n + 2){M_H}}} \times 100\%  = 17,24\% \\
 \Rightarrow \dfrac{{2n + 2}}{{14n + 2}} = 0,1724\\
 \Rightarrow n = 4\\
 \Rightarrow CTPT:{C_4}{H_{10}}\\
d)\\
CTDGN:{({C_3}{H_7})_x}\\
 \Rightarrow x = 2\\
 \Rightarrow CTPT:{C_6}{H_{14}}
\end{array}\)