Bài 20: Mở đầu về hóa học hữu cơ

\(n_X=n_{O_2}=\dfrac{0,224}{32}=0,007\left(mol\right)\)

\(M_X=\dfrac{0,42}{0,007}=60\) \((g/mol)\)

\(PTHH:X+O_2\rightarrow\left(t^o\right)CO_2+H_2O\)

\(m_{O_2}=32.\dfrac{10,08}{22,4}=14,4\left(g\right)\)

\(BTKL:m_X+m_{O_2}=m_{CO_2}+m_{H_2O}\)

\(\Rightarrow m_{CO_2}+m_{H_2O}=6+14,4=20,4\left(g\right)\) (1)

Ta có: \(\dfrac{m_{CO_2}}{m_{H_2O}}=\dfrac{11}{6}\)  \(\Leftrightarrow6m_{CO_2}-11m_{H_2O}=0\) (2)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}m_{CO_2}=13,2\\m_{H_2O}=7,2\end{matrix}\right.\)

Bảo toàn C: \(n_C=n_{CO_2}=\dfrac{13,2}{44}=0,3\left(mol\right)\)

Bảo toàn H: \(n_H=2.n_{H_2O}=2.\dfrac{7,2}{18}=0,8\left(mol\right)\)

\(n_O=\dfrac{6-\left(0,3.12+0,8.1\right)}{16}=0,1\left(mol\right)\)

Đặt CTTQ X: \(C_xH_yO_z\)

\(x:y:z=0,3:0,8:0,1=3:8:1\)

\(CTĐG:\left(C_3H_8O\right)_n=60\)

                 \(\Rightarrow n=1\)

\(\Rightarrow CTPTX:C_3H_8O\)

$\rm m_{CO_2} + m_{H_2O} = m_{A} + m_{O_2} = 6 = 14,4 = 20,4 (g)$
Vì $\rm m_{CO_2} : m_{H_2O} = 11 : 6$

$\rightarrow \begin{cases} \rm m_{CO_2} = \dfrac{11}{11 + 6} . 20,4 = 13,2 (g) \\ \rm m_{H_2O} = 20,4 - 13,2 = 7,2 (g) \end{cases}$

$\rightarrow \begin{cases} \rm n_{CO_2} = \dfrac{13,2}{44} = 0,3 (mol) \\ \rm n_{H_2O} = \dfrac{7,2}{18} = 0,4 (mol) \end{cases}$

Theo BTNT:
$\rightarrow \begin{cases} \rm n_{C} = n_{CO_2} = 0,3 (mol) \\ \rm n_{H} = 2n_{H_2O} = 0,8 (mol) \\ \rm n_{O(X)} = 2n_{CO_2} + n_{H_2O} - 2n_{O_2} = 0,1 (mol) \end{cases}$

$\rm \rightarrow n_C : n_H : n_O = 0,3 : 0,8 : 0,1 = 3 : 8 : 1$

$\rm \rightarrow X: (C_3H_8O)_n$
Mà $\rm M_X = \dfrac{0,42}{\dfrac{0,224}{32}} = 60 (g/mol)$
$\rm \rightarrow n = \dfrac{60}{60} = 1$

$\rm \rightarrow X:C_3H_8O$
CTCT của X:

$\rm CH_3-CH_2-CH_2-OH$
$\rm CH_3-CH(OH)-CH_3$
$\rm C_2H_5-O-CH_3$