BTNT `C:`

`n_C = n_{CO_2} = (44)/(44) = 1` mol

BTNT `H:`

`n_H = 2n_{H_2O} = 2 . (27)/(18) = 3` mol

Xét

`m_C + m_H = 1 . 12 + 3 . 1 = 15g 23g`

`=> A` chứa `C, H, O`

`=> n_O = (23 - 15)/(16) = 0,5` mol

`n_C : n_H : n_O = 1 : 3 : 0,5 = 2 : 6 : 1`

`=>`CT Đơn giản của `A` là `(C_2H_6O)_n`

Mà `M_A = 23 . 2 = 46g//mol`

`=> (12 . 2 + 1 . 6 + 16)n = 46`

`=> n = 1`

`=>`CTPT của `A: C_2H_6O`

CTCT:

$\rm \hspace{1cm}H\hspace{,55cm}H\\\hspace{1,1cm}|\hspace{,75cm}|\\H-C-C-O-H\\\hspace{1,1cm}|\hspace{,75cm}|\\\hspace{1cm}H\hspace{,55cm}H$

$\rm \hspace{1cm}H\hspace{1,5cm}H\\\hspace{1,1cm}|\hspace{1,75cm}|\\H-C-O-C-H\\\hspace{1,1cm}|\hspace{1,75cm}|\\\hspace{1cm}H\hspace{1,5cm}H$

`A` `+` $O_{2}$ㅤ $\xrightarrow[]{t^{0}}$ㅤ $CO_{2}$ `+` $H_{2}$`O
$n_{CO_{2}}$ `=` $\dfrac{44}{44}$ `=` `1` `(mol)`
$n_{H_{2}O}$ `=` $\dfrac{27}{18}$ `=` `1,5` `(mol)`
Theo định luật bảo toàn nguyên tố`:`
$n_{C}$ `=` $n_{CO_{2}}$ `=` `1` `(mol)` `⇒` $m_{C}$ `=` `12` `(g)`
$n_{O}$ `=` $2n_{H_{2}O}$ `=` `2` `.` `1,5` `=` `3` `(mol)` `⇒` $m_{H}$ `=` `3` `(g)`
Ta có`:` $m_{C}$ `+` $m_{H}$ `=` `12` `+` `3` `=` `15` `(g)` `;` $m_{A}$ `=` `23` `(g)`
`⇒` $m_{C}$ `+` $m_{H}$ `<` $m_{A}$ ㅤ`(15 < 23)`
Vậy chứng tỏ trong `A` có nguyên tố `O`
$m_{O}$ `=` $m_{A}$ `-` $m_{C}$ `-` $m_{H}$ `=` `23` `-` `12` `-` `3` `=` `8` `(g)`
$n_{O}$ `=` $\dfrac{8}{16}$ `=` `0,5` `(mol)`
`CT``TQ:` $C_{x}$$H_{y}$$O_{z}$ ㅤ`(` `x; y; z ∈ `$Z^{+}$`)` 
Ta có`:` `x` `:` `y` `:` `z` `=` $n_{C}$ `:`  $n_{H}$ `:`  $n_{O}$ `=` `1` `:` `3` `:` `0,5` `=` `2` `:` `6` `:` `1` 
`CTĐG:` $(C2H6O)_{n}$  `(` `n ∈ `$Z^{+}$`)` 
Theo đề bài ta có`:` $d_{A/H_{2}}$ `=` $\dfrac{M_{A}}{M_{H_{2}}}$ `=` `23`
`⇒` $M_{A}$ `=` `46` `(g``/mol)` 
`⇒` `(12 ` `.` `2` `+` `1` `.` `6` `+` `16)``.` `n` `=` `46`
`⇔` `46n` `=` `46`
`⇔` `n` `=` `1` `(TMĐK)`
`⇒` `CTPT` của `A` là`:` $C_{2}$$H_{6}$`O`
`CTCT` của `A` là`:` $CH_{3}$ `-`  $CH_{2}$ `-` `OH` hoặc $CH_{3}$ `-` `O` `-` $CH_{3}$

$\text{#KL0112}$