2 câu trả lời
$M_A=12.2+6+16=46(g/mol)\\ \%_C=\dfrac{2.12}{46}.100\%\approx 52,17\%\\ \%_H=\dfrac{6}{46}.100\%\approx 13,04\%\\ \%_O=100\%-13,04\%-52,17\%=34,79\%$
$M_{C_2H_6O}=46(g/mol)$
$\text{%}C=\dfrac{12.2}{46}.100=52$$\text{%}$
$\text{%}H=\dfrac{1.6}{46}.100=13$$\text{%}$
$\text{%}O=100-52-14=35$$\text{%}$
