Tìm số nguyên tử C, H, O có trong a, 0,08 mol benzyl axetat ( có mùi hoa nhài ) CH3COOCH2C6H5 b, 16,2 gam canxi hidrocacbonat Ca(HCO3)2 c, 3,42 gam đường kính C12H22O11 d, 3,36 lít (đktc) fomandehit HCHO

Em tham khảo nha :

\(\begin{array}{l}
a)\\
CTHH:C{H_3}COOHC{H_2}{C_6}{H_5}\\
{n_A} = 0,08mol\\
{n_C} = 9{n_A} = 9 \times 0,08 = 0,72mol\\
{N_C} = {n_C} \times 6 \times {10^{23}} = 0,72 \times 6 \times {10^{23}} = 4,32 \times {10^{23}}\\
{n_O} = 2{n_A} = 2 \times 0,08 = 0,16mol\\
{N_O} = {n_O} \times 6 \times {10^{23}} = 0,16 \times 6 \times {10^{23}} = 9,6 \times {10^{22}}\\
{n_H} = 11{n_A} = 11 \times 0,08 = 0,88mol\\
{N_H} = {n_H} \times 6 \times {10^{23}} = 0,88 \times 6 \times {10^{23}} = 5,28 \times {10^{23}}\\
b)\\
CTHH:Ca{(HC{O_3})_2}\\
{n_B} = \dfrac{m}{M} = \dfrac{{16,2}}{{162}} = 0,1mol\\
{n_C} = 2{n_B} = 2 \times 0,1 = 0,2mol\\
{N_C} = {n_C} \times 6 \times {10^{23}} = 0,2 \times 6 \times {10^{23}} = 1,2 \times {10^{23}}\\
{n_O} = 6{n_B} = 6 \times 0,1 = 0,6mol\\
{N_O} = {n_O} \times 6 \times {10^{23}} = 0,6 \times 6 \times {10^{23}} = 3,6 \times {10^{23}}\\
{n_H} = 2{n_B} = 2 \times 0,1 = 0,2mol\\
{N_H} = {n_H} \times 6 \times {10^{23}} = 0,2 \times 6 \times {10^{23}} = 1,2 \times {10^{23}}\\
c)\\
CTHH:{C_{12}}{H_{22}}{O_{11}}\\
{n_3} = \dfrac{m}{M} = \dfrac{{3,42}}{{342}} = 0,01mol\\
{n_C} = 12{n_3} = 12 \times 0,01 = 0,12mol\\
{N_C} = {n_C} \times 6 \times {10^{23}} = 0,12 \times 6 \times {10^{23}} = 0,72 \times {10^{23}}\\
{n_O} = 11{n_3} = 11 \times 0,01 = 0,11mol\\
{N_O} = {n_O} \times 6 \times {10^{23}} = 0,11 \times 6 \times {10^{23}} = 6,6 \times {10^{22}}\\
{n_H} = 22{n_3} = 22 \times 0,01 = 0,22mol\\
{N_H} = {n_H} \times 6 \times {10^{23}} = 0,22 \times 6 \times {10^{23}} = 1,32 \times {10^{23}}\\
d)\\
CTHH:HCHO\\
{n_D} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_C} = {n_D} = 0,15mol\\
{N_C} = {n_C} \times 6 \times {10^{23}} = 0,15 \times 6 \times {10^{23}} = 0,9 \times {10^{23}}\\
{n_O} = {n_D} = 0,15mol\\
{N_O} = {n_O} \times 6 \times {10^{23}} = 0,15 \times 6 \times {10^{23}} = 9 \times {10^{22}}\\
{n_H} = 2{n_D} = 2 \times 0,15 = 0,3mol\\
{N_H} = {n_H} \times 6 \times {10^{23}} = 0,3 \times 6 \times {10^{23}} = 1,8 \times {10^{23}}
\end{array}\)