Bài 1:

$n_{C_4H_8}=\dfrac{20,86}{12.4+8}=0,3725(mol)$

a,

Số phân tử $C_4H_8=0,3725.6.10^{23}= 2,235.10^{23}$ 

b,

$V_{C_4H_8}=0,3725.24=8,94l$

c,

* Cách 1:

$n_C=4n_{C_4H_8}=1,49(mol)$

$\to m_C=1,49.12=17,88g$

$\to m_H=20,86-17,88=2,98g$

* Cách 2:

$\%m_{C(C_4H_8)}=\dfrac{12.4.100}{12.4+1.8}=\dfrac{600}{7}\%$

$\to m_C=20,86.\dfrac{600}{7}\%=17,88g$

$\to m_H=20,86-17,88=2,98g$

Bài 2:

$\%m_{Cu(CuO)}=\dfrac{64.100}{64+16}=80\%$

$\%m_{Cu(CuS)}=\dfrac{64.100}{64+32}=66,67\%$

$\%m_{Cu(CuS_2)}=\dfrac{64.100}{64+32.2}=50\%$

$\%m_{Cu(CuSO_3)}=\dfrac{64.100}{64+32+16.3}=44,44\%$

$\%m_{Cu(CuSO_4)}=\dfrac{64.100}{64+32+16.4}=40\%$

$\%m_{Cu(Cu_2O)}=\dfrac{64.2.100}{64.2+16}=88,89\%$ 

Vậy $Cu_2O$ giàu $Cu$ nhất, $CuSO_4$ nghèo $Cu$ nhất