Trong 1 mol hợp chất:

$n_H=\dfrac{98.2,04\%}{1}\approx 2(mol)$

$n_S=\dfrac{98.32,65\%}{32}\approx 1(mol)$

$n_O=\dfrac{98-2-32}{16}=4(mol)$

$\to CTHH:H_2SO_4$

a) CTHH: Cl_xO_y

Có \(\dfrac{m_{Cl}}{m_O}=\dfrac{38,8\%}{61,2\%}\)

=> \(\dfrac{35,5.n_{Cl}}{16.n_O}=\dfrac{97}{153}=>\dfrac{n_{Cl}}{n_O}=\dfrac{2}{7}\)

=> CTHH: Cl₂O₇

PTK = 35,5.2 + 16.7 = 183 (đvC)

b) CTHH: Ca_xH_yP_zO_t

Có m_Ca : m_H : m_P : m_O = 17,09% : 1,71% : 26,5% : 54,7%

=> 40.n_Ca : 1.n_H : 31.n_P : 16.n_O = 17,09 : 1,71 : 26,5 : 54,7

=> n_Ca : n_H : n_P : n_O = 0,42725 : 1,71 : 0,8545 : 3,41875

= 1:4:2:8

=> CTHH: CaH₄P₂O₈ hay Ca(H₂PO₄)₂

PTK = 40.1 + (1.2 + 31.1 + 16.4).2 = 234 (đvC)

a)\(Al_2O_3\)

b)\(ZnCl_2\)

c)\(H_2S\)

d)\(Fe\left(OH\right)_3\)

e)\(H_2S\)

f)\(H_2SO_4\)

g)\(SO_3\)

a) Al₂O₃

B) ZnCl₂

C) H₂S

D) Fe(OH)₃

e) H₂S

F) H₂SO₄

G) SO₃

2) \(\%m_{\dfrac{O}{FeO}}=\dfrac{16}{72}.100\approx22,222\%\\ \%m_{\dfrac{O}{Fe2O3}}=\dfrac{3.16}{160}.100=30\%\\ \%m_{\dfrac{O}{Fe3O4}}=\dfrac{64}{232}.100\approx27,586\%\)

Bài 1:

%mO=48%

M(phân tử)= (2.56)/28%=400(g/mol)

Số nguyên tử S: (24% . 400)/32= 3(nguyên tử)

Số nguyên tử O: (48% . 400)/16= 12(nguyên tử)

=> CTHH: Fe2(SO4)3

Trong 1 mol X:

$n_K=\dfrac{101.38,8\%}{39}\approx 1(mol)$

$n_N=\dfrac{101.13,9\%}{14}\approx 1(mol)$

$n_O=\dfrac{101-14-39}{16}=3(mol)$

Vậy CTHH là $KNO_3$

$\to$ Chọn A 

CTHH là : K_xN_yO_z

\(\%K=\dfrac{39x}{101}\cdot100\%=38.8\%\)

\(\Rightarrow y=1\)

\(\%N=\dfrac{14x}{101}\cdot100\%=13.9\%\)

\(z=\dfrac{101-39-14}{16}=3\)

\(KNO_3\)

Gọi CTHH của X là: \(H_xS_y\)

\(\Rightarrow x:y=\dfrac{2,04\%}{1}:\dfrac{32,65\%}{32}=2,04:1,02=2:1\)

\(\Rightarrow CTHH:H_2S\)

a,Ta có:M_X = 23.2 = 46 (g/mol)

b,Gọi CTHH của X là N_yO_z

=> m_N = 46.30,43%=14 (g)=>y=14:14=1

=> m_O = 46-14 = 32 (g) => z=32:16=2

Vậy CTHH: NO₂