Giải thích các bước giải:

`a) m_{Cu} = {40%.160}/{100%} = 64g`

`=>n_{Cu} = {64}/{64} = 1 mol`

`m_S = {20%.160}/{100%} = 32g`

`=>n_S = {32}/{32} = 1 mol`

`m_O = {40%.160}/{100%} = 64g`

`=>n_O = {64}/{16} = 4 mol`

Vậy CTHH là: `CuSO_4`

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`b) m_N = {82,35%.17}/{100%} = 14g`

`=>n_N = {14}/{14} = 1 mol`

`m_H = {17,65%.17}/{100%} = 3g`

`=>nH = 3/1 = 3 mol`

Vậy CTHH là: `NH_3`

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`c) %O = 100 - 32,39 - 22,53 = 45,08%`

`m_{Na} = {32,39%.142}/{100%} = 46g`

`=>n_{Na} = {46}/{23} = 2 mol`

`m_S = {22,53%.142}/{100%} = 32g`

`=>n_S = {32}/{32} = 1 mol`

`m_O = {45,08%.142}/{100%} = 64g`

`=>n_O = {64}/{16} = 4 mol`

Vậy CTHH là: `Na_2SO_4`

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`d) %O = 100 - 36,8 - 21 = 42,2%`

`m_{Fe} = {36,8%.152}/{100%} = 56g`

`=>n_{Fe} = {56}/{56} = 1 mol`

`m_S = {21%.152}/{100%} = 32g`

`=>n_S = {32}/{32} = 1 mol`

`m_O = {42,2%.152}/{100%} = 64g`

`=>n_O = {64}/{16} = 4 mol`

Vậy CTHH là `FeSO_4`

a)

`m_{Cu}`=`160×40%=64 gam`

`m_S`=`160×20%=32 gam`

`m_O`=`160-(64+32)`=`64gam`

⇒`n_{Cu}`=$\dfrac{64}{64}$=`1mol`

⇒`n_S`=$\dfrac{32}{32}$=`1mol`

⇒`n_O`=$\dfrac{64}{16}$=`4mol`

⇒CTHH: `CuSO_4`

b)

`m_N`=`17×82,35%`≈`14 gam`

`m_H`=`17-13,9995`≈`3 gam`

⇒`n_N`=$\dfrac{14}{14}$=`1mol`

⇒`n_H`=$\dfrac{3}{1}$=`3mol`

⇒CTHH: `NH_3`

c)

`m_{Na}`=`142×32,39%≈`46 gam`

`m_S`=`142×22,53%`=`32 gam`

`m_O`=`142-(46+32)`=64 gam`

`n_{Na}`=$\dfrac{46}{23}$≈`2 mol`

`n_S`=$\dfrac{32}{32}$=`1mol`

`n_O`=$\dfrac{64}{16}$=`4mol`

⇒CTHH `Na_2SO_4`

d)

`m_{Fe}`=`152×36,8%≈56 gam`

`m_S`=`152×21%≈32 gam`

`m_O`=`152-(56+32)=64 gam`

`n_{Fe}`=$\dfrac{56}{56}$=`1mol`

`n_S`=$\dfrac{32}{32}$=`1mol`

`n_O`=$\dfrac{64}{16}$=`4mol`

⇒CTHH `FeSO_4`