Đáp án+ Giải thích các bước giải:

`1)`
`a)`
$\text{$PTK$ = $NTK_O$ = 16(đvC)}$
`\text{CTTQ:}\ XH_4`
`<=>NTK_X+1.4=16`
`=>NTK_X=16-4=12(đvC)`
`\text{X : Cacbon (C)}`
`b)`
`%m_C={12}/{16}.100%=75%`
`%m_H=100%-75%=25%`
`2)`
`a)`
`\text{CTTQ:}\ XO_2`
`%m_O=50%`
`<=>{16.2}/{NTK_X+16.2}.100%=50%`
`<=>3200=50.NTK_X+1600`
`<=>1600=50.NTK_X`
`=>NTK_X=32(đvC)`
`\text{ X : Lưu huỳnh (S)}`
`b)`
`PTK=32+16.2=64(đvC)`
`PTK_{\text{hợp chất}}=NTK_{Cu}=64(đvC)`

`1)`

`a)`

`NTK_X = 16 -4 = 12(đvC)`

` X: `cacbon `(C)`

`b)`

`%C = {12}/{16} .100 = 75%`

`%H = 100-75=25%`

`2)`

`a)`

Hợp chất: `XO_2`

`%O = {16 . 2}/{X + 16.2} .100 = 50%`

` X = 32(đvC)`

`X` là lưu huỳnh `(S)`

`b)`

`PTK = {16 . 2}/{50%} = 64(đvC)`

`PTK = NTK_{Cu} (=64đvC)`