Bài giải :
`a.CO` và `CO_2`
`-CO:`
`PTK_{CO}=12+16=28(đvC)`
`⇒%m_C=\frac{12}{28}.100%≈42,857%`
`⇒%m_O=100%-%m_C=100%-42,857%= 57,143%`
`-CO_2:`
`PTK_{CO}=12+16.2=44(đvC)`
`⇒%m_C=\frac{12}{44}.100%≈27,28%`
`⇒%m_O=100%-%m_C=100%-27,28%= 72,72%`
`b.Fe_2O_3` và `Fe_3O_4`
`-Fe_2O_3:`
`PTK_{Fe_2O_3}=56.2+16.3=160(đvC)`
`⇒%m_{Fe}=\frac{56.2}{160}.100%7%`
`⇒%m_O=100%-%m_{Fe}=100%-70%= 30%`
`-Fe_3O_4:`
`PTK_{Fe_3O_4}=56.3+16.4=232(đvC)`
`⇒%m_{Fe}=\frac{56.3}{232}.100%≈72,4170%`
`⇒%m_O=100%-%m_{Fe}=100%-72,41%=27,59%`
