Oxit có dạng: `S_xO_y`
`M_{S_2O_y}=40.2=80` $(g/mol)$
`%S=(32x)/(80).100%=40%`
`=>x=1`
`M_{SO_y}=80`
`=>32+16y=80`
`=>y=3`
`->` Oxit: `SO_3`
Ta đặt CTTQ :SxOy
M Oxit =40.2=80đvC
=>40%=\(\dfrac{32.x}{80}.100\)
=>x=1
->60%=\(\dfrac{16.y}{80}.100\)
=>y=3
Vậy CTHH :SO3