Suppose the mass of the compound is 100g,
Mass of sodium in the compound rArr 100*32.5% = 32.5 g⇒100⋅32.5%=32.5g
Mass of sulphur in the compound rArr 100*22.6% = 22.6 g⇒100⋅22.6%=22.6g
Mass of oxygen in the compound rArr 100*44.9% = 44.9 g⇒100⋅44.9%=44.9g
No. of moles of sodium in the compound rArr 32.5/23 = 1.4 mol⇒32.523=1.4mol
No. of moles of sulphur in the compound rArr 22.6/32 = 0.7 mol⇒22.632=0.7mol
No. of moles of oxygen in the compound rArr 44.9/16 = 2.8 mol⇒44.916=2.8mol
Mole ratio rArr color(white)(xxxx) Nacolor(white)(xxxx) : color(white)(xxxx)S color(white)(xxxxx) : color(white)(xxxx)O⇒××Na××:××S××x:××O
color(white)(xxx.xxx) rArr color(white)(xxx)1.4mol color(white)(xxx) : color(white)(xxx) 0.7mol color(white)(xxx) : color(white)(xxx) 2.8mol×x.×x⇒×x1.4mol×x:×x0.7mol×x:×x2.8mol
|Dividing through by the smallest ratio..|
color(white)(xxx.xxx) rArr color(white)(xxx)(1.4mol)/(0.7mol) color(white)(xxx) : color(white)(xxx) (0.7mol)/(0.7mol) color(white)(xxx) : color(white)(xxx) (2.8mol)/(0.7mol)×x.×x⇒×x1.4mol0.7mol×x:×x0.7mol0.7mol×x:×x2.8mol0.7mol
color(white)(xxx.xxx) rArr color(white)(xx.xxx) 2 color(white)(xxxxx) : color(white)(xxxxxx) 1 color(white)(xxxx) : color(white)(xxxxx) 4×x.×x⇒×.×x2××x:×××1××:××x4
Thus, the empirical formula of the compound is Na_2SO_4Na2SO4.
