Suppose the mass of the compound is 100g,
Mass of sodium in the compound #rArr 100*32.5% = 32.5 g#
Mass of sulphur in the compound #rArr 100*22.6% = 22.6 g#
Mass of oxygen in the compound # rArr 100*44.9% = 44.9 g#
No. of moles of sodium in the compound # rArr 32.5/23 = 1.4 mol#
No. of moles of sulphur in the compound # rArr 22.6/32 = 0.7 mol#
No. of moles of oxygen in the compound # rArr 44.9/16 = 2.8 mol#
Mole ratio #rArr color(white)(xxxx) Nacolor(white)(xxxx) : color(white)(xxxx)S color(white)(xxxxx) : color(white)(xxxx)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#
|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)#
#color(white)(xxx.xxx) rArr color(white)(xx.xxx) 2 color(white)(xxxxx) : color(white)(xxxxxx) 1 color(white)(xxxx) : color(white)(xxxxx) 4#
Thus, the empirical formula of the compound is #Na_2SO_4#.
