Question

0.2 g atom of silicon combine with 21.3 g of chlorine. Find the empirical formula of the compound formed.

Answer 1

Since 0.2g atom of Si = given mass/mol. Mass

so, given mass = 0.2 × 28 = 5.6 g

Element mass /At. mass = Gram atom Ratio

Si 5.6/28 = 0.21

Cl 21.3/35.5 = 0.63

Si:Cl = 0.21:0.63 = 1:3

now we see that one part of silicon reacts with 3 parts of chlorine so, the empirical formula is SiCl₃

Answer 2

0.2 g atom of silicon = 0.2 × 28

= 5.6 g of Si

Mass of chlorine (Cl₂₎ = 21.3 g

∴ Total mass of compound = 5.6 + 21.3 = 26.9 g

Percentage of silicon = `5.6/26.9 xx 100` = 20.28%

Percentage of (Cl₂) in compound = 100 − 20.82 = 79.18%

Atomic ratio of chlorine = `79.18/35.5` = 2.23

∴ Simplest ratio of Si = `0.74/0.74` = 1

Simplest ratio of Cl = `2.23/0.74` = 3

∴ The empirical formula of the compound formed is SiCl₃