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