Question

The percentage composition of a compound is
K = 28.16%, CI = 25.63% and O = 46.21%. The
simplest formula of compound is

Answer 1

Let the compound contains $\tt{a}$ atoms of K, $\tt{b}$ atoms of Cl and $\tt{c}$ atoms of O.

• Atomic mass of K $\tt{=39\ g\,mol^{-1}}$

• Atomic mass of Cl $\tt{=35.5\ g\,mol^{-1}}$

• Atomic mass of O $\tt{=16\ g\,mol^{-1}}$

Percentage composition of K,

$\tt{\longrightarrow \dfrac{39a}{39a+35.5b+16c}=\dfrac{28.16}{100}}$

$\tt{\longrightarrow39a=28.16\times\dfrac{39a+35.5b+16c}{100}}$

Percentage composition of Cl,

$\tt{\longrightarrow \dfrac{35.5b}{39a+35.5b+16c}=\dfrac{25.63}{100}}$

$\tt{\longrightarrow35.5b=25.63\times\dfrac{39a+35.5b+16c}{100}}$

Percentage composition of O,

$\tt{\longrightarrow \dfrac{16c}{39a+35.5b+16c}=\dfrac{46.21}{100}}$

$\tt{\longrightarrow16c=46.21\times\dfrac{39a+35.5b+16c}{100}}$

We take all these in a ratio and we get,

$\tt{\longrightarrow39a:35.5b:16c=28.16:25.63:46.21}$

$\tt{\longrightarrow a:b:c=\dfrac{28.16}{39}:\dfrac{25.63}{35.5}:\dfrac{46.21}{16}}$

$\tt{\longrightarrow a:b:c=0.72:0.72:2.88}$

$\tt{\longrightarrow a:b:c=1:1:4}$

Hence simplest / empirical formula of the compound is KClO₄.

Answer 2

Answer:

Let the compound contains \tt{a}a atoms of K, \tt{b}b atoms of Cl and \tt{c}c atoms of O.

Atomic mass of K \tt{=39\ g\,mol^{-1}}=39 gmol

−1

Atomic mass of Cl \tt{=35.5\ g\,mol^{-1}}=35.5 gmol

−1

Atomic mass of O \tt{=16\ g\,mol^{-1}}=16 gmol

−1

Percentage composition of K,

\tt{\longrightarrow \dfrac{39a}{39a+35.5b+16c}=\dfrac{28.16}{100}}⟶

39a+35.5b+16c

39a

=

100

28.16

\tt{\longrightarrow39a=28.16\times\dfrac{39a+35.5b+16c}{100}}⟶39a=28.16×

100

39a+35.5b+16c

Percentage composition of Cl,

\tt{\longrightarrow \dfrac{35.5b}{39a+35.5b+16c}=\dfrac{25.63}{100}}⟶

39a+35.5b+16c

35.5b

=

100

25.63

\tt{\longrightarrow35.5b=25.63\times\dfrac{39a+35.5b+16c}{100}}⟶35.5b=25.63×

100

39a+35.5b+16c

Percentage composition of O,

\tt{\longrightarrow \dfrac{16c}{39a+35.5b+16c}=\dfrac{46.21}{100}}⟶

39a+35.5b+16c

16c

=

100

46.21

\tt{\longrightarrow16c=46.21\times\dfrac{39a+35.5b+16c}{100}}⟶16c=46.21×

100

39a+35.5b+16c

We take all these in a ratio and we get,

\tt{\longrightarrow39a:35.5b:16c=28.16:25.63:46.21}⟶39a:35.5b:16c=28.16:25.63:46.21

\tt{\longrightarrow a:b:c=\dfrac{28.16}{39}:\dfrac{25.63}{35.5}:\dfrac{46.21}{16}}⟶a:b:c=

39

28.16

:

35.5

25.63

:

16

46.21

\tt{\longrightarrow a:b:c=0.72:0.72:2.88}⟶a:b:c=0.72:0.72:2.88

\tt{\longrightarrow a:b:c=1:1:4}⟶a:b:c=1:1:4

Hence simplest / empirical formula of the compound is KClO₄.