Question

Two oxides of metal M has 27. 6% and 30% oxygen by weight. If the formula of the first oxide is M3O4, What is the formula of II oxide ?

Answer 1

Consider the oxide $\sf{M_3O_4.}$ It has 27.6% oxygen by mass.

$\longrightarrow\sf{\dfrac{4\times16}{3M+4\times16}=\dfrac{27.6}{100}}$

$\longrightarrow\sf{\dfrac{64}{3M+64}=\dfrac{27.6}{72.4+27.6}}$

By rule of componendo,

$\longrightarrow\sf{\dfrac{3M}{64}=\dfrac{72.4}{27.6}}$

$\longrightarrow\sf{M=\dfrac{72.4}{27.6}\times\dfrac{64}{3}}$

$\longrightarrow\sf{M=56\ g}$

Let the formula of second oxide be $\sf{M_aO_b.}$ It has 30% oxygen by mass.

$\longrightarrow\sf{\dfrac{16b}{56a+16b}=\dfrac{30}{100}}$

$\longrightarrow\sf{\dfrac{2b}{7a+2b}=\dfrac{3}{7+3}}$

Again by rule of componendo,

$\longrightarrow\sf{\dfrac{7a}{2b}=\dfrac{7}{3}}$

$\longrightarrow\sf{\dfrac{a}{b}=\dfrac{2}{3}}$

$\longrightarrow\sf{a:b=2:3}$

Hence the formula of second oxide is $\bf{M_2O_3.}$

Answer 2

Answer:

Consider the oxide \sf{M_3O_4.}M

3

O

4

. It has 27.6% oxygen by mass.

\longrightarrow\sf{\dfrac{4\times16}{3M+4\times16}=\dfrac{27.6}{100}}⟶

3M+4×16

4×16

=

100

27.6

\longrightarrow\sf{\dfrac{64}{3M+64}=\dfrac{27.6}{72.4+27.6}}⟶

3M+64

64

=

72.4+27.6

27.6

By rule of componendo,

\longrightarrow\sf{\dfrac{3M}{64}=\dfrac{72.4}{27.6}}⟶

64

3M

=

27.6

72.4

\longrightarrow\sf{M=\dfrac{72.4}{27.6}\times\dfrac{64}{3}}⟶M=

27.6

72.4

×

3

64

\longrightarrow\sf{M=56\ g}⟶M=56 g

Let the formula of second oxide be \sf{M_aO_b.}M

a

O

b

. It has 30% oxygen by mass.

\longrightarrow\sf{\dfrac{16b}{56a+16b}=\dfrac{30}{100}}⟶

56a+16b

16b

=

100

30

\longrightarrow\sf{\dfrac{2b}{7a+2b}=\dfrac{3}{7+3}}⟶

7a+2b

2b

=

7+3

3

Again by rule of componendo,

\longrightarrow\sf{\dfrac{7a}{2b}=\dfrac{7}{3}}⟶

2b

7a

=

3

7

\longrightarrow\sf{\dfrac{a}{b}=\dfrac{2}{3}}⟶

b

a

=

3

2

\longrightarrow\sf{a:b=2:3}⟶a:b=2:3

Hence the formula of second oxide is \bf{M_2O_3.}M

2

O

3

.