Chemistry, asked by tomarlalit5078, 7 months ago

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 ?

Answers

Answered by shadowsabers03
12

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.}

Answered by KrishnaKumar01
0

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

.

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