Question

Two oxides of metal were found to contain 31.6% &48% of oxygen respectively.The formula of first oxide is represented by M2O3. Calculate the formula of the second oxide

Answer 1

The formula of the second oxide is $MO_3$

Explanation:

Let atomic weight of metal M be m:

moles of M=$\frac{\text {given mass}}{\text{ Molar mass}}=\frac{68.4}{m}$

moles of O=$\frac{\text {given mass}}{\text{ Molar mass}}=\frac{31.6}{16}$

Given the formula is $M_2O_3$

$\frac{\frac{68.4}{m}}{\frac{31.6}{16}}=\frac{2}{3}$

m= 52

Now the second oxide contains 48 % of oxygen:

Thus moles of M= $\frac{\text {given mass}}{\text{ Molar mass}}=\frac{52}{52}=1$

moles of O=$\frac{\text {given mass}}{\text{ Molar mass}}=\frac{48}{16}=3$

Thus formula of oxide is M:O= 1:3. The formula is $MO_3$

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Answer 2

Answer:

The formula of the second oxide is MO_3MO

3

Explanation:

Let atomic weight of metal M be m:

moles of M=\frac{\text {given mass}}{\text{ Molar mass}}=\frac{68.4}{m}

Molar mass

given mass

=

m

68.4

moles of O=\frac{\text {given mass}}{\text{ Molar mass}}=\frac{31.6}{16}

Molar mass

given mass

=

16

31.6

Given the formula is M_2O_3M

2

O

3

\frac{\frac{68.4}{m}}{\frac{31.6}{16}}=\frac{2}{3}

16

31.6

m

68.4

=

3

2

m= 52

Now the second oxide contains 48 % of oxygen:

Thus moles of M= \frac{\text {given mass}}{\text{ Molar mass}}=\frac{52}{52}=1

Molar mass

given mass

=

52

52

=1

moles of O=\frac{\text {given mass}}{\text{ Molar mass}}=\frac{48}{16}=3

Molar mass

given mass

=

16

48

=3

Thus formula of oxide is M:O= 1:3. The formula is MO_3MO

3

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