Question

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[SUBJECT - CHEMISTRY]

QUESTION:-

Two oxides of a metal , one contain 27.6% and 30% oxygen respictively . if the formula of the first oxide is X3O4 , find that of second.


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Answer 1

By percent I assume you're saying by mass.

So in X3O4 there is 27.6% oxygen.

Wt of O4 = 64 g/moles

This is 27.6% Hence the molecular wt of the compound itself is

$64 \times 100 \div 27.6$
Which gives us 231.884057971 g/moles

Hence the wt of X3 is 167.884057971 g/moles
Thus atomic weight of X is 55.961352657

(Bonus: such a molecular is possessed by Iron(Fe))

Hence we take 70% of X by mass in the next compound and 30% of O by mass.

Let's say we have 1kg of new compound. Thus 700g is X and 300g is O.

Thus number of atoms of

X = 700/55.961352657 = 12.5086325967

O = 300/16 = 18.75

Hence you can see the ratio of no. of atoms of X and O is in the ratio:-

= 12.5 (approx) : 18.75
= 2:3

Hence in each molecule of the new compound there are 2 atoms of X and 3 atoms of O. Which makes our new compound as

X2O3

(Bonus: Both the compounds mentioned in the question, if X is replaced by Fe, turn out to be real compounds where Fe3O4 is a mixed oxide and Fe2O3 is an Iron(III) oxide.)

Answer 2

By percent I assume you're saying by mass.

So in X3O4 there is 27.6% oxygen.

Wt of O4 = 64 g/moles

This is 27.6% Hence the molecular wt of the compound itself is  

64 \times 100  \div 27.6

Which gives us 231.884057971 g/moles

Hence the wt of X3 is 167.884057971 g/moles

Thus atomic weight of X is 55.961352657

(Bonus: such a molecular is possessed by Iron(Fe))

Hence we take 70% of X by mass in the next compound and 30% of O by mass.

Let's say we have 1kg of new compound. Thus 700g is X and 300g is O.

Thus number of atoms of

X = 700/55.961352657 = 12.5086325967

O = 300/16 = 18.75

Hence you can see the ratio of no. of atoms of X and O is in the ratio:-

= 12.5 (approx) : 18.75

= 2:3

Hence in each molecule of the new compound there are 2 atoms of X and 3 atoms of O. Which makes our new compound as  

X2O3

(Bonus: Both the compounds mentioned in the question, if X is replaced by Fe, turn out to be real compounds where Fe3O4 is a mixed oxide and Fe2O3 is an Iron(III) oxide.)