Question

a compound contains 70% by mass of iron and 30% by mass of oxygen. what is its empirical formula​

Answer 1

Explanation:

Fe2O3

70% of Fe = 70g of Fe

30%of O = 30g of O

use the molar mass of two elements

70× 1mole Fe/55.845

= 1.2535

for o: 30 × 1 mole o/ 16

=1.875

mole ratio of two elements

for Fe: 1.2535/1.2535 = 1

for o : 1.875/ 1.2535= 1.5

(Fe1 O1.5)2

Fe2O3

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Answer 2

LET THE TOTAL MASS OF THE COMPOUND BE 'X'

MASS OF IRON=56g

MASS OF OXYGEN=16g

NO. OF ATOMS OF IRON=70/56=1.25

NO. OF ATOMS OF OXYGEN=30/16=1.875

DIVIDING 1.25 DOESNT CHANGE THE VALUE IT COULD BE ANY NUMBER FOR FINDING THE ANSWER I TOOK 1.25

1.25/1.25=1

1.875/1.25=1.5

NOW MULTIPLY 2

1×2=2

1.5×2=3

NO.OF IRON ATOMS=2

NO.OF OXYGEN ATOMS=3

EMPIRICAL FORMULA MEANS THE THERE SHOULD NOT BE ANY INTEGRAL COMMON FACTOR BETWEEN THE NO.OF ATOMS OF EACH ELEMENT

THEREFORE THE COMPOUND IS:

Fe2O3