Choose the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively and its molecular mass is 160:

FeO

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F e_{3} O_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F e_{2} O_{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

F e O_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $F e_{2} O_{3}$
Let the molecule be $F e_{a} O_{b}$
$∴$ mass % of iron $\times$ molar mass of $F e_{x} O_{y}$ = Atomic mass of iron $\times$ a
$\frac{69.9}{100} \times 160 = 56 a$
$a = 2$
Similarly for oxygen
$\frac{30.1}{100} \times 160 = 16 b$
$b = 3$
Molecular formula is $F e_{2} O_{3}$ .

Suggest Corrections

Similar questions

Q. Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are

69.9

and

30.1

respectively.

Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69 g mol^–1.

The molecular formula of the oxide of iron in which mass percentage of iron and oxygen are 69.0 and 30.1 respectively ?

Q. Determine the empirical formula of an oxide of iron, which has

69.9 %

iron and

30.1 %

dioxygen by mass.

Join BYJU'S Learning Program

Choose the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively and its molecular mass is 160:

The correct option is C Fe2O3Let the molecule be FeaOb∴ mass % of iron × molar mass of FexOy= Atomic mass of iron × a69.9100×160=56aa=2Similarly for oxygen30.1100×160=16bb=3Molecular formula is Fe2O3.