Question

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

Answer 1

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Answer 2

From the available data Percentage of iron = 69.9
Percentage of oxygen= 30.1
Total percentage of iron & oxygen= 69.9+30.1= 100%

Step 1 calculation of simplest whole number ratios of the elements

Element
Percentage
Atomic mass
Atomic ratio
Simplest ratio
Simplest whole no ratio

Fe
69.9
55.84
69.9/55.84=1.25
1.25= 1
2

O
30.1
16
30.1/16 = 1.88
1.88=1.5
3

Step 2 Writing the empirical formula of the compound
The empirical formula of the compound = Fe2 O3

Step 3 determination of molecular formula of the compound

Empirical formula mass = 2 X69.9 + 3 X16=187.8 amu

Molecular mass of oxide= 159.69g/mol(given)
Now we know molecular formula = n x Empirical formula

And n= molecular mass / empirical formula mass= 159.69/187.8 = 0.85 = approx 1

Therefore molecular formula = n x empirical formula

=1 x(Fe2O3) = Fe2O3

The molecular formula of the oxide is Fe2O3


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