Question

Determine the molecular formula of an oxide of iron, in which the mass per cent
How much copper can
of iron and oxygen are 69.9 and 30.1, respectively.
(overage) of chlorine using the following dat
10​

Answer 1

Answer

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

Explanation:

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Answer 2

Explanation:

Given that the molar mass of the oxide is 159.69 g mol-1. Answer Answer Mass % of iron = 69.9 % [Given] Mass % of oxygen = 30.1 % [Given] Fe : O = 1.25 : 1.88 Convert in simple ratio we get Fe : O = 2 : 3 The empirical formula of the iron oxide is Fe2O3.