Question

$\mathfrak{\huge{\purple{\underline{\underline{Question :}}}}}$
Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.​

Answer 1

ANSWER

The iron oxide has 69.9% iron and 30.1% dioxygen by mass.

Thus, 100 g of iron oxide contains 69.9 g iron and 30.1 g dioxygen.

The number of moles of iron present in 100 g of iron oxide are

55.8

69.9

=1.25.

The number of moles of dioxygen present in 100 g of iron oxide are

32

30.1

=0.94.

The ratio of the number of oxygen atoms to the number of carbon atoms present in one formula unit of iron oxide is

1.25

2×0.94

=1.5:1=3:2.

Hence, the formula of the iron oxide is Fe

2

O

3

Answer 2

Your answer sis

In 100 g of the compound there will be: 69.9 g of Iron and 30.1 g of oxygen

Now, Divide the composition of each element by the molar mass of the molar mass of each element

Molar mass of iron= 56 g/mol

Molar mass of oxygen= 32 g/mol

$\frac{69.9}{56}$ = 1.2

$\frac{30.1}{16}$ = 1.8

Now, take the ratio of both the composition of the elements;

1.2 : 1.8

You'll get: 2 : 3 as the ratio

This means that there will be 2 atoms of Iron and 3 atoms of oxygen

The empirical formula is: Fe₂O₃

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