Question

Determine the empirical formula of an oxide of an iron which has 69.9% iron and 30.1% oxygen by mass. fe=58.85 and o=16 amu

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

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Answer 2

Given :-

• Percentage of iron ( Fe ) = 69.9 %

• Percentage of oxygen ( O ) = 30.1 %

To Find :-

• Empirical formula

Solution :-

We know,

Atomic mass of iron ( Fe ) = 55.84 g

Atomic mass of oxygen ( O ) = 16 g

$\bullet \sf \ No \ of \ mole \ of \ iron \ ( Fe)= \dfrac{69.9}{55.84}= 1.25 \\\\\bullet \ No \ of \ mole \ of \ oxygen \ (O) = \dfrac{30.1}{16}= 1.88$

Simplest ratio :-

$\longrightarrow \sf \dfrac{1.25}{1.25} \ : \ \dfrac{1.88}{1.25} \\\\\longrightarrow 1 \ : \ 1.5$

Simplest whole number ratio :-

$\longrightarrow \sf 1\times 2 \ : \ 1.5 \times 2 \\\\\longrightarrow 2 \ : \ 3$

$\boxed{\bf Empirical \ formula = Fe_2 O_3}$