Question

determine the molecular formula of an oxide of iron ,in which the mass percent of iron and oxygen sre 69.9 and 30.1,respectively​

Answer 1

Fe%=69.9

O%=30.1

1. mass of Fe=56
2. mass of O=16

divide mass percent with mass of element

1. 69.9÷56=1.24
2. 30.1÷16=1.88

divide smallest no.

1. 1.24÷1.24=1
2. 1.88÷1.24=1.5

For finding simplest ratio multiply 2 .

we get 1×2=2 and 1.5×2=3

So

• Fe simplest ratio is 2

&

• O simplest ratio is 3

Hence the molecular formula of the compound is Fe2O3