Question

Please help with this problem

Answer 1

Answer : The empirical formula of the iron oxide is
`Fe_2O_3`.

Solution : Given,

Mass of iron oxide = 0.450 g

Mass of iron = 0.315 g

Molar mass of iron = 56 g/mole

Molar mass of oxygen = 16 g/mole

First we have to calculate the Mass of oxygen.

Mass of iron oxide = Mass of iron + Mass of Oxygen

0.450 g = 0.315 g + Mass of oxygen

Mass of oxygen = 0.450 - 0.315 = 0.315 g

Step 1 : convert given mass into moles.

Moles of Fe =
`\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}= (0.315g)/(56g/mole)=0.005625moles`

Moles of O =
`\frac{\text{ Mass of O}}{\text{ Molar mass of O}}= (0.135g)/(16g/mole)=0.0084375moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Fe =
`(0.005625)/(0.005625)=1`

For O =
`(0.0084375)/(0.005625)=1.5`

The ratio of Fe : O = 1 : 1.5

To make the ratio as a whole number multiply numerator and denominator by 2.

The ratio of Fe : O =
`(1* 2)/(1.5* 2)=2:3`

The mole ratio of the element is represented by subscripts in empirical formula.

Therefore, the empirical formula =
`Fe_(2)O_(3)`.