Question

What is the empirical formula of a compound composed of 3.25% hydrogen ( H ), 19.36% carbon ( C ), and 77.39% oxygen ( O ) by mass? Insert subscripts as needed.

Answer 1

Answer : The empirical of the compound is,
`C_1H_2O_3`

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 19.36 g

Mass of H = 3.25 g

Mass of O = 77.39 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (19.36g)/(12g/mole)=1.613moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (3.25g)/(1g/mole)=3.25moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (77.39g)/(16g/mole)=4.837moles`

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

For C =
`(1.613)/(1.613)=1`

For H =
`(3.25)/(1.613)=2.01\approx 2`

For o =
`(4.837)/(1.613)=2.99\approx 3`

The ratio of C : H : O = 1 : 2 : 3

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

The Empirical formula =
`C_1H_2O_3`

Therefore, the empirical of the compound is,
`C_1H_2O_3`