Question

An alcohol is 62.04 % C and 10.43 % H by mass. The rest is oxygen. What is the empirical formula of the alcohol

Answer 1

Answer: The empirical formula is
`C_3H_6O`

Step-by-step explanation:

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 = 62.04 g

Mass of H = 10.43

Mass of O = (100-(62.04+10.43)) g = 27.53 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (62.04g)/(12g/mole)=5.17`moles

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (10.43g)/(1g/mole)=10.43moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (27.53g)/(16g/mole)=1.72moles`

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

For C =
`(5.17)/(1.72)=3`

For H =
`(10.43)/(1.72)=6`

For O =
`(1.72)/(1.72)=1`

The ratio of C :H : O= 3: 6 : 1

Hence the empirical formula is
`C_3H_6O`