Question

A compound is 52.14% C, 13.13% H, and 34.73% O. What is the empirical formula of the compound?

Answer 1

% H = 100 - ( 52.14 + 34.73 )=13.13 %

assume 100 g of this compound
mass H = 13.13 g
moles H = 13.13 g / 1.008 g/mol=13

mass C = 52.14 g
moles C = 52.14 g/ / 12.011 g/mol=4

mass O = 34.73 g
moles O = 34.73 g/ 15.999 g/mol=2

the empirical formula is C4H13O2

Answer 2

Answer: The empirical formula is
`C_2H_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 = 52.14 g

Mass of H = 13.13 g

Mass of O = 34.73 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (52.14g)/(12g/mole)=4.35moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (13.13g)/(1g/mole)=13.13moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (34.73g)/(16g/mole)=2.17moles`

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

For C =
`(4.35)/(2.17)=2`

For H =
`(13.13)/(2.17)=6`

For O =
`(2.17)/(2.17)=1`

The ratio of C: H: O = 2: 6: 1

Hence the empirical formula is
`C_2H_6O`.