Question

What is the empirical formula of a molecule containing 65.5% carbon 5.5% hydrogen and 29% oxygen?

Answer 1

I assume C₃H₃O.

Hope I helped! Tell me if I'm wrong!

Answer 2

Answer: The empirical formula is
`C_(3)H_(3)O_1`

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 = 65.5 g

Mass of H = 5.5 g

Mass of O = 29.0 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (65.5g)/(12g/mole)=5.5moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (5.5g)/(1g/mole)=5.5moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (29g)/(16g/mole)=1.8moles\approx 1moles`

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

For C =
`(5.5)/(1.8)=3`

For H =
`(5.5)/(1.8)=3`

For O =
`(1.8)/(1.8)=1`

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

Hence the empirical formula is
`C_(3)H_(3)O_1`