Question

A compound is composed of 58.8% c, 9.8% h, and 31.4% o, and the molar mass is 102 g/mol. what is the molecular formula for this compound?

Answer 1

Answer: Thus the molecular formula is
`C_(5)H_(10)O_2`

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 = 58.8 g

Mass of H = 9.8 g

Mass of O = 31.4 g

Step 1 : convert given masses into moles.

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (58.8g)/(12g/mole)=4.9moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (9.8g)/(1g/mole)=9.8moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (31.4g)/(16g/mole)=1.96moles`

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

For C =
`(4.9)/(1.96)=2.5`

For H =
`(9.8)/(1.96)=5`

For O =
`(1.96)/(1.96)=1`

The ratio of C : H : O= 2.5 : 5 : 1

Converting them into whole number ratios:

Hence the empirical formula is
`C_(5)H_(10)O_2`

The empirical weight of
`C_(5)H_(10)O_2` = 5(12) + 10(1) + 2(16)= 102g.

Given :The molecular weight = 102 g/mole

Now we have to calculate the molecular formula.

`n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}`

`n=(102g/mole)/(102g/eq)=1`

Thus the molecular formula is
`C_(5)H_(10)O_2`