Question

Kethoxal, a substance used to combat viruses, is 48.6% carbon, 8.17% hydrogen, and 43.2% oxygen. It has a molar mass of 148 g / mol. What is the molecular formula of kethoxal?

Answer 1

Answer : The molecular formula of kethoxal is,
`C_6H_(12)O_4`

Solution :

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 = 48.6 g

Mass of H = 8.17 g

Mass of O = 43.2 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}}= (48.6g)/(12g/mole)=4.05moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (8.17g)/(1g/mole)=8.17moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (43.2g)/(16g/mole)=2.7moles`

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

For C =
`(4.05)/(2.7)=1.5`

For H =
`(8.17)/(2.7)=3.02\approx 3`

For O =
`(2.7)/(2.7)=1`

The ratio of C : H : O = 1.5 : 3 : 1

To make a whole number we multiplying the ratio by 2, we get:

The ratio of C : H : O = 3 : 6 : 2

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

The Empirical formula =
`C_3H_6O_2`

The empirical formula weight = 3(12) + 6(1) + 2(16) = 74 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

`n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}`

`n=(148)/(74)=2`

Molecular formula =
`(C_3H_6O_2)_n=(C_3H_6O_2)_2=C_6H_(12)O_4`

Therefore, the molecular of the kethoxal is,
`C_6H_(12)O_4`