Question

Oxalic acid is an organic substance. Its composition is 26.7% , 2.2% , and 71.1% (by mass), and its molecular weight is 90 amu. What is its molecular formula?

Answer 1

This is an incomplete question, here is a complete question.

Oxalic acid is an organic substance. Its composition is 26.7% C, 2.2% H, and 71.1% O (by mass), and its molecular weight is 90 amu. What is its molecular formula?

Answer : The molecular formula of a compound is,
`C_2H_2O_4`

Solution : Given,

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 = 26.7 g

Mass of H = 2.2 g

Mass of O = 71.1 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}}= (26.7g)/(12g/mole)=2.225moles`

Moles of H =
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (2.2g)/(1g/mole)=2.2moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (71.1g)/(16g/mole)=4.444moles`

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

For C =
`(2.225)/(2.2)=1.0\aaprox 1`

For H =
`(2.2)/(2.2)=1`

For O =
`(4.444)/(2.2)=2.02\approx 2`

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

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

The Empirical formula =
`C_1H_1O_2=CHO_2`

The empirical formula weight = 1(12) + 1(1) + 2(16) = 45 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=(90)/(45)=2`

Molecular formula =
`(C_1H_1O_2)_n=(C_1H_1O_2)_2=C_2H_2O_4`

Therefore, the molecular of the compound is,
`C_2H_2O_4`