Question

An organic compound has 68.8% C , 5% H
and 26.2% O.Calculate its empirical formula

Answer 1

Answer:

C₇H₆O₂

Explanation:

Let's first convert the percentages to masses by considering the mass of each element in 100g of this compound

C = 68.8g                                 H = 5g                                      O = 26.2g

Now, to find the number of moles of each element, let's divide the above values by the molar masses of the respective elements

(equation used: no. of moles = mass of the element x molar mass of the element)

(I will be using the respective molar masses rounded off to the nearest whole number as we are told to use the rounded off values for calculations. But please use the value you were told to use by your teachers)

C= 68.8g / 12gmol⁻¹ = 5.733mol

H= 5g /1gmol⁻¹ = 5mol

O= 26.2g / 16gmol⁻¹ = 1.637mol

Now let's divide the above values by the smallest value among them

C= 5.733mol / 1.637mol =3.5

H= 5mol / 1.637mol = 3.0

O= 1.637mol / 1.637mol = 1

The final step of the calculation is to multiply each term by 2 as 3.5 is at the exact middle of 3 and 4 ( this is a step taught by my teacher. Please follow this method only if you were taught to do so)

C=3.5 x 2 = 7

H= 3 x 2 = 6

O= 1 x 2 = 2

Then the empirical formula = C₇H₆O₂

If you weren't taught to do as above,

C=4                                      H=3                                             O=1

Then the empirical formula = C₄H₃O₂

For the above calculations, i used the molar masses of C, H and O as  12gmol⁻¹,  1gmol⁻¹  and  16gmol⁻¹ respectively.

Hope this answer helps.