Question

87.48 g of an unknown compound composed of carbon, hydrogen and oxygen contains 41.17 g C and 5.18 g H. Give the empirical formula of this compound.

C4H5O2

C3H5O2

C3H5O4

C3H5O3

C4H6O3

Answer 1

Let's get the composition of Oxygen:

mass O = 87.48 - 41.17 - 5.18 = 41.13 g

Now, let's get the moles:

C = 41.17 / 12 = 3.43

H = 5.18 / 1 = 5.18

O = 41.13 / 16 = 2.57

Now, the relation between then to get the number of atoms:

C = 3.43 / 2.57 = 1.33

H = 5.18 / 2.57 = 2.01

O = 2.57 / 2.57 = 1

The empirical formula should be:

C_1.33H₂O

However this does not match the options...at first.

However, if we just have the molecular weight of this compound, we can calculate the molecular formula (which I think is the real question here).

the composition of C, H and O in this is:

%C = 41.17/87.48 * 100 = 47.1%

%H = 5.18/87.48 * 100 = 5.9%

%O = 41.13/87.48 * 100 = 47%

So, if we calculate the percent for every option, the only one that match would be the last compound, which mean that the formula is C₄H₆O₃ but this would be molecular formula. The empirical is the above.

Hope this helps