Question

Compute the molecular formula for a compound of molecular weight 90 g/mol which is shown to contain 40% C, 6.67% H, and 53.33% oxygen. Group of answer choicesC1H2O1C2H1O2C2H4O2C3H6O3C4H8O3

Answer 1

In the statement, it was said that the substance was formed by hydrogen, carbon and oxygen, which allows us to write the following:

CxHyOz

We need to find the value of x, y and z to determine the molecular formula of this substance. We know the atomic masses of each element, so we have:

Cx ---> 12x

Hy ---> 1y

Oz ---> 16z

Now just use rules of three, the centesimal proportion given in the statement (40% C, 6.67% H, and 53.33% oxygen) and its molar mass (90 g/mol):

C

100% substance ----- 40% C

90 g of substance ----- 12 x g of C

1,200x = 3,600

x = 3,600/1,200 = 3

H

100% substance ----- 6,67% H

90 g of substance ----- 1y g of H

100y = 600.3

y = 6

O

100% substance ---- 53.33% O

90 g of substance ----- 16z of O

1,600z = 4,799.7

z = 3

Answer: C3H6O3