Question

A compound is 40.0% c, 6.70% h, and 53.3% o by mass. assume that we have a 100.-g sample of this compound. the molecular formula mass of this compound is 240 amu . what are the subscripts in the actual molecular formula?

Answer 1

When you have 100 g of compound, then based on the percentages given, there are 40.0 g C, 6.70 g H, and 53.3 g O. Convert those to moles:

C: 40.0 g / 12.0 = 3.33 moles of C 
H: 6.70 g / 1.01 = 6.63 moles of H 
O: 53.3 / 16.0 = 3.33 moles of O 

Dividing by the smallest (3.33), we get a C:H:O mole ratio of 1:2:1
So, The empirical formula is CH2O.
Now, That formula has a molar mass of [12.0 + 2(1.0) + 16.0] = 30.0

And we are given it's molar mass is = 240

So, no. of units of CH2O = 240 / 30 = 8

8 x CH2O = C8H16O8, and that is the molecular formula.

$C_8H_{16}O_8$