Answer:
You need the mass composition of the compound.
This composition works fine for the formula mass 150 amu.
C: 40.0%
H: 6.7%
O: 53.3%
From that, you can solve the problem following theses steps:
1) Convert mass composition fo molar ratios by dividing ech element by its atomic mass =>
C: 40.0 / 12 = 3.33
H: 6.70 / 1 = 6.70
O: 53.3 / 16 = 3.33
2) Divide all the numbers by the smallest one =>
C: 3.33 / 3.33 = 1
H: 6.70 / 3.33 = 2
O: 3.33 / 3.33 = 1
3) Write the empirial formula and calculate its mass:
C H2 O => 1 * 12.0 g/mol + 2 * 1.0 g/mol + 1 * 16.0 g/mol = 30 g/mol
4) Calculate how many times the empirical mass is contained in the molecular mass:
150 / 30 = 5
5) Conclusion:
The molecular formula is C5 H10 O5, i.e. the number of each atom in the molecular formula are:
C =5, H = 10, O = 5
