Question

The molecular formula mass of this compound is 150 amu . what are the subscripts in the actual molecular formula? enter the subscripts for c, h, and o, respectively, separated by commas (e.g., 5,6,7).

Answer 1

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.