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).

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.