Anabolic steroids are sometimes used illegally by athletes to increase muscle strength. a forensic chemist analyzes some tablets suspected of being a popular steroid.
he determines the substance contains only c, h and o and has a molar mass of 300.42g/mol.

when a 1.200g sample is subjected to combustion analysis, 3.516g of co2 and 1.007g h2o are collected. a) what is the empirical formula of the unknown substance in the tablets? b)what is the molecular formula of the unknown substance in the tablets?

The empirical formula is =

The formula of steroid =

Explanation:

Mass of water obtained = 1.007 g

Molar mass of water = 18 g/mol

Moles of = 1.007 g /18 g/mol = 0.0559 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.0559 = 0.1119 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.1119 x 1.008 = 0.1128 g

Mass of carbon dioxide obtained = 3.516 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of = 3.516 g  /44.01 g/mol = 0.0799 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.0799 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.0799 x 12.0107 = 0.9595 g

Given that the steroids only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = 1.200 g

Mass of O in sample = 1.200 - 0.9595 - 0.1128 = 0.1277 g  

Molar mass of O = 15.999 g/mol

Moles of O  = 0.1277 / 15.999  = 0.00798 moles

Taking the simplest ratio for H, O and C as:

0.1119 : 0.00798 : 0.0799

= 14 : 1 : 10

The empirical formula is =

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 10×12 + 14×1 + 16= 150 g/mol

Molar mass = 300.42 g/mol

So,  

Molecular mass = n × Empirical mass

300.42 = n × 150

⇒ n = 2

The formula of steroid =