a 0.1014 g sample of a purified cho compound was burned in a combustion apparatus and produced 0.1486 g co2 and 0.0609 g of h2o. mass spectrometry analysis revealed that the cho compound had a molar mass of 180 g/mol. what is the molecular formula for this compound?

Explanation:

Mass of water obtained = 0.0609 g

Molar mass of water = 18 g/mol

Moles of = 0.0609 g /18 g/mol = 0.00338 moles

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

Moles of H = 2 x 0.00338 = 0.00676 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.00676 x 1.008 = 0.00681 g

Mass of carbon dioxide obtained = 0.1486 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of = 0.1486 g  /44.01 g/mol = 0.00337 moles

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

Moles of C = 0.00337 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.00337 x 12.0107 = 0.04047 g

Given that the compound 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 = 0.1014 g

Mass of O in sample = 0.1014 - 0.04047 - 0.00681 = 0.05412 g  

Molar mass of O = 15.999 g/mol

Moles of O  = 0.05412  / 15.999  = 0.00338 moles

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

0.00676 : 0.00338 : 0.00337

= 2 : 1 : 1

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 = 1×12 + 2×1 + 16= 30 g/mol

Molar mass = 180 g/mol

So,  

Molecular mass = n × Empirical mass

180 = n × 30

⇒ n = 6

The formula of compound =

2b2t

explanation:

2b2t