Question

When a 0.860 g sample of an organic compound containing C, H, and O was burned completely in oxygen, 1.64 g of CO₂ and 1.01 g of H₂O were produced. What is the empirical formula of the compound?

Answer 1

Given that combustion of 0.86 g of organic compound produced 1.64 g CO2 and 1.01 g H2O.

Mass of carbon present in 1.64 g of CO2 = 1.64g x molar mass of carbon/ molar mass of CO2

                                                                         = 1.64g x 12 (g/mol) / 44 (g/mol)

                                                                         = 0.447 g

Mass of hydrogen present in 1.01 g of H2O = 1.01g x molar mass of hydrogen/ molar mass of H2O

                                                                         = 1.01g x2 (g/mol) / 18 (g/mol)

                                                                         = 0.112 g

Mass of C + Mass of H + Mass of O = Mass of ognanic compound

0.447 g + 0.112 g + Mass of O = 0.86 g

Mass of O = 0.86g – ( 0.447 g + 0.112g )

                   = 0.301 g

Therefore, Mass of O = 0.301 g

                     Mass of carbon = 0.447 g

                    Mass of carbon = 0.112 g

Dividing each mass by elemental molar mass tells us how many moles of each element present in the sample of butyric acid.

For C, 0.447 g/ 12 (g/mol) = 0.037 mol C

For H, 0.112 g/ 1 (g/mol) = 0.112 mol H

For O, 0.301 g/ 16 (g/mol) = 0.0188 mol O

Divide each by smallest among them i.e. 0.0188

For C, 0.037/ 0.0188 = 1.97 mol C

For H, 0.112/ 0.0188 = 5.95 mol H

For O, 0.0188/ 0.0188 = 1 mol O

Round each value to the nearest integer,

C = 2

H = 6

O =1

Therefore, empirical formula of organic compound = C₂H₆O