Question

Combustion analysis of toluene, a common organic solvent, gives 5.27 mg of co2 and 1.23 mg of h2o. if the compound contains only carbon and hydrogen, what is its empirical formula?

Answer 1

C7H8 First, determine the number of relative moles of each element we have and the molar masses of the products. atomic mass of carbon = 12.0107 atomic mass of hydrogen = 1.00794 atomic mass of oxygen = 15.999 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 We have 5.27 mg of CO2, so 5.27 / 44.0087 = 0.119749 milli moles of CO2 And we have 1.23 mg of H2O, so 1.23 / 18.01488 = 0.068277 milli moles of H2O Since there's 1 carbon atom per CO2 molecule, we have 0.119749 milli moles of carbon. Since there's 2 hydrogen atoms per H2O molecules, we have 2 * 0.068277 = 0.136554 milli moles of hydrogen atoms. Now we need to find a simple integer ratio that's close to 0.119749 / 0.136554 = 0.876937 Looking at all fractions n/m where n ranges from 1 to 10 and m ranges from 1 to 10, I find a closest match at 7/8 = 0.875 with an error of only 0.001937, the next closest match has an error over 6 times larger. So let's go with the 7/8 ratio. The numerator in the ratio was for carbon atoms, and the denominator was for hydrogen. So the empirical formula for toluene is C7H8.