Question

(a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of 7.23 mg of ethyl butyrate produces16.43 mg of CO2 and 6.71 mg of H2O. What is the empirical formula of the compound? (Type your answer using the format CO2 for CO2.)

(b) Nicotine, a component of tobacco, is composed of C, H, and N. A 4.725 mg sample of nicotine was combusted, producing 12.818 mg of CO2 and 3.675 mg of H2O. What isthe empirical formula for nicotine?

If the substance has a molecular mass of 160 +/- 5 g/mol, what is its molecular formula?

Answer 1

(a) Mass of C = molar mass of C/molar mass of CO2 x mass of CO2

= 12.01/44.01 x 16.43 = 4.484 mg

Mass of H = 2 x molar mass of H/molar mass of H2O x mass of H2O

= 2 x 1.008/18.02 x 6.74 = 0.754 mg

Mass of O = mass of sample - mass of C - mass of H

= 7.23 - 4.484 - 0.754 = 1.992 mg

Since moles = mass/molar mass

Moles of C : H : O = 4.484/12.01 : 0.754/1.008 : 1.992/16.00

= 0.3733 : 0.7480 : 0.1245

= 3 : 6 : 1

Empirical formula is C3H6O

(b) Mass of C = molar mass of C/molar mass of CO2 x mass of CO2

= 12.01/44.01 x 12.818 = 3.498 mg

Mass of H = 2 x molar mass of H/molar mass of H2O x mass of H2O

= 2 x 1.008/18.02 x 3.675 = 0.411 mg

Mass of N = mass of sample - mass of C - mass of H

= 4.725 - 3.498 - 0.411 = 0.816 mg

Since moles = mass/molar mass

Moles of C : H : N = 3.498/12.01 : 0.411/1.008 : 0.816/14.01

= 0.2913 : 0.4077 : 0.05824

= 5 : 7: 1

Empirical formula is C5H7N

Let the molecular formula be C_5aH_7aN_a

Molecular mass = 12.01 x 5a + 1.008 x 7a + 14.01 x a = 160

81.116a = 160 => a = 2

Molecular formula is C10H14N2