(a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of 2.78 mg of ethyl butyrate produces 6.32 \(\mathrm{mg}\) of \(\mathrm{CO}_{2}\) and 2.58 \(\mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of the compound? (b) Nicotine, a component of tobacco, is composed of \(C, H,\) and \(N .\) A 5.250 -mg sample of nicotine was combusted, producing 14.242 \(\mathrm{mg}\) of \(\mathrm{CO}_{2}\) and 4.083 \(\mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for nicotine? If nicotine has a molar mass of \(160 \pm 5 \mathrm{g} / \mathrm{mol},\) what is its molecular formula?

Step by step solution

01

Convert mass of product to mass of element

Using the given masses of CO₂ and H₂O from the combustion of ethyl butyrate: Mass of C: 6.32 mg CO₂ × \(\frac{12.01\ g/mol}{44.01\ g/mol}\) = 1.72 mg C Mass of H: 2.58 mg H₂O × \(\frac{2.02\ g/mol}{18.02\ g/mol}\) = 0.289 mg H Mass of O (by subtraction): 2.78 mg (Ethyl Butyrate) - 1.72 mg C - 0.289 mg H = 0.771 mg O

02

Convert mass of element to moles

For Carbon, Hydrogen, and Oxygen: n(C) = \(\frac{1.72\ mg}{12.01\ \mathrm{mg}\,/\,\mathrm{mol}}\) = 0.143 mol C n(H) = \(\frac{0.289\ mg}{1.01\ \mathrm{mg}\,/\,\mathrm{mol}}\) = 0.286 mol H n(O) = \(\frac{0.771\ mg}{16.00\ \mathrm{mg}\,/\,\mathrm{mol}}\) = 0.0482 mol O

03

Determine the mole ratio and empirical formula

Mole ratio: C : H : O = 0.143 : 0.286 : 0.0482 ≈ 3 : 6 : 1 Empirical formula: C3H6O1 or simply C₃H₆O (b) Finding the empirical formula for nicotine:

04

Convert mass of product to mass of element

Using the given masses of CO₂ and H₂O from the combustion of nicotine: Mass of C: 14.242 mg CO₂ × \(\frac{12.01\ g/mol}{44.01\ g/mol}\) = 3.881 mg C Mass of H: 4.083 mg H₂O × \(\frac{2.02\ g/mol}{18.02\ g/mol}\) = 0.462 mg H Mass of N (by subtraction): 5.250 mg (Nicotine) - 3.881 mg C - 0.462 mg H = 0.907 mg N

05

Convert mass of element to moles

For Carbon, Hydrogen, and Nitrogen: n(C) = \(\frac{3.881\,mg}{12.01\,\mathrm{mg}\,/\,\mathrm{mol}}\) = 0.323 mol C n(H) = \(\frac{0.462\,mg}{1.01\,\mathrm{mg}\,/\,\mathrm{mol}}\) = 0.458 mol H n(N) = \(\frac{0.907\,mg}{14.01\,\mathrm{mg}\,/\,\mathrm{mol}}\) = 0.0647 mol N

06

Determine the mole ratio and empirical formula

Mole ratio: C : H : N = 0.323 : 0.458 : 0.0647 ≈ 5 : 7 : 1 Empirical formula: C5H7N

07

Determine the molecular formula

Molar mass of empirical formula: M(C5H7N) = 5 × 12.01 g/mol + 7 × 1.01 g/mol + 14.01 g/mol ≈ 79.1 g/mol Molar mass of nicotine: 160 g/mol (given) Multiplication factor: \(\frac{160\,g/mol}{79.1\,g/mol}\) ≈ 2 (to the nearest whole number) Molecular formula: 2 × C5H7N = C10H14N2

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