Question

(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?

Short answer

The empirical formula of ethyl butyrate is C₃H₆O. For nicotine, the empirical formula is C5H7N and its molecular formula is C10H14N2.

Step-by-step solution

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

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

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:

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

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

Determine the mole ratio and empirical formula

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

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

Key concepts

Combustion Analysis

Combustion analysis is a common method in chemistry to determine the composition of an unknown organic compound. This process involves burning the substance in pure oxygen, where it converts to combustion products—typically carbon dioxide (\(\mathrm{CO}_2\)) and water vapor (\(\mathrm{H}_2\mathrm{O}\)). By measuring the amounts of these products, we can back-calculate to find the amounts of each element in the original compound.
The primary goal is to identify how much of each element is present by mass. Using stoichiometry, the masses of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\) are converted to the masses of carbon and hydrogen. Then, the mass of any additional elements, like oxygen, is typically determined by subtraction from the initial mass of the compound.
This information helps derive the empirical formula, reflecting the simplest whole-number ratio of atoms in the compound.

Elemental Composition

Determining elemental composition refers to calculating the proportions of each element within a compound. After you've ascertained how much of each element is present from combustion analysis, these quantities are converted into moles, using the respective atomic masses.

• Carbon (C) contributes to \(\mathrm{CO}_2\), where each molecule contains one carbon atom.
• Hydrogen (H) is inferred from \(\mathrm{H}_2\mathrm{O}\), with each molecule containing two hydrogen atoms.

Understanding these contributions allows you to know how the mass is distributed among the elements in the compound. From the molar quantity of each element, determined through conversion with atomic masses, you establish the empirical formula.
This process requires careful subtraction to determine any other elements' presence, like nitrogen or oxygen, enhancing comprehension of the compound's complete elemental makeup.

Molecular Formula Determination

The molecular formula provides the actual number of atoms of each element in a molecule, and it can be determined once the empirical formula and molar mass are known.
The determination involves comparing the calculated molar mass of the empirical formula with the given molar mass of the compound. The molecular formula is often a simple multiple of the empirical formula, which can be deduced by dividing the compound's molar mass by that of the empirical formula.
For instance, if the empirical formula's molar mass is obtained and then finds the compound has a molar mass approximately double, the molecular formula would be the empirical formula repeated twice. Understanding this concept ensures you know precisely how many atoms are present in each molecule and not just the simplest ratio—this is quintessential in providing full insight into the compound's actual chemical structure.