Problem 56 - Chemistry: The Central Science

Question

(a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of \(2.78 \mathrm{mg}\) 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

Answer

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

Step 1: Convert the mass of CO2 and H2O into moles.

Calculate the moles of CO2 and H2O using their molar masses. Moles of CO2 = Mass of CO2/Molar mass of CO2 Mmoles of CO2 = \( \frac{6.32\ mg}{44.01\ mg/mol}\) = 0.1435 moles Moles of H2O = Mass of H2O/Molar mass of H2O Mmoles of H2O = \( \frac{2.58\ mg}{18.02\ mg/mol}\) = 0.1432 moles

Step 2: Calculate the moles of C and H in ethyl butyrate.

Use the molar ratio (1:1) between CO2 and C; H2O and H. Moles of C = Moles of CO2 = 0.1435 moles Moles of H = Moles of H2O × 2 = 0.1432 moles × 2 = 0.2864 moles

Step 3: Calculate the mass of C and H in ethyl butyrate.

Find the mass of C and H using their molar masses. Mass of C = (0.1435 moles) × \( 12.01\ \frac{mg}{mole}\) = 1.72 mg Mass of H = (0.2864 moles) × \( 1.0\ \frac{mg}{mole}\) = 0.286 mg

Step 4: Find the mass of O in ethyl butyrate.

Subtract the mass of C and H from the total mass of ethyl butyrate. Mass of O = Mass of ethyl butyrate - Mass of C - Mass of H Mass of O = 2.78 mg - 1.72 mg - 0.286 mg = 0.774 mg

Step 5: Calculate the moles of O in ethyl butyrate.

Moles of O = Mass of O/Molar mass of O Mmoles of O = \( \frac{0.774\ mg}{16.0\ mg/mol}\) = 0.04833 moles

Step 6: Calculate the empirical formula.

Find the whole number ratio of moles of C, H, and O by dividing by the smallest quantity. Divide moles of each element by 0.04833: C: \(\frac{0.1435}{0.04833}\) ≈ 2.97 H: \(\frac{0.2864}{0.04833}\) ≈ 5.93 O: \(\frac{0.04833}{0.04833}\) ≈ 1 The empirical formula of ethyl butyrate is C₃H₆O. Part (b): Nicotine

Step 1: Calculate the moles of CO2, H2O, C, and H from the given masses.

Repeat Steps 1-3 from Part (a) for nicotine. Moles of CO2 = \( \frac{14.242\ mg}{44.01\ mg/mol}\) = 0.3237 moles Moles of H2O = \( \frac{4.083\ mg}{18.02\ mg/mol}\) = 0.2267 moles Moles of C = Moles of CO2 = 0.3237 moles Moles of H = Moles of H2O × 2 = 0.2267 moles × 2 = 0.4534 moles

Step 2: Find mass of N in nicotine.

Subtract the mass of C and H from the total mass of nicotine. Mass of C = (0.3237 moles) × \(12.01\ \frac{mg}{mole}\) ≈ 3.885 mg Mass of H = (0.4534 moles) × \(1.0\ \frac{mg}{mole}\) ≈ 0.453 mg Mass of N = Mass of nicotine - Mass of C - Mass of H Mass of N = 5.250 mg - 3.885 mg - 0.453 mg = 0.912 mg

Step 3: Calculate the moles of N in nicotine.

Moles of N = Mass of N/Molar mass of N Mmoles of N = \( \frac{0.912\ mg}{14.0\ mg/mol}\) = 0.0651 moles

Step 4: Calculate the empirical formula.

Find the whole number ratio of moles of C, H, and N by dividing by the smallest quantity. Divide moles of each element by 0.0651: C: \(\frac{0.3237}{0.0651}\) ≈ 4.97 H: \(\frac{0.4534}{0.0651}\) ≈ 6.95 N: \(\frac{0.0651}{0.0651}\) ≈ 1 The empirical formula of nicotine is C₅H₇N.

Step 5: Determine the molecular formula of nicotine.

Calculate the molar mass of the empirical formula and compare it with the molar mass of nicotine. Molar mass of C5H7N = 5 × 12.01 + 7 × 1.0 + 14.0 = 73.09 g/mol Divide the known molar mass of nicotine by the molar mass of the empirical formula: Molar mass ratio = \( \frac{160}{73.09}\) ≈ 2.19 ≈ 2 Multiply the empirical formula by the molar mass ratio to find the molecular formula: Nicotine molecular formula = C5H7N × 2 = C₁₀H₁₄N₂

Key Concepts

Chemical Combustion

Chemical combustion refers to the process of burning a substance in the presence of oxygen, resulting in the release of energy, usually in the form of heat and light. When a substance combusts completely, it typically forms specific products, such as carbon dioxide (CO2) and water (H2O) when hydrocarbons are burned.
In the context of calculating empirical formulas, chemical combustion analysis is often used. By determining the amounts of CO2 and H2O produced during the combustion of an unknown compound, scientists can deduce the amounts of carbon and hydrogen in the original compound.

• The carbon in the compound is converted to CO2.
• The hydrogen is converted to H2O.

This enables the steps necessary for determining the empirical formula of the compound through the information obtained about the moles and mass of these combustion products. This method is an effective approach for identifying a compound's composition exploring the proportions of its constituent elements.

Molar Mass Calculation

Molar mass calculation is a vital tool in chemistry that involves determining the mass of one mole of a substance, often expressed in grams per mole (g/mol).
In the context of this exercise, calculating the molar mass is essential for converting between grams and moles, as well as for determining how much of an element is present in a compound.
For example, during the combustion of a compound:

• Moles of CO2 are calculated using its molar mass of 44.01 g/mol.
• Moles of H2O are calculated using its molar mass of 18.02 g/mol.

To find the empirical formula, it's necessary to determine the mass and number of moles of each element in a compound. The molar masses of carbon (C), hydrogen (H), and often nitrogen (N) are typically used for such conversions. Understanding these values allows for the calculation of how much of each element corresponds to a certain mass of the compound, providing the ratio of elements, which is fundamental in establishing the compound's empirical formula.

Molecular Formula Determination

The molecular formula determination starts with the empirical formula and extends to find the actual number of atoms of each element in a molecule. While the empirical formula represents the simplest ratio of elements, it doesn't indicate the true number of atoms. For that, the molecular formula is required and can be calculated using the molar mass of the compound.
In nicotine's case from the example:

• The empirical formula is determined to be C₅H₇N.
• The empirical formula's molar mass is 73.09 g/mol.

By comparing the given molar mass of nicotine (160 g/mol) with that of the empirical formula, we find the molecular formula.

• The molecular formula is obtained by multiplying the empirical formula by the ratio obtained from dividing the compound's molar mass by the empirical formula's molar mass.
• If the ratio is about 2, as in nicotine, the molecular formula is C₁₀H₁₄N₂.

This process ensures that the molecular formula is reflective of the actual substance, providing precise information about the compound's composition.