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 \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). \(\mathrm{A}\) 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 CH₂O. The empirical formula of nicotine is C₅H₇N, and its molecular formula is C₁₀H₁₄N₂.

Step-by-step solution

Determine moles of each element

We can calculate the moles of carbon and hydrogen using the amounts of CO₂ and H₂O formed. Firstly, we need to convert the mass of CO₂ and H₂O into moles: Moles of CO₂ = \(\frac{6.32 \text{ mg}}{44.01 \text{ mg/mol}} = 0.0001436 \text{ mol}\) Moles of H₂O = \(\frac{2.58 \text{ mg}}{18.02 \text{ mg/mol}} = 0.0001431 \text{ mol}\) Now for each product, we can find the moles of C and H: Moles of C = moles of CO₂ = \( 0.0001436 \text{ mol}\) Moles of H = 2 × moles of H₂O = \( 2 \times 0.0001431 = 0.0002862 \text{ mol}\)

Determine the mole ratio of each element

Divide each mole value by the smallest mole value: Mole ratio of C = \(\frac{0.0001436}{0.0001431} \approx 1 \) Mole ratio of H = \(\frac{0.0002862}{0.0001431} \approx 2 \)

Write the empirical formula

Based on the mole ratios, the empirical formula of ethyl butyrate is CH₂O. ### Part B - Empirical formula and Molecular formula of Nicotine ###

Determine moles of each element

We can calculate the moles of carbon and hydrogen using the amounts of CO₂ and H₂O formed. Firstly, we need to convert the mass of CO₂ and H₂O into moles: Moles of CO₂ = \(\frac{14.242 \text{ mg}}{44.01 \text{ mg/mol}} = 0.0003236 \text{ mol}\) Moles of H₂O = \(\frac{4.083 \text{ mg}}{18.02 \text{ mg/mol}} = 0.0002266 \text{ mol}\) Now for each product, we can find the moles of C and H: Moles of C = moles of CO₂ = \( 0.0003236 \text{ mol}\) Moles of H = 2 × moles of H₂O = \( 2 \times 0.0002266 = 0.0004532 \text{ mol}\) To determine moles of N, subtract the mass of C and H from the total mass of nicotine, then divide by the molar mass of N: Mass of C = 0.0003236 mol × 12.01 mg/mol = 3.881 mg Mass of H = 0.0004532 mol × 1.008 mg/mol = 0.4566 mg Mass of N = 5.250 mg - (3.881 mg + 0.4566 mg) = 0.9124 mg Moles of N = \(\frac{0.9124 \text{ mg}}{14.01 \text{ mg/mol}} = 0.0000651 \text{ mol}\)

Determine the mole ratio of each element

Divide each mole value by the smallest mole value: Mole ratio of C = \(\frac{0.0003236}{0.0000651} \approx 5 \) Mole ratio of H = \(\frac{0.0004532}{0.0000651} \approx 7 \) Mole ratio of N = \(\frac{0.0000651}{0.0000651} \approx 1 \)

Write the empirical formula

Based on the mole ratios, the empirical formula of nicotine is C₅H₇N.

Calculate the empirical formula mass

Empirical formula mass = 5(12.01) + 7(1.008) + 14.01 = 80.09 g/mol

Divide the molar mass by the empirical formula mass

Molar mass is given as 160 ± 5 g/mol. We will use the average value of 160 g/mol \(\frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} = \frac{160}{80.09} \approx 2\)

Write the molecular formula

Multiply the empirical formula by the integer found in step 5: Molecular formula of nicotine = (C₅H₇N) × 2 = C₁₀H₁₄N₂

Key concepts

Understanding Stoichiometry

Stoichiometry is the mathematical relationship between the relative quantities of reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of moles, making it possible to predict the amounts of substances consumed and produced in a reaction.

When performing stoichiometric calculations, the first step is often to balance the chemical equation to ensure that the same number of atoms of each element appears on both sides. In the context of empirical formula determination, stoichiometry helps us to determine the mole ratio of each element in a compound by using the mass of products formed during a reaction, such as CO₂ and H₂O in a combustion reaction.

Importance in Empirical Formula Determination

Stoichiometry is essential when deriving an empirical formula. It involves converting the masses of substances to moles, using molar masses, and then finding the simplest whole-number ratio between moles of each element. This ratio reflects the relative number of atoms of each element in the compound.

A strong grasp of stoichiometry is crucial for students to accurately perform these conversions and to understand the underlying principles guiding these calculations.

Combustion Analysis Explained

Combustion analysis is a laboratory technique often used to determine the empirical formula of an organic compound. In this process, a known mass of a compound undergoes complete combustion in excess oxygen to produce CO₂ and H₂O.

The masses of these products are measured, allowing us to calculate the moles of carbon and hydrogen in the original compound. If there are other non-metal elements like nitrogen or sulfur, their presence can also be determined using further chemical analysis.

Role in Empirical Formula Determination

Through combustion analysis, students can learn how to isolate and quantify the carbon and hydrogen components of organic molecules. This is a critical step in working out the empirical formula, as shown in the provided exercise with ethyl butyrate and nicotine. Their corresponding CO₂ and H₂O outputs indicate carbon and hydrogen content, which then can be articulated into a formula reflecting the simplest whole-number ratio of atoms in the compound.

Molecular Formula Calculation

The molecular formula of a compound represents the actual number of atoms of each element present in a molecule, which might be an integer multiple of the empirical formula. Determining the molecular formula involves calculating the empirical formula mass and then comparing it with the experimentally determined molar mass of the compound.

Once the empirical formula is found, the mass of this formula is summed up from the atomic masses of each element. This sum is then divided into the known molar mass of the compound. The result is typically a whole number, which indicates how many empirical formula units are present in the molecular formula.

Calculating Nicotine's Molecular Formula

In the example of nicotine, after determining the empirical formula as C₅H₇N, you would calculate its mass. Dividing nicotine's molar mass by the empirical formula mass gives a ratio used to multiply the subscripts in the empirical formula, yielding the molecule's actual atomic composition. This final step is crucial for understanding the compound's true structure and for predicting its chemical properties.