Problem 56 - Chemistry The Central Science In Si Units

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 \(\mathrm{N}\). A \(5.250-\mathrm{mg}\) 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

(a) Ethyl butyrate: \( C_3H_6O \). (b) Nicotine: Empirical: \( C_5H_7N \), Molecular: \( C_{10}H_{14}N_2 \).

Step 1: Calculate Moles of Carbon from CO2 in Ethyl Butyrate

The mass of carbon in CO2 is determined by using the molar mass of CO2 (44.01 g/mol) and knowing that each mole of CO2 corresponds to one mole of carbon atoms.\[ \text{Mass of } \text{CO}_2 = 6.32 \text{ mg} = 0.00632 \text{ g} \]\[ n(\text{C in } \text{CO}_2) = \frac{0.00632 \text{ g}}{44.01 \text{ g/mol}} = 0.0001436 \text{ mol of CO}_2 \]\[ \text{Mass of C} = 0.0001436 \times 12.01 \text{ g/mol} = 0.001724 \text{ g of C} \]

Step 2: Calculate Moles of Hydrogen from H2O in Ethyl Butyrate

The mass of hydrogen in water is calculated using the molar mass of water (18.02 g/mol) where each mole of water contains two moles of hydrogen atoms.\[ \text{Mass of } \text{H}_2\text{O} = 2.58 \text{ mg} = 0.00258 \text{ g} \]\[ n(\text{H in } \text{H}_2\text{O}) = \frac{0.00258 \text{ g}}{18.02 \text{ g/mol}} = 0.0001432 \text{ mol of H}_2\text{O} \]\[ \text{Mass of H} = 0.0001432 \times 2 \times 1.008 \text{ g/mol} = 0.0002887 \text{ g of H} \]

Step 3: Calculate Mass of Oxygen in Ethyl Butyrate

The mass of oxygen in ethyl butyrate is found by subtracting the mass of carbon and hydrogen from the total mass of the compound.\[ \text{Total mass of ethyl butyrate} = 2.78 \text{ mg} = 0.00278 \text{ g} \]\[ \text{Mass of O} = 0.00278 \text{ g} - (0.001724 \text{ g of C} + 0.0002887 \text{ g of H}) = 0.0007673 \text{ g of O} \]

Step 4: Determine Empirical Formula of Ethyl Butyrate

Convert the masses of each element to moles and find the simplest whole-number ratio.\[ n(\text{C}) = \frac{0.001724}{12.01} = 0.0001436 \text{ mol} \]\[ n(\text{H}) = \frac{0.0002887}{1.008} = 0.0002865 \text{ mol} \]\[ n(\text{O}) = \frac{0.0007673}{16.00} = 0.00004796 \text{ mol} \]Divide by the smallest number of moles:\[ C: \frac{0.0001436}{0.00004796} \approx 3, \quad H: \frac{0.0002865}{0.00004796} \approx 6, \quad O: \frac{0.00004796}{0.00004796} = 1 \]Thus, the empirical formula is \( C_3H_6O \).

Step 5: Calculate Moles of Carbon from CO2 in Nicotine

Following a similar calculation for nicotine's carbon content in CO2.\[ \text{Mass of } \text{CO}_2 = 14.242 \text{ mg} = 0.014242 \text{ g} \]\[ n(\text{C}) = \frac{0.014242}{44.01} = 0.0003234 \text{ mol of C} \]\[ \text{Mass of C} = 0.0003234 \times 12.01 = 0.003882 \text{ g of C} \]

Step 6: Calculate Moles of Hydrogen from H2O in Nicotine

Determine the hydrogen contribution from the water produced by nicotine combustion.\[ \text{Mass of } \text{H}_2\text{O} = 4.083 \text{ mg} = 0.004083 \text{ g} \]\[ n(\text{H}) = \frac{0.004083}{18.02} \times 2 = 0.0004527 \text{ mol of H} \]\[ \text{Mass of H} = 0.0004527 \times 1.008 = 0.0004562 \text{ g of H} \]

Step 7: Calculate Mass and Moles of Nitrogen in Nicotine

Find the mass of nitrogen by subtracting the carbon and hydrogen masses from the total sample mass, and then calculate moles.\[ \text{Total mass of nicotine} = 5.250 \text{ mg} = 0.005250 \text{ g} \]\[ \text{Mass of N} = 0.005250 \text{ g} - (0.003882 \text{ g} + 0.0004562 \text{ g}) = 0.000911 \text{ g of N} \]\[ n(\text{N}) = \frac{0.000911}{14.01} = 0.000065 \text{ mol of N} \]

Step 8: Determine Empirical Formula of Nicotine

Calculate the simplest whole-number ratio of the elements.\[ n(\text{C}) = \frac{0.003882}{12.01} = 0.0003 \text{ mol} \]\[ n(\text{H}) = \frac{0.0004562}{1.008} = 0.000452 \text{ mol} \]\[ n(\text{N}) = 0.000065 \text{ mol} \]Ratios (approximately after scaling):\[ C: 5, \quad H: 7, \quad N: 1 \]Empirical formula: \( C_5H_7N \).

Step 9: Determine Molecular Formula for Nicotine

Comparing the mass of the empirical formula unit with the given molar mass:Empirical formula mass: \[ 5 \times 12.01 + 7 \times 1.008 + 14.01 = 81.13 \text{ g/mol} \]Molecular formula: \[ n = \frac{160}{81.13} \approx 2 \]Thus, the molecular formula is \( C_{10}H_{14}N_2 \).

Key Concepts

ethyl butyrate combustion

Understanding the combustion of ethyl butyrate is not just about flames and heat. It’s about chemistry in action! Ethyl butyrate is an ester that contributes to the delightful scent of pineapples. When it combusts, it reacts with oxygen to form carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)).
When performing the combustion, we start by measuring the amount of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) produced. Here's why:

• Carbon calculation: Each molecule of \(\text{CO}_2\) represents a carbon atom from the original compound. By knowing how much \(\text{CO}_2\) was produced, we can calculate how much carbon was in the original sample.
• Hydrogen calculation: Similarly, each \(\text{H}_2\text{O}\) molecule gives information about hydrogen atoms. Since water (\(\text{H}_2\text{O}\)) has two hydrogen atoms, we use this to determine the hydrogen content.

After figuring out how much carbon and hydrogen are in ethyl butyrate, we look at the total original mass of the compound to determine the rest, which must be oxygen. By finding the simplest mole ratio of carbon, hydrogen, and oxygen, we deduce the empirical formula. In the case of ethyl butyrate from the exercise, the formula is \(C_3H_6O\), expressing the simplest whole number ratio of its atoms.

nicotine combustion

Nicotine is most well-known for being a potent and addictive component of tobacco. Chemically, it contains carbon, hydrogen, and nitrogen. When nicotine combusts, it forms carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)), just like ethyl butyrate, along with nitrous gases which are typically not measured. The combustion data allows us to trace back the original elemental composition of nicotine.
To do so, you begin by determining:

• Carbon Content: The amount of \(\text{CO}_2\) produced tells you how many moles of carbon were present.
• Hydrogen Content: The \(\text{H}_2\text{O}\) indicates the number of hydrogen atoms initially there.
• Nitrogen Calculation: The leftover mass after accounting for carbon and hydrogen is attributed to nitrogen, which is unique for nicotine.

The calculation results in the empirical formula \(C_5H_7N\). However, to find out the actual molecular formula, we need to compare the empirical formula mass with the actual molar mass of nicotine. Given that the molar mass of nicotine is about 160 g/mol, the molecular formula is deduced to be \(C_{10}H_{14}N_2\), by scaling up the empirical formula.

molecular formula determination

Determining the molecular formula from an empirical formula involves a few more steps and requires a bit of detective work with molar mass. Simply put, while the empirical formula gives the smallest whole number ratio of atoms in a compound, the molecular formula tells us the actual number of atoms.
Here’s how to bridge the gap from empirical to molecular:

• Empirical Formula Mass: Calculate the mass based on the empirical formula by adding up the atomic masses of the constituent elements. For instance, in nicotine, the empirical formula \(C_5H_7N\) adds up to about 81 g/mol.
• Comparing Molar Masses: Divide the actual molar mass by the empirical formula mass. This quotient indicates how many times we multiply the empirical formula to get the molecular formula. So, for nicotine, with a molar mass of roughly 160 g/mol, we find it fits nearly two empirical units.
• Molecular Formula Scaling: Multiply the subscripts in the empirical formula by the result from the comparison step. With nicotine, this results in a molecular formula of \(C_{10}H_{14}N_2\).

Thus, understanding this process helps unravel the complex but fascinating world of molecular compositions.