Problem 92 - Chemistry and Chemical Reactivity

Question

Nicotine, a poisonous compound found in tobacco leaves, is \(74.0 \%\) C, \(8.65 \%\) H, and \(17.35 \%\) N. Its molar mass is \(162 \mathrm{g} / \mathrm{mol}\). What are the empirical and molecular formulas of nicotine?

Step-by-Step Solution

Answer

Empirical formula: \( \text{C}_5\text{H}_7\text{N} \); Molecular formula: \( \text{C}_{10}\text{H}_{14}\text{N}_2 \).

Step 1: Determine the moles of each element

Assume we have 100 grams of nicotine, which gives us the percentages as direct masses: 74.0 grams of carbon (C), 8.65 grams of hydrogen (H), and 17.35 grams of nitrogen (N). Calculate the moles of each element:1. For carbon: \[ \frac{74.0 \, \text{grams C}}{12.01 \, \text{grams/mol}} = 6.16 \, \text{mol C} \].2. For hydrogen: \[ \frac{8.65 \, \text{grams H}}{1.008 \, \text{grams/mol}} = 8.58 \, \text{mol H} \].3. For nitrogen: \[ \frac{17.35 \, \text{grams N}}{14.01 \, \text{grams/mol}} = 1.24 \, \text{mol N} \].

Step 2: Calculate the simplest whole-number mole ratio

Divide the moles of each element by the smallest number of moles calculated to find the simplest whole-number ratio:1. For carbon: \( \frac{6.16}{1.24} = 4.97 \approx 5 \).2. For hydrogen: \( \frac{8.58}{1.24} = 6.92 \approx 7 \).3. For nitrogen: \( \frac{1.24}{1.24} = 1 \).The empirical formula is therefore \( \text{C}_5\text{H}_7\text{N} \).

Step 3: Determine the empirical formula mass

Calculate the molar mass of the empirical formula \( \text{C}_5\text{H}_7\text{N} \):\[ 5(12.01) + 7(1.008) + 1(14.01) = 81.13 \, \text{g/mol} \].

Step 4: Calculate the molecular formula

Divide the given molar mass of nicotine by the empirical formula mass to find the multiple:\[ \frac{162 \, \text{g/mol}}{81.13 \, \text{g/mol}} = 2 \].Since the molecular formula is a multiple of the empirical formula by 2, the molecular formula is \( \text{C}_{10}\text{H}_{14}\text{N}_2 \).

Key Concepts

Mole Calculations

Understanding mole calculations is key to finding the composition of a compound and determining its formula. The concept of a mole allows chemists to count atoms, molecules, and other chemical entities in a given mass of a substance because it provides a bridge between the atomic scale and real-world quantities.

First, calculate the number of moles from a given mass and molar mass (the mass of one mole of a substance). For example, suppose you have 74.0 grams of carbon. You'll need its molar mass, which is 12.01 grams/mole for carbon. Using these values, you can calculate the number of moles by dividing the grams of carbon by its molar mass:

• For carbon: \( \frac{74.0 \, \text{grams of C}}{12.01 \, \text{grams/mol}} = 6.16 \, \text{moles of C} \)
• For hydrogen: \( \frac{8.65 \, \text{grams of H}}{1.008 \, \text{grams/mol}} = 8.58 \, \text{moles of H} \)
• For nitrogen: \( \frac{17.35 \, \text{grams of N}}{14.01 \, \text{grams/mol}} = 1.24 \, \text{moles of N} \)

This crucial step sets the stage for deriving the empirical formula, which reflects the simplest whole-number ratio of these elements in the compound.

Molecular Formula Determination

After finding the empirical formula, we need to determine the molecular formula, which represents the actual numbers of atoms in a molecule of a compound. The molecular formula can be a multiple of the empirical formula.

To find the molecular formula, compare the empirical formula mass to the experimental molar mass of the compound. First, calculate the empirical formula mass by summing the molar masses of the constituent atoms in the empirical formula. For example, with nicotine's empirical formula \( \text{C}_5\text{H}_7\text{N} \), calculate:

• Carbon: 5 atoms \( \times 12.01 \, \text{g/mol} = 60.05 \, \text{g/mol} \)
• Hydrogen: 7 atoms \( \times 1.008 \, \text{g/mol} = 7.056 \, \text{g/mol} \)
• Nitrogen: 1 atom \( \times 14.01 \, \text{g/mol} = 14.01 \, \text{g/mol} \)

Total empirical formula mass = 81.11 g/mol.

Next, divide the compound's molar mass by the empirical formula mass to find a multiplication factor. For nicotine, \( \frac{162 \, \text{g/mol}}{81.11 \, \text{g/mol}} = 2 \). Multiply the empirical formula by this result to obtain the molecular formula: \( \text{C}_{10}\text{H}_{14}\text{N}_2 \). This shows the compound contains twice as many atoms of each element than indicated in the empirical formula.

Empirical Formula

The empirical formula is the simplest whole-number ratio of elements in a compound. Understanding how to determine it is fundamental in chemistry.

Start by calculating the moles of each element from their masses or percentages: for instance, if nicotine is 74.0% carbon, in a 100 g sample, that's 74.0 grams of carbon. Then convert to moles using the molar mass of each element, as demonstrated in the section on mole calculations.

Once you have the moles, derive the simplest ratio of moles by dividing all by the smallest mole quantity calculated. For example, consider the moles calculated for nicotine — carbon, hydrogen, nitrogen:

• Carbon: \( \frac{6.16}{1.24} = 4.97 \approx 5 \)
• Hydrogen: \( \frac{8.58}{1.24} = 6.92 \approx 7 \)
• Nitrogen: \( \frac{1.24}{1.24} = 1 \)

This results in an empirical formula of \( \text{C}_5\text{H}_7\text{N} \). Understanding the empirical formula is crucial as it forms the basis for determining the molecular formula of compounds, helping chemists understand the compound's composition.