Problem 92 - Chemistry and Chemical Reactivity

Question

Nicotine, a poisonous compound found in tobacco leaves, is \(74.0 \% \mathrm{C}, 8.65 \% \mathrm{H},\) and \(17.35 \% \mathrm{N} .\) 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: \( C_5H_7N \); Molecular formula: \( C_{10}H_{14}N_2 \).

Step 1: Convert Percentages to Grams

Assume we have a 100 g sample of nicotine. This means we have 74.0 g of C, 8.65 g of H, and 17.35 g of N.

Step 2: Convert Grams to Moles

Use the atomic masses of the elements to convert grams to moles. For carbon: \( 74.0 \text{ g C} \times \frac{1 \text{ mol C}}{12.01 \text{ g}} = 6.16 \text{ mol C} \). For hydrogen: \( 8.65 \text{ g H} \times \frac{1 \text{ mol H}}{1.008 \text{ g}} = 8.58 \text{ mol H} \). For nitrogen: \( 17.35 \text{ g N} \times \frac{1 \text{ mol N}}{14.01 \text{ g}} = 1.24 \text{ mol N} \).

Step 3: Determine the Molecular Ratio

Divide each mole quantity by the smallest number of moles calculated to find the simplest whole-number ratio. For carbon: \( \frac{6.16}{1.24} = 4.97 \approx 5 \). For hydrogen: \( \frac{8.58}{1.24} = 6.92 \approx 7 \). For nitrogen: \( \frac{1.24}{1.24} = 1 \).

Step 4: Write the Empirical Formula

Based on the mole ratio, the empirical formula is \( C_5H_7N \).

Step 5: Calculate Molar Mass of Empirical Formula

Calculate the molar mass of the empirical formula \( C_5H_7N \). \(5 \times 12.01 + 7 \times 1.008 + 14.01 \approx 81.13 \text{ g/mol} \).

Step 6: Determine Molecular Formula

Divide the given molar mass by the empirical molar mass: \( \frac{162 \text{ g/mol}}{81.13 \text{ g/mol}} \approx 2 \). Thus, the molecular formula is \( C_{10}H_{14}N_2 \).

Key Concepts

Molar Mass Calculation

To understand the composition of a compound like nicotine, calculating the molar mass is crucial. Molar mass is the weight of one mole of a substance and is typically expressed in grams per mole (g/mol). For calculating the molar mass of a molecule, you need to know the atomic masses of each element involved. Simply multiply the atomic mass of each element by the number of atoms present in the molecule and sum them up.

In the given exercise, the empirical formula for nicotine is determined to be \(C_5H_7N\). To calculate its molar mass, consider the atomic masses: carbon (C) is approximately 12.01 g/mol, hydrogen (H) is approximately 1.008 g/mol, and nitrogen (N) is about 14.01 g/mol.

We multiply the number of each type of atom by its atomic mass and add them together:

• Carbon: \(5 \times 12.01 = 60.05\) g/mol
• Hydrogen: \(7 \times 1.008 = 7.056\) g/mol
• Nitrogen: \(1 \times 14.01 = 14.01\) g/mol

The total molar mass of the empirical formula \(C_5H_7N\) is approximately 81.116 g/mol, which helps in determining the molecular formula when compared to the known molar mass of 162 g/mol for nicotine.

Chemical Composition Analysis

Analyzing the chemical composition is essential to deduce the empirical formula from percent composition. In this process, we assume a sample size of 100 grams for simplicity. The given percentages are then directly converted into grams.

• Carbon: 74.0% translates to 74.0 grams of carbon.
• Hydrogen: 8.65% becomes 8.65 grams of hydrogen.
• Nitrogen: 17.35% translates to 17.35 grams of nitrogen.

After converting these masses into moles using the atomic masses of the elements, we obtain the number of moles of each element present in the compound.

• Carbon: \( \frac{74.0 \text{ g C}}{12.01 \text{ g/mol}} = 6.16 \text{ mol C}\)
• Hydrogen: \( \frac{8.65 \text{ g H}}{1.008 \text{ g/mol}} = 8.58 \text{ mol H}\)
• Nitrogen: \( \frac{17.35 \text{ g N}}{14.01 \text{ g/mol}} = 1.24 \text{ mol N}\)

Next, the mole quantities are converted into the simplest whole-number ratio, with the smallest mole quantity as the divisor. This ratio gives us the empirical formula of the compound, which is used as a stepping-stone to the molecular formula.

Percent Composition Conversion

Converting percent composition into a practical empirical formula requires precision and understanding of mole concepts.

To carry out a percent composition conversion effectively, assume a hypothetical sample size that simplifies calculations, such as 100 grams. This converts percentage composition directly into gram measurements.

Once you have the percentage-by-mass, they need to be converted into moles using their respective atomic weights, which allows for a direct comparison of the number of moles.

• A percentage of 100 g makes it easy to say "74 g of carbon gives 6.16 moles of carbon" based on its atomic weight. This conversion method is essential for both empirical and molecular formula calculations.

In the case of nicotine, understanding and accurately converting the percent composition ensures that you can reliably calculate both empirical and the extended molecular formulas. This step-by-step conversion and calculation teach significant skills involved in general chemical equation balancing and compounds analysis.