Problem 56 - 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

The empirical formula is \(C_5H_7N\) and the molecular formula is \(C_{10}H_{14}N_2\).

Step 1: Convert Percentages to Grams

Assume you have 100 grams of nicotine. This suggests you have 74 grams of carbon (C), 8.65 grams of hydrogen (H), and 17.35 grams of nitrogen (N).

Step 2: Convert Grams to Moles

To find moles, divide the mass of each element by its atomic weight:- Moles of C: \( \frac{74.0}{12.01} \approx 6.16 \text{ moles} \)- Moles of H: \( \frac{8.65}{1.008} \approx 8.58 \text{ moles} \)- Moles of N: \( \frac{17.35}{14.01} \approx 1.24 \text{ moles} \)

Step 3: Determine the Simplest Mole Ratio

Divide each mole value by the smallest number of moles to determine the simplest whole number ratio:- Ratio for C: \( \frac{6.16}{1.24} \approx 4.97 \approx 5 \)- Ratio for H: \( \frac{8.58}{1.24} \approx 6.92 \approx 7 \)- Ratio for N: \( \frac{1.24}{1.24} = 1 \)Thus, the empirical formula is \( C_5H_7N \).

Step 4: Calculate the Empirical Formula Mass

Calculate the mass of the empirical formula \( C_5H_7N \):- Carbon: \( 5 \times 12.01 = 60.05 \)- Hydrogen: \( 7 \times 1.008 = 7.056 \)- Nitrogen: \( 1 \times 14.01 = 14.01 \)Total = \( 60.05 + 7.056 + 14.01 = 81.116 \text{ g/mol} \)

Step 5: Determine the Molecular Formula

Find the ratio of the molar mass to the empirical mass: \( \frac{162}{81.116} \approx 2 \). The molecular formula is twice the empirical formula, therefore it is \( C_{10}H_{14}N_2 \).

Key Concepts

Molecular Formula

A molecular formula shows the exact number of each type of atom in a molecule. It's like the blueprint of a compound, providing precise information about its composition. For nicotine, scientific analysis delivers its molecular formula as \(C_{10}H_{14}N_2\). This means each nicotine molecule includes:

• 10 carbon atoms
• 14 hydrogen atoms
• 2 nitrogen atoms

Compared to the empirical formula, which is the simplest representation, the molecular formula signifies the actual whole-number ratios in a molecule. Understanding this formula allows chemists to delve deep into the compound’s structure and functionality.

Molar Mass

The molar mass is a crucial concept in chemistry that refers to the mass of one mole of a substance, typically calculated in grams. For nicotine, the molar mass is given as \(162\, \mathrm{g/mol}\). Calculating molar mass involves summing the masses of the individual atoms in a molecule. For nicotine, this is performed using its molecular formula \(C_{10}H_{14}N_2\):

• Carbon: \(10 \times 12.01\, \text{g/mol}\)
• Hydrogen: \(14 \times 1.008\, \text{g/mol}\)
• Nitrogen: \(2 \times 14.01\, \text{g/mol}\)

Knowing the molar mass helps in determining the amount of each element present in a given sample of a compound.

Percent Composition

Percent composition provides insight into the makeup of a compound by indicating the percentage by mass of each element present. For nicotine, it's composed of 74.0% carbon \( (C) \), 8.65% hydrogen \( (H) \), and 17.35% nitrogen \( (N) \). This analysis is crucial for determining:

• The empirical formula: It uses these percentages to estimate the simplest whole-number ratio of atoms.
• The purity of a sample: By comparing known percent compositions, scientists can identify and evaluate the purity of samples.

This concept aids in understanding a compound’s elemental makeup and serves as the foundation for further molecular analysis.

Elemental Analysis

Elemental analysis breaks down a compound to reveal the types and quantities of atoms present. In the case of nicotine, the elemental analysis shows that it contains carbon, hydrogen, and nitrogen. Through processes such as combustion analysis or spectroscopic methods, chemists can meticulously assess:

• The identity of elements: Confirming which elements are part of the compound's structure.
• The quantitative composition: Measuring how much of each element is present, which is crucial for empirical and molecular formula determination.

By utilizing elemental analysis, scientists can accurately piece together detailed profiles of chemical compounds, facilitating research, product development, and quality control efforts.