Question

A sample of the poisonous compound nicotine extracted from cigarette smoke was found to contain \(74.0 \%\) by weight of carbon (C, atomic weight \(=12.0 \mathrm{~g} / \mathrm{mole}), 8.65 \%\) by weight of hydrogen \((\mathrm{H}\), atomic weight \(=1.01 \mathrm{~g} / \mathrm{mole})\), and \(17.3 \%\) by weight of nitrogen \((\mathrm{N}\), atomic weight \(=14.0\) \(\mathrm{g} /\) mole). What is the empirical formula of nicotine?

Short answer

The empirical formula of nicotine is C5H7N.

Step-by-step solution

Convert the percentage (by weight) into grams

Assume we have 100 grams of nicotine. This makes the calculations easier since the percentages directly represent grams. - Carbon: 74.0% => 74.0 grams - Hydrogen: 8.65% => 8.65 grams - Nitrogen: 17.3% => 17.3 grams

Convert grams into moles for each element

To do this, divide the grams by their respective atomic weights: - Carbon: 74.0 g / 12.0 g/mol = 6.17 mol - Hydrogen: 8.65 g / 1.01 g/mol = 8.56 mol - Nitrogen: 17.3 g / 14.0 g/mol = 1.24 mol

Find the simplest ratio of moles

To determine the simplest whole-number ratio of the elements, divide each mole value by the smallest mole value. In this case, the smallest mole value is 1.24 mol (nitrogen). - Carbon: 6.17 mol / 1.24 mol = 4.97 ≈ 5 - Hydrogen: 8.56 mol / 1.24 mol = 6.90 ≈ 7 - Nitrogen: 1.24 mol / 1.24 mol = 1

Write the empirical formula

Using the simplest whole-number ratios obtained in step 3, the empirical formula for nicotine compound is C5H7N1. The empirical formula can be written as C5H7N.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us understand the quantitative relationships between substances in chemical reactions. It's all about measuring and analyzing the amounts of reactants and products. In the context of calculating an empirical formula, stoichiometry is used to balance the ratios of atoms in a compound based on the number of moles.
This involves several key steps:

• First, convert the percentage composition of each element in a compound to grams. Assuming a 100g sample simplifies this step since the percentages align directly with grams.
• Next, calculate the number of moles of each element. This is done by dividing the mass of each element by its atomic weight.
• Finally, find the simplest whole number ratio of moles, which reveals the formula of the compound in its simplest form.

By understanding stoichiometry, we can accurately determine the relationships between the elements, which is crucial for formulating the empirical formula.

Elemental Analysis

Elemental analysis gives us insight into the composition of substances through quantitative examination of each element within the compound. It essentially tells us what percentage of the compound each element makes up.

For instance, in nicotine, elemental analysis shows us that it is composed of 74.0% carbon, 8.65% hydrogen, and 17.3% nitrogen by weight. Each of these percentages reflects the mass contributions of the individual elements in a compound.
The process involves:

• Analyzing a sample to determine the weight percentage of each element.
• Using this data to convert the percentage of each element to grams, as if the sample weighed exactly 100 grams.
• Sigificantly aiding in determining the compound's empirical formula by illustrating the composition in terms of elemental weights.

This analysis is crucial because it feeds directly into calculations needed for empirical formula determination.

Chemical Composition

Chemical composition refers to the proportion of different elements in a compound. Understanding this concept helps chemists determine the empirical and molecular formulas of a compound, which describe the types and numbers of atoms in the compound.

In the exercise with nicotine, the chemical composition was given in terms of percentages by weight—74.0% carbon, 8.65% hydrogen, and 17.3% nitrogen. These percentages imply the makeup of nicotine in terms of these elements. To simplify, we work with the assumption of 100 grams to directly translate percentages to grams, making our initial calculations much simpler for determining empirical formulas.
Knowing the chemical composition, one can calculate the empirical formula by finding the simplest whole number mole ratio of each element. This tells us not just the presence of specific elements in a compound but also how vastly each one contributes to the overall structure.

Molar Mass Calculation

Calculating the molar mass is a crucial step in many stoichiometric computations and helps us transition from mass values to mole values. This calculation is essential when converting the mass of an element in grams to the number of moles, which is a vital step in determining empirical formulas.

An element's molar mass is the mass of one mole of its atoms, usually in grams per mole (\[g/mol\]). For example, carbon has a molar mass of 12.0 g/mol, hydrogen 1.01 g/mol, and nitrogen 14.0 g/mol.
To find the molar mass:

• Divide the mass of each element by its atomic weight. For example, if you have 74 grams of carbon and the atomic weight is 12.0 g/mol, you calculate 74 / 12 = 6.17 moles of carbon.
• Apply this to hydrogen and nitrogen similarly, converting masses to moles using their respective atomic weights.

This process confirms how many moles of each element are in the compound, paving the way for calculating the empirical formula. It's a simple yet powerful tool that translates the language of mass into moles, which is the key currency in chemistry.