Question

Nicotine is a poisonous, addictive compound found in tobacco. A sample of nicotine contains \(6.16 \mathrm{mmol}\) of \(\mathrm{C}, 8.56 \mathrm{mmol}\) $$ \text { of } \mathrm{H} \text { , and } 1.23 \mathrm{mmol} \text { of } \left.\mathrm{N}[1 \mathrm{mmol} \text { ( } 1 \text { millimole })=10^{-3} \mathrm{~mol}\right] \text { . } $$ $$ \text { What is the empirical formula of nicotine? } $$

Short answer

The empirical formula of nicotine is \(C_{5}H_{7}N\).

Step-by-step solution

Identify the moles of each element

Given: \(6.16 \text{ mmol of } \text{C}\), \(8.56 \text{ mmol of } \text{H}\), and \(1.23 \text{ mmol of } \text{N}\). Convert these from millimoles to moles: \(1 \text{ mmol} = 10^{-3} \text{ mol}\). Thus: \(6.16 \text{ mmol of } \text{C} = 6.16 \times 10^{-3} \text{ mol}\), \(8.56 \text{ mmol of } \text{H} = 8.56 \times 10^{-3} \text{ mol}\), \(1.23 \text{ mmol of } \text{N} = 1.23 \times 10^{-3} \text{ mol}\).

Determine the smallest number of moles among the elements

\(1.23 \times 10^{-3} \text{ mol of } \text{N}\) is the smallest number of moles.

Calculate the mole ratio of each element

Divide the number of moles of each element by the smallest number of moles: \(\frac{6.16 \times 10^{-3}}{1.23 \times 10^{-3}} = 5.01\), \(\frac{8.56 \times 10^{-3}}{1.23 \times 10^{-3}} = 6.96\), and \(\frac{1.23 \times 10^{-3}}{1.23 \times 10^{-3}} = 1\).

Round the mole ratios to the nearest whole numbers

The ratios are approximately 5 for C, 7 for H, and 1 for N.

Write the empirical formula

The empirical formula can be written by using the whole number mole ratios: \(C_{5}H_{7}N\).

Key concepts

Mole Ratios

Mole ratios are fundamental in chemistry when dealing with chemical compositions. They help us understand the proportion of elements in a compound. In the given problem, we are provided with the amount of each element in millimoles. First, we convert these millimoles to moles by multiplying by \(10^{-3}\). This step is crucial because it normalizes the quantities for easier comparison. Once converted, the smallest value is identified; this becomes the reference point for determining mole ratios.

By dividing each mole value by this smallest quantity, we obtain ratios that tell us how many times more of each element is present relative to the smallest quantity. For instance, in our nicotine example, carbon (\text{C}) and hydrogen (\text{H}) have mole ratios derived by dividing their values by that of nitrogen (\text{N}). These ratios often result in decimal numbers, which are then rounded to the nearest whole numbers to simplify the empirical formula.

Chemical Formulas

A chemical formula represents the types and numbers of atoms in a molecule. In our example, we aim to find the empirical formula of nicotine, the simplest whole-number ratio of elements in the compound. Start from the individual moles of each atom and convert these to whole numbers via mole ratios.

For nicotine, we find 6.16 mmol of C, 8.56 mmol of H, and 1.23 mmol of N, which convert to respective moles. Using the smallest mole value, the ratios are discovered to be approximately 5:7:1. Therefore, the empirical formula for nicotine becomes \(C_5H_7N\). This simplified representation reveals the fundamental building blocks of nicotine, essential for further chemical analysis and reactions.

Stoichiometry

Stoichiometry is the area of chemistry that involves quantifying the relationships between reactants and products in chemical reactions. It relies heavily on the principles of mole ratios and chemical formulas. For solving the empirical formula of nicotine, stoichiometry guided us through converting millimoles to moles, finding the smallest quantity, and calculating the mole ratios.

These steps—converted through clearly defined stoichiometric relationships—help ensure that our chemical equations and formulas are balanced and accurate. Stoichiometry not only assists in finding empirical formulas but also plays a crucial role in predicting the outcomes of reactions and synthesizing new compounds. By understanding the stoichiometric relationships, students can master more complex chemical interactions and appreciate the balanced nature of chemical equations.