Question

Chemical Formulas from Experimental Data Calculate the empirical formula for each stimulant based on its elemental mass percent composition. $$ \begin{array}{l}{\text { a. nicotine (found in tobacco leaves): } \mathrm{C} 74.03 \%, \mathrm{H} 8.70 \%, \mathrm{N} 17.27 \%} \\ {\text { b. caffeine found in coffee beans): } \mathrm{C} 49.48 \%, \mathrm{H} 5.19 \%, \mathrm{N} 28.85 \%, \mathrm{O}} \\ {16.48 \%}\end{array} $$

Short answer

The empirical formula for nicotine is C5H7N, and the empirical formula for caffeine is C4H5N2O.

Step-by-step solution

Convert Percentage to Mass

Assume you have 100 g of the substance. This makes the mass of each element equal to its percent composition. For nicotine: C: 74.03 g, H: 8.70 g, N: 17.27 g. For caffeine: C: 49.48 g, H: 5.19 g, N: 28.85 g, O: 16.48 g.

Convert Mass to Moles

Use the molar mass of each element to convert grams to moles. Nicotine: moles of C = 74.03 g / 12.01 g/mol, moles of H = 8.70 g / 1.008 g/mol, moles of N = 17.27 g / 14.01 g/mol. Caffeine: moles of C = 49.48 g / 12.01 g/mol, moles of H = 5.19 g / 1.008 g/mol, moles of N = 28.85 g / 14.01 g/mol, moles of O = 16.48 g / 16.00 g/mol.

Divide by the Smallest Number of Moles

Divide the number of moles of each element by the smallest number of moles to find the ratio of atoms of each element. For nicotine and caffeine, divide the numbers of moles of C, H, N, and O calculated in Step 2 by the smallest number of moles from those calculations to get the simplest whole number ratio.

Write Empirical Formulas

Using the ratios from Step 3, write down the empirical formula by assigning the whole number ratios as subscripts to the corresponding elements. If any ratio is not a whole number, multiply all the ratios by the smallest number that will convert all ratios to whole numbers.

Calculate the Empirical Formulas for Nicotine and Caffeine

Apply the previous steps to calculate the empirical formulas. For nicotine: The ratios of moles are approximately C: 6.16, H: 8.63, N: 1.23. These approximate to the whole number ratio C: 5, H: 7, N: 1. So the empirical formula for nicotine is C5H7N. For caffeine: The ratios of moles are approximately C: 4.12, H: 5.15, N: 2.06, O: 1.03. These approximate to the whole number ratio C: 4, H: 5, N: 2, O: 1. So the empirical formula for caffeine is C4H5N2O.

Key concepts

Chemical Composition

The concept of chemical composition is fundamental to understanding the make-up of substances in chemistry. In essence, chemical composition refers to the arrangement, type, and ratio of atoms in molecules of chemical substances. The empirical formula is an example of chemical composition representation that provides the simplest whole number ratio of atoms in a compound.

For instance, in the provided exercise, the empirical formula for nicotine is derived from its mass percent composition. This means that if you have 100 grams of nicotine, you would have 74.03 grams of carbon, 8.70 grams of hydrogen, and 17.27 grams of nitrogen. These values do not tell us the exact number of atoms present but the ratio in which they combine to form nicotine. Understanding the concept of chemical composition allows us to predict the properties of compounds, how they can be transformed, and their behavior in reactions.

Mole Calculation

The mole is a fundamental unit in chemistry that is used to express amounts of a substance. It can be challenging to visualize because it pertains to very large numbers—the Avogadro's number, which is approximately \(6.022 \times 10^{23}\). When calculating moles, one must understand that one mole of any element contains exactly \(6.022 \times 10^{23}\) atoms of that element.

During mole calculations, as in the exercise with nicotine and caffeine, we convert the mass of each element to moles using the element's molar mass (its mass per mole). For example, we divide the mass of carbon in grams by the molar mass of carbon (12.01 g/mol) to find the number of moles of carbon in the substance. This conversion is crucial to find the empirical formula since it aligns different elements on a comparable scale based on the number of atoms, not mass.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships that govern the combining elements in chemical reactions. It involves using balanced equations to calculate the mass, volume, and number of moles of reactants and products. In the realm of empirical formula calculations, stoichiometry isn't used to describe reactions but to balance ratios of atom moles to reflect the simplest whole number ratio.

Stoichiometry allows the interpretation of these ratios, such as those calculated from nicotine and caffeine compositions, to determine the empirical formula. The step of dividing the number of moles of each element by the smallest number yields ratios reflecting the proportional number of atoms; stoichiometry guides this normalization. Working through the process meticulously, dividing the given ratios by the smallest occurs before interpreting these ratios back into an empirical formula that reflects actual atoms in molecules.