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Ch.3 - Chemical Reactions and Reaction Stoichiometry
Brown - Chemistry: The Central Science 15th Edition
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Tutti i libri di testoBrown 15th EditionCh.3 - Chemical Reactions and Reaction StoichiometryProblema 56b
Capitolo 3, Problema 56b

(b) Nicotine, a component of tobacco, is composed of C, H, and N. A 5.250-mg sample of nicotine was combusted, producing 14.242 mg of CO2 and 4.083 mg of H2O. What is the empirical formula for nicotine? If nicotine has a molar mass of 160 ± 5 g/mol, what is its molecular formula?

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Determine the moles of carbon in CO2 and the moles of hydrogen in H2O. Use the molar mass of CO2 (44.01 g/mol) and H2O (18.02 g/mol) to convert the mass of CO2 and H2O to moles. Remember that each mole of CO2 contains one mole of carbon and each mole of H2O contains two moles of hydrogen.
Calculate the mass of carbon and hydrogen in the sample by multiplying the moles of carbon and hydrogen by their respective atomic masses (12.01 g/mol for C and 1.008 g/mol for H).
Subtract the total mass of carbon and hydrogen from the initial mass of the nicotine sample to find the mass of nitrogen. Use the atomic mass of nitrogen (14.01 g/mol) to convert this mass to moles of nitrogen.
Find the simplest whole number ratio of moles of C, H, and N to determine the empirical formula. Divide the moles of each element by the smallest number of moles calculated among C, H, and N.
To find the molecular formula, divide the molar mass of nicotine (given as 160 ± 5 g/mol) by the molar mass of the empirical formula unit. Multiply the subscripts in the empirical formula by this ratio to get the molecular formula.

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Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of organic compounds. In this process, a sample is burned in excess oxygen, producing carbon dioxide (CO2) and water (H2O). The masses of these products are measured to calculate the amounts of carbon, hydrogen, and other elements in the original sample, which is essential for determining the empirical formula.
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Combustion of Non-Hydrocarbons

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. It is derived from the moles of each element calculated from the combustion products. For nicotine, the empirical formula can be determined by converting the masses of CO2 and H2O produced into moles of carbon and hydrogen, and then finding the ratio of these moles to establish the formula.
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Empirical vs Molecular Formula

Molecular Formula

The molecular formula indicates the actual number of atoms of each element in a molecule of a compound. It can be derived from the empirical formula by comparing the molar mass of the compound to the molar mass of the empirical formula. If the molar mass of nicotine is known, the molecular formula can be calculated by determining how many times the empirical formula fits into the molar mass.
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Determining Molecular Formulas
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