Question

Methyl salicylate is a common "active ingredient" in liniments such as Ben-Gay \(^{\mathrm{TM}} .\) It is also known as oil of wintergreen. It is made up of carbon, hydrogen, and oxygen atoms. When a sample of methyl salicylate weighing \(5.287 \mathrm{~g}\) is burned in excess oxygen, \(12.24 \mathrm{~g}\) of carbon dioxide and \(2.505 \mathrm{~g}\) of water are formed. What is the simplest formula for oil of wintergreen?

Short answer

Answer: The simplest formula for oil of wintergreen is C2H2O.

Step-by-step solution

Find moles of carbon atoms in the compound

Calculate the moles of carbon atoms in the compound by dividing the mass of CO2 formed by the molar mass of CO2, then multiplying by the stoichiometric ratio of carbon atoms to CO2 molecules: moles_C = (mass_CO2 / molar mass_CO2) * (1 mol C / 1 mol CO2) moles_C = (12.24 g / 44.01 g/mol) * (1 mol C / 1 mol CO2) = 0.278 moles_C

Find moles of hydrogen atoms in the compound

Calculate the moles of hydrogen atoms in the compound by dividing the mass of H2O formed by the molar mass of H2O, then multiplying by the stoichiometric ratio of hydrogen atoms to H2O molecules: moles_H = (mass_H2O / molar mass_H2O) * (2 mol H / 1 mol H2O) moles_H = (2.505 g / 18.02 g/mol) * (2 mol H / 1 mol H2O) = 0.278 moles_H

Find the mass of oxygen atoms in the compound

Subtract the masses of carbon and hydrogen atoms from the total mass of the methyl salicylate sample to find the mass of oxygen atoms: mass_O = total mass of sample - mass_C - mass_H mass_O = 5.287 g - (0.278 moles_C * 12.01 g/mol) - (0.278 moles_H * 1.01 g/mol) = 2.107 g

Find moles of oxygen atoms in the compound

Calculate the moles of oxygen atoms in the compound by dividing the mass of oxygen atoms by the molar mass of oxygen: moles_O = mass_O / molar mass_O moles_O = 2.107 g / 16.00 g/mol = 0.132 moles_O

Determine the simplest ratio of C, H, and O atoms

Divide each amount of moles by the smallest amount of moles (in this case, moles of O, which is the smallest one) to get the simplest whole number ratios of each element to one another: simplest_mole_ratio_C = moles_C / moles_O = 0.278 / 0.132 = 2.1 ≈ 2 simplest_mole_ratio_H = moles_H / moles_O = 0.278 / 0.132 = 2.1 ≈ 2 simplest_mole_ratio_O = moles_O / moles_O = 0.132 / 0.132 = 1

Write the simplest formula for oil of wintergreen

Combine the elements and their ratios to form the simplest formula for the compound: Methyl salicylate simplest formula: C_2H_2O_1, or C2H2O

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. By understanding the concept of stoichiometry, students can predict the amounts of substances consumed and produced in a reaction. A fundamental aspect of stoichiometry is the mole concept, where a mole represents Avogadro's number (\(6.022 \times 10^{23}\) particles) of a substance. It allows chemists to count particles by weighing them.

When solving stoichiometric problems, such as finding the empirical formula of a compound, it is essential to determine the number of moles of each element involved. This involves converting the mass of each element (or compound) to moles using the molar mass, understanding the mole ratio in which elements combine, and balancing these ratios to reflect the law of conservation of mass. The ability to perform these conversions and utilize mole ratios is crucial to mastering stoichiometry.

Molar Mass

Molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). It corresponds to the combined atomic weight of all atoms in a molecular formula. By referring to the periodic table, students can add up the atomic weights of each element in a molecule to find its molar mass. For example, the molar mass of carbon dioxide (CO2) is calculated by adding the molar mass of one carbon atom (12.01 g/mol) to the molar mass of two oxygen atoms (2 x 16.00 g/mol), resulting in 44.01 g/mol.

The concept of molar mass serves as a bridge between the microscopic world of atoms and the macroscopic world of grams. It is pertinent for converting mass measurements into moles, which is a stepping stone for calculating the empirical formula of a compound. Having a clear understanding of molar mass allows students to accurately calculate the number of moles of each element in a compound, as demonstrated in the problem under consideration.

Combustion Analysis

Combustion analysis is a laboratory technique used to determine the empirical formula of organic compounds, typically consisting of carbon, hydrogen, and sometimes oxygen and nitrogen. In this method, a known mass of a compound is burned in excess oxygen, and the masses of the resulting carbon dioxide (CO2) and water (H2O) are measured. From these values, one can calculate the number of moles of carbon (from CO2) and hydrogen (from H2O) in the original sample.

To further analyze the exercise, the student must note that the oxygen in CO2 and H2O comes from both the organic compound and the excess oxygen used in the combustion. Therefore, to find the amount of oxygen in the original compound, it is necessary to subtract the mass of carbon and hydrogen from the total mass of the sample. This step is crucial in accurately determining the compound's simplest formula. Combustion analysis is a practical application of stoichiometry and understanding its process helps in solving real-world chemical analysis problems.

Mole Ratio

The mole ratio is the ratio of moles of one substance to the moles of another substance in a balanced chemical reaction. It is derived from the coefficients of a balanced equation. For empirical formula determination, the mole ratio is used to convert the moles of each element in a compound to the simplest whole-number ratio.

In the methyl salicylate problem, after finding the moles of carbon, hydrogen, and oxygen, the next step is to divide the number of moles of each element by the smallest number of moles obtained. This gives the simplest mole ratio for each element. In some cases, these ratios may not be whole numbers, and thus they may need to be multiplied by a common factor to achieve whole-number ratios, as empirical formulas must be expressed in whole numbers. To convey this more clearly in educational material, illustrative examples or visual aids showing how to round and adjust ratios can greatly aid student comprehension.