Question

Vanillin, the dominant flavoring in vanilla, contains C, H, and O. When \(1.05 \mathrm{~g}\) of this substance is completely combusted, \(2.43 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.50 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. What is the empirical formula of vanillin?

Short answer

The empirical formula of vanillin is \(C_3H_3O\).

Step-by-step solution

Calculate moles of CO₂ and H₂O produced

The balanced equation for the complete combustion of a hydrocarbon is: \(C_xH_yO_z + O_2 \rightarrow CO_2 + H_2O\) We are given the mass of CO₂ and H₂O produced. We will convert the mass of each product to moles using their respective molar masses: Molar mass of CO₂: \(1 C + 2 O = 12.01 + (2 × 16.00) = 44.01 \frac{g}{mol}\) Molar mass of H₂O: \(2 H + 1 O = (2 × 1.01) + 16.00 = 18.02 \frac{g}{mol}\) Now, we'll convert the mass of CO₂ and H₂O to moles: Moles of CO₂ produced = \(\frac{2.43 g}{44.01 g/mol} = 0.0552 mol\) Moles of H₂O produced = \(\frac{0.50 g}{18.02 g/mol} = 0.0277 mol\)

Calculate moles of C and H in vanillin

From the combustion reaction, we know that each mole of CO₂ produced corresponds to one mole of C in vanillin, and each mole of H₂O produced corresponds to two moles of H in vanillin: Moles of C in vanillin = Moles of CO₂ produced = \(0.0552 mol\) Moles of H in vanillin = 2 × Moles of H₂O produced = \(2 × 0.0277 mol = 0.0554 mol\)

Calculate moles of O in vanillin

We can now determine the moles of O in vanillin by using mass conservation. We are given the mass of vanillin, and now we know the mass of C and H in vanillin from the moles we calculated: Mass of C in vanillin = Moles of C × Molar mass of C = \(0.0552 mol × 12.01 \frac{g}{mol} = 0.662 g\) Mass of H in vanillin = Moles of H × Molar mass of H = \(0.0554 mol × 1.01 \frac{g}{mol} = 0.056 g\) Mass of O in vanillin = Total mass of vanillin - Mass of C - Mass of H = \(1.05 g - 0.662 g - 0.056 g = 0.332 g\) Now, we can calculate the moles of O: Moles of O in vanillin = \(\frac{0.332 g}{16.00 g/mol} = 0.0208 mol\)

Find the empirical formula of vanillin

Using the moles of C, H, and O in vanillin, we will now find the empirical formula by dividing each mol value by the smallest mol value: Mole ratio of C : H : O = \(\frac{0.0552}{0.0208} : \frac{0.0554}{0.0208} : \frac{0.0208}{0.0208} = 2.65 : 2.66 : 1\) Since the ratio is very close to the whole numbers 3:3:1, the empirical formula of vanillin is: \(C_3H_3O_1\) or simply \(C_3H_3O\).

Key concepts

Complete Combustion of Hydrocarbons

Complete combustion of hydrocarbons is a chemical process where a hydrocarbon reacts with oxygen to produce carbon dioxide (CO₂), water (H₂O), and energy. This process requires an adequate supply of oxygen to ensure that the hydrocarbon burns completely. During complete combustion, all carbon atoms in the hydrocarbon are converted to CO₂, and all hydrogen atoms are converted to H₂O.

Understanding this principle is fundamental when analyzing combustion reactions in chemistry, such as those involving the determination of an unknown compound's empirical formula. For instance, in the vanillin example, knowing that each CO₂ molecule produced in the combustion represents one carbon atom and each H₂O molecule represents two hydrogen atoms in the original compound is crucial to deducing the empirical formula.

When a student struggles with the concept of complete combustion, a valuable exercise improvement advice would be to visualize or model the reaction. They could balance a simplified equation of a hydrocarbon's combustion or use molecular models to see the transformation of reactants to products.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. It allows chemists to predict the amount of substances consumed and produced in a reaction, ensuring that the chemical equations are balanced in terms of both mass and charge.

In practice, stoichiometry involves calculations using the mole concept, molar mass, and Avogadro’s number to convert between mass, moles, and molecular counts. For a student facing challenges with stoichiometry, focusing on the mole concept is imperative. This often entails practice in converting grams to moles and vice versa, as seen in the example where calculations were made to find the moles of CO₂ and H₂O produced from the combustion of vanillin.

Additionally, it can be helpful for students to work on a variety of problems, especially ones that require using stoichiometry to determine the composition of a substance, in different contexts to build confidence and understanding.

Molar Mass

The molar mass of a substance is the weight of one mole of that substance, typically expressed in grams per mole (g/mol). It is a critical concept in chemistry because it acts as a conversion factor between the mass of a substance and the number of moles. This allows chemists to count molecules by weighing them, which is more practical than counting individual atoms or molecules.

To determine the molar mass, one needs to sum the atomic masses of all the atoms in a molecule, as seen in the vanillin example. Learning to calculate molar mass effectively is a foundational skill in chemistry. If a student finds it difficult, it might be beneficial to start by calculating the molar mass of simple molecules and gradually progress to more complex compounds.

Interactive tools, such as online molar mass calculators, can also aid in understanding this concept more deeply, as it reinforces the connection between the mass of a sample and the quantity of particles within that sample.