Problem 95 - Chemistry The Central Science

Question

Vanillin, the dominant flavoring in vanilla, contains \(\mathrm{C}, \mathrm{H}\) , and \(\mathrm{O} .\) 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?

Step-by-Step Solution

Answer

The empirical formula of Vanillin is \(\textrm{C}_3\textrm{H}_3\textrm{O}\).

Step 1: Find moles of Carbon and Hydrogen from CO2 and H2O

To find the moles of Carbon and Hydrogen, we need to use the mass and molar mass of CO2 and H2O. Let's find the moles of Carbon and Hydrogen using the given data. Molar mass of CO2 = 12.01 (1 Carbon) + 16.00 (2 Oxygen) = 44.01 g/mol Moles of CO2 = Mass of CO2 / Molar mass of CO2 = 2.43 g / 44.01 g/mol = 0.0552 mol Molar mass of H2O = 1.008 (2 Hydrogen) + 16.00 (1 Oxygen) = 18.016 g/mol Moles of H2O = Mass of H2O / Molar mass of H2O = 0.50 g / 18.016 g/mol = 0.0278 mol Since there is 1 mole of Carbon in 1 mole of CO2 and 2 moles of Hydrogen in 1 mole of H2O: Moles of Carbon = 0.0552 mol Moles of Hydrogen = 0.0278 mol * 2 = 0.0556 mol

Step 2: Calculate the mass of Oxygen in Vanillin

To find the mass of Oxygen present in Vanillin, we can subtract the masses of Carbon and Hydrogen from the initial mass of Vanillin. Mass of Carbon = Moles of Carbon * Molar mass of Carbon = 0.0552 mol * 12.01 g/mol = 0.662 g Mass of Hydrogen = Moles of Hydrogen * Molar mass of Hydrogen = 0.0556 mol * 1.008 g/mol = 0.056 g Mass of Oxygen = Initial mass of Vanillin - Mass of Carbon - Mass of Hydrogen = 1.05 g - 0.662 g - 0.056 g = 0.332 g

Step 3: Find moles of Oxygen in Vanillin

Now that we have the mass of Oxygen, we can find the number of moles. Molar mass of Oxygen = 16.00 g/mol Moles of Oxygen = Mass of Oxygen / Molar mass of Oxygen = 0.332 g / 16.00 g/mol = 0.0208 mol

Step 4: Determine the empirical formula

In order to find the empirical formula, we need to divide the moles of each element by the smallest number of moles. Smallest number of moles = 0.0208 mol (Oxygen) Dividing each number of moles by the smallest number of moles: C: 0.0552 mol / 0.0208 mol = 2.65 ≈ 3 H: 0.0556 mol / 0.0208 mol = 2.68 ≈ 3 O: 0.0208 mol / 0.0208 mol = 1 Therefore, the empirical formula of Vanillin is \(\textrm{C}_3\textrm{H}_3\textrm{O}\).

Key Concepts

Understanding Stoichiometry

Stoichiometry is a core concept in chemistry that involves calculating the amounts of reactants and products in chemical reactions. It is based on the principle that matter is neither created nor destroyed during a chemical reaction, which is known as the law of conservation of mass.

Using stoichiometry, chemists can determine the relative quantities of substances needed to react with one another. For example, in combustion reactions like the one involving vanillin in the exercise, stoichiometry allows for the calculation of how much carbon dioxide (CO₂) and water (H₂O) will form from a known mass of vanillin. By relating the moles of reactants to the moles of products, scientists can ensure that equations are balanced and predict the yields of products.

For students trying to solve stoichiometric problems, a balanced chemical equation is the starting point. From there, it's a matter of converting masses to moles, using molar ratios from the balanced equation, and then, if necessary, converting moles back to masses. Through practice, determining the empirical formula of a compound from combustion data can become an intuitive process.

Role of Combustion Analysis

Combustion analysis is an analytical technique often used to determine the empirical formula of an organic compound. Through combustion, a substance is reacted with excess oxygen to produce known compounds—namely CO₂ and H₂O—whose masses can be measured to infer the amounts of C and H in the original compound.

This technique is particularly useful when the compound's composition is unknown, such as the case with vanillin in the exercise. By measuring the masses of CO₂ and H₂O produced, it's possible to work backwards to determine the moles, and thus the mass, of carbon and hydrogen present in the original sample. With the mass of the original sample and the masses of C and H accounted for, the mass of oxygen can be deduced by subtraction. Knowing the masses of each element allows us to calculate the ratio of atoms present, and from there, the empirical formula.

Significance of Molar Mass

Molar mass, the mass of one mole of a substance, is a fundamental property used in stoichiometry to convert between mass and moles. The molar mass of a compound is determined by adding together the molar masses of its constituent elements, which can be found on the periodic table.

Understanding molar mass is vital when conducting combustion analysis, as it helps translate the mass of CO₂ and H₂O produced back to the moles of C and H in the original compound, as demonstrated in the vanillin exercise. This is crucial for calculating the empirical formula, as it eventually allows us to determine the simplest whole number ratio of atoms within a compound.

In exercises like the given example, students will benefit significantly from familiarity with molar mass calculations. Often, readability can be improved by breaking down each step and associating it with its conceptual basis, providing a logical flow from the mass of a substance to its molar composition.