Question

Vanillin, the dominant flavoring in vanilla, contains \(\mathrm{C}, \mathrm{H}\), 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 C8H8O3.

Step-by-step solution

Calculate the moles of CO2 and H2O produced

To find the moles of each product, we'll use the molar mass of CO2 (44.01 g/mol) and H2O (18.02 g/mol) and the mass of each product: Moles of CO2 = mass of CO2 ÷ molar mass of CO2 Moles of CO2 = \( \frac{2.43 \mathrm{~g}}{44.01 \mathrm{~g/mol}} \) = 0.0552 mol Moles of H2O = mass of H2O ÷ molar mass of H2O Moles of H2O = \( \frac{0.50 \mathrm{~g}}{18.02 \mathrm{~g/mol}} \) = 0.0277 mol

Determine moles of C and H from the products

Knowing that each mole of CO2 contains one mole of C, and each mole of H2O contains two moles of H, we can find the moles of C and H in vanillin: Moles of C = 0.0552 mol Moles of H = 0.0277 mol × 2 = 0.0554 mol

Calculate the mass of O in vanillin

We can find the mass of O by subtracting the masses of C and H from the total mass of vanillin: Mass of C = moles of C × molar mass of C Mass of C = 0.0552 mol × 12.01 g/mol = 0.662 g Mass of H = moles of H × molar mass of H Mass of H = 0.0554 mol × 1.01 g/mol = 0.056 g Mass of O in vanillin = mass of vanillin - mass of C - mass of H Mass of O = 1.05 g - 0.662 g - 0.056 g = 0.332 g

Determine moles of O in vanillin

Use the molar mass of O (16.00 g/mol) to find the moles of O: Moles of O = mass of O ÷ molar mass of O Moles of O = \( \frac{0.332 \mathrm{~g}}{16.00 \mathrm{~g/mol}} \) = 0.02075 mol

Find the ratio of moles of C, H, and O

Divide the moles of each element by the smallest number of moles to find the empirical formula: Mole ratio of C : \( \frac{0.0552}{0.02075} \) ≈ 2.66 Mole ratio of H : \( \frac{0.0554}{0.02075} \) ≈ 2.67 Mole ratio of O : \( \frac{0.02075}{0.02075} \) = 1 The mole ratios are very close to integer values, so we can round them to get the empirical formula: Empirical formula of vanillin: C8H8O3

Key concepts

Combustion Analysis

Combustion analysis is a traditional laboratory method used to determine the elemental composition of a substance by completely burning it and measuring the masses of the products. For organic compounds containing carbon (C) and hydrogen (H), the combustion process typically results in the formation of carbon dioxide (CO2) and water (H2O), which can be collected and weighed.

The key steps involve burning the substance in an excess of oxygen to ensure full combustion, capturing and measuring the CO2 and H2O produced, and then using stoichiometry to back-calculate the amounts of carbon and hydrogen present in the original substance. If the compound also contains oxygen, the amount of oxygen can be deduced by subtracting the masses of carbon and hydrogen from the initial mass of the compound.

To improve student understanding, it's important to emphasize the indirect nature of this analysis—since we don't measure the elements directly but infer them from the combustion products.

Mole Concept

The mole concept is foundational to understanding chemistry, as it links the micro world of atoms and molecules to the macro world that we can measure. One mole represents Avogadro's number (\(6.022 \times 10^{23}\) entities), and by extension, one mole of any substance contains the same number of units—be they atoms, molecules, ions, or others.

When determining the empirical formula, the mole concept allows us to convert the mass of each element to the number of moles, thus helping to identify the ratio between elements in a compound. By dividing the moles of each element by the smallest number of moles obtained, students can determine the simplest whole-number ratio of atoms in the compound.

Visual aids such as diagrams can be very beneficial in helping students understand the mole concept. Also, using analogies like a 'dozen' helps to convey that 'mole' is just another way to count particles, much like a dozen is a way to count eggs.

Molar Mass

Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). This concept is crucial for converting between the mass of a substance and the number of moles, a standard practice in chemical calculations.

Each element's molar mass is found on the periodic table, corresponding to the atomic weight of the element. For a molecule, the molar mass is the sum of the atomic weights of all the atoms in a single molecule of that substance. Understanding how to use molar mass is key to solving empirical formula problems.

In instructional material, it's beneficial to walk through the process of calculating molar masses for compounds step-by-step to solidify the students' understanding and confidence in using this value for further calculations.

Stoichiometry

Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. It is based on the conservation of mass and the balanced chemical equation for the reaction.

Using stoichiometry, we can predict the amounts of products that will form during a reaction, or the amounts of reactants needed to form a certain amount of product. This is essential for determining the empirical formula from combustion analysis since the process relies on understanding the stoichiometric relationships between the substance combusted and the products formed.

To strengthen comprehension, it's useful to work through several stoichiometric calculations with different scenarios, reinforcing the concept that stoichiometry is simply a way to 'count' atoms and molecules by mass instead of by individual particles.