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?

Short answer

The empirical formula of vanillin is \(\mathrm{C_2 H O}\).

Step-by-step solution

Calculate the moles of carbon

First, we need to find the moles of carbon in vanillin. Since the combustion of vanillin produces CO₂, the mass of carbon in CO₂ can be used to determine the moles of carbon. The given mass of CO₂ is 2.43 g. Given, molecular weight of carbon (C) is 12.01 g/mol and molecular weight of oxygen (O) is 16.00 g/mol. So the molecular weight of CO₂ is 12.01 + (16.00 × 2) = 44.01 g/mol. To find the moles of carbon in the sample, divide the mass of CO₂ produced by its molecular weight: Moles of carbon = (mass of carbon dioxide) / (molecular weight of carbon dioxide) Moles of carbon = 2.43 g / 44.01 g/mol = 0.0552 mol

Calculate the moles of hydrogen

Next, we need to find the moles of hydrogen in vanillin. Since the combustion of vanillin produces H₂O, the mass of hydrogen in H₂O can be used to determine the moles of hydrogen. The given mass of H₂O is 0.50 g. Given, molecular weight of hydrogen (H) is 1.008 g/mol. So the molecular weight of H₂O is (1.008 × 2) + 16.00 = 18.016 g/mol. To find the moles of hydrogen in the sample, divide the mass of H₂O produced by its molecular weight: Moles of hydrogen = (mass of water) / (molecular weight of water) Moles of hydrogen = 0.50 g / 18.016 g/mol = 0.0277 mol

Calculate the moles of oxygen

To find the moles of oxygen in vanillin, we need to subtract the mass of carbon and hydrogen from the total mass of the vanillin compound: Mass of oxygen = total mass of vanillin - (mass of carbon + mass of hydrogen) Mass of oxygen = 1.05 g - (2.43 g + 0.50 g) = -1.88 g Since this result is mathematically incorrect, we need to re-check our calculation in Step 1 and Step 2. It appears we've made an error in calculating the mass of carbon and the mass of hydrogen. We need to find the mass of each element instead of the mass of CO₂ and H₂O. Mass of carbon in CO₂ = moles of carbon × molecular weight of carbon Mass of carbon in CO₂ = 0.0552 mol × 12.01 g/mol = 0.662 g Mass of hydrogen in H₂O = moles of hydrogen × molecular weight of hydrogen Mass of hydrogen in H₂O = 0.0277 mol × 1.008 g/mol = 0.0279 g Now, let's recalculate the mass of oxygen in vanillin: Mass of oxygen = total mass of vanillin - (mass of carbon + mass of hydrogen) Mass of oxygen = 1.05 g - (0.662 g + 0.0279 g) = 0.3601 g Now, determine the moles of oxygen: Moles of oxygen = mass of oxygen / molecular weight of oxygen Moles of oxygen = 0.3601 g / 16.00 g/mol = 0.0225 mol

Calculate the empirical formula

To find the empirical formula, we will divide the moles of each element by the smallest number of moles, and then round the resulting ratios to the nearest whole number. Divide moles by the smallest moles (in this case, moles of oxygen = 0.0225 mol): C: 0.0552 mol / 0.0225 mol = 2.454 H: 0.0277 mol / 0.0225 mol = 1.231 O: 0.0225 mol / 0.0225 mol = 1 Rounding each ratio to the nearest whole number, we get: C: 2 H: 1 O: 1 Therefore, the empirical formula of vanillin is \(\mathrm{C_2 H O}\).

Key concepts

Combustion Analysis

Combustion analysis is a key method used in chemistry to determine the elemental composition of a compound. It involves burning a sample in excess oxygen to produce combustion products like carbon dioxide and water, which can then be analyzed to find the amounts of carbon and hydrogen in the original sample. This method is especially useful for organic compounds. In our exercise, vanillin was completely combusted, producing carbon dioxide and water. By measuring the weight of these products, we can backtrack to find out how much carbon and hydrogen the vanillin contained originally. This is done by converting the mass of carbon dioxide and water produced into moles, which then gives us the moles of carbon and hydrogen in the original substance. All of these steps are important to accurately determine the empirical formula of the compound.

Moles of Elements

The concept of moles is central in chemistry, as it provides a bridge between the atomic world and the macroscopic quantities we can measure and observe. The mole is a standard unit that represents a quantity same as Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities (atoms, molecules, etc.). To find the moles of elements in the combustion of vanillin, we started by using the masses of carbon dioxide and water produced during the reaction.

• For carbon, the molecular weight of \( \mathrm{CO}_2 \) is needed to calculate how many moles of \( \mathrm{CO}_2 \) and hence carbon it contains.
• For hydrogen, using the molecular weight of \( \mathrm{H}_2\mathrm{O} \) enables us to find out how many moles of hydrogen are present.
• The calculation involves dividing the mass of the compound by its molecular weight, which yields the moles.

This step-by-step conversion from mass to moles is crucial for finding out the relative amounts of each element in the original vanillin compound.

Molecular Weight Calculation

Calculating molecular weight is essential when converting between mass and moles, which further helps in determining empirical formulas. Molecular weight refers to the sum of the atomic masses of all atoms in a molecule. In the context of our vanillin combustion analysis:

• The molecular weight of \( \mathrm{CO}_2 \) was determined by adding the atomic masses of one carbon atom and two oxygen atoms, totaling \(44.01 \ \mathrm{g/mol}\).
• For \( \mathrm{H}_2\mathrm{O} \), it's the sum of twice the atomic mass of hydrogen plus the atomic mass of oxygen, giving \(18.016 \ \mathrm{g/mol}\).

The accurate calculation of molecular weight enables the correct determination of moles of each element, which are crucial for finding out the empirical formula. Missing the right molecular weight can lead to significant discrepancies in finding the composition of the compound, as seen in the initial incorrect calculation of oxygen in the example. Each accurate step in molecular weight calculation ensures proper interpretation of experimental data and successful derivation of theoretical conditions.