Problem 171 - Introductory Chemistry Atoms First

Question

The flavoring agent vanillin contains carbon, hydrogen, and possibly oxygen. When \(0.450 \mathrm{~g}\) of vanillin is subjected to combustion analysis, the results are \(63.08 \% \mathrm{C}\) and \(5.30 \% \mathrm{H}\). If the molar mass is approximately \(152 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula of vanillin?

Step-by-Step Solution

Answer

The molecular formula of vanillin is \(\mathrm{C_9H_8O_3}\).

Step 1: Convert percentages to grams

We will directly use the given percentage of carbon and hydrogen to convert them to grams since the mass of vanillin combusted is given as \(0.450 \mathrm{~g}\): \[0.6308 \times 0.450 \mathrm{~g} = 0.28386 \mathrm{~g~C}\] \[0.0530 \times 0.450 \mathrm{~g} = 0.02385 \mathrm{~g~H}\]

Step 2: Convert grams to moles

Now, let's convert the grams of carbon and hydrogen to moles by dividing by their molar masses: \[\frac{0.28386 \mathrm{~g~C}}{12.01 \mathrm{~g/mol}} = 0.02364 \mathrm{~mol~C}\] \[\frac{0.02385 \mathrm{~g~H}}{1.01 \mathrm{~g/mol}} = 0.02361 \mathrm{~mol~H}\]

Step 3: Find the mole ratio

Divide the moles of carbon and hydrogen by the smaller number of moles to get the mole ratio: \[\frac{0.02364 \mathrm{~mol~C}}{0.02361} = 1\] \[\frac{0.02361 \mathrm{~mol~H}}{0.02361} = 1\]

Step 4: Determine the empirical formula

The empirical formula is the simplest whole number ratio of elements in a compound. From step 3, the mole ratio of carbon and hydrogen is both 1. Since we don't have the percentage composition of oxygen, we can assume the remainder is oxygen: \[\% ~\mathrm{O} = 100\% - \% ~\mathrm{C} - \% ~\mathrm{H} = 100\% - 63.08\% - 5.30\% = 31.62\%\] \[0.3162 \times 0.450 \mathrm{~g} = 0.14229 \mathrm{~g~O}\] \[\frac{0.14229 \mathrm{~g~O}}{16.00 \mathrm{~g/mol}} = 0.00889 \mathrm{~mol~O}\] Divide the moles of oxygen by the smallest number of moles: \[\frac{0.00889 \mathrm{~mol~O}}{0.02361} = 0.376 \approx \frac{1}{3}\] Thus the empirical formula is \(\mathrm{C_3H_3O}\).

Step 5: Calculate the empirical formula mass

The empirical formula mass is the sum of the molar masses of elements in the empirical formula: \[\mathrm{Empirical~Formula~Mass} = 3 \times 12.01 \mathrm{~g/mol} + 3 \times 1.01 \mathrm{~g/mol} + 16.00 \mathrm{~g/mol} = 51.09 \mathrm{~g/mol}\]

Step 6: Determine the molecular formula

Now, divide the molar mass by the empirical formula mass and round the result to the nearest whole number to obtain the molecular formula: \[\frac{152 \mathrm{~g/mol}}{51.09 \mathrm{~g/mol}} \approx 3\] Multiply the empirical formula by this whole number to arrive at the molecular formula: \[\mathrm{Molecular~Formula} = 3 \times (\mathrm{C_3H_3O}) = \mathrm{C_9H_8O_3}\] Therefore, the molecular formula of vanillin is \(\mathrm{C_9H_8O_3}\).

Key Concepts

Understanding Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of a compound. It involves burning a sample of a substance and analyzing the resulting products. When the compound combusts, it reacts with oxygen to produce carbon dioxide and water, among other possible gases. These products can then be measured to determine the amount of carbon and hydrogen in the original compound.

In vanillin's case, the combustion analysis showed percentages for carbon and hydrogen. These percentages are crucial because they help to identify the molecular structure of the compound when combined with the molar mass. This method is particularly useful for organic compounds, as it efficiently provides the ratios of different elements present.

• By knowing only the percentages of certain elements, we can work backwards to find the empirical formula.
• Combustion analysis is often the first step in discovering the empirical formula of organic substances.

Determining the Empirical Formula

The empirical formula of a compound is the simplest ratio of its constituent elements. After obtaining elemental percentages from combustion analysis, we convert these percentages to moles. This gives us the basic mole ratio of the elements.

For vanillin, the percentages of carbon, hydrogen, and the derived oxygen gave us the empirical formula \( \mathrm{C_3H_3O} \). This indicates the simplest whole number ratio. It's crucial because it forms the foundation for determining the molecular formula.

• The empirical formula may not represent the actual number of atoms in a molecule, but rather the simplest ratio of elements.
• Converting mass to moles accurately reflects the ratios of each element in the compound.

Finding the Mole Ratio

The mole ratio is a vital step in determining the empirical formula. By converting the percentage composition into moles, we determine the simplest whole number ratio of each element. This ratio allows us to understand how elements are proportionally combined in the compound.

In our vanillin analysis, we calculated the moles of carbon, hydrogen, and oxygen. We then divided by the smallest number of moles to arrive at the simplest whole number ratio. This was done to ensure our empirical formula truly reflects the nature of the compound, leading to \( \mathrm{C_3H_3O} \) as the empirical formula.

• Dividing by the smallest mole value normalizes the ratios, simplifying the formula.
• Mole ratios help visualize the composition of a compound at an atomic level, showing how atoms relate to each other in size and number.

Calculating Molar Mass for Molecular Formula

Once we have an empirical formula, we use molar mass to determine the molecular formula. The molar mass is an essential step since it tells us whether the empirical formula must be multiplied to match the true molecular formula.

In the exercise, given that the approximate molar mass of vanillin is \(152 \mathrm{~g/mol}\), we computed the empirical formula mass of \( \mathrm{C_3H_3O} \) to be \(51.09 \mathrm{~g/mol} \). Then, by dividing the actual molar mass by the empirical formula mass, we found that the molecular formula is three times the empirical formula, giving us\( \mathrm{C_9H_8O_3} \).

• The molar mass ensures the empirical formula correctly replicates the actual molecular structure when multiplied appropriately.
• It is a verification tool to align calculations with real-world measurements of the compound's mass.