Question

Cinnamaldehyde is a compound that is responsible for the characteristic aroma of cinnamon. It contains \(81.79 \% \mathrm{C}\), \(6.10 \% \mathrm{H},\) and the remaining is oxygen. Its molar mass is \(132 \mathrm{~g} / \mathrm{mol} .\) Determine its molecular formula.

Short answer

The molecular formula of cinnamaldehyde is C₉H₈O.

Step-by-step solution

Determine the moles of each element in the empirical formula.

Since we have percentages of each element, suppose that we have 100g of cinnamaldehyde. This way, we can immediately consider the percentages of each element as grams and calculate the moles of C, H, and O. C: \( \frac{81.79 g}{12.01 g/mol} = 6.81 mol \) H: \( \frac{6.10 g}{1.008 g/mol} = 6.05 mol \) Since we do not have the percentage of oxygen, we can calculate the amount of oxygen in grams by subtracting the amounts (grams) of carbon and hydrogen from the total. Hence, Oxygen in grams = \(100 g - (81.79 g + 6.10 g) = 12.11 g \) Then, the moles of O: O: \( \frac{12.11 g}{16.00 g/mol} = 0.757 mol \)

Calculate the lowest whole number ratio.

In order to find the lowest whole number ratio for each element, we need to divide all the moles by the smallest mole value we have: Smallest moles: 0.757 mol (O) C: \( \frac{6.81}{0.757} \approx 9 \) H: \( \frac{6.05}{0.757} \approx 8 \) O: \( \frac{0.757}{0.757} = 1 \) Empirical formula: C₉H₈O

Determine the molecular formula from the empirical formula.

Calculate the molar mass of the empirical formula: Molar mass of C₉H₈O: \(9 * 12.01 g/mol + 8 * 1.008 g/mol + 1 * 16.00 g/mol = 108.09 g/mol \) Now, divide the given molar mass by the molar mass of the empirical formula to find the factor that we need to multiply the empirical formula: Factor: \( \frac{132 g/mol}{108.09 g/mol} \approx 1.22 \) Since the multiplication factor is close to 1 (with some deviation due to rounding errors in our calculations), we can assume that the molecular formula is the same as the empirical formula. Molecular formula: C₉H₈O

Key concepts

Empirical Formula

The empirical formula represents the simplest whole number ratio of the atoms of each element present in a compound. It is determined through elemental analysis, which provides the percentages of each element. By assuming a 100g sample, these percentages convert directly to grams. Once you have grams, calculate the number of moles for each element using their respective atomic masses:

• For Carbon (C), use the atomic mass of 12.01 g/mol.
• For Hydrogen (H), use 1.008 g/mol.
• For Oxygen (O), use 16.00 g/mol.

In our example, you found 6.81 moles of carbon, 6.05 moles of hydrogen, and 0.757 moles of oxygen. The next step is to divide all moles by the smallest mole value present, which normalizes the ratios to their simplest whole numbers. In our exercise, this allows us to form the empirical formula of C₉H₈O, by dividing each mole value by 0.757, the smallest mole number. This step is crucial, as it lays the foundation for determining the molecular formula.

Molar Mass

Molar mass is the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). It is a critical factor in determining both empirical and molecular formulas. For a compound, the molar mass is the sum of the atomic masses of all the atoms in the empirical formula. In the case of Cinnamaldehyde with an empirical formula of C₉H₈O, calculate it as follows:

• Carbon: 9 atoms × 12.01 g/mol = 108.09 g/mol
• Hydrogen: 8 atoms × 1.008 g/mol = 8.064 g/mol
• Oxygen: 1 atom × 16.00 g/mol = 16.00 g/mol

Adding these values gives a molar mass of 132 g/mol for the empirical formula. In our context, we used this empirical molar mass to find how it relates to the actual or given molar mass of the compound.

Elemental Analysis

Elemental analysis is the process of determining the percentage composition of each element within a compound. It is often the first step in figuring out a compound's empirical formula. Knowing how much of each element is present, typically reported as a percentage, helps chemists make an initial assessment of the compound's identity. By treating these percentages as grams in a 100 g sample, you can simplify the conversions needed to find the number of moles of each element:

• The given percentages are used directly as weights in grams.
• The gram weights are divided by each element’s atomic mass to find moles.

Once you have calculated the moles, you convert them to the lowest whole number ratios to determine the empirical formula. In our example, an analysis showing 81.79% C, 6.10% H, and the rest as oxygen provided a clear pathway to finding C₉H₈O as the empirical formula.