Question

Vanillin, the active component of the flavoring vanilla, has the composition \(63.15 \% \mathrm{C}, 5.30 \% \mathrm{H}\), and \(31.55 \% \mathrm{O}\). (a) What is its empirical formula? (b) If the molar mass of vanillin is \(152.14 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

Short answer

The empirical formula of vanillin is \(C_3H_3O\) and the molecular formula is \(C_9H_9O_3\).

Step-by-step solution

Conversion of Percentages to Moles

Convert the given percentages of each element into moles. This is done by assuming there is 100g of the compound, hence the % composition would equal the mass composition. The atomic masses of \(C\), \(H\), and \(O\) are approximately 12.01g/mol, 1.01g/mol, and 16.00g/mol respectively. Therefore, the number of moles are: \(C = \frac{63.15g}{12.01g/mol} = 5.25mol\) \(H = \frac{5.3g}{1.01g/mol} = 5.25mol\) \(O = \frac{31.55g}{16g/mol} = 1.97mol\)

Calculate the Empirical Formula

Divide each mole value by the smallest number of moles calculated (In this case, 1.97mol) to get the mole ratio of each element:\(C = \frac{5.25}{1.97} = 2.67 \approx 3\) \(H = \frac{5.25}{1.97} = 2.67 \approx 3\) \(O = \frac{1.97}{1.97} = 1\) Therefore, the empirical formula is \(C_3H_3O\).

Find the Molecular Formula

Calculate the molar mass of the empirical formula (\(C_3H_3O\)):Molar mass = 3*(12.01g/mol) + 3*(1.01g/mol) + 1*(16.00g/mol) = 50.1g/mol. Divide the given molar mass of the compound by the molar mass of the empirical formula:Molecular formula multiplier = \(\frac{152.14g/mol}{50.1g/mol} = 3\) Therefore, the molecular formula of vanillin is \(C_9H_9O_3\) (multiplying the subscripts of the empirical formula by 3).

Key concepts

Understanding Molecular Formula

The molecular formula of a compound reveals the actual number of atoms of each element present in a molecule. It can be different from the empirical formula, which only gives the simplest whole number ratio of the elements. For instance, with vanillin, the empirical formula is \( C_3H_3O \), which expresses the ratio as 3:3:1. However, this is not the whole story, as the molecular formula provides more detailed information.

In our example, the molecular formula of vanillin is \( C_9H_9O_3 \). This means there are actually nine carbon atoms, nine hydrogen atoms, and three oxygen atoms in each molecule of vanillin. This formula provides more clarity about the true composition of the molecule.

• It is determined by multiplying the subscripts in the empirical formula by a whole number.
• This number is found by dividing the molar mass of the compound by the molar mass of the empirical formula.

Understanding the molecular formula helps chemists and students alike to comprehend the actual structure and quantity of atoms in a compound.

The Basics of Percent Composition

Percent composition is a handy way to express how much of each element is present in a compound in terms of percentage by mass. For vanillin, the composition is given as \(63.15\%\) carbon, \(5.30\%\) hydrogen, and \(31.55\%\) oxygen, indicating the mass fractions contributed by each of these elements to the 100g sample.

When working with percent compositions:

• Assume a 100 g sample, which simplifies calculations as percentages directly translate to grams.
• Convert gram quantities to moles to further explore their significance within the compound.
• Calculate moles by dividing the mass of each element by its atomic mass.

This process helps in determining the empirical formula, a crucial step in finding the molecular formula.

The Process of Moles Conversion

Converting mass or percent composition to moles is a fundamental step in chemistry. This conversion provides a bridge between the atomic and macroscopic worlds. In the case of vanillin and its composition, moles conversion allows us to link the given percent or mass of each element to the number of moles it represents.

This involves the following steps:

• To convert mass to moles, divide the mass by the atomic mass of the element, often found on the periodic table:
• For example, converting the percent of carbon in vanillin (\(63.15\text{g}\)) to moles involves \( \frac{63.15}{12.01} \approx 5.25 \text{ moles of carbon}\).
• This provides a way to compare the amounts of each element directly.

Converting to moles is akin to finding how many pieces of each type we have in the chemical puzzle, where each mole represents \(6.022 \times 10^{23}\) atoms. It prepares us for further steps like determining the empirical formula.