Question

Butyric acid is a compound that is present in butter. It has a molar mass of \(88.11 \mathrm{~g} / \mathrm{mol}\) and is composed of \(54.5 \% \mathrm{C}\), \(9.2 \% \mathrm{H}\), and \(36.3 \% \mathrm{O}\). What is the molecular formula for this substance?

Short answer

The molecular formula for butyric acid is \(\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_2\).

Step-by-step solution

Determine Masses Based on Percentage Composition

For simplicity, consider a 100 g sample of butyric acid. This means it contains 54.5 g of carbon (C), 9.2 g of hydrogen (H), and 36.3 g of oxygen (O).

Convert Masses to Moles

Use the molar mass of each element to convert grams to moles:o For carbon (C): \ \[ \text{{Moles of C}} = \frac{{54.5 \text{{ g}}}}{{12.01 \text{{ g/mol}}}} \approx 4.537 \text{{ moles}} \]o For hydrogen (H): \ \[ \text{{Moles of H}} = \frac{{9.2 \text{{ g}}}}{{1.008 \text{{ g/mol}}}} \approx 9.127 \text{{ moles}} \]o For oxygen (O): \ \[ \text{{Moles of O}} = \frac{{36.3 \text{{ g}}}}{{16.00 \text{{ g/mol}}}} \approx 2.269 \text{{ moles}} \]

Determine the Simplest Whole Number Ratio

Divide each mole value by the smallest number of moles calculated:o For carbon (C): \ \[ \frac{{4.537}}{{2.269}} \approx 2.0 \]o For hydrogen (H): \ \[ \frac{{9.127}}{{2.269}} \approx 4.0 \]o For oxygen (O): \ \[ \frac{{2.269}}{{2.269}} \approx 1.0 \]

Write the Empirical Formula

With the simplest whole number ratio determined, the empirical formula is \ \[ \mathrm{C}_2 \mathrm{H}_4 \mathrm{O} \].

Calculate the Empirical Formula Molar Mass

Calculate the molar mass of the empirical formula \ \[ \mathrm{C}_2 \mathrm{H}_4 \mathrm{O} \]:o Carbon (C): \ \[ 2 \times 12.01 = 24.02 \text{ g/mol} \]o Hydrogen (H): \ \[ 4 \times 1.008 = 4.032 \text{ g/mol} \]o Oxygen (O): \ \[ 1 \times 16.00 = 16.00 \text{ g/mol} \]o Total molar mass: \ \[ 24.02 + 4.032 + 16.00 = 44.052 \text{ g/mol} \]

Determine the Molecular Formula

Divide the given molar mass of butyric acid by the empirical formula molar mass to find the ratio:\[ \text{Ratio} = \frac{88.11}{44.052} \approx 2 \]Multiply each subscript in the empirical formula by this ratio:\[ \text{Molecular formula} = (\mathrm{C}_2 \mathrm{H}_4 \mathrm{O})_2 = \mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_2 \]

Key concepts

Percent Composition

Understanding percent composition is crucial in determining the amount of each element in a compound. This value tells you the percentage, by mass, of each element within the compound.

For butyric acid, we have the following percent composition:

• 54.5% Carbon (C)
• 9.2% Hydrogen (H)
• 36.3% Oxygen (O)

A practical way to work with these percentages is to consider a 100 g sample of the compound. This simplifies calculations because the percentages can be directly converted to grams (e.g., 54.5% becomes 54.5 g of Carbon). This clear linkage between percentages and mass helps us in the next steps of converting these masses into moles.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms in a molecule. It is derived from experimental data, such as the percent composition of each element in a compound.

To find the empirical formula for butyric acid from the given composition, follow these steps:

• First, convert the mass percentages to moles by dividing by the atomic masses:
\begin{itemize}
• Moles of Carbon: \[ \frac{54.5 \text{ g}}{12.01 \text{ g/mol}} \rightarrow 4.537 \text{ moles} \]
• Moles of Hydrogen: \[ \frac{9.2 \text{ g}}{1.008 \text{ g/mol}} \rightarrow 9.127 \text{ moles} \]
• Moles of Oxygen: \[ \frac{36.3 \text{ g}}{16.00 \text{ g/mol}} \rightarrow 2.269 \text{ moles} \]


• Then, determine the simplest whole-number ratio by dividing each element's mole quantity by the smallest value obtained:
\begin{itemize}
• Ratio for Carbon: \[ \frac{4.537}{2.269} \rightarrow 2.0 \]
• Ratio for Hydrogen: \[ \frac{9.127}{2.269} \rightarrow 4.0 \]
• Ratio for Oxygen: \[ \frac{2.269}{2.269} \rightarrow 1.0 \]


• Using these ratios, the empirical formula of butyric acid is \[ \text{C}_2 \text{H}_4 \text{O}\].

Molar Mass Calculation

The molar mass of a compound is the mass of one mole of that compound. It is calculated by summing the atomic masses of all atoms in the compound's empirical formula.

To find the molar mass of the empirical formula for butyric acid (\text{C}_2 \text{H}_4 \text{O}), consider the contribution of each element:

• Carbon: \[ 2 \times 12.01 \text{ g/mol} = 24.02 \text{ g/mol} \]
• Hydrogen: \[ 4 \times 1.008 \text{ g/mol} = 4.032 \text{ g/mol} \]
• Oxygen: \[ 1 \times 16.00 \text{ g/mol} = 16.00 \text{ g/mol} \]
• Total molar mass: \[ 24.02 + 4.032 + 16.00 = 44.052 \text{ g/mol} \]

Next, to determine the molecular formula, divide the given molar mass of butyric acid (88.11 g/mol) by the molar mass of the empirical formula (44.052 g/mol): \[ \text{Ratio} = \frac{88.11 \text{ g/mol}}{44.052 \text{ g/mol}} \rightarrow 2 \]
Finally, multiply the subscripts in the empirical formula by this ratio to obtain the molecular formula: \[ (\text{C}_2 \text{H}_4 \text{O})_2 = \text{C}_4 \text{H}_8 \text{O}_2 \]. This is the molecular formula of butyric acid.