Question

Sorbic acid consists of \(\mathrm{C}=64.3 \%, \mathrm{H}=7.2 \%\) and \(28.5 \%\) oxygen. Find the empirical formula. (a) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\) (d) \(\mathrm{CH}_{2} \mathrm{O}\)

Short answer

(c) \(\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{O}\) is the empirical formula.

Step-by-step solution

Convert Percentages to Masses

Assume you have a 100 g sample of sorbic acid. This means you have 64.3 g of carbon, 7.2 g of hydrogen, and 28.5 g of oxygen.

Convert Masses to Moles

Convert the mass of each element to moles using their respective atomic masses: \( \text{C: } \frac{64.3 \, \text{g}}{12.01 \, \text{g/mol}} \approx 5.36 \, \text{mol}; \text{H: } \frac{7.2 \, \text{g}}{1.008 \, \text{g/mol}} \approx 7.14 \, \text{mol}; \text{O: } \frac{28.5 \, \text{g}}{16.00 \, \text{g/mol}} \approx 1.78 \, \text{mol}.\)

Determine the Simplest Mole Ratio

Divide the number of moles of each element by the smallest number of moles calculated (which is for oxygen, 1.78): \( \text{C: } \frac{5.36}{1.78} \approx 3.01; \text{H: } \frac{7.14}{1.78} \approx 4.01; \text{O: } \frac{1.78}{1.78} = 1.00.\)

Write the Empirical Formula

The simplest ratio of atoms is approximately 3:4:1, which corresponds to the empirical formula \(\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{O}\).

Key concepts

Conversion of Percentages to Masses

Understanding how to convert percentages to masses is an essential initial step in determining an empirical formula. This method is straightforward when you imagine working with a 100-gram sample of the compound. Why 100 grams? Simply because it makes the math easy and intuitive. With a 100-gram sample, the percentage of each element directly translates into grams. For example, if you have a compound that is 64.3% carbon, 7.2% hydrogen, and 28.5% oxygen (like in our sorbic acid problem), you treat these values as gram masses:

• Carbon: 64.3% of 100g = 64.3g
• Hydrogen: 7.2% of 100g = 7.2g
• Oxygen: 28.5% of 100g = 28.5g

By assuming a 100-gram sample, you eliminate the percentage and directly work with mass, which is a necessary step before you can move onto calculating moles.

Mole Ratio Determination

After converting masses to moles, determining the mole ratio is the next crucial step. A mole ratio helps you understand the proportion of each element in a compound. But how do you find this ratio?First, determine the number of moles of each element by using their masses and corresponding atomic masses. For instance, from the masses derived in the previous step:

• Carbon: 64.3g / 12.01 g/mol ≈ 5.36 mol
• Hydrogen: 7.2g / 1.008 g/mol ≈ 7.14 mol
• Oxygen: 28.5g / 16.00 g/mol ≈ 1.78 mol

Now, take the smallest mole number, which here is for oxygen (1.78 mol), to find the simplest mole ratio. Divide the moles of all elements by this smallest number:

• Carbon: 5.36 ÷ 1.78 ≈ 3.01
• Hydrogen: 7.14 ÷ 1.78 ≈ 4.01
• Oxygen: 1.78 ÷ 1.78 = 1.00

These simplifications result in a ratio of approximately 3:4:1. This tells you the simplest molecular structure: \[\text{C}_3\text{H}_4\text{O}\]Each quotient corresponds to the subscript in the empirical formula, giving a clear picture of how the elements combine in the compound.

Atomic Masses

Atomic masses are fundamental to converting an element's mass to moles, an indispensable step in empirical formula calculations. Each element has a unique atomic mass, sometimes called atomic weight, which is the average mass of atoms naturally occurring on Earth.Using these atomic masses, chemists can convert grams into moles because 1 mole of any atom is defined as its atomic mass in grams. For example:

• Carbon: Atomic Mass ≈ 12.01 g/mol
• Hydrogen: Atomic Mass ≈ 1.008 g/mol
• Oxygen: Atomic Mass ≈ 16.00 g/mol

In our specific example with sorbic acid:

• From the mass of carbon (64.3 g), we calculate moles as \(64.3 \text{ g} / 12.01 \text{ g/mol} \approx 5.36 \text{ mol}\).
• From the mass of hydrogen (7.2 g), we calculate moles as \(7.2 \text{ g} / 1.008 \text{ g/mol} \approx 7.14 \text{ mol}\).
• From the mass of oxygen (28.5 g), we calculate moles as \(28.5 \text{ g} / 16.00 \text{ g/mol} \approx 1.78 \text{ mol}\).

Understanding and applying atomic masses accurately ensures that conversions from mass to moles are correct, which directly impacts the outcome of the empirical formula, highlighting the essential role of atomic masses in chemical calculations.