Problem 65 - A Complete Resource Book in Chemistry for JEE Main

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) \(\mathrm{CH}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The empirical formula is C\(_3\)H\(_4\)O.

Step 1: Assume 100g Total Mass

Assume that the total mass of sorbic acid is 100g. This means there are 64.3g of Carbon (C), 7.2g of Hydrogen (H), and 28.5g of Oxygen (O).

Step 2: Calculate Moles of Each Element

Convert the mass of each element to moles using their atomic masses.For Carbon: \[\text{Moles of C} = \frac{64.3 \, \text{g}}{12.01 \, \text{g/mol}} \approx 5.35 \, \text{mol}\]For Hydrogen: \[\text{Moles of H} = \frac{7.2 \, \text{g}}{1.008 \, \text{g/mol}} \approx 7.14 \, \text{mol}\]For Oxygen: \[\text{Moles of O} = \frac{28.5 \, \text{g}}{16.00 \, \text{g/mol}} \approx 1.78 \, \text{mol}\]

Step 3: Find the Simplest Whole Number Ratio

Divide each mole value by the smallest number of moles calculated in the previous step.For Carbon: \[\frac{5.35}{1.78} \approx 3.01\]For Hydrogen: \[\frac{7.14}{1.78} \approx 4.01\]For Oxygen: \[\frac{1.78}{1.78} = 1\]These approximate to 3, 4, and 1 respectively, revealing the whole number ratio of C : H : O to be 3 : 4 : 1.

Step 4: Determine the Empirical Formula

Using the ratios from the previous step, derive the empirical formula. The formula is determined by the ratio of moles: C\(_3\)H\(_4\)O.

Key Concepts

Molecular Composition

In chemistry, understanding the molecular composition of a compound is crucial. Molecular composition refers to the specific elements that make up a molecule, including their proportion by mass. For instance, in sorbic acid, the composition is given by the percentage of each element present: 64.3% Carbon (C), 7.2% Hydrogen (H), and 28.5% Oxygen (O). This indicates that in every 100 grams of sorbic acid:

• 64.3 grams are Carbon
• 7.2 grams are Hydrogen
• 28.5 grams are Oxygen

Using these percentages, chemists can find the empirical formula, which shows the simplest integer ratio of the elements in a compound. Understanding molecular composition is fundamental as it provides the initial clue about the chemical formula. It helps us determine how various elements combine to form chemical substances and offers insights into their possible structures and properties.

Mole Ratio

The mole ratio is a comparison of the amount of each element in a compound, expressed in moles. This ratio is pivotal for determining the empirical formula of a compound. By converting the mass of each element to moles using the atomic masses, we can establish the ratio of the components:

• For Carbon: given as 64.3 g, converted to moles using the atomic mass of 12.01 g/mol
• For Hydrogen: given as 7.2 g, converted using 1.008 g/mol
• For Oxygen: given as 28.5 g, converted using 16.00 g/mol

After calculating the moles:

• Carbon: approximately 5.35 mol
• Hydrogen: approximately 7.14 mol
• Oxygen: approximately 1.78 mol

Next, divide each mole value by the smallest mole value (1.78 mol for Oxygen in this case). This process helps to find the simplest whole number ratio, which is 3:4:1 for Carbon, Hydrogen, and Oxygen respectively. This ratio translates directly into the empirical formula of the compound.

Chemical Calculations

Chemical calculations are an integral part of solving chemistry problems, such as finding the empirical formula of a compound. These calculations involve converting between the different units of chemical substance amounts: grams, moles, and molecules, using known atomic weights and Avogadro's number.To compute the empirical formula of sorbic acid, start by assuming a total of 100 grams of the compound to simplify the percentage-to-grams conversion. Calculate moles for each element to reveal their amounts on a common ground:

• Carbon calculation: \[\text{Moles of C} = \frac{64.3 \text{ g}}{12.01 \text{ g/mol}} \approx 5.35 \text{ mol}\]
• Hydrogen calculation: \[\text{Moles of H} = \frac{7.2 \text{ g}}{1.008 \text{ g/mol}} \approx 7.14 \text{ mol}\]
• Oxygen calculation: \[\text{Moles of O} = \frac{28.5 \text{ g}}{16.00 \text{ g/mol}} \approx 1.78 \text{ mol}\]

Each element's moles are then compared by dividing through by the smallest number of moles obtained (Oxygen's 1.78 moles here). This yields the whole number coefficient for each element in the empirical formula.These chemical calculations not only serve in determining empirical formulas but are foundational skills for any further complex chemical analysis or molecular studies.