Problem 140 - Chemistry The Molecular Science

Question

There are four binary potassium compounds of oxygen. They contain these mass percents of potassium: Compound I, \(83.0 \%\); Compound II, \(55.0 \%\); Compound III, \(44.9 \%\); and Compound IV, \(71.0 \%\). One compound has molar mass equal to \(110.2 \mathrm{~g} / \mathrm{mol}\). Use this information to determine the chemical formula of each compound.

Step-by-Step Solution

Answer

Compounds: I - K₂O, II - KO₃, III - KO₂, IV - K₂O₂; IV matches 110.2 g/mol.

Step 1: Define the Problem and Known Values

We have four compounds of potassium (K) and oxygen (O), with known percentages of potassium in each: Compound I (83.0%), Compound II (55.0%), Compound III (44.9%), and Compound IV (71.0%). Additionally, we know one compound has a molar mass of 110.2 g/mol. Our goal is to determine the chemical formula of each compound.

Step 2: Calculate the Mass Percent of Oxygen

Using the mass percent of potassium, calculate the mass percent of oxygen for each compound by using the equation: \[ \text{Percent of O} = 100\% - \text{Percent of K} \]- Compound I: 17.0%- Compound II: 45.0%- Compound III: 55.1%- Compound IV: 29.0%

Step 3: Convert Mass Percent to Moles

Convert the mass percent of potassium and oxygen to moles. Use the molar masses, where potassium (K) is 39.1 g/mol and oxygen (O) is 16.0 g/mol.- Compound I: \[ \text{Moles of K} = \frac{83.0}{39.1} \approx 2.12 \ \text{Moles of O} = \frac{17.0}{16.0} \approx 1.06 \]- Compound II: \[ \text{Moles of K} = \frac{55.0}{39.1} \approx 1.41 \ \text{Moles of O} = \frac{45.0}{16.0} \approx 2.81 \]- Compound III: \[ \text{Moles of K} = \frac{44.9}{39.1} \approx 1.15 \ \text{Moles of O} = \frac{55.1}{16.0} \approx 3.44 \]- Compound IV: \[ \text{Moles of K} = \frac{71.0}{39.1} \approx 1.82 \ \text{Moles of O} = \frac{29.0}{16.0} \approx 1.81 \]

Step 4: Determine Empirical Formulas

Divide the moles of each element by the smallest number of moles calculated for each compound to simplify them into whole numbers representing subscripts in empirical formulas:- Compound I: Approximate ratio \((2.12:1.06)\) simplifies to K₂O.- Compound II: Approximate ratio \((1.41:2.81)\) simplifies to KO₂.- Compound III: Approximate ratio \((1.15:3.44)\) simplifies to KO₃.- Compound IV: Approximate ratio \((1.82:1.81)\) simplifies to K₂O₂.

Step 5: Match the Molar Mass with Formulas

Calculate or approximate the molar mass of each empirical formula and match it with the given molar mass (110.2 g/mol) to identify the correct compound:- K₂O: Calculated molar mass is \(2(39.1) + 16.0 = 94.2 \text{ g/mol}\)- KO₂: Calculated molar mass is \(39.1 + 2(16.0) = 71.1 \text{ g/mol}\)- KO₃: Calculated molar mass is \(39.1 + 3(16.0) = 87.1 \text{ g/mol}\)- K₂O₂: Calculated molar mass is \(2(39.1) + 2(16.0) = 110.2 \text{ g/mol}\) - Matches given molar mass.

Step 6: Finalize the Formulas

Now that we have identified which compound corresponds to the provided molar mass, finalize the chemical formulas: - Compound I is K₂O. - Compound II is KO₃. - Compound III is KO₂. - Compound IV is K₂O₂.

Key Concepts

Mass Percent Composition

Mass percent composition is a way to express the concentration of an element within a compound. It tells us the percentage, by mass, of each element present in the compound. This value is vital when you need to know the amount of a specific element you are working with in a chemical reaction.

To determine the mass percent of an element, you can use the formula: \[ \text{Mass percent of } X = \left( \frac{ \text{mass of } X}{ \text{total mass of compound}} \right) \times 100 \% \]Let's break this process down using binary potassium compounds as an example:

• For Compound I with 83.0% potassium, subtract this from 100% to find that oxygen makes up 17.0% of the compound.
• Similarly, Compound II has 55.0% potassium and 45.0% oxygen.
• In Compound III, potassium is 44.9%, leaving 55.1% for oxygen.
• Finally, Compound IV contains 71.0% potassium and 29.0% oxygen.

Understanding mass percent composition allows for deeper insights into the make-up of a compound and is essential for tasks like empirical formula determination.

Empirical Formula Determination

Empirical formulas reveal the simplest ratio of atoms within a compound. These formulas are crucial when experimental data on elemental composition is available, but the molecular structure remains unknown.

To determine the empirical formula, follow these general steps:

• Convert the mass percent of each element to moles using their respective atomic masses.
• For example, given 83.0% potassium in Compound I, compute moles of potassium as \( \frac{83.0}{39.1} \approx 2.12 \). For oxygen, with 17.0%, compute \( \frac{17.0}{16.0} \approx 1.06 \).


• Simplify the mole ratios by dividing each number of moles by the smallest mole value in that set. For Compound I, dividing both by 1.06 gives K₂O.

Simplifying mole ratios helps predict the empirical formula. This formula reflects the simplest whole-number ratio of the elements in the compound, aiding in comprehending its basic framework.

Molar Mass Calculation

Calculating molar mass helps identify compounds by providing a link between a compound's mass and its corresponding detailed formula. The molar mass is the sum of the atomic masses of every atom in its empirical or molecular formula. This concept is crucial in confirming the identity of compounds when given a molar mass.

For instance, determine the molar mass of an empirical formula by:- Adding the atomic masses of all elements: - K₂O: \(2 \times 39.1 + 1 \times 16.0 = 94.2\;\text{g/mol}\) - KO₂: \(39.1 + 2 \times 16.0 = 71.1\;\text{g/mol}\) - KO₃: \(39.1 + 3 \times 16.0 = 87.1\;\text{g/mol}\) - K₂O₂: \(2 \times 39.1 + 2 \times 16.0 = 110.2\;\text{g/mol}\)

If the calculated molar mass matches the given molar mass data (110.2 g/mol), it confirms the compound's identity. Thus, K₂O₂ is our match. Mastering molar mass calculations ensures you correctly interpret or predict chemical formula identities.