Problem 10 - Chemistry in 30 days

Question

A \(0.242 \mathrm{~g}\) sample of potassium is heated in oxygen. The result is \(0.440 \mathrm{~g}\) of a crystalline compound. What is the formula of this compound? (a) \(\mathrm{KO}\) (b) \(\mathrm{K}_{2} \mathrm{O}\) (c) \(\mathrm{KO}_{2}\) (d) \(\mathrm{KO}_{3}\)

Step-by-Step Solution

Answer

The compound formula is \( \text{K}_2\text{O} \).

Step 1: Identify Mass of Oxygen

Calculate the mass of oxygen that reacted with potassium. Subtract the mass of potassium from the mass of the compound: \( 0.440 \, \text{g} - 0.242 \, \text{g} = 0.198 \, \text{g} \) of oxygen.

Step 2: Convert Mass to Moles for Potassium and Oxygen

Convert the mass of potassium and oxygen to moles using their molar masses (Potassium: \( 39.1 \, \text{g/mol} \), Oxygen: \( 16.0 \, \text{g/mol} \)). Calculate:- Moles of Potassium: \( \frac{0.242 \, \text{g}}{39.1 \, \text{g/mol}} \approx 0.00619 \, \text{mol} \)- Moles of Oxygen: \( \frac{0.198 \, \text{g}}{16.0 \, \text{g/mol}} \approx 0.0124 \, \text{mol} \) (requires two oxygen atoms, convert for \( \text{O}_2 \), so divide by 2: \( \approx 0.0062 \, \text{mol} \))

Step 3: Determine Molar Ratio

Compare the moles of potassium to moles of oxygen. Since the moles of potassium (\( 0.00619 \, \text{mol} \)) and moles of \( \text{O}_2 \) (\( 0.0062 \, \text{mol} \)) are approximately equal, the ratio of potassium to dioxygen is roughly 1:0.5, corresponding to a 2:1 ratio of K:O.

Step 4: Infer Chemical Formula

Based on the molar ratio of 2:1 (K:O), the compound formula matches that of \( \text{K}_2\text{O} \), where two potassium atoms combine with one oxygen atom.

Key Concepts

Molar Mass Calculation

The molar mass is essential for converting between grams and moles, crucial in chemical reactions. To find an element's molar mass (in grams per mole), look at the atomic mass on the periodic table.
For example, potassium (K), with an atomic mass of approximately 39.1 g/mol, and oxygen (O), approximately 16.0 g/mol. To calculate the moles from a given mass, use the formula:\[\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}\]Using this, you can find moles of any component to help in reactions and calculations. Remember, accurate molar mass calculations are a key step in determining other chemical properties and behavior in reactions.

Molar Ratio Determination

Determining the molar ratio involves comparing the amounts in moles of each element in a compound. This concept is crucial for finding how elements combine in chemical reactions.
In the example exercise, we found the molar amounts for potassium and oxygen. The moles of potassium (0.00619) and oxygen (0.0062) are about equal, signifying a close ratio. To determine the molar ratio:

• Divide the number of moles of each element by the smallest number of moles calculated.
• Convert to whole numbers to get the simplest ratio.

Here, the ratio of potassium to oxygen simplifies to 2:1 after considering both elements, crucial for formula inference.

Chemical Formula Inference

In chemistry, inferring the chemical formula is about translating the molar ratio into a meaningful representation of the molecule. Each element's subscript in the formula is based on its molar ratio from calculations.
For example, if you have a K:O ratio of 2:1, that aligns with the compound \( \text{K}_2\text{O} \). Chemical formulas provide quick information about:

• The number of each element's atoms in the compound.
• How elements interact and stabilize in compounds.

These formulas are foundational in understanding reactivity, properties, and the behavior of compounds in different chemical environments.

Mass to Mole Conversion

The mass to mole conversion is about translating a substance's mass into manageable, comparable units of quantity (moles). Moles provide a consistent way to measure and work with substances in chemical reactions based on the number of constituent particles.
To convert mass to moles and vice versa:

• Divide mass by molar mass (to get moles).
• Multiply moles by molar mass (to return to mass).

This is vital in the initial step of many calculations in chemistry. For example, converting the mass of reactants to moles allows for efficient use of stoichiometry in reactions, ensuring accurate predictions of product formation.