Problem 14 - Chemistry in 30 days

Question

The percent loss in weight after heating a pure sample of potassium chlorate (mol. wt. = 122.5) will be (a) \(12.25\) (b) \(24.50\) (c) \(39.18\) (d) \(49.0\)

Step-by-Step Solution

Answer

The percent loss in weight after heating is approximately 39.18% (option c).

Step 1: Understanding Potassium Chlorate Decomposition

When potassium chlorate (KClO₃) is heated, it decomposes into potassium chloride (KCl) and oxygen gas (O₂). The reaction can be represented as follows:\[ 2 ext{KClO}_3 ightarrow 2 ext{KCl} + 3 ext{O}_2 \] This means that for every 2 moles of KClO₃, 3 moles of O₂ are released.

Step 2: Determine Molar Weights

Calculate the molar weight of KClO₃ and the products: - Molar mass of KClO₃ = 122.5 g/mol - Molar mass of KCl = 74.5 g/mol - Molar mass of O₂ = 32 g/mol Since 2 moles of KClO₃ decompose, we need the combined molar mass of 3 moles of O₂, which is 3 × 32 = 96 g/mol.

Step 3: Calculate Weight Loss

When 2 moles of KClO₃ decompose: - Original mass = 2 × 122.5 = 245 g - Mass of KCl left after decomposition = 2 × 74.5 = 149 g - Thus, the weight loss due to O₂ is 245 - 149 = 96 g

Step 4: Percent Loss Calculation

Calculate the percent loss in weight:\[ ext{Percent Loss} = rac{ ext{Weight Loss}}{ ext{Original Weight}} imes 100 = rac{96}{245} imes 100 \]This calculation yields approximately 39.18%.

Key Concepts

Understanding Potassium Chlorate Decomposition

When you heat potassium chlorate, represented as \( \text{KClO}_3 \), an interesting reaction occurs. It breaks down or decomposes into two products: potassium chloride (\( \text{KCl} \)) and oxygen gas (\( \text{O}_2 \)). This is a fairly common chemical process where a compound is split into simpler substances. The decomposition reaction is especially important in understanding weight loss in chemical reactions involving gas release. In our case, the balanced chemical equation looks like this:\[2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2\]This equation tells us a few things:

• 2 moles of \( \text{KClO}_3 \) break down to produce 2 moles of \( \text{KCl} \)
• 3 moles of \( \text{O}_2 \) are released as a gas

Understanding this process is crucial as it explains why there's a change in specimen mass after heating, which is key for calculating weight loss.

Mastering Molar Mass Calculations

Molar mass is essentially the weight of one mole of a substance, and it’s expressed in grams per mole (g/mol). In the context of potassium chlorate decomposition, knowing the molar masses of all substances involved helps determine how much each substance contributes to the overall weight of the compound before and after the reaction.Here's how we calculate it for the elements involved:

• KClO₃: The molar mass is given as 122.5 g/mol. It already incorporates Potassium (K), Chlorine (Cl), and Oxygen (O) in its composition.
• KCl: With potassium and chlorine, its molar mass is 74.5 g/mol.
• O₂: Oxygen naturally forms a diatomic molecule; the molar mass for oxygen as \( \text{O}_2 \) is 32 g/mol. Thus for 3 moles, it's 3 × 32= 96 g/mol.

By understanding and accurately calculating these values, we can proceed to perform percent weight loss calculations during the decomposition of potassium chlorate.

Analyzing Chemical Reactions for Weight Loss

Analyzing chemical reactions, particularly for processes like decomposition, can illustrate changes in mass. Here, we're interested in the percent weight loss that occurs when potassium chlorate decomposes. Let's break down the calculation:

• First, calculate the total initial mass of the decomposing compound: 2 moles of KClO₃ have a combined mass of 245 g (as 2 × 122.5 g/mol = 245 g).
• After decomposition, we calculate the mass of the residual product, KCl: 2 moles again, with a mass of 149 g (2 × 74.5 g/mol = 149 g).
• Weight loss is thus attributed to the release of oxygen gas (O₂): original mass - final mass, or 245 g - 149 g = 96 g.
• The percent weight loss is determined by comparing the weight loss to the original weight using the formula:\[\text{Percent Loss} = \frac{\text{Weight Loss}}{\text{Original Weight}} \times 100\]Substituting the values, \[\frac{96}{245} \times 100 \approx 39.18\%\]

Analyzing these steps shows how the conversion of matter affects overall weight, emphasizing the role of gas formation in weight loss calculations.