Question

A mixture of potassium chloride and potassium bromide weighing \(3.595 \mathrm{~g}\) is heated with chlorine, which converts the mixture completely to potassium chloride. The total mass of potassium chloride after the reaction is \(3.129 \mathrm{~g}\). What percentage of the original mixture was potassium bromide?

Short answer

Answer: The percentage of potassium bromide (KBr) in the original mixture is approximately 19.3%.

Step-by-step solution

Calculate the mass difference before and after chlorination

As the mass of the mixture before the reaction is given as \(3.595 \mathrm{~g}\), and after the reaction, it is \(3.129 \mathrm{~g}\), we can find the mass difference by subtracting the mass after from the mass before: $$ \text{Mass difference} = 3.595 - 3.129 = 0.466 \mathrm{~g} $$ Now, we know that the mass difference (\(0.466 \mathrm{~g}\)) is the mass of bromine (Br) replaced by chlorine (Cl) during the reaction.

Calculate the number of moles of bromine (Br) replaced

We can now calculate the number of moles of bromine (Br) that were replaced by chlorine (Cl). We will use the molar mass of bromine (Br), which is approximately \(79.90 \mathrm{~g/mol}\): $$ \text{Moles of Br} = \frac{\text{Mass of Br}}{\text{Molar mass of Br}} = \frac{0.466}{79.90} \approx 0.00583 \mathrm{~mol} $$ So, \(0.00583\) moles of bromine (Br) were replaced during the reaction.

Calculate the mass of potassium bromide (KBr) in the original mixture

Now, we can calculate the mass of potassium bromide (KBr) in the original mixture. We will use the molar mass of potassium bromide (KBr), which is approximately \(39.10 \mathrm{~g/mol} + 79.90 \mathrm{~g/mol} = 119.00 \mathrm{~g/mol}\): $$ \text{Mass of KBr in the original mixture} = \text{Moles of Br} \times \text{Molar mass of KBr} \approx 0.00583 \times 119.00 \approx 0.693 \mathrm{~g} $$ So, there are \(0.693 \mathrm{~g}\) of potassium bromide (KBr) in the original mixture.

Calculate the percentage of potassium bromide (KBr) in the original mixture

Finally, we can calculate the percentage of potassium bromide (KBr) in the original mixture, using the mass of potassium bromide (KBr) we found in Step 3 and the total initial mass of the mixture: $$ \text{Percentage of KBr} = \frac{\text{Mass of KBr}}{\text{Initial mass of the mixture}} \times 100 \%= \frac{0.693}{3.595} \times 100 \% \approx 19.3 \% $$ So, approximately \(19.3 \%\) of the original mixture was potassium bromide (KBr).

Key concepts

Chemical Reactions

When substances undergo chemical reactions, they rearrange their atoms to form new substances. In this exercise, the mixture of potassium chloride (KCl) and potassium bromide (KBr) reacts with chlorine (Cl₂), resulting in the formation of potassium chloride (KCl) and the replacement of bromine (Br) with chlorine. This is a classic example of a substitution reaction where chlorine gas is more reactive and displaces bromine from its compound.

• The reactants are the substances you start with: KCl, KBr, and Cl₂.
• The products are what you get after the reaction: KCl formed from both the original KCl and converted KBr.

Understanding the transformation that takes place in a chemical reaction is key to solving problems like this, where you need to account for changes in substance composition.

Mass Conservation

The Law of Mass Conservation, also known as the Law of Conservation of Mass, states that mass cannot be created or destroyed in a chemical reaction. This principle tells us that the total mass of the reactants before a reaction must equal the total mass of the products after the reaction.
In this problem, the mass difference between the mixture before and after the reaction (0.466 g) represents the mass of bromine lost and replaced by chlorine. This demonstrates the balance maintained in the chemical equation, where the mass replaced is exactly accounted for by the new element taking its place. So, no atoms are lost; they are simply rearranged. This helps us calculate the exact amount of potassium bromide that was there initially.

Molar Mass

Molar mass is a measure of the mass of one mole of a substance. It is expressed in grams per mole (g/mol) and crucial for converting between mass and moles in chemical equations.
To tackle this exercise, knowledge of the molar mass of bromine (Br, about 79.90 g/mol) and potassium bromide (KBr, approximately 119.00 g/mol) is critical.

• We use the molar mass of bromine to find out how many moles were replaced by chlorine.
• Then, the moles are multiplied by the molar mass of potassium bromide to deduce its original mass.

By mastering these conversions, you can seamlessly transition between different quantities and better understand chemical formulas and reactions.

Percentage Composition

Percentage composition refers to the percentage by mass of each element in a compound. In the context of this exercise, it involves determining how much of the initial mixture was constituted by potassium bromide (KBr).
We calculate this by dividing the mass of potassium bromide (0.693 g), as determined earlier, by the total mass of the mixture (3.595 g). Then, we multiply the resulting fraction by 100 to get the percentage:

• Percentage of KBr = \( \left(\frac{0.693}{3.595}\right) \times 100 \approx 19.3\% \)

This calculation helps us understand the relative proportions of components in mixtures, which is fundamental when dealing with complex chemical mixtures.