Problem 52 - Chemistry: An Atoms-Focused Approach

Question

Oxygen gas may be prepared in the laboratory by decomposing a compound of potassium, chlorine, and oxygen. The other product of the decomposition is \(\mathrm{KCl}(s)\) Complete decomposition of \(2.917 \mathrm{g}\) of the compound produces \(1.143 \mathrm{g}\) of oxygen gas. What is the empirical formula of the compound?

Step-by-Step Solution

Answer

Answer: The empirical formula of the compound is K2Cl3O6.

Step 1: Determine the mass of potassium chloride (KCl)

Since we know the mass of the compound before decomposition and the mass of oxygen gas produced after decomposition, we can calculate the mass of KCl produced as the difference between them: Mass of KCl = Mass of compound - Mass of oxygen gas Mass of KCl = 2.917 g - 1.143 g Mass of KCl = 1.774 g

Step 2: Calculate the moles of each element

To determine the empirical formula, we need to find the mole ratio between the elements in the compound. First, let's convert the mass of KCl and O2 to moles using their molar masses: Moles of KCl = Mass of KCl / Molar mass of KCl Moles of KCl = 1.774 g / (39.10 g/mol + 35.45 g/mol) = 0.02512 mol Moles of O2 = Mass of O2 / Molar mass of O2 Moles of O2 = 1.143 g / (32.0 g/mol) = 0.03572 mol

Step 3: Find the mole ratio between KCl and O2

We need to find the ratio between moles of KCl and moles of O2. To do this, we divide the moles of KCl by the moles of O2: Mole ratio of KCl to O2 = Moles of KCl / Moles of O2 Mole ratio of KCl to O2 = 0.02512 mol / 0.03572 mol = 0.703

Step 4: Determine the empirical formula

We know that the compound contains potassium (K), chlorine (Cl), and oxygen (O). Based on the mole ratio between KCl and O2, the empirical formula can be found. Since the mole ratio is approximately 0.7, it closely corresponds to a 2:3 ratio. Therefore, the empirical formula would be K2Cl3O6 (if 2 moles of KCl are used for every 3 moles of O2).

Key Concepts

Moles Calculation

Understanding moles is fundamental in chemistry for quantifying the amount of a substance. A mole is a unit that measures an amount of substance, and is based on Avogadro's number, which is approximately \(6.022 \times 10^{23}\). This makes it easier to relate mass to the number of atoms or molecules present.

• To calculate moles, you need the mass of the element or compound and its molar mass, which is the mass of one mole of that substance.
• The formula for moles is: \(\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}\).
• This helps in converting a physical mass we can measure into a more usable form for scientific calculations.

By transforming given masses into moles, we can effectively analyze and compare the amounts of different materials involved in chemical reactions.

Mass-Mole Conversion

Converting between mass and moles is essential for understanding reactions at molecular levels. It bridges the macroscopic world of grams and kilograms to the atomic world of moles.To perform a mass-mole conversion, follow these steps:

• Identify the molar mass of the substance, which is the mass of \(1\) mole of that substance measured in grams per mole \(\text{(g/mol)}\). This can often be found on the periodic table.
• Use the conversion formula: \(\text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass in g/mol}}\).

For example, in our decomposition exercise, we converted the mass of potassium chloride and oxygen into moles to find their respective quantities in a more meaningful way that allows for analysis and comparison.
This conversion is particularly useful in chemical reactions and formulas since reactions often occur on a molecule-to-molecule basis rather than on a gram-to-gram basis.

Chemical Decomposition

Chemical decomposition involves breaking down a compound into simpler substances. It is a type of chemical reaction where one reactant yields two or more products.In the given exercise, decomposition is shown by a single compound breaking into potassium chloride (\(\text{KCl}\)) and oxygen gas (\(\text{O}_2\)). This process is crucial for understanding how compounds behave when subjected to various conditions such as heat or electricity.

• Decomposition allows us to determine the composition of the original compound by analyzing the products.
• Specifically, we use the masses of the resulting products to backtrack and figure out what the original formula might have been.

By measuring the masses of \(\text{KCl}\) and \(\text{O}_2\) produced, the reaction completes a narrative telling us about the elemental make-up of the original compound.

Mole Ratio Analysis

Mole ratio analysis is key to determining empirical formulas. It involves calculating the simplest whole-number ratio of the moles of each element in a compound. To perform a mole ratio analysis:

• First, compute the moles of each element involved in the reaction, as shown in our step-by-step solution.
• Next, divide the number of moles of each element by the smallest number of moles calculated among the elements to find the ratio.
• The resulting numbers should present themselves as near whole numbers, indicating the simplest ratio of atoms in the compound.

For the decomposition exercise, by determining the mole ratio between potassium chloride and oxygen, we concluded on an empirical formula reflecting the basic composition of the initial compound.
This ratio lets chemists understand and represent the proportional relationship between atoms in a compound, leading to insights into both its structure and reactivity.