Problem 90 - Chemistry: The Central Science

Question

An oxybromate compound, \(\mathrm{KBrO}_{x}\), where \(x\) is unknown, is analyzed and found to contain \(52.92 \%\) Br. What is the value of \(x\) ?

Step-by-Step Solution

Answer

The value of \(x\) in the oxybromate compound, \(\mathrm{KBrO}_{x}\), is approximately 2, making the chemical formula \(\mathrm{KBrO}_{2}\).

Step 1: Determine the molar mass of the unknown compound

The general formula for an oxybromate compound is \(\mathrm{KBrO}_{x}\). To find the molar mass, we need the molar masses of each element in the compound: potassium (K), bromine (Br), and oxygen (O). From the periodic table, we have: Molar mass of K = 39 g/mol Molar mass of Br = 80 g/mol Molar mass of O = 16 g/mol The molar mass of the unknown compound will be \(\mathrm{K} + \mathrm{Br} + x \cdot \mathrm{O} = 39 + 80 + 16x\) g/mol.

Step 2: Calculate the mass percentage of each element in the compound

We are given that the mass percentage of Br in the compound is \(52.92\%\). Let the mass percentage of K be \(A \%\) and the mass percentage of O be \(B \%\). We know that the sum of all mass percentages should be equal to 100%, so we can write the equation: \(A + 52.92 + B = 100\)

Step 3: Express mass percentages in terms of molar mass

The mass percentages of each element can also be expressed in terms of their molar masses: \(A \% = \frac{39}{39 + 80 + 16x} \cdot 100\%\) \(B \% = \frac{16x}{39 + 80 + 16x} \cdot 100\%\) We already have the mass percentage of Br as \(52.92\%\): \(52.92 \% = \frac{80}{39 + 80 + 16x} \cdot 100\%\)

Step 4: Solve for x

Now let's solve the equation in terms of x: \(52.92 = \frac{80}{39 + 80 + 16x} \cdot 100\) Divide both sides by 100: \(0.5292 = \frac{80}{39 + 80 + 16x}\) Now, cross-multiply: \(0.5292(39 + 80 + 16x) = 80\) Expand and simplify the equation: \(0.5292(119 + 16x) = 80\) \(62.9752 + 8.4672x = 80\) Now, isolate x: \(8.4672x = 80 - 62.9752\) \(8.4672x = 17.0248\) Finally, divide by \(8.4672\) to find the value of x: \(x = \frac{17.0248}{8.4672}\) \(x \approx 2\)

Step 5: Conclusion

Based on our calculations, the value of x in the oxybromate compound, \(\mathrm{KBrO}_{x}\), is approximately 2. Therefore, the chemical formula of the compound is \(\mathrm{KBrO}_{2}\).

Key Concepts

Molar Mass Calculation

Understanding the molar mass of a compound is fundamental in chemistry, as it helps determine the mass of one mole of that substance. It's calculated by adding together the atomic masses of each element in a compound, multiplied by the number of atoms of that element in the formula. The atomic mass values are sourced from the periodic table and are usually expressed in grams per mole (g/mol).

For example, in our oxybromate compound \(\mathrm{KBrO}_{x}\), to calculate the molar mass, we need to consider the molar masses of potassium (K), bromine (Br), and oxygen (O), each represented by their respective atomic weight: K = 39 g/mol, Br = 80 g/mol, and O = 16 g/mol. We don't know the value of x yet, so the molar mass is expressed as \(39 + 80 + 16x\) g/mol.

Mass Percent Composition

The mass percent composition of an element within a compound reflects the percentage by mass that the element contributes to the total mass of the compound. This is an important concept since it allows chemists to understand the proportion of each element in a compound. The mass percent is calculated using the formula:\[\text{Mass percent of element} = \left(\frac{\text{Mass of element in compound}}{\text{Molar mass of compound}}\right) \times 100\%\]

In the problem provided, the mass percent of bromine (Br) in the compound \(\mathrm{KBrO}_{x}\) is given as 52.92%. Using this percentage, along with the molar mass calculation, we can determine the value of x, the number of oxygen atoms, by setting up equations that reflect the mass percent composition for each element.

Stoichiometric Calculations

Stoichiometry involves calculations that relate the quantities of reactants and products in a chemical reaction. It is rooted in the conservation of mass and the concept of moles. While our current problem doesn't involve a reaction per se, stoichiometric principles are still applied to find the ratio of elements in a compound.

By using the given mass percent of bromine and the molar masses of each element involved, we can set up stoichiometric equations to deduce the formula of the compound. In our example, the stoichiometric equation relating the given mass percent to the molar mass allowed us to isolate and solve for x, which denotes the number of oxygen atoms associated with each molecule of potassium bromate.

Empirical Formula Determination

The empirical formula represents the simplest whole number ratio of atoms of each element in a compound. Determining this formula is a process that often involves calculating the molar ratios of elements from mass percent composition data.

To determine the empirical formula of our oxybromate compound, we figured out the molar ratio of the elements to each other, using the mass percent of bromine and the atomic masses. After solving for x in the stoichiometric equation and establishing that x is approximately equal to 2, we established the empirical formula as \(\mathrm{KBrO}_{2}\).

The empirical formula is crucial because it provides a base for understanding the actual molecular composition and reacting ratios in a substance, and in this case, it allowed the exact identification of the compound being studied.