Question

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

Short answer

The value of \(x\) in the compound \(\mathrm{KBrO}_{x}\) is 1.

Step-by-step solution

Calculate the Percentage of K and O in the Compound

Br makes up 52.92% of the compound. Since the percentages of the elements in a compound should add up to 100%, we can calculate the percentage of K and O in the compound as follows: Percentage of K and O = 100% - 52.92% = 47.08%

Convert the Percentages to Grams

For simplicity reasons, let's assume we have 100 grams of the compound. This means we have 52.92 grams of Br and 47.08 grams of (K and O).

Determine the Moles of K, Br, and O

Now, we will convert the grams to moles using the molar mass of each element: (1) Moles of K = \(\frac{\text{mass of K}}{\text{molar mass of K}}\) (2) Moles of Br = \(\frac{\text{mass of Br}}{\text{molar mass of Br}}\) (3) Moles of O = \(\frac{\text{mass of O}}{\text{molar mass of O}}\) Let's assume there is 1 mole of K in the compound, which means that there are 39.10 grams of K (since the molar mass of K is 39.10 g/mol). We already know there are 52.92 grams of Br. The molar mass of Br (Br) is 79.90 g/mol. (2) Moles of Br = \(\frac{52.92}{79.90} = 0.662\) We can now find the mass of O in the compound by subtracting the masses of K and Br (39.10 + 52.92 = 92.02 g) from the total mass of 100 g. This gives us 7.98 g of O. Divide this by the molar mass of O (16.00 g/mol) to find the moles of O: (3) Moles of O = \(\frac{7.98}{16.00} = 0.499\)

Determine the Ratio of Moles

To find the empirical formula of the compound, we need to determine the ratio of moles of K, Br, and O. We can divide the number of moles of each element by the smallest number of moles to find the whole number ratio. Since the number of moles of O is the smallest, we will divide all moles by 0.499: Mole ratio for K = \(\frac{1}{0.499}\) ≈ 2 Mole ratio for Br = \(\frac{0.662}{0.499}\) ≈ 1 Mole ratio for O = \(\frac{0.499}{0.499}\) ≈ 1

Determine the Empirical Formula and the Value of \(x\)

Now that we have the mole ratios for K, Br, and O, we can write the empirical formula for the compound: \(\mathrm{K}_{2}\mathrm{Br}\mathrm{O}\) To find the value of \(x\), compare the empirical formula to the original formula (\(\mathrm{KBrO}_{x}\)). The mole ratio of Br to O in the empirical formula is 1:1, which confirms that the value of x is: \(x = 1\) Thus, the compound is \(\mathrm{K}_{2}\mathrm{Br}\mathrm{O}\), and the value of \(x\) is 1.

Key concepts

Percentage Composition

Understanding the percentage composition of a compound is crucial for determining its empirical formula. Percentage composition represents the percent by mass of each element in a compound. It illustrates how much of a compound is made up by a particular element and is calculated using the formula:

Percentage composition = \(\frac{{\text{mass of the element}}}{{\text{total mass of the compound}}} \times 100\)%

When analyzing the compound \(\mathrm{KBrO}_{x}\), it was found to contain 52.92% Br by mass. To find the percentage composition of K and O, it was essential to subtract this value from 100%, assuming that the sum of the percentage compositions of all elements in a compound equals 100%. This approach simplified the problem and laid the groundwork for further calculations by partitioning the compound into its elemental components based on their mass percentages.

Mole Concept

The mole concept is a fundamental principle in chemistry that relates the mass of substances to the amount of entities, such as atoms, molecules, or ions. A mole is defined as the amount of substance that contains as many entities as there are atoms in 12 grams of carbon-12, which is approximately \(6.022 \times 10^{23}\) entities and is known as Avogadro's number.

In the given exercise, the mole concept was employed to convert grams to moles for each element in \(\mathrm{KBrO}_{x}\). This step was necessary to find the empirical formula of the compound. For instance, knowing that the molar mass of K is 39.10 g/mol, the number of moles of K present in the compound was calculated. Similarly, the mass of Br and O were converted to moles using their respective molar masses. These calculations allow chemists to compare the element's proportions on a mole-to-mole basis, rather than mass-to-mass, leading to a clearer understanding of the compound's composition.

Molar Mass

Molar mass is the mass of one mole of a given substance and is expressed in units of grams per mole (g/mol). It is an intrinsic property that combines the atomic masses of the elements in a compound as listed on the periodic table. The molar mass is a bridge between the atomic scale and the real-world scale, enabling us to count out a specific number of molecules by measuring a specific mass of a substance.

Determining the molar mass of each element in \(\mathrm{KBrO}_{x}\) allowed us to convert the mass of each component to moles. For example, by using the molar mass of Br (79.90 g/mol), the 52.92 grams of Br in the compound was converted into moles, which is a necessary step when determining the empirical formula. An accurate molar mass is critical for this conversion and, subsequently, for calculating the correct mole ratio of elements, which ultimately reveals the simplest whole-number ratio of atoms in a compound and solves for the empirical formula.