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

In conclusion, the value of \(x\) in the compound \(\mathrm{KBrO}_{x}\) is 2, making the compound \(\mathrm{KBrO}_{2}\).

Step-by-step solution

Find the molecular weight of KBrOx

We know the molecular weight of K, Br, and O: K (Potassium) = 39.1 g/mol Br (Bromine) = 79.9 g/mol O (Oxygen) = 16.0 g/mol Since we don't know the value of x yet, the molecular weight of KBrOx can be expressed as: KBrOx = 39.1 + 79.9 + 16x

Calculate the moles of Br and K in a 100 g sample

As the percentage of Bromine in the compound is given, we can assume a 100 g sample, so 52.92 g of Br will be present. moles of Br = mass of Br / molecular weight of Br moles of Br = 52.92 g / 79.9 g/mol = 0.6621 molesBr Since there's exactly one potassium atom (K) for each bromine atom (Br) in KBrOx, the amount of potassium (K) in the sample is the same as the amount of Br: moles of K = 0.6621 molesK

Determine the moles of O

As we now know the moles of both K and Br, we can find the moles of O by subtracting these from the total weight of the sample. mass of K = (moles of K) * (molecular weight of K) mass of K = 0.6621 molesK * 39.1 g/mol = 25.88 g mass of sample = 100 g total mass of K and Br = mass of K + mass of Br total mass of K and Br = 25.88 g + 52.92 g = 78.8 g mass of O = mass of sample - total mass of K and Br mass of O = 100g - 78.8 g = 21.2 g moles of O = mass of O / molecular weight of O moles of O = 21.2 g / 16.0 g/mol = 1.325 molesO

Find the mole ratio between Br and O

We can now get the mole ratio between Br and O to determine the value of x. mole ratio between Br and O = moles of Br / moles of O mole ratio between Br and O = 0.6621 molesBr / 1.325 molesO = 0.5

Calculate the value of x

Since the mole ratio of Br and O is 0.5, that means for every one mole of Br, there are two moles of O. Therefore, the value of x is 2. In conclusion, the value of x in the compound KBrOx is 2, making the compound KBrO2.

Key concepts

Molecular Weight Calculation

Understanding the molecular weight calculation is crucial when analyzing a chemical compound's composition. The molecular weight, also referred to as the molar mass, is the sum of the atomic weights of all the atoms in a molecule. Each element's atomic weight can be found on the periodic table and is expressed in atomic mass units (amu), or grams per mole (g/mol) when dealing with moles.

For example, a compound like KBrO_x, where K stands for Potassium, Br for Bromine, and O for Oxygen, has a molecular weight that is the combined weights of all these elements. In our exercise, this would be calculated as the sum of the weight of K (39.1 g/mol), Br (79.9 g/mol), and O multiplied by x (16 g/mol each). When calculating, keep in mind to multiply the atomic weight of elements times the number of atoms of that element present in the compound. This is fundamental in predicting how much of each element is present in a given sample of the compound, leading to the mole concept.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and the macroscopic world we live in. One mole is defined as Avogadro's number (approximately 6.022 x 10²³) of entities, whether they be atoms, molecules, ions, or other particles.

In the given exercise, by assuming a 100 g sample, we simplify the problem and directly relate the percent composition to grams, which then allows us to calculate the moles of each element. The moles of an element are computed by dividing the mass of the element in the sample by its molecular weight. Knowing the moles gives us a count of how many formulas of the compound are present in the sample, and this count is critical when determining the proportion of elements (stoichiometry) and solving for unknown variables, such as x in KBrO_x.

Stoichiometry

The field of stoichiometry is a quantitative study of the relationships that exist between substances in a chemical reaction. It builds on the mole concept to allow chemists to predict the amounts of substances consumed and produced in a given reaction.

In this instance, stoichiometry comes into play when we calculate the mole ratio of Br to O in KBrO_x. We can then apply this mole ratio to determine the correct stoichiometric coefficient for oxygen, which is represented by x in our chemical formula. By dividing the moles of Br by the moles of O found in our sample, the ratio provides insight into the structure of the compound. With stoichiometry, it's possible to discover that in every molecule of KBrO_x, there are twice as many atoms of oxygen as there are of bromine, concluding that the compound is actually KBrO₂. This kind of analysis is vital for chemists to understand the exact proportions of elements within compounds and to manipulate chemical reactions predictably.