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 \(x ?\)

Short answer

The value of \(x\) in the potassium oxybromate compound, KBrO\(_x\), is approximately 1. Thus, the compound's formula is KBrO.

Step-by-step solution

Calculate the mass of each element in 100 g of KBrOx

Assuming we have 100 g of the potassium oxybromate compound, we can calculate the mass of each element within the compound. Bromine accounts for 52.92% of the mass, and since potassium and oxygen make up the remainder, we have: Mass of Bromine (Br) = 52.92 g Mass of Potassium (K) + Oxygen (O) = 100 g - 52.92 g = 47.08 g

Calculate the moles of each element in the compound

Next, we will calculate the moles of each element in the compound by dividing the mass of the element by its molar mass: Moles of Br = \(\frac{52.92 \ \text{g}}{79.904 \ \text{g/mol}}\) = 0.66229 mol Moles of K + O = \(\frac{47.08 \ \text{g}}{39.098 + 15.999x }\text{g/mol}\) (\(15.999x\) represents the mass of x oxygen atoms)

Divide moles by the smallest mole value to find the empirical formula

We will find the ratio of K to Br to O by dividing the calculated moles of each substance by the smallest mole value (0.66229 mol): Ratio of K/Br = \(\frac{0.66229 \ \text{mol}}{0.66229 \ \text{mol}}\) = 1 Ratio of O/Br = \(\frac{0.66229}{\frac{47.08 \ \text{g}}{39.098 + 15.999x }\text{g/mol}}\) = \(\frac{0.66229}{\frac{47.08}{(39.098 + 15.999x)}}\)

Solve for x

To find the value of x, we can set up a proportionality equation using the ratio formulas derived above and solve for x: \(\frac{\text{O}}{\text{K}} = \frac{0.66229}{\frac{47.08}{(39.098 + 15.999x)}} = 1\) Now, solve for x: \((39.098 + 15.999x) = 47.08\) \(15.999x = 7.982\) \(x = \frac{7.982}{15.999} = 0.499\) Since the value of x must be a whole number, and given that 0.499 is very close to 0.5, we can multiply x by 2 to obtain the whole number: \(x = 0.499 * 2 = 0.998 \approx 1\) So, the value of x is approximately 1. Therefore, the formula for the potassium oxybromate compound is KBrO.

Key concepts

Empirical Formula

An empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. In chemistry, it is crucial because it gives us insight into the basic composition of a compound without detailing its entire molecular structure. To find an empirical formula, follow these general steps:
First, determine the percentage composition by mass of each element present in the compound. Then assume a convenient amount of the compound (like 100 g) to simplify the math, as seen in our oxybromate example. The percent by mass directly indicates the grams of each element due to this assumption.
Next, convert these masses into moles using the atomic masses from the periodic table. This involves dividing the mass of each element by its respective atomic mass. Finally, calculate the simplest whole-number mole ratio of the elements. You achieve this by dividing the number of moles of each element by the smallest number of moles present.

• Key point: The empirical formula is not necessarily the same as the molecular formula, but it is generally directly related or a simple multiple of it.

Molar Mass Calculation

Molar mass is a central concept in chemistry, representing the mass in grams of one mole of a substance. It combines all the atomic masses of the individual elements that constitute a substance's chemical formula. Calculating this value is essential for conversion between moles and grams.
To calculate the molar mass of a compound, identify the atomic mass of each element from the periodic table and multiply that by the number of atoms of that element in the formula. Then, sum these masses for all elements in the compound.
For example, if a compound's formula is KBrO, we would calculate its molar mass by:

• Potassium (K): Atomic mass = 39.098 g/mol
• Bromine (Br): Atomic mass = 79.904 g/mol
• Oxygen (O): Atomic mass = 15.999 g/mol
• Total = 39.098 + 79.904 + 15.999 = 135.001 g/mol

These calculations allow chemists to understand the relationship between mass, molecules, and moles in a substance, crucial for balancing equations and determining proportions in reactions.

Elemental Analysis

Elemental analysis is a method used to determine the relative quantity of each element within a compound. This analysis is fundamental in chemistry as it helps in deducing the empirical formula from the percent composition by mass.
Most analysis techniques involve combustion or similar reactions to decompose the compound. The elements are measured by mass spectrometry, gas chromatography, or similar instruments. In our problem, where an element like bromine comprises 52.92% of the compound, this value is directly used to infer the grams of bromine per 100 grams of the compound.

• Understanding this composition is important when identifying unknown substances or verifying the purity of a sample.
• Such results are often the first step in constructing an empirical formula, leading synthesizing and applying chemicals in various applications.

These processes help assure the accuracy and effectiveness of chemical manufacturing and research, contributing to advancements in technology, healthcare, and industrial applications.