Question

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

Short answer

The value of \(x\) in the compound \(\mathrm{KBrO}_{x}\) is approximately \(3\). The compound can be written as \(\mathrm{KBrO}_3\).

Step-by-step solution

Calculate Molar Mass of Atoms in the Compound

To find the value of \(x\), we first need to find the molar mass of the atoms present in the compound. The molar mass of an element is the mass of 1 mole of that element. Using the periodic table, we can find the molar masses for each element: - Molar mass of Potassium (K) = 39.10 g/mol - Molar mass of Bromine (Br) = 79.90 g/mol - Molar mass of Oxygen (O) = 16.00 g/mol

Express Molar Mass as an Equation with Unknown \(x\)

Now we can represent the molar mass of the compound \(\mathrm{KBrO}_{x}\), with unknown \(x\), as an equation. In the compound, there is one atom of Potassium (K), one atom of Bromine (Br), and \(x\) atoms of Oxygen (O). So, the molar mass of the compound can be expressed as: Molar Mass = Molar Mass of K + Molar Mass of Br + (\(x\) * Molar Mass of O)

Calculate Mass Fraction of Br

We are given that the compound contains \(52.92 \% \mathrm{Br}\). This can be expressed as a fraction of the molar mass of Bromine (Br) to the molar mass of the whole compound: Mass Fraction of Br = \(\frac{\text{Molar Mass of Br}}{\text{Molar Mass of the compound}}\) We plug in the given value of \(52.92 \%\) and the molar masses we found in step 1: \(\frac{79.90}{39.10+79.90+x(16.00)} = 0.5292\)

Solve the Equation for \(x\)

Now we will solve the equation to find the value of \(x\). Multiply both sides by the denominator: \(79.90 = 0.5292 (39.10+79.90+x(16.00))\) Divide both sides by \(0.5292\) to isolate the expression in parentheses: \(\frac{79.90}{0.5292} = 39.10+79.90+x(16.00)\) Subtract the constant terms from both sides: \(\frac{79.90}{0.5292} -39.10 -79.90 = x (16.00)\) Then, divide both sides by \(16.00\) to isolate \(x\): \(x = \frac{\frac{79.90}{0.5292} - 39.10 -79.90}{16.00} \approx 3\) So, the value of \(x\) in the compound \(\mathrm{KBrO}_{x}\) is approximately \(3\). The compound can be written as \(\mathrm{KBrO}_3\).

Key concepts

Molar Mass

To determine the value of \( x \) in the compound \( \mathrm{KBrO}_x \), we first need to understand what molar mass is. Molar mass is the mass of one mole of a substance. One mole is Avogadro's number of atoms or molecules, which is approximately \( 6.022 \times 10^{23} \). This value allows us to convert between the mass of a substance and the number of moles.
To find the molar mass of our compound, we add together the molar masses of all the atoms present:

• Potassium (K) has a molar mass of 39.10 g/mol.
• Bromine (Br) has a molar mass of 79.90 g/mol.
• Oxygen (O), with its molar mass of 16.00 g/mol, must be multiplied by \( x \), as there are \( x \) oxygen atoms in \( \mathrm{KBrO}_x \).

The total molar mass becomes:
\[ \text{Molar Mass of } \mathrm{KBrO}_x = 39.10 + 79.90 + x \times 16.00 \] grams per mole. This expression allows us to find the molar mass of the whole compound and is essential for solving further parts of the problem.

Mass Fraction

Understanding mass fraction is crucial in stoichiometry. Mass fraction is the ratio of the mass of a single component within a compound to the total mass of the compound. It's typically expressed as a percentage.
In our exercise, we know that bromine (Br) makes up 52.92% of the mass of \( \mathrm{KBrO}_x \). To use this information, we can write:

• Mass fraction of Br = \( \frac{\text{Mass of Br}}{\text{Total mass of the compound}} \)

Given that Bromine has a molar mass of 79.90 g/mol, we can set up the equation:
\[ \frac{79.90}{39.10 + 79.90 + x \times 16.00} = 0.5292 \]
By setting up this equation, we're expressing the part-to-whole relationship in terms of molar masses. This form of expression makes it possible to isolate and solve for the unknown \( x \), which represents the number of oxygen atoms in the molecule.

Chemical Formula Determination

Chemical formula determination involves figuring out the exact number of each type of atom in a compound. After establishing the molar mass and mass fraction in our example, we solve for \( x \), the unknown number of oxygen atoms in \( \mathrm{KBrO}_x \).
Starting with the mass fraction equation for Br, \( \frac{79.90}{39.10 + 79.90 + x \times 16.00} = 0.5292 \), we proceed to solve for \( x \):

• Multiply both sides by the denominator to eliminate the fraction:
  \( 79.90 = 0.5292 \times (39.10 + 79.90 + x \times 16.00) \)
• Divide by 0.5292 to isolate the terms involving \( x \):
  \( \frac{79.90}{0.5292} = 39.10 + 79.90 + x \times 16.00 \)
• Simplify further by moving constant terms:
  \( \frac{79.90}{0.5292} - 39.10 - 79.90 = x \times 16.00 \)
• Finally, solve for \( x \):
  \( x = \frac{\frac{79.90}{0.5292} - 39.10 - 79.90}{16.00} \approx 3 \)

Thus, the unknown \( x \) in our compound \( \mathrm{KBrO}_x \) is 3. Therefore, the chemical formula is \( \mathrm{KBrO}_3 \), showing that there are three oxygen atoms in the compound alongside one bromine and one potassium.