Question

When solid \(\mathrm{KClO}_{3}\) is heated strongly, it decomposes to form solid potassium chloride, \(\mathrm{KCl}\), and \(\mathrm{O}_{2}\) gas. \(\mathrm{A}\) \(0.415 \mathrm{g}\) sample of impure \(\mathrm{KClO}_{3}\) is heated strongly and the \(\mathrm{O}_{2}\) gas produced by the decomposition is collected over water. When the wet \(\mathrm{O}_{2}\) gas is cooled back to \(26^{\circ} \mathrm{C}\), the total volume is \(229 \mathrm{mL}\) and the total pressure is 323 Torr. What is the mass percentage of \(\mathrm{KClO}_{3}\) in the original sample? Assume that none of the impurities produce oxygen on heating. The vapor pressure of water is 25.22 Torr at \(26^{\circ} \mathrm{C}\).

Short answer

To find the mass percentage of \(\mathrm{KClO}_{3}\) in the original sample, calculate the mass of \(\mathrm{KClO}_{3}\) obtained through the decomposition process and divide it by the initial sample mass, then multiply by 100% to convert to a percentage.

Step-by-step solution

Mole Calculation for Oxygen

We need to figure out the number of moles of oxygen gas collected. To do this, we need to correct the total pressure given to the pressure of oxygen by subtracting the vapor pressure of the water. Therefore, the pressure of oxygen, \(P_{O_{2}} = \text{total pressure} - \text{vapor pressure of water} = 323 \text{Torr} - 25.22 \text{Torr} = 297.78 \text{Torr}\). Next, we convert the pressure to atmosphere (1 Torr = 1/760 atm), so \(P_{O_{2}} = 297.78/760 \text{atm}\). Now, we convert the temperature from Celsius to Kelvin scale (K = \(^{\circ}C + 273.15)\), which gives T = 26 + 273.15 = 299.15 K. Given that the volume V = 229 ml = 0.229 L, we can use the ideal gas law equation \(PV = nRT\) (where n is the number of moles we are seeking, R is the ideal gas constant) to calculate the number of moles of oxygen (\(n_{O_{2}})\). Therefore, \(n_{O_{2}}= \frac{P_{O_{2}}V}{RT} = \frac{(297.78/760) * 0.229}{0.0821 * 299.15}\).

Mole Calculation for \(\mathrm{KClO}_{3}\)

From the reaction equation, we know that for every 2 moles of \(O_{2}\) produced, 2 moles of \(\mathrm{KClO}_{3}\) decomposes. Hence, the number of moles of \(\mathrm{KClO}_{3}\) is equal to the number of moles of \(O_{2}\). So, \(n_{\mathrm{KClO}_{3}} = n_{O_{2}}\).

Find Mass of \(\mathrm{KClO}_{3}\) and calculate mass percentage

The molar mass of \(\mathrm{KClO}_{3}\) is 39.1 (K) + 35.5 (Cl) + 16*3 (O) = 122.6 g/mol. Thus, the mass of \(\mathrm{KClO}_{3}\) is equal to the molar mass of \(\mathrm{KClO}_{3}\) times the number of moles, and the mass percentage of \(\mathrm{KClO}_{3}\) in the original sample is \(\frac{\text{mass of } \mathrm{KClO_{3}}}{\text{original mass of sample}} \times 100\%\).

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental principle used to relate the pressure, volume, temperature, and amount of a gas. This relationship is described by the formula \( PV = nRT \), where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume of the gas,
• \( n \) is the number of moles of the gas,
• \( R \) is the ideal gas constant (approximately 0.0821 L atm/mol K),
• \( T \) is the temperature in Kelvin.

To solve problems with the ideal gas law, it’s crucial to ensure all units are compatible, especially temperature in Kelvin and pressure in atmospheres. In the decomposition of potassium chlorate, for example, we need to account for the collected oxygen gas by using the ideal gas law, which helps us find the number of moles based on measured physical conditions like temperature and pressure. Here, correcting pressure to account for water vapor pressure is crucial to isolating the oxygen gas's contribution.

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. Each gas in a mixture behaves independently, contributing to the total pressure according to its own concentration. The total pressure of the gas mixture is the sum of the partial pressures of each component gas.
In our exercise, oxygen gas (\( O_2 \)) is collected over water, meaning the measured pressure is due to both the oxygen and the water vapor. To find the true pressure of the oxygen, the vapor pressure of water (which changes with temperature) must be subtracted from the total measured pressure. This is essential because the ideal gas law calculations require the pressure of the gas being solved for, which in this case is oxygen alone, not the mixture.
By accurately determining partial pressures, we ensure that we calculate the correct number of moles of gas, thereby leading to precise chemical decomposition analysis.

Molar Mass Calculation

Molar mass is the mass of one mole of a given substance, used primarily to convert between grams of a substance and the amount in moles. It acts as a bridge in various chemical calculations. Determining the molar mass correctly is essential in the decomposition process of potassium chlorate (\( KClO_3 \)), which is calculated as 39.1 (for K) + 35.5 (for Cl) + 16*3 (for three O atoms), totaling 122.6 g/mol.
With the molar mass known, we can convert the moles of a substance, calculated using the ideal gas law, back into grams. This is vital when trying to find the mass percentage of pure potassium chlorate in an impure sample. The calculated mass of \( KClO_3 \) divided by the total sample mass gives the percentage composition, crucial in assessing the purity or effectiveness of a chemical decomposition reaction. Understanding and applying molar mass in conversions ensures that the data and results of chemical reactions are both meaningful and accurate.