Question

Oxygen gas generated by the decomposition of potassium chlorate is collected over water. The volume of oxygen collected at \(24^{\circ} \mathrm{C}\) and atmospheric pressure of \(760 \mathrm{mmHg}\) is \(128 \mathrm{~mL}\). Calculate the mass of oxygen gas obtained. The pressure of the water vapour at \(24^{\circ} \mathrm{C}\) is \(22.4 \mathrm{~mm} \mathrm{Hg}\) (a) \(1.36 \mathrm{~g}\) (b) \(1.52 \mathrm{~g}\) (c) \(0.163 \mathrm{~g}\) (d) \(1.63 \mathrm{~g}\)

Short answer

The mass of oxygen gas obtained is approximately 1.63 g.

Step-by-step solution

Determine the partial pressure of oxygen gas

Subtract the vapor pressure of water from the atmospheric pressure to obtain the partial pressure of the oxygen gas. Since the atmospheric pressure is 760 mmHg and the vapor pressure of water at 24°C is 22.4 mmHg, the partial pressure of oxygen is given by: \[ P_{\text{O}_2} = P_{\text{atm}} - P_{\text{H}_2\text{O}} = 760 \mathrm{~mmHg} - 22.4 \mathrm{~mmHg} \]

Convert the pressure to atmospheres

The pressure is given in mmHg and must be converted to atmospheres, since the gas constant (R) value when using mL and grams for volume and mass is commonly given in L·atm/(mol·K). The conversion is: \[ 1 \text{ atm} = 760 \text{ mmHg} \] Thus, the pressure of O2 in atmospheres is: \[ P_{\text{O}_2}(\text{atm}) = \frac{P_{\text{O}_2}(\text{mmHg})}{760} \]

Calculate the number of moles of oxygen using the ideal gas law

Use the ideal gas law: \[ PV = nRT \] where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in kelvin (24°C + 273.15). Solve for n to find: \[ n = \frac{PV}{RT} \] Remember to convert the volume of oxygen from mL to L by dividing by 1000.

Convert moles of oxygen to grams

Using the molar mass of O2 (32.0 g/mol), convert the moles of oxygen gas to mass in grams using: \[ \text{mass} = n \times \text{molar mass} \] The molar mass of oxygen (O2) is 32.0 g/mol.

Perform the calculations

Follow the calculations based on the previous steps to obtain the mass of oxygen gas collected. This involves plug in the values into the ideal gas equation to find moles, then multiply the result by the molar mass to find the mass in grams.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that correlates the pressure (P), volume (V), temperature (T), and amount in moles (n) of an ideal gas. It is represented by the formula: \[ PV = nRT \]where R is the universal gas constant with a value of 0.0821 L·atm/(mol·K). To apply this law effectively, it's important to ensure that all units are consistent—pressure should be in atmospheres, volume in liters, and temperature in kelvins.
In practical problems, like the one provided, we often have to manipulate the units and conditions to match these standard units. For example, pressures might need to be converted from mmHg to atm, and volumes from mL to L, to apply the ideal gas law effectively. Temperature must always be in kelvins, which simply requires adding 273.15 to the Celsius temperature.
The ideal gas law is the foundation for solving many types of gas-related problems in chemistry, including determining the amount of gas produced or consumed in a reaction.

Partial Pressure

Partial pressure refers to the pressure each gas in a mixture would exert if it occupied the entire volume alone at the same temperature. The total pressure exerted by the gas mixture is the sum of the partial pressures of each individual gas. This is defined by Dalton's Law of Partial Pressures.
In the exercise problem, we consider the partial pressure of oxygen, which we calculate by subtracting the water vapor pressure from the atmospheric pressure. This adjustment is crucial because the presence of water vapor affects the total pressure measurement in our system. By focusing on the oxygen's partial pressure, we can isolate the effect of the gas of interest and apply the ideal gas law correctly to calculate the amount of gas collected.

Molar Mass

Molar mass is a physical property that represents the mass of one mole of a substance. It is typically expressed in units of grams per mole (g/mol). For elemental molecules like oxygen (O2), which exist as diatomic molecules in nature, the molar mass is the sum of the atomic masses of each atom in the molecule.
In the context of the given exercise, knowing the molar mass of oxygen (32.0 g/mol) allows us to convert the number of moles calculated from the ideal gas law into the measurable mass of oxygen gas collected. This conversion is a critical step that bridges the gap between the abstract concept of 'moles' and the tangible amount of substance that we can weigh.
Understanding and utilizing molar mass is essential in stoichiometry for performing calculations between the mass of substances and mole quantities.

Stoichiometry

Stoichiometry is the aspect of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. It involves calculations that are based on the balanced chemical equation and the conservation of mass and atoms.
In stoichiometry, the coefficients in the chemical equation dictate the ratios of moles of each substance involved. For instance, if a reaction states that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water, it establishes a fixed proportion that must be maintained.
To apply stoichiometry to real-world problems, like the one given, you would first calculate the number of moles of oxygen produced (from the ideal gas law), then use the stoichiometry from the balanced equation of the reaction to find the amounts of other reactants or products, if necessary. The beauty of stoichiometry lies in its ability to predict the outcomes of chemical reactions and to ensure that experiments are conducted efficiently and safely.