Problem 209 - Introductory Chemistry Atoms First

Question

Air is approximately \(21 \%\) by volume oxygen, \(\mathrm{O}_{2}\). How many grams of oxygen are present in \(200.0 \mathrm{~L}\) of air at \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) ? (Hint: Use the ideal gas equation from Chapter \(11 .\) )

Step-by-Step Solution

Answer

The mass of oxygen present in 200.0 L of air at 1.00 atm and 25°C is approximately 54.72 grams.

Step 1: Find the number of moles of air

Use the Ideal Gas Equation to find the number of moles of air, n_air: PV = nRT Where: - P is the pressure (1.00 atm) - V is the volume (200.0 L) - n is the number of moles - R is the ideal gas constant (0.0821 L⋅atm/mol⋅K) - T is the temperature (25°C = 298.15 K, after converting to Kelvin) Rearrange the equation to solve for n_air: n_air = PV / RT

Step 2: Calculate the number of moles of air

Plug the values into the equation and calculate n_air: n_air = (1.00 atm × 200.0 L) / (0.0821 L⋅atm/mol⋅K × 298.15 K) n_air ≈ 8.143 mol

Step 3: Find the number of moles of oxygen

Given that the air is approximately 21% by volume oxygen, we can find the number of moles of oxygen (n_oxygen) by multiplying n_air by the volume percentage: n_oxygen = n_air × 0.21 n_oxygen ≈ 8.143 mol × 0.21 ≈ 1.71 mol

Step 4: Calculate the mass of oxygen

Now, we need to convert the number of moles of oxygen to grams. To do this, we will use the molecular weight of oxygen, which is 32 g/mol: mass_oxygen = n_oxygen × (32 g/mol) mass_oxygen ≈ 1.71 mol × 32 g/mol ≈ 54.72 g

Step 5: Present the final answer

The mass of oxygen present in 200.0 L of air at 1.00 atm and 25°C is approximately 54.72 grams.

Key Concepts

Moles of Gas

In chemistry, the concept of a "mole" is crucial for understanding amounts of substances. A mole is a unit that measures the amount of a substance. Specifically, it equals Avogadro's number, which is about \(6.022 \times 10^{23}\) of whatever the molecule or atom you are counting. In gas calculations, we use the ideal gas equation to find the number of moles. The ideal gas equation is given by:\[PV = nRT\]- **P** stands for pressure in atmospheres (atm).- **V** is volume in liters (L).- **n** is the number of moles.- **R** is the ideal gas constant, approximately 0.0821 L⋅atm/mol⋅K.- **T** is the temperature in Kelvin (K).
By rearranging the equation to \(n = \frac{PV}{RT}\), we can directly compute the moles of air. This allows us to advance in exercises like the one provided and is foundational to predicting gas behavior under different conditions. Always ensure temperature is in Kelvin to maintain unit consistency.

Volume Percentage

Volume percentage is a way of expressing the concentration of a component in a mixture. It is defined as the volume of a component divided by the total volume of the mixture multiplied by 100%. In the problem, we know that "air" is composed of about 21% oxygen by volume. This simply means that in every 100 liters of air, approximately 21 liters are oxygen.

• When working with gases, especially mixtures like air, knowing the percentage by volume helps to break down the problem. It simplifies determining how much of one component gas contributed to the total volume.
• Once we find the total moles of the air mixture, we can use the volume percentage to find the moles of a specific gas, such as oxygen.

In solving our problem, the 21% oxygen translated directly into finding the moles of oxygen from the total moles of air as \(n_{oxygen} = n_{air} \times 0.21\). This cascading of knowledge greatly reduces complexity in multi-component systems.

Molecular Weight

The molecular weight (or molar mass) of a substance tells us the mass of one mole of its molecules. Molecular weight is usually measured in grams per mole (g/mol). Oxygen, a diatomic molecule represented by \(\text{O}_2\), has a molecular weight of 32 g/mol (16 g/mol for each oxygen atom multiplied by two).
Why is this important? Because when we have the amount of gas in moles, it can be converted to mass using the molecular weight. The conversion is straightforward: multiply the number of moles by the molecular weight: \[ \text{mass} = n \times \text{Molecular Weight} \]

• For example, in our exercise, once we knew there were 1.71 moles of oxygen, multiplying by 32 g/mol gave the gram amount of oxygen.

Understanding molecular weight is essential in translating between quantitative (mass) and qualitative (moles) chemical data, making it a core aspect of chemistry calculations.

Pressure and Temperature

Pressure and temperature profoundly influence gas behavior. They are two crucial parameters in the ideal gas equation. To handle gas problems accurately, understanding and converting units properly is vital.
**Pressure** is often given in atmospheres (atm) in chemical calculations:

• 1 atm is the standard pressure at sea level.
• In solving gas problems, ensure all inputs match the ideal gas constant's units.

**Temperature** must always be in Kelvin (K) for ideal gas calculations:

• To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.
• Kelvin keeps the temperature scale positive in calculations, avoiding inaccuracies.

For example, converting 25°C to Kelvin by adding 273.15 results in 298.15 K. Understanding how pressure and temperature shift the number and behavior of gas particles helps you predict outcomes in chemical reactions and energy dynamics.