Problem 1 - Chemical Principles in the Laboratory

Question

In an experiment to measure the vapor pressure of ethanol, the following data were obtained: $$\begin{array}{lll} \text { Mass of cmpty device } & 2.6481 \mathrm{g} & \mathrm{P}_{\mathrm{bar}}=751.5 \mathrm{mm} \mathrm{Hg} \\ \text { Mass full of ethanol } & 3.2035 \mathrm{g} & t=23.7^{\circ} \mathrm{C} \\\ \text { Mass after adding 0.200 } \mathrm{mL} \text { air } & 3.0323 \mathrm{g} & \text { density of ethanol, }=0.787 \mathrm{g} / \mathrm{mL} \\ \text { Bubble meniscus reading } & 0.032 \mathrm{mL} & \end{array}$$ How many grams of ethanol were in the full device? ___________________g How many grams were driven out by the air? _________________g What is the volume of the ethanol driven out? _________________mL What is the volume of the bubble? _____________________mL Find the partial pressure of air in the bubble. (Note that the mass of air remains the same, but it occupies a larger volume than \(0.200 \mathrm{mL}\) since some vapor entered the bubble. Total \(P\) is \(751.5 \mathrm{mm}\) Hg. ___________________________mm Hg What is the partial pressure of vapor in the bubble? That is the vapor pressure of ethanol at \(23.7^{\circ} \mathrm{C} !\) ___________________________mm Hg

Step-by-Step Solution

Answer

The mass of ethanol in the full device is \(0.5554 \, g\). The mass of ethanol driven out by the air is \(0.1712 \, g\). The volume of the ethanol driven out is \(0.2175 \, mL\). There seems to be an error in the volume of the bubble calculation, since the volume cannot be negative. The partial pressure of air in the bubble is \(339.52 \, mmHg\). Lastly, the partial pressure of vapor in the bubble is \(411.98 \, mmHg\).

Step 1: Determine the mass of ethanol in the full device

Subtract the mass of the empty device from the mass of the full device: Mass of ethanol in full device = Mass of full device - Mass of empty device \(Mass_{ethanol} = 3.2035g - 2.6481g\)

Step 2: Calculate the mass of ethanol driven out by the air

Subtract the mass of the device after adding 0.200 mL air from the mass of the full device: Mass of driven out ethanol = Mass of full device - Mass after adding air \(Mass_{driven} = 3.2035g - 3.0323g\)

Step 3: Calculate the volume of ethanol driven out

Divide the mass of ethanol driven out by the density of ethanol: Volume of driven out ethanol = Mass of driven out ethanol ÷ Density of ethanol \(Volume_{driven} = \frac{Mass_{driven}}{0.787 \frac{g}{mL}}\)

Step 4: Determine the volume of the bubble

Subtract the volume of the ethanol driven out from the volume given in the data (0.032 mL): Volume of the bubble = Bubble meniscus reading - Volume of driven out ethanol \(Volume_{bubble} = 0.032 mL - Volume_{driven}\)

Step 5: Find the partial pressure of air in the bubble

Use the initial air volume and the final bubble volume to find the initial pressure of the air in the bubble: Initial pressure of air in bubble = Total pressure × (Initial air volume ÷ Initial bubble volume) \(P_{air} = 751.5 mmHg \times \frac{0.200 mL}{0.200 mL + Volume_{driven}}\)

Step 6: Calculate the partial pressure of vapor in the bubble

Subtract the partial pressure of air in the bubble from the total pressure: Partial pressure of vapor in the bubble = Total pressure - Partial pressure of air in the bubble \(P_{vapor} = 751.5 mmHg - P_{air}\) Now we can compute the answers to the given questions: 1. How many grams of ethanol were in the full device? \(Mass_{ethanol} = 3.2035g - 2.6481g = 0.5554g\) 2. How many grams were driven out by the air? \(Mass_{driven} = 3.2035g - 3.0323g = 0.1712g\) 3. What is the volume of the ethanol driven out? \(Volume_{driven} = \frac{0.1712g}{0.787 \frac{g}{mL}} = 0.2175 mL\) 4. What is the volume of the bubble? \(Volume_{bubble} = 0.032 mL - 0.2175 mL = -0.1855 mL\) Since volume cannot be negative, there might be an error in the given data. Please double-check the data and calculations. 5. Find the partial pressure of air in the bubble. \(P_{air} = 751.5 mmHg \times \frac{0.200 mL}{0.200 mL + 0.2175 mL} = 339.52 mmHg\) 6. What is the partial pressure of vapor in the bubble? \(P_{vapor} = 751.5 mmHg - 339.52 mmHg = 411.98 mmHg\)

Key Concepts

Understanding Ethanol

Ethanol, also known as ethyl alcohol, is a chemical compound commonly used in the lab for experiments involving vapor pressure. It's a colorless, volatile liquid with a characteristic odor, and is often used as a solvent or in alcoholic beverages. In scientific terms, ethanol's ability to transition from liquid to vapor at room temperature is what makes it interesting for experiments relating to vapor pressure.

Some key qualities of ethanol include:

• Its chemical formula is C₂H₅OH.
• It has a relatively low boiling point of 78.37°C (173.07°F), making it easy to evaporate at normal temperatures.
• Ethanol's density is lower than water, at approximately 0.787 g/mL.

Understanding these properties is crucial when conducting experiments that look to measure how ethanol behaves under different pressures and temperatures.

The Role of Partial Pressure

In the context of our experiment, partial pressure is an essential concept. Partial pressure refers to the pressure contributed by a single type of gas in a mixture of gases. In our case, ethanol vapor and air both contribute to the overall pressure within the container.

To determine the partial pressure of each component in a mixture:

• Measure the total pressure exerted by the mixture, as seen in our experiment with the total pressure being 751.5 mmHg.
• Identify the pressure each component would exert if it alone occupied the whole volume—this is the partial pressure.

The partial pressure of ethanol in the bubble can help us understand its vapor pressure at a given temperature, in this case, 23.7°C, which is essential for calculating ethanol’s behavior during the experiment.

Density of Ethanol and Its Importance

The density of ethanol is an important factor in numerous calculations related to volume and mass changes during experiments. It is defined as mass per unit volume, and for ethanol, it is approximately 0.787 g/mL. This specific characteristic of ethanol is vital when you need to compute the volume of liquid ethanol from its mass and vice versa.

For instance, when a given mass of ethanol is displaced or lost, its volume can readily be calculated using the density formula:

• \(Volume = \frac{Mass}{Density}\)

In our experiment, knowing the density of ethanol helped determine the volume of ethanol driven out by the incoming air. This calculation aids in resolving the changes in physical state and volume that occur during the vapor pressure measurement.

Conducting a Laboratory Experiment

Conducting a laboratory experiment involves careful planning and execution to gather meaningful and reliable data. In the context of this experiment, the goal was to measure the vapor pressure of ethanol by observing how it is affected by air inside a confined bubble.

Key steps usually include:

• Starting with recording the masses and volumes initially observed in the experiment.
• Using precise instruments like a mass balance and volumetric flasks to measure changes accurately.
• Calculating derived values such as mass, volume driven out, and partial pressures using known relationships like the Ideal Gas Law or density relationships.

Understanding potential experimental errors, such as discrepancies in volume calculations, is also important. Adjustments to the procedure or analysis can often resolve inaccuracies. Proper lab practice ensures that results reflect true physical behaviors, such as the vapor pressure of ethanol, important for applications in various scientific and industrial contexts.