2015 AP Chemistry free response 2a (part 1 of 2) | Chemistry | Khan Academy

Ethine (C₂H₄) molar mass of 28.1 g per mole may be prepared by the dehydration of ethanol (C₂H₅OH) molar mass 46.1 g per mole using a solid catalyst. A setup for the lab synthesis is shown in the diagram above. The equation for the dehydration reaction is given below.

So we have the ethanol, and then in the presence of a catalyst, we're going to yield after our reaction some ethine and some water. We have some metrics on the actual reaction.

A student added a 0.200 g sample of ethanol (C₂H₅OH) to a test tube using the setup shown above. So this is the glass wool with ethanol. The ethanol is right over here; there's a solid catalyst right over there. The student heated the test tube gently with a Bunsen burner until all of the ethanol evaporated, and gas generation stopped.

When the reaction stopped, the volume of gas collected was 0.854 L at 0.822 atmospheres and 305 Kelvin. The vapor pressure of water at 305 Kelvin is 35.7 torr.

In the presence of the catalyst, the ethanol reacts in a dehydration reaction to produce ethine and water vapor. After cooling, the ethine gets captured at the top in a gaseous state, while the water condenses into liquid water.

Now, let's try to answer the questions:

1. Calculate the number of moles of ethine that are actually produced in the experiment and measured in the gas collection tube.
2. Calculate the percent yield of ethine in the experiment.

Alright, let's tackle part one. We have to figure out the actual moles produced in the experiment and measured in the gas collection tube. They said until all of the ethanol evaporated, gas generation stopped.

The volume of gas was 0.854 L at 0.822 atmospheres. The temperature is 305 Kelvin, and the vapor pressure of water is 35.7 torr.

To find the partial pressure of ethine, we will subtract the partial pressure of water from the total pressure. The total pressure given is 0.822 atmospheres. The partial pressure of water must be converted from torr to atmospheres:

Vapor pressure of water is 35.7 torr.

Since 1 atmosphere is 760 torr, we can convert it:

[
\text{Partial Pressure of Water} = \frac{35.7 \text{ torr}}{760 \text{ torr/atm}} \approx 0.0470 \text{ atm}
]

Now we can subtract the vapor pressure of water from the total pressure:

[
\text{Partial Pressure of Ethine} = \text{Total Pressure} - \text{Partial Pressure of Water}
]

Thus:

[
\text{Partial Pressure of Ethine} = 0.822 \text{ atm} - 0.0470 \text{ atm} \approx 0.775 \text{ atm}
]

Now we can use the ideal gas law to find the number of moles of ethine produced. The ideal gas law is:

[
PV = nRT
]

Rearranging gives:

[
n = \frac{PV}{RT}
]

Substituting the known values:

[
n = \frac{(0.775 \text{ atm})(0.854 \text{ L})}{(0.08206 \text{ L atm/(mol K)})(305 \text{ K})}
]

Calculating the right side:

[
n = \frac{0.775 \times 0.854}{0.08206 \times 305}
]

Calculating step by step:

1. (0.775 \times 0.854 \approx 0.66285)
2. (0.08206 \times 305 \approx 25.0783)
3. (\frac{0.66285}{25.0783} \approx 0.0264)

So the approximate number of moles of ethine that are produced is:

[
n \approx 0.0264 \text{ moles}
]

That’s part one: the number of moles of ethine that are actually produced in the experiment and measured in the gas collection tube.