## Procedure:

1.) Various steps were taken in order to determine the molar mass of a gas in a lighter by collecting the gas over water. The initial mass of the lighter was measured to help determine how much gas, in grams, was released under water into the jar. A triple beam balance was used to determine the mass of the lighter.

2.) A transparent, plastic container was filled 3/4 of the way with water from a sink. The temperature of the water was taken using a thermometer and left submerged in the water until the temperature did not change. The initial temperature was needed to help find the vapor pressure of water that corresponds with the temperature.

3.) A clear, glass jar was filled completely with water to help find the volume of the gas released in the jar. For accuracy, the water was measured with a 100 mL graduated cylinder and repeatedly filled to find the initial volume of the jar.

4.) The water from the jar was dumped in the plastic container and the glass jar was submerged under water until all of the air bubbles escaped. The jar was tilted and tapped to help the bubbles escape. A pipet was used to remove the remaining visible bubbles in the jar while completely submerged under water.

5.) The jar was angled downwards in the water and the lighter was put at the opening of the jar, so that the gas went into the jar. To release the gas from the lighter, the lever was pushed down and held down until the gas filled 1/4 of the jar. As the gas was going into the jar the water was being pushed out. After 1/4 of the jar was filled with the gas, the lid was twisted on while under water to make sure not to let any of the gas escape from the jar.

6.) The jar was taken out from the container and placed on the table with the lid facing up. The lid was opened to let the gas escape from the jar in order to measure the amount of water left in the jar. The remaining water was measured with a 100 mL graduated cylinder. The difference in the initial and final volume of water was the volume of gas released.

7.) The lighter laid out to dry for 30 minutes before weighing it. If water was retained in the lighter the mass of the gas would have been lower than the amount that truly escaped.

8.) At the end of the experiment, enough data was collected to plug in the experimental values into the PV=nRT equation. This equation was able to be used because it was assumed that the gas behaves like an ideal gas. The pressure was given at the beginning of the experiment because it was assumed that the atmospheric pressure in the room is standard pressure, which is 1 atm or 760 mmHg. The volume was collected when the difference from the initial and final volume of the water in the jar was measured. The gas constant is known and the temperature was measured previously. Therefore, "n" could be solved by plugging in the known values.

2.) A transparent, plastic container was filled 3/4 of the way with water from a sink. The temperature of the water was taken using a thermometer and left submerged in the water until the temperature did not change. The initial temperature was needed to help find the vapor pressure of water that corresponds with the temperature.

3.) A clear, glass jar was filled completely with water to help find the volume of the gas released in the jar. For accuracy, the water was measured with a 100 mL graduated cylinder and repeatedly filled to find the initial volume of the jar.

4.) The water from the jar was dumped in the plastic container and the glass jar was submerged under water until all of the air bubbles escaped. The jar was tilted and tapped to help the bubbles escape. A pipet was used to remove the remaining visible bubbles in the jar while completely submerged under water.

5.) The jar was angled downwards in the water and the lighter was put at the opening of the jar, so that the gas went into the jar. To release the gas from the lighter, the lever was pushed down and held down until the gas filled 1/4 of the jar. As the gas was going into the jar the water was being pushed out. After 1/4 of the jar was filled with the gas, the lid was twisted on while under water to make sure not to let any of the gas escape from the jar.

6.) The jar was taken out from the container and placed on the table with the lid facing up. The lid was opened to let the gas escape from the jar in order to measure the amount of water left in the jar. The remaining water was measured with a 100 mL graduated cylinder. The difference in the initial and final volume of water was the volume of gas released.

7.) The lighter laid out to dry for 30 minutes before weighing it. If water was retained in the lighter the mass of the gas would have been lower than the amount that truly escaped.

8.) At the end of the experiment, enough data was collected to plug in the experimental values into the PV=nRT equation. This equation was able to be used because it was assumed that the gas behaves like an ideal gas. The pressure was given at the beginning of the experiment because it was assumed that the atmospheric pressure in the room is standard pressure, which is 1 atm or 760 mmHg. The volume was collected when the difference from the initial and final volume of the water in the jar was measured. The gas constant is known and the temperature was measured previously. Therefore, "n" could be solved by plugging in the known values.

## Data:

Mass of lighter before
Mass of lighter after Initial volume of jar filled up (mL) Final volume of jar with gas (mL) Temperature of water (°C) Atmospheric pressure (mmHg) Water vapor pressure at 23°C |
16.5 g
16.25 g 260 mL 208.5 mL 23°C 760 mmHg 21 mmHg |

## Calculations:

**Calculating the mass of the gas:**

In order to calculate the mass of the gas in the lighter the mass of the lighter after the gas was released needs to be subtracted from the mass of the gas before the gas was released

Initial mass of lighter - Final mass of the lighter = The mass of the gas released

16.5 g - 16.25 g = .25 g

**Calculating the pressure of the gas:**

In order to calculate the pressure of the gas the equation P (atm) = P (gas) + P (water vapor) was used. The water vapor pressure needs to be subtracted from the atmospheric pressure, which is standard pressure. The water vapor pressure depends on the temperature of the water. The water vapor pressure for 23°C is 21 mmHg. The atmospheric pressure is 760 mmHg.

**Atmospheric pressure - Water vapor pressure = Pressure of the gas**

760 mmHg = P gas + 21 mmHg

- 21 mmHg - 21 mmHg

P (gas) = 739 mmHg

**Calculating the temperature:**

In order to use the temperature in the PV=nRT equation, the temperature must be in Kelvin.

°C + 273K = temperature in Kelvin

23°C + 273K = 296K

**Calculating the volume of the gas:**

In order to find the volume of the gas the initial volume of water in the jar needs to be subtracted by the final volume of water in the jar.

Initial volume of water - Final volume of water = Volume of the gas

260 mL - 208.5 mL = 51.5 mL

In order to plug the volume into the PV=nRT equation, the volume must be in liter.

51.5 mL 1 L

x ------------ = .0515 L of the gas

1000 mL

**Finding the moles of the gas:**

Plug in the values obtained from the experiment and the given values. The number of moles if the unknown values. The number of moles is multiplied by the gas constant and temperature. The pressure and volume is divided by this value.

PV=nRT (The values are all about the gas in the lighter)

(739 mmHg)(.0515L) = (n)(62.4 (L x torr)/(mol x K))(296K)

(38.05 mmhg x L) = (18,470.4n (L x torr)/mol)

n = .0021 moles of the gas

**Calculating the molar mass:**

The molar mass is the mass given on the period table. It is the mass of the element per mole.

**mass (g) /moles = mass (g) /mole**

.25g/.002 moles = 125 g/mole

## Conclusion:

The lab concluded that the molar mass of butane gas, C4H10, is 125 g/mol. The molar mass was calculated from the experimental values obtained from the lab and given values and constants. The ideal gas law equation was able to be used because it was assumed that the gas behaves like an ideal gas. The equation was used to plug in the values and obtain the number of moles of butane. The mass of butane released was divided by the number of moles to find the molar mass. The molar mass, given on the periodic tables, is 58.12 g/mol. The percent error in the lab was 115%. There was a large percent error in this lab because of the assumptions made in the beginning of the experiment. The atmospheric pressure was assume to be standard pressure and the gas was assume to behave as an ideal gas, in order to use the ideal gas law equation. Also, the butane gas was mixed with other gases inside the lighter. Not only was butane released into the jar, but other gases as well. The mass of the other alkanes were included in the nominator or the g/mol ratio. If the other gases were not mixed in with the butane the mass would have been lower and the overall molar mass would have been lower as well.

## Analysis:

**1. What is the molar mass of the gas, based on your calculations?**

Based on the calculations, the molar mass of butane is 125 g/mol

2. The gas is an alkane (contains only carbon and hydrogen with single bonds). Based on your molar mass, suggest a possible formula for the gas.

2. The gas is an alkane (contains only carbon and hydrogen with single bonds). Based on your molar mass, suggest a possible formula for the gas.

A possible formula for the gas is C9H16. The mass of C9H16 is 125 g. This formula has the closest mass that we obtained from the lab.

3. Most of the gas in the lighter is butane: C4H10. Based on that, what is your percent error? Show your work.

3. Most of the gas in the lighter is butane: C4H10. Based on that, what is your percent error? Show your work.

The percent error is 115 percent.

| Experimental Value - Theoretical Value |

-------------------------------------------------- × 100% = 115%

| Theoretical Value |

4. What effect would each of the following errors have had on the value of the molar mass you calculated? (Higher, lower, no effect) Explain why.

4. What effect would each of the following errors have had on the value of the molar mass you calculated? (Higher, lower, no effect) Explain why.

a.) You forgot to change Celsius temperature to Kelvin for your calculations.

The molar mass would be lower. The number of moles would increase, so when you divide the mass by the number of moles it decreases.

b.) You forgot to correct the pressure for the vapor pressure of water

The molar mass would be lower. The number of moles would increase, so the molar mass would decrease because the mass is being divided by a larger number.

c.) You had air bubbles in your flask before you released the lighter gas into it

The molar mass would be lower because there would be more volume of the gas accounted for than the actual amount. This will increase the number of moles resulted from the ideal gas law equation. When the mass is divided by this larger number (the number of moles) the molar mass will decrease.

d.) The lighter was not completely dry when weighed the second time.

The molar mass would be lower because the difference in the mass of butane released into the jar will be appear to be less. If the water does not evaporate the mass would be higher so when you subtract the initial mass by the final mass it will be smaller. Although the overall mass of the lighter will be heavier because of the water, the difference in the initial and final is what we use as the mass. Therefore, the mass would be smaller not larger. When it is divided by the number of moles the molar mass will decrease. A smaller nominator results in an decrease in number if the denominator is kept constant.

**5. There are really several gases mixed together in lighter fluid. Based on your results, do the other gases have a higher or lower molar mass than butane? Explain.**

The other gases have a lower molar mass because it there are multiple gases mixed in there and butane alone accounts for 47% of the molar mass, they must have a lower molar mass. However, added all together, the other gases have a higher molar mass. The molar mass we obtained during the lab was 125 g/mol and butane is 58.12 g/mol. That leaves 66.88 g/mol for the other gases mixed together with butane.

6. In this lab, we have referred to the gas inside the lighter. However, observations of a transparent lighter clearly show that the substance inside the lighter is a liquid. Explain.

6. In this lab, we have referred to the gas inside the lighter. However, observations of a transparent lighter clearly show that the substance inside the lighter is a liquid. Explain.

The substance inside the lighter is a liquid because the alkane is placed under very high pressure inside the lighter. In standard temperature and pressure, the alkane would be a gas, but because it is not in standard pressure it causes a phase change inside the lighter. Therefore, it appear to be a liquid. However, when the lever is pushed down, the pressure change cause the alkane to be released in a form of a gas and not a liquid.