## Procedure

1. In this lab, gas from a lighter was collected over water. To determine the molar mass of the chemical in the lighter, the inital mass of the lighter first had to be measured. This was done using a balance. It was important that none of the gas from the lighter was released after the mass was measured prior to the experiment being completed.

2. A glass jar was then filled completely with water. Water was first measured into a graduated cylinder and then poured into the jar to keep count of the volume of water that was placed into the jar.

3. A plastic bin was then filled about half-way with water. After the water had sat for a short amount of time, the temperature was measured using a thermometer. The glass jar was then submereged into the tub of water. It was important that all of the air bubbles be removed from the jar using various methods. The jar was wiggled and shifted to remove air bubbles. A pipette was also carefully used to remove the remaining air bubbles.

4. After all the excess air bubbles were removed, the jar was positioned at an angle with the bottom of the jar facing upward. The lighter was quickly submerged into the water and placed into the mouth of the jar. The button on the lighter was then pushed to release the gas. It was crucial that none of the gas escaped the glass jar.

5. The gas was released into the jar until about a fourth of the jar was filled with the gas from the lighter. After the jar was filled with a sufficient amount of gas, the jar was capped tightly while still underwater. None of the gas escaped during this process.

6. After the lighter was removed from the water, it was placed on a paper towel to allow the water to drain form the lighter. The lighter had time to dry and then was placed on the balance to be weighed again. The final mass of the lighter was recorded for later use.

7. Finally, the top of the jar was removed and the gas was allowed to escape the jar. After the gas had escpaed, the remaining water was poured into a graduated cylinder to be measured.

8. After each of the processes were complete, the lab area was cleaned up and each item was returned to its home. The data collected was then plugged into the Pv=nRT equation to begin to find the molar mass of the gas collected.

2. A glass jar was then filled completely with water. Water was first measured into a graduated cylinder and then poured into the jar to keep count of the volume of water that was placed into the jar.

3. A plastic bin was then filled about half-way with water. After the water had sat for a short amount of time, the temperature was measured using a thermometer. The glass jar was then submereged into the tub of water. It was important that all of the air bubbles be removed from the jar using various methods. The jar was wiggled and shifted to remove air bubbles. A pipette was also carefully used to remove the remaining air bubbles.

4. After all the excess air bubbles were removed, the jar was positioned at an angle with the bottom of the jar facing upward. The lighter was quickly submerged into the water and placed into the mouth of the jar. The button on the lighter was then pushed to release the gas. It was crucial that none of the gas escaped the glass jar.

5. The gas was released into the jar until about a fourth of the jar was filled with the gas from the lighter. After the jar was filled with a sufficient amount of gas, the jar was capped tightly while still underwater. None of the gas escaped during this process.

6. After the lighter was removed from the water, it was placed on a paper towel to allow the water to drain form the lighter. The lighter had time to dry and then was placed on the balance to be weighed again. The final mass of the lighter was recorded for later use.

7. Finally, the top of the jar was removed and the gas was allowed to escape the jar. After the gas had escpaed, the remaining water was poured into a graduated cylinder to be measured.

8. After each of the processes were complete, the lab area was cleaned up and each item was returned to its home. The data collected was then plugged into the Pv=nRT equation to begin to find the molar mass of the gas collected.

## Data

During this lab, the following data was collected:

Initial mass of the lighter 16.5 g

Final mass of the lighter 16.25 g

Initial mL of water in the jar 260.0 mL

Final mL of water in the jar 208.5 mL

Temperature of water 23.0 °C

At the beginning of the lab, the following data was provided:

Pressure of water vapor 21 mm Hg (at 23.0 °C - measured temperature)

Pressure of atm 1 atm or 760 mm Hg

Initial mass of the lighter 16.5 g

Final mass of the lighter 16.25 g

Initial mL of water in the jar 260.0 mL

Final mL of water in the jar 208.5 mL

Temperature of water 23.0 °C

At the beginning of the lab, the following data was provided:

Pressure of water vapor 21 mm Hg (at 23.0 °C - measured temperature)

Pressure of atm 1 atm or 760 mm Hg

## Calculations

Calculating the pressure of the gas:

Since the pressure of the atmosphere was given (1 atm) and the pressure of the water vapor at 23 °C was given (21 mm Hg), we

began by converting the pressure of the atmosphere in atm into mm Hg. This was done by setting up a fence post with a

conversion factor. Next, each part was placed into the equaiton of collecting gas over water. The equation is P atm= P gas + P water

vapor. The ending number resulted in the pressure of the gas.

1 atm x 760 mm Hg

_______________ = 760 mm Hg (pressure of atmosphere)

1 atm

P atm = P gas + P water vapor

760 mm Hg = P gas + 21 mm Hg

- 21 mm Hg - 21 mm Hg

Calculating the mass of the gas (in grams) :

The mass of the gas was calculated by subtracting the final mass of the lighter from the initial mass of the lighter. The resulting

number was the mass, in grams, of the gas.

initial mass of the lighter - final mass of the lighter = mass in grams of the gas

16.5 g - 16.25 g =

Calculating the volume of the gas (in L) :

The volume of the gas was calculated by subtracting the final volume of the water in the jar from the initial volume of the water in

the jar. This volume was in milliliters, so it was then converted into liters using a fence post and conversion factor.

Initial volume of water in jar - final mass of water in jar = volume of gas in mL

260.0 mL - 208.5 mL = 51.5 mL of gas

51.5 mL of gas x 1 L

______________ =

1000 mL

Calculating the temperature:

Temperature was measured with a thermometer in degrees Celsius (°C). This value was then converted into Kelvin (K) using

the conversion factor.

°C + 273 = K

23.0 °C + 273 =

Calculating moles of gas:

The moles of gas were calculated using the equation Pv=nRT. This equation is pressure times volume is equal to number of

moles times a constant value times temperature. The values calculated above for pressure, volume, and temperature were

plugged into this equation. The constant R value used was 62.4 L x torr/ mol x K. The equation was solved for n, number of moles

of gas.

Pv=nRT

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

n=

Calculating molar mass of the gas:

The molar mass of the gas was calculated by dividing grams by moles. This was done because the units for molar mass are

grams per mole. The grams of the gas found above were placed on top, and the moles of the gas found above were placed

on the bottom. The resulting number was the molar mass of the gas.

molar mass = grams/mol

molar mass of the gas= .25 g of gas

________ =

.002mol of gas

Since the pressure of the atmosphere was given (1 atm) and the pressure of the water vapor at 23 °C was given (21 mm Hg), we

began by converting the pressure of the atmosphere in atm into mm Hg. This was done by setting up a fence post with a

conversion factor. Next, each part was placed into the equaiton of collecting gas over water. The equation is P atm= P gas + P water

vapor. The ending number resulted in the pressure of the gas.

1 atm x 760 mm Hg

_______________ = 760 mm Hg (pressure of atmosphere)

1 atm

P atm = P gas + P water vapor

760 mm Hg = P gas + 21 mm Hg

- 21 mm Hg - 21 mm Hg

__P gas = 739 mm Hg__Calculating the mass of the gas (in grams) :

The mass of the gas was calculated by subtracting the final mass of the lighter from the initial mass of the lighter. The resulting

number was the mass, in grams, of the gas.

initial mass of the lighter - final mass of the lighter = mass in grams of the gas

16.5 g - 16.25 g =

__.25 g of gas__Calculating the volume of the gas (in L) :

The volume of the gas was calculated by subtracting the final volume of the water in the jar from the initial volume of the water in

the jar. This volume was in milliliters, so it was then converted into liters using a fence post and conversion factor.

Initial volume of water in jar - final mass of water in jar = volume of gas in mL

260.0 mL - 208.5 mL = 51.5 mL of gas

51.5 mL of gas x 1 L

______________ =

__0.015 L gas__1000 mL

Calculating the temperature:

Temperature was measured with a thermometer in degrees Celsius (°C). This value was then converted into Kelvin (K) using

the conversion factor.

°C + 273 = K

23.0 °C + 273 =

__296 K__Calculating moles of gas:

The moles of gas were calculated using the equation Pv=nRT. This equation is pressure times volume is equal to number of

moles times a constant value times temperature. The values calculated above for pressure, volume, and temperature were

plugged into this equation. The constant R value used was 62.4 L x torr/ mol x K. The equation was solved for n, number of moles

of gas.

Pv=nRT

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

n=

__0.002060513 mol of gas__Calculating molar mass of the gas:

The molar mass of the gas was calculated by dividing grams by moles. This was done because the units for molar mass are

grams per mole. The grams of the gas found above were placed on top, and the moles of the gas found above were placed

on the bottom. The resulting number was the molar mass of the gas.

molar mass = grams/mol

molar mass of the gas= .25 g of gas

________ =

__125 g/mol of gas__.002mol of gas

## Conclusion

From this lab, the molar mass of the Butane gas , C4H10, was determined to be 125 grams per mole. This number was found by using tools to measure the values for volume, temperature, and mass. The other values were either provided or constant numbers. The values gathered were then plugged into formulas. The molar mass that was calculated was quite far off from the molar mass of Butane that comes form the periodic table. There were many sources of error that accounted for the mass that was too high. Some of those sources of error were other gasses in the lighter fluid, the pressure of the atmpshere not being at STP, and errors while measuring values. The results showed that it is hard to get extremely close to the mass on the periodic table by collecting gas over water, and that even the slightest change can have a large effect on the final value.

## Analysis

1. Based on our calculations, the molar mass of the gas was 125 grams per mole.

2. According to the molar mass we calculated, a possible formula for the gas would be C9H20. This formula would result in the mass closest to the value that we calculated for the molar mass.

3. Based on the fact that most of the gas in the lighter is Butane, our percent error was 393%. This was calculated by putting actual over theoretical and then multiplying by one hundred.

Actual

__________ x 100 = percent error C4H10 = (12 x 4) + (1 x 10) = 58 g (theoretical)

theoretical

125 g

_________ x 100 = 393 % error

58 g

4.

a. Forgetting to convert Celsius to Kelvin would resuld in a lower molar mass because it would make the value for n larger and make the fraction

smaller.

b. Forgetting to correct the pressure for the vapor pressure o fthe water would result in a lower molar mass because it would increase the value

of n and make the fraction smaller, resulting in a smaller final value.

c. Having air bubbles in the flask before releasing lighter gas into the flask would result in a lower molar mass because the increased volume would

result in the number of moles being larger. This would make the fraction smaller and result in a smaller molar mass.

d. If the lighter was not completely dry when it was weighed the second time, it would cause the molar mass to be higher because the grams

would be a larger value. When grams are placed over moles, the bigger value for grams results in a larger fraction and therefore a larger molar

mass.

5. Based on my results, the other gases in the lighter would collectively have a larger mass than the Butane because the mass that I got was much

larger than the typical mass of Butane. These gasses had to weigh more to take up the excess space in the jar and account for the excess mass.

6. The substance inside the lighter is seen in liquid form when it is inside the lighter because it is at a very high pressure inside the lighter. Butane would typically be a gas at STP, so it is placed in an environment with a high pressure to keep it stored as liquid.When the Butane is released, it is pushed through a very small hole and released. As the pressure changes from high to low the Butane goes through a phase change and becomes a gas so that it can be ignighted and used in the flame. The difference in pressure explains the different states that the Butane is in.

2. According to the molar mass we calculated, a possible formula for the gas would be C9H20. This formula would result in the mass closest to the value that we calculated for the molar mass.

3. Based on the fact that most of the gas in the lighter is Butane, our percent error was 393%. This was calculated by putting actual over theoretical and then multiplying by one hundred.

Actual

__________ x 100 = percent error C4H10 = (12 x 4) + (1 x 10) = 58 g (theoretical)

theoretical

125 g

_________ x 100 = 393 % error

58 g

4.

a. Forgetting to convert Celsius to Kelvin would resuld in a lower molar mass because it would make the value for n larger and make the fraction

smaller.

b. Forgetting to correct the pressure for the vapor pressure o fthe water would result in a lower molar mass because it would increase the value

of n and make the fraction smaller, resulting in a smaller final value.

c. Having air bubbles in the flask before releasing lighter gas into the flask would result in a lower molar mass because the increased volume would

result in the number of moles being larger. This would make the fraction smaller and result in a smaller molar mass.

d. If the lighter was not completely dry when it was weighed the second time, it would cause the molar mass to be higher because the grams

would be a larger value. When grams are placed over moles, the bigger value for grams results in a larger fraction and therefore a larger molar

mass.

5. Based on my results, the other gases in the lighter would collectively have a larger mass than the Butane because the mass that I got was much

larger than the typical mass of Butane. These gasses had to weigh more to take up the excess space in the jar and account for the excess mass.

6. The substance inside the lighter is seen in liquid form when it is inside the lighter because it is at a very high pressure inside the lighter. Butane would typically be a gas at STP, so it is placed in an environment with a high pressure to keep it stored as liquid.When the Butane is released, it is pushed through a very small hole and released. As the pressure changes from high to low the Butane goes through a phase change and becomes a gas so that it can be ignighted and used in the flame. The difference in pressure explains the different states that the Butane is in.