## Procedure

1. The mass of the glass jar was measured using a triple beam balance when the jar was empty, and this data was recorded.

2. The mass of the lighter before releasing the gas was measured using a triple beam balance, and this data was recorded.

3. A plastic bin was filled about three quarters full with water from the sink, and the temperature of the water was measured using a thermometer and then recorded.

4. The volume of the jar before the gas was released into it was measured. This was performed by using a graduated cylinder to measure the maximum volume of water, in milliliters, that could be contained in the jar.

5. The pressure of the gas was calculated by subtracting the pressure of the water vapor, found by using the temperature of the water, from the atmospheric pressure. This was possible due to the assumption given in the beginning of the lab that the atmospheric pressure in the room is standard pressure. This assumption allowed one to use Dalton's Law of Partial Pressures which states that the pressure of the water vapor added to the pressure of the gas collected over the water is equal to the standard, or total, pressure. The pressure of the water vapor was found on the chart that was given at the start of the lab, and the pressure correlated with the temperature of the water.

6. The jar was submerged in the water in the plastic bin, and air bubbles left in the jar were removed.

7. The gas was released into the jar by submerging the lighter in the water and directing the gas bubbles into the jar. Gas was released into the jar until the volume of the gas was approximately one third of the volume of the jar.

8. The mass of the lighter was measured after the gas was released using a triple beam balance.

9. The final mass of the gas was calculated by subtracting the mass of the lighter after releasing the gas from the initial mass of the lighter before releasing the gas found in step 2.

10. The volume of water in the jar containing the gas was calculated by pouring the water from the jar into a graduated cylinder and measuring this volume of water in mL.

11. The volume of the gas in the jar was calculated by subtracting the volume of the water in the jar with the gas (found in step 10) from the initial volume of the jar (found in step 4).

12. Due to the assumption that the gas behaves like an ideal gas, given in the beginning of the lab, the Ideal Gas Law, PV=nRT, was used to calculate the number of moles of gas in the jar. The P was the pressure of the gas which was found in step 5. The V was the volume of the gas which was found in step 11. The n was the number of moles of the gas. The R was the gas constant, and the T was the temperature of the gas in Kelvin.

13. The molar mass of the gas was calculated by dividing the number of moles of butane gas (calculated in step 12) by the grams of butane gas in the jar (calculated in step 9).

## Data

Mass of empty glass jar
214.30g Volume of water in glass jar containing gas 174mL |
Mass of lighter before releasing gas
16.50g |
Temperature of water
23 degrees Celcius |
Volume of glass jar
262mL |
Mass of lighter after releasing gas
16.3g |

## Calculations

## The pressure of the gas:

The pressure of the gas was calculated using Dalton's Law of Partial Pressures by subtracting the pressure of the water vapor from the total pressure, which in this case was the atmospheric pressure, 760mm Hg. The pressure of the water vapor was found by using the temperature of the water. Once the pressure of the water vapor was found, this was subtracted from the atmospheric pressure, which in this case was 1 atm.

The pressure of the water vapor that corresponds with the water temperature of 23 degrees Celcius was 21 mm Hg. 1 atm is equal to 760 mm Hg.

760mm Hg-21mm Hg= 739 mm Hg

The pressure of the gas in the gas jar was 739 mm Hg.

The pressure of the water vapor that corresponds with the water temperature of 23 degrees Celcius was 21 mm Hg. 1 atm is equal to 760 mm Hg.

760mm Hg-21mm Hg= 739 mm Hg

The pressure of the gas in the gas jar was 739 mm Hg.

## The volume of the gas:

The volume of the gas was calculated by subtracting the volume of water in the jar with the gas (174mL) from the initial volume of water in the jar when the jar was completely filled with water (262 mL)

262mL-174mL= 88mL

The volume of gas was needed in liters, not milliliters, to be used in the PV=nRT equation, so the volume in milliliters was divided by 1000mL to convert to liters.

88mL/1000mL= 0.088 L butane gas

The volume of the gas in the glass jar was calculated to be 0.088 L.

262mL-174mL= 88mL

The volume of gas was needed in liters, not milliliters, to be used in the PV=nRT equation, so the volume in milliliters was divided by 1000mL to convert to liters.

88mL/1000mL= 0.088 L butane gas

The volume of the gas in the glass jar was calculated to be 0.088 L.

## Temperature of the water in Kelvin:

The temperature of the water was needed in degrees Kelvin to be used in the PV=nRT formula. The temperature of the water was converted from degrees celcius to degrees kelvin by adding 273 to the temperature in degrees Kelvin.

23 degrees celcius + 273= 296 K

The temperature of the water was 296 degrees Kelvin.

23 degrees celcius + 273= 296 K

The temperature of the water was 296 degrees Kelvin.

## Moles of gas:

The Ideal Gas Law, PV=nRT, was used to calculate the number of mole of butane gas. The P in the equation is the pressure of the gas, the V is the volume of the gas, the n is the number of moles of gas, the R is the gas constant, and the T is the temperature of the gas in Kelvin. The pressure of the gas was calculated to be 739mm Hg. The volume of the gas was calculated to be 88 mL. The gas constant in this equation is the gas constant for mm Hg which is 62.4mm Hg. The temperature of the gas can be assumed to be the same as the temperature of the water because the gas is passing through the water so the temperature of the gas is 296 K.

(739mm Hg)(0.088L)=n(62.4mm Hg)(296K)

n=((739mm Hg)(0.088L)) / ((62.4mm Hg)(296K)= 0.00352 moles butane gas

The moles of butane gas was calculated to be 0.00352 moles.

(739mm Hg)(0.088L)=n(62.4mm Hg)(296K)

n=((739mm Hg)(0.088L)) / ((62.4mm Hg)(296K)= 0.00352 moles butane gas

The moles of butane gas was calculated to be 0.00352 moles.

## Mass of the gas in grams:

The mass of the gas in the glass jar was calculated by subtracting the mass of the lighter after releasing the gas(16.30g) from the initial mass of the lighter before releasing the gas(16.50g).

16.50g-16.30g=0.20g

The mass of the gas in the glass jar was calculated to be 0.20g.

16.50g-16.30g=0.20g

The mass of the gas in the glass jar was calculated to be 0.20g.

## Molar mass of the gas:

The molar mass of any element or compound is the mass per one mole, and is in the units g/mol. The molar mass of the butane gas was calculated by dividing the mass of butane gas in grams(0.20g) by the moles of butane gas(0.00352 moles).

0.20g/0.00352moles= 56.81 g/mole= molar mass of butane gas.

The molar mass of butane gas was calculated to be 56.81 g/mole.

0.20g/0.00352moles= 56.81 g/mole= molar mass of butane gas.

The molar mass of butane gas was calculated to be 56.81 g/mole.

## Conclusion

The objective of this lab was to find the molar mass of a chemical used for combustion in a lighter. This was done by finding the number of moles in a sample of this chemical and also finding the mass of the same chemical sample then dividing the mass in grams by the number of moles since the units of molar mass is grams per mole. By using the data gathered through the procedures and calculations above and the three assumptions given at the beginning of the lab, the number of moles of the gas was found, and the mass of the gas sample was calculated by finding the difference between the mass of the lighter before releasing the gas and after releasing the gas. To calculate the number of moles of the gas, the Ideal Gas Law, which states that the pressure of the gas multiplied by the volume of that gas is equal to the number of moles multiplied by the gas constant multiplied by the temperature in Kelvin, was used after the assumption that the gas behaved like an ideal gas was given. Another law that was required to find the pressure of the gas to use as P in the Ideal Gas Law was Dalton's Law of partial pressures, which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each individual gas. In this lab, the total pressure was equal to the standard atmospheric pressure, and the pressure of the water vapor was subtracted from the total pressure to calculate the pressure of the gas. The number of moles found was 0.00352 and the mass of the gas sample was 0.20g. When the mass of the sample was divided by the number of moles, a molar mass of 56.81 g/mole was calculated. There were also many sources of error in this lab which influenced the value of the molar mass that was calculated. Sources of error include: balances not being calibrated, lighter containing water from the previous class, water present in the jar before being submerged, graduated cylinder read incorrectly, graduated cylinder systemic error related to the markings, incorrect temperature measurement due to random and/or systemic error causing an incorrect water vapor pressure, oxygen bubbles left in the jar after it was submerged, gas released from the lighter but not going into the jar, water remaining in the lighter when it was weighed the second time, water drops stuck to the inside of the jar when finding the volume of water in the jar when gas was in the jar, and incorrect calculations.

## Analysis

1. Based on the calculations, the molar mass of the gas was 56.81 grams per mole.

2. The gas is an alkane, so it contains only carbon and hydrogen with single covalent bonds. The molar mass of carbon is 12.01g/mol and the molar mass of hydrogen is 1.008 g/mol. The molar mass of the chemical in this lab was 56.81 g/mol, but for the sake of this question and due to many sources of error the molar mass of the chemical can be rounded to 57 g/mol. The chemical formula of this chemical was found by dividing the molar mass by 12 to find the number of moles of carbon. When 57 was divided by 12, the result was 4.75 moles of carbon, but since there has to be a whole number of moles in the chemical formula, it was assumed that there were 4 moles of carbon in the chemical formula. The mass of these 4 moles of carbon is 48 grams so this can be subtracted from the 57 grams/mol of the compound, which resulted in a mass of 9 grams. Since the chemical formula can only contain carbon and hydrogen, these 9 grams must be made up of hydrogen moles. Due to the fact that the molar mass of hydrogen is approximately 1 g/mol, there must be 9 moles of hydrogen in this chemical formula. All in all, the chemical formula contains 4 moles of carbon and 9 moles of hydrogen, resulting in a chemical formula of C4H9.

3. The chemical formula of the gas used in this lab, butane, is C4H10, and its molar mass is approximately 58 g/mol. The percent error was calculated by finding the difference between the actual molar mass of butane and the molar mass of butane that was calculated in the lab, dividing this difference by the actual molar mass of butane then multiplying this by 100 to calculate the result as a percent.

(58-56.81)/58 X 100 = 2.0%

The percent error calculated was 2.0%.

(58-56.81)/58 X 100 = 2.0%

The percent error calculated was 2.0%.

4. A) If I forgot to change the temperature of the gas from Celcius to Kelvin, the molar mass I calculated would have been lower. The ideal gas law, PV=nRT, requires the temperature to be in units Kelvin. If the temperature was left in Celcius, which is 273 degrees lesser than Kelvin, the number of moles calculated would have been significantly greater because the PV was divided by a lesser number than if the temperature was in Kelvin. Due to the mass of the gas in grams being divided by a greater number, the molar mass of the gas in this lab would have been lower.

B) If one forgot to correct the pressure for the vapor pressure of water, the molar mas one calculated would be lower. Since the pressure was not corrected for the pressure of the water vapor, there is no need to use Dalton's Law of Partial Pressures and it can be assumed that the pressure of the gas is equal to the total pressure. The total pressure in this lab was assumed to be the same as the atmospheric pressure, 760 mm Hg, so the pressure of the gas is equal to 760mm Hg. Since the pressure used in the Ideal Gas law will be greater, the number of moles would be greater since the value of PV would be greater than it should be. Due to the greater number of moles, the molar mass would be lower because the mass of the gas in the lab would be divided by a greater number.

C) If there were air bubbles in the jar before one released the lighter gas into it, the molar mass one calculated would have been lower. These air bubbles would have affected the volume of the gas that was calculated. Since there were air bubbles in the jar, the volume of the gas calculated would have been greater than the actual volume of the gas in the jar, so the value of PV in the Ideal Gas Law, PV=nRT, would have been greater. This greater value of PV would have lead to a greater number of moles calculated, thereby decreasing the molar mass because the mass of the gas would have been divided by a greater number of moles.

D) If the lighter was not completely dried the second time it was weighed, this would result in a lower molar mass being calculated. If the mass of the lighter after the gas was released was greater than it should have been this would have lead to a lesser (and possibly negative) mass of the gas when the mass of the lighter after releasing the gas was subtracted from the initial mass of the lighter, This lesser value would result in a lesser molar mass because there would be a lesser mass of the gas being divided by the number of moles.

B) If one forgot to correct the pressure for the vapor pressure of water, the molar mas one calculated would be lower. Since the pressure was not corrected for the pressure of the water vapor, there is no need to use Dalton's Law of Partial Pressures and it can be assumed that the pressure of the gas is equal to the total pressure. The total pressure in this lab was assumed to be the same as the atmospheric pressure, 760 mm Hg, so the pressure of the gas is equal to 760mm Hg. Since the pressure used in the Ideal Gas law will be greater, the number of moles would be greater since the value of PV would be greater than it should be. Due to the greater number of moles, the molar mass would be lower because the mass of the gas in the lab would be divided by a greater number.

C) If there were air bubbles in the jar before one released the lighter gas into it, the molar mass one calculated would have been lower. These air bubbles would have affected the volume of the gas that was calculated. Since there were air bubbles in the jar, the volume of the gas calculated would have been greater than the actual volume of the gas in the jar, so the value of PV in the Ideal Gas Law, PV=nRT, would have been greater. This greater value of PV would have lead to a greater number of moles calculated, thereby decreasing the molar mass because the mass of the gas would have been divided by a greater number of moles.

D) If the lighter was not completely dried the second time it was weighed, this would result in a lower molar mass being calculated. If the mass of the lighter after the gas was released was greater than it should have been this would have lead to a lesser (and possibly negative) mass of the gas when the mass of the lighter after releasing the gas was subtracted from the initial mass of the lighter, This lesser value would result in a lesser molar mass because there would be a lesser mass of the gas being divided by the number of moles.

5. Based on the calculations, there may be some mixture of gases inside the lighter, but it is impossible to quantify it. The molar mass that was calculated was so close to the actual molar mass of butane that it fell within the range of error. If it had to be said whether the molar masses of these other gases were higher or lower than the molar mass of butane, I would say that they are lower because the molar mass that we calculated was lesser than the actual molar mass of butane, so these gases may be decreasing the value of butane's molar mass that was calculated.

6. The lighter fluid inside the lighter is mostly butane which is an alkane and a gas at standard atmospheric pressure and temperature. The reason why the butane appears as a liquid inside the lighter is due to the fact that the gas is placed in its containing body under very high pressure which causes a phase change in the butane from a gas to a liquid. The opposite occurs as the butane is released. The high pressure of the butane causes it to go through the tiny port of the lighter and as it emerges into atmospheric pressure, the butane undergoes an immediate phase change from a liquid to a gas.