Butane Lab
Procedure
The goal of the lab was to determine the molar mass and the plausible formula of the chemical, using a glass bottle, a lighter, and a handful of laboratory equipments. The weight of the glass bottle, of the lighter, and of the bottle cap was measured. The box was filled with water. The glass bottle was filled with water and then submerged in the box of water. All of the air bubble was taken out of the bottle. The lighter was also submerged in the water and its gas was released. The bottle trapped the unknown gas and the gas filled the bottle for about one-third. The inside contents of the bottle was locked with the bottle cap and the weight of the bottle filled with water and gas was measured. Using the data and other given information, the molar mass and chemical formula of the gas was calculated.
To find the molar mass of the unknown gaseous compound, the mole of the gas was needed, and to find the mole, PV = nRT equation was used. Because the unit of molar mass is g/mole, finding the number of moles is not sufficient. The mass of the gas also had to be founded.
To find the temperature, a thermostat was used. It showed that the temperature of the gas was 23 degrees Celsius. Celsius was converted to Kelvin by adding 273 to 23.
To find the pressure, a given chart was used. Using the measured temperature, 23 degrees Celsius, the chart showed that the vapor pressure of water in 23 degrees Celsius was 21 mmHg. The atmospheric pressure, 760 mmHg, was subtracted by the pressure of the gas, which was 21 mmHg. Then mmHg was converted to standard atm by dividing the pressured with 760 mmHg.
To find the volume of the mixture of the gas, the volume of glass bottle was subtracted by the volume of water left after adding gas to the bottle. The volume of the glass bottle was 262 mL and the volume of water after adding gas mixture was 187 mL. The mL was converted to L.
0.0821 atm*L/moles*K was the constant.
The numbers were plugged in for the PV = nRT equation to find the moles of the gas.
To find the mass of the gas mixture, the lighter's mass was weighted for both before and after releasing of the gas. Before the release, the lighter weighted 17.6 g. After the release, the lighter weighted 17.8 g. The difference between the two measurements show the mass of the gas.
The mass of the gas mixture was divided by the number of moles of the gas mixture to find the molar mass of the gas mixture.
For finding plausible chemical compound for the molar mass required mix-n-match. The molar mass of carbon was fitted into molar mass of gas mixture as much as it could without resulting in non-integer number. 5 units of carbon could fit into the molar mass of the gas mixture. The rest of the molar mass of the mixture was fitted by molar mass of hydrogen.
To find the molar mass of the unknown gaseous compound, the mole of the gas was needed, and to find the mole, PV = nRT equation was used. Because the unit of molar mass is g/mole, finding the number of moles is not sufficient. The mass of the gas also had to be founded.
To find the temperature, a thermostat was used. It showed that the temperature of the gas was 23 degrees Celsius. Celsius was converted to Kelvin by adding 273 to 23.
To find the pressure, a given chart was used. Using the measured temperature, 23 degrees Celsius, the chart showed that the vapor pressure of water in 23 degrees Celsius was 21 mmHg. The atmospheric pressure, 760 mmHg, was subtracted by the pressure of the gas, which was 21 mmHg. Then mmHg was converted to standard atm by dividing the pressured with 760 mmHg.
To find the volume of the mixture of the gas, the volume of glass bottle was subtracted by the volume of water left after adding gas to the bottle. The volume of the glass bottle was 262 mL and the volume of water after adding gas mixture was 187 mL. The mL was converted to L.
0.0821 atm*L/moles*K was the constant.
The numbers were plugged in for the PV = nRT equation to find the moles of the gas.
To find the mass of the gas mixture, the lighter's mass was weighted for both before and after releasing of the gas. Before the release, the lighter weighted 17.6 g. After the release, the lighter weighted 17.8 g. The difference between the two measurements show the mass of the gas.
The mass of the gas mixture was divided by the number of moles of the gas mixture to find the molar mass of the gas mixture.
For finding plausible chemical compound for the molar mass required mix-n-match. The molar mass of carbon was fitted into molar mass of gas mixture as much as it could without resulting in non-integer number. 5 units of carbon could fit into the molar mass of the gas mixture. The rest of the molar mass of the mixture was fitted by molar mass of hydrogen.
Data
Weight of lighter before releasing gas- 17.6 g
Weight of lighter after releasing gas- 17.8 g
Weight of glass bottle- 205.8 g
Weight of cap- 10.2 g
Weight of glass bottle of water, gas, and cap- 474.7 g
Volume of glass bottle- 262 mL
Volume of water left in the bottle of water and of gas- 187 mL
Weight of lighter after releasing gas- 17.8 g
Weight of glass bottle- 205.8 g
Weight of cap- 10.2 g
Weight of glass bottle of water, gas, and cap- 474.7 g
Volume of glass bottle- 262 mL
Volume of water left in the bottle of water and of gas- 187 mL
Calculations
Temperature
(temperature of water vapor in Celsius) + (273k) = temperature of water vapor in Kelvin
23*C + 273k => 296k
The temperature of the gas in Kelvin was 296k.
Pressure
(atmospheric Pressure) - (pressure of water vapor) = pressure of the gas
760 mmHg - 21mmHg = 739 mmHg
(pressure of the gas in mmHg) / 760 mmHg = pressure of the gas in atm
739 mmHg / 760 mmHg = 0.97 atm
The pressure of the gas in atm was 0.97 atm.
Volume
(volume of full glass bottle in mL) - (volume of water left after release of gas in mL) = volume of gas after release of gas in mL
262 mL - 187 mL = 75 mL
(volume of gas after the release of gas in mL) / 1000 mL = .075 L
75mL / 1000 mL = .075 L
The volume of the gas in L was .075 L.
Constant
The R of PV = nRT was 0.0821 atm*L/moles*K.
Plug-in PV = nRT
(Pressure in atm)(volume in L) = n(gas constant in L*atm/mol*K)(Temperature in K)
(0.97 atm)(0.075 L) = n(0.0821 L*atm/mol*K)(296K)
n= (0.97 atm)(0.075 L)/(0.0821L*atm/mol*K)(296K)
n= 0.0030 mole of the gas
The mole of the gas mixture was found to be 0.0030 mole.
Mass of the gas mixture
l(mass of lighter before the release of gas) - (mass of lighter after the release of gas)l = mass of the gas
l17.6 g - 17.8 gl = .2 g
The mass of the gas mixture was found to be .2 g.
Finding Molar Mass
(mass of the gas) / (mole of the gas) = molar mass of the gas
0.2 g / 0.0030 mole = 66.6 g/mole
The molar mass of the gas mixture was found to be 66.6 g/mole.
Finding Plausible Chemical Compound
molar mass of the gas = 12g(carbon ratio) + 1g(hydrogen ratio) = one plausible chemical compound of the gas mixture
For the ratio of carbon and hydrogen, plug-in method was used to determine a plausible chemical compound of the gas mixture.
66.6 g = (12 g)*5 + (1 g)*7 = C5H7
One plausible chemical compound of the gas mixture could be C5H7.
(temperature of water vapor in Celsius) + (273k) = temperature of water vapor in Kelvin
23*C + 273k => 296k
The temperature of the gas in Kelvin was 296k.
Pressure
(atmospheric Pressure) - (pressure of water vapor) = pressure of the gas
760 mmHg - 21mmHg = 739 mmHg
(pressure of the gas in mmHg) / 760 mmHg = pressure of the gas in atm
739 mmHg / 760 mmHg = 0.97 atm
The pressure of the gas in atm was 0.97 atm.
Volume
(volume of full glass bottle in mL) - (volume of water left after release of gas in mL) = volume of gas after release of gas in mL
262 mL - 187 mL = 75 mL
(volume of gas after the release of gas in mL) / 1000 mL = .075 L
75mL / 1000 mL = .075 L
The volume of the gas in L was .075 L.
Constant
The R of PV = nRT was 0.0821 atm*L/moles*K.
Plug-in PV = nRT
(Pressure in atm)(volume in L) = n(gas constant in L*atm/mol*K)(Temperature in K)
(0.97 atm)(0.075 L) = n(0.0821 L*atm/mol*K)(296K)
n= (0.97 atm)(0.075 L)/(0.0821L*atm/mol*K)(296K)
n= 0.0030 mole of the gas
The mole of the gas mixture was found to be 0.0030 mole.
Mass of the gas mixture
l(mass of lighter before the release of gas) - (mass of lighter after the release of gas)l = mass of the gas
l17.6 g - 17.8 gl = .2 g
The mass of the gas mixture was found to be .2 g.
Finding Molar Mass
(mass of the gas) / (mole of the gas) = molar mass of the gas
0.2 g / 0.0030 mole = 66.6 g/mole
The molar mass of the gas mixture was found to be 66.6 g/mole.
Finding Plausible Chemical Compound
molar mass of the gas = 12g(carbon ratio) + 1g(hydrogen ratio) = one plausible chemical compound of the gas mixture
For the ratio of carbon and hydrogen, plug-in method was used to determine a plausible chemical compound of the gas mixture.
66.6 g = (12 g)*5 + (1 g)*7 = C5H7
One plausible chemical compound of the gas mixture could be C5H7.
Conclusion
Using the PV = nRT equation, the molar mass of the unknown gas was calculated to be 66.6 g/mole. To find the molar mass of the gas, moles- along with pressure, volume, and temperature- were founded. The pressure, volume, and temperature of the gas was calculated to be 0.97 atm, 0.075 L, and 296k, respectively. The gas had 0.0030 moles. The mass of the gas was also needed to find the molar mass of the gas. The mass of the gas was found to be 0.2 g. The mass of the gas was divided by the moles of the gas to find the molar mass, which was 66.6 g/mole. Using the molar mass, combinations of carbon and hydrogen were made to propose plausible gaseous compounds in the lighter. C5H7 was proposed to be one plausible gaseous compound, because its molar mass closely matched up with the molar mass from the calculations.
Analysis
1. What is the molar mass of the gas, based on your calculations?
The molar mass of the gas was found to be 66.6 g/mole.
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.
Since the molar mass of the gas is calculated to be 66.6 g/mole, the possible formula of the gas could be C5H7.
3. Most of the gas in the lighter is butane: C4H10. Based on that, what is your percent error? Show your work!!
Percent Error: ((molar mass of C5H7 - molar mass of C4H10) / mass of C4 H10) x 100%
(67 g - 58g) / 58g x 100% = 16 %
The percent error of the lab is 16%.
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.
In the PV = nRT, the temperature number would be 273 lower than the Kelvin number. The mole number would be higher, because of the lower temperature number. The molar mass would become lower, because the increasing mole number is on the denominator side and higher divider results in lower overall number.
b. You forgot to correct the pressure for the vapor pressure of water.
The pressure in PV = nRT would be higher, because it would not be subtracted by 21 mmHg. Since pressure is on the denominator side, number of moles would be high as well. Because higher mole is in the denominator of g/mole, the molar mass of the gas mixture would be lower than the actual molar mass.
c. You had air bubbles in your flask before you released the lighter gas into it.
The air bubble would have mixed in with the gas, changing the result of the lab. The volume of the gas would have been recorded to be higher than the actual volume. The higher volume in PV = nRT equation would result in higher mole number. More number of moles results in lower molar mass number, because mole is on the denominator side. Higher denominator leads to lower number as a whole.
d. The lighter was not completely dry when weighted the second time.
Wet lighter would have affected the weight of the lighter. The mass of the lighter would have been bigger than the actual mass. Mass of the gas in the nominator side of the g/mole, so bigger mass of the gas would result in bigger molar mass. However, the amount of water stuck to the lighter would have been insignificant and probably would not have impacted the overall result much.
5. There are really several gases mixed together in lighter fluid. Bawsed on your results, do the other gases have a higher or lower molar mass than butane? Explain!
Assuming the calculation of the molar mass of the unknown gas is relatively accurate, other gases should have a higher molar mass than butane. Butane’s molar mass is only 58 g/mole, which is lower than the calculated molar mass of the gas, which is 66.6 g/mole. The calculated molar mass of the gas should have the average molar mass of the entire gas, including butane. Therefore, other gases should have higher molar mass to balance butane’s molar mass of 58 g/mole to 66.6 g/mole.
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 lighter fuel is in a high pressure, which allows the fuel to stay in liquid form. When the button is pressed, small leaks are formed. The atmosphere outside the lighter has much lower pressure, so the liquid fuel turns into gas immediately as it leaks out of the container.
The molar mass of the gas was found to be 66.6 g/mole.
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.
Since the molar mass of the gas is calculated to be 66.6 g/mole, the possible formula of the gas could be C5H7.
3. Most of the gas in the lighter is butane: C4H10. Based on that, what is your percent error? Show your work!!
Percent Error: ((molar mass of C5H7 - molar mass of C4H10) / mass of C4 H10) x 100%
(67 g - 58g) / 58g x 100% = 16 %
The percent error of the lab is 16%.
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.
In the PV = nRT, the temperature number would be 273 lower than the Kelvin number. The mole number would be higher, because of the lower temperature number. The molar mass would become lower, because the increasing mole number is on the denominator side and higher divider results in lower overall number.
b. You forgot to correct the pressure for the vapor pressure of water.
The pressure in PV = nRT would be higher, because it would not be subtracted by 21 mmHg. Since pressure is on the denominator side, number of moles would be high as well. Because higher mole is in the denominator of g/mole, the molar mass of the gas mixture would be lower than the actual molar mass.
c. You had air bubbles in your flask before you released the lighter gas into it.
The air bubble would have mixed in with the gas, changing the result of the lab. The volume of the gas would have been recorded to be higher than the actual volume. The higher volume in PV = nRT equation would result in higher mole number. More number of moles results in lower molar mass number, because mole is on the denominator side. Higher denominator leads to lower number as a whole.
d. The lighter was not completely dry when weighted the second time.
Wet lighter would have affected the weight of the lighter. The mass of the lighter would have been bigger than the actual mass. Mass of the gas in the nominator side of the g/mole, so bigger mass of the gas would result in bigger molar mass. However, the amount of water stuck to the lighter would have been insignificant and probably would not have impacted the overall result much.
5. There are really several gases mixed together in lighter fluid. Bawsed on your results, do the other gases have a higher or lower molar mass than butane? Explain!
Assuming the calculation of the molar mass of the unknown gas is relatively accurate, other gases should have a higher molar mass than butane. Butane’s molar mass is only 58 g/mole, which is lower than the calculated molar mass of the gas, which is 66.6 g/mole. The calculated molar mass of the gas should have the average molar mass of the entire gas, including butane. Therefore, other gases should have higher molar mass to balance butane’s molar mass of 58 g/mole to 66.6 g/mole.
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 lighter fuel is in a high pressure, which allows the fuel to stay in liquid form. When the button is pressed, small leaks are formed. The atmosphere outside the lighter has much lower pressure, so the liquid fuel turns into gas immediately as it leaks out of the container.