Procedure
1. Fill up the tub with a decent amount of water.
2. Find out the amount of water the glass bottle holds in milliliters by filling the bottle up with water using graduated cylinders.
3. Measure the mass of the lighter in grams with a triple beam balance.
4. Using a thermometer, find out the temperature of the water in the tub.
5. After recording all the data above, proceed to placing the glass bottle in the tub without the cap.
6. Fill up the glass bottle with water. Make sure to tap the glass gently with fingers to get all the air bubbles out. It may take awhile.
7. Once that is accomplished, get the lighter and release gas into the glass bottle by pressing the trigger.
8. Bubbles will form into the bottle, so try to not let them escape from the bottle. Continue to release gas into the bottle until it gets up to about a third of the bottle.
9. Secure the bottle with the cap tightly while it's still submerged in the water.
10. Take out the bottle and dry it up with a paper towel. Dry the lighter as well completely (make sure to tap it a few times to drain the water out of the lighter.
11. After drying the bottle, open the bottle and pour what water is left into the graduated cylinders to calculate the volume of the water without gas.
11. Plug in volume and temperature into PV=nRT. For temperature, assume the temperature of the water. For pressure, assume the atmospheric pressure in the room is standard pressure (1 atm or 760 mmHg) and subtract the water vapor pressure from it to find the pressure of gas. (Patm = Pwatervapor + Pgas). For the volume, take the initial volume of the water in the bottle and subtract it from the amount of water without the gas inside the jar. Make sure to convert the milliliters to liters. This would solve for moles.
12. Subtract the mass of the lighter after submersion from the mass of the lighter before submersion. This solves for grams of gas.
13. Plug in the difference in mass over the amount of moles in order to calculate the molar mass of butane. (gives g/mol)
2. Find out the amount of water the glass bottle holds in milliliters by filling the bottle up with water using graduated cylinders.
3. Measure the mass of the lighter in grams with a triple beam balance.
4. Using a thermometer, find out the temperature of the water in the tub.
5. After recording all the data above, proceed to placing the glass bottle in the tub without the cap.
6. Fill up the glass bottle with water. Make sure to tap the glass gently with fingers to get all the air bubbles out. It may take awhile.
7. Once that is accomplished, get the lighter and release gas into the glass bottle by pressing the trigger.
8. Bubbles will form into the bottle, so try to not let them escape from the bottle. Continue to release gas into the bottle until it gets up to about a third of the bottle.
9. Secure the bottle with the cap tightly while it's still submerged in the water.
10. Take out the bottle and dry it up with a paper towel. Dry the lighter as well completely (make sure to tap it a few times to drain the water out of the lighter.
11. After drying the bottle, open the bottle and pour what water is left into the graduated cylinders to calculate the volume of the water without gas.
11. Plug in volume and temperature into PV=nRT. For temperature, assume the temperature of the water. For pressure, assume the atmospheric pressure in the room is standard pressure (1 atm or 760 mmHg) and subtract the water vapor pressure from it to find the pressure of gas. (Patm = Pwatervapor + Pgas). For the volume, take the initial volume of the water in the bottle and subtract it from the amount of water without the gas inside the jar. Make sure to convert the milliliters to liters. This would solve for moles.
12. Subtract the mass of the lighter after submersion from the mass of the lighter before submersion. This solves for grams of gas.
13. Plug in the difference in mass over the amount of moles in order to calculate the molar mass of butane. (gives g/mol)
Conclusion
In this lab, the purpose was to figure out the molar mass of butane in a lighter by collecting the gas over water. Using Patm = Pwatervapor + Pgas, one was able to solve for the pressure of gas. Taking that pressure and the subtracted volume of water and the temperature of the water, plug it into the formula PV=nRT to solve for moles. To find the grams of gas, take the mass of the lighter without gas and subtract it from the lighter with gas. Afterwards, set it up as grams over moles with the calculations done and that should give the molar mass. With that molar mass, there could be plausible formulas for the chemical.
Analysis Questions
1. The molar mass of the gas based on calculations is 141.1 grams per mole.
2. Based on the molar mass, a possible formula for the gas is C11H4.
3. The percent error is 143%
2. Based on the molar mass, a possible formula for the gas is C11H4.
3. The percent error is 143%
4. A. If one forgot to change Celcius temperature to Kelvin for calculations, the mole value would be too high. This is because in the formula PV=nRT, the temperature required to solve correctly is in Kelvin. Kelvin is 273 degrees greater than Celcius. With the number of grams divided by a larger number of moles, the molar mass would be lower.
B. If one forgot to correct the pressure for the vapor pressure of water, the value of the pressure of the gas would be incorrect and would be higher. This is because nothing else is being added to equal the total pressure in Dalton's Law of Partial Pressure.
C. If there were air bubbles in the flask before lighter gas was released into it, the volume calculated from the water would be too high compared to the true value.
D. If the lighter was not completely dry when weighed the second time, the mass would be greater than the true value because it measure the mass of the water as well.
5. Based on the results, the other gases have a higher molar mass than butane because according to calculations of this experiment, it produced a molar mass of 141 g/mol of butane. The true value of butane is 58 g/mol. The experimental value of gas compared to the accepted value of the gas has a difference of 83 g/mol which means the other gases has a higher molar mass.
6. when the substance inside the lighter is released, it releases a gas because when the trigger is released, the liquid goes through a phase change due to the heat going into the system (endothermic).
B. If one forgot to correct the pressure for the vapor pressure of water, the value of the pressure of the gas would be incorrect and would be higher. This is because nothing else is being added to equal the total pressure in Dalton's Law of Partial Pressure.
C. If there were air bubbles in the flask before lighter gas was released into it, the volume calculated from the water would be too high compared to the true value.
D. If the lighter was not completely dry when weighed the second time, the mass would be greater than the true value because it measure the mass of the water as well.
5. Based on the results, the other gases have a higher molar mass than butane because according to calculations of this experiment, it produced a molar mass of 141 g/mol of butane. The true value of butane is 58 g/mol. The experimental value of gas compared to the accepted value of the gas has a difference of 83 g/mol which means the other gases has a higher molar mass.
6. when the substance inside the lighter is released, it releases a gas because when the trigger is released, the liquid goes through a phase change due to the heat going into the system (endothermic).