Lab Report Help

Created By: Andrew Weeldreyer
Mrs. Wolf Period:3

AP Chem
Lab Report: Determination of the Molar Volume of a Gas

Objective: To react Mg and HCL and form hydrogen gas, then after collecting lab data, determine the hydrogen gas’s molar volume at STP through calculation. Materials: -LabQuest data collector - 3.0 M of HCl solution - Ribbon of Mg metal - Gas pressure sensor - Temperature probe - 600 mL beaker - 125 mL Erlenmeyer flask - 20 mL gas syringe - Rubber stopper with two-way valve - Plastic tubing - Analytical balance Procedure: 1.Retrieve an Erlenmeyer flask then determine its volume. 2. Obtain a Mg ribbon from teacher then make sure to polish until sample is a metallic silver then measure ribbon on a analytical balance then place in flask, afterwards record mass in your data. 3. Fill a large beaker with room temperature water. 4. Start up LabQuest and attach the pressure and temperature probe. 5. Connect a rubber stopper to the pressure probe and later to the flask with valve closed so that absolutely no hydrogen gas will escape. 6. Posses a five mL sample 3.0 M HCl solution and place in a syringe. Connect syringe to the two-way valve on stopper 7. Place the Erlenmeyer flask in the water bath, and begin to collect your data. 8. After a handful of seconds open up your two-way valve and add the HCl solution to the flask, then immediately close the valve. 9. Swirl flask with hands in the water bath until reaction and data collection has concluded. Data Table: | Trail One | Trail Two | Mass of Mg (g) | 0.012 g | 0.014 g | Volume of flask (mL) | 146.0 mL | 146.8 mL | Maximum Pressure (atm) | 1.07 atm | 1.09 atm | Initial Pressure (atm) | 0.991 atm | 0.990 atm | Change in Pressure (atm) | 0.079 atm | 0.1 atm | Temperature (K) | 296 K | 295 K | Vapor pressure in water (atm) | 0.0277 atm | 0.0261 atm |

Analysis: I) Pre-Lab Questions A. Use Dalton’s law and the vapor pressure of water at 22 C to calculate the partial pressure of hydrogen gas collected. (746 mmHg)-(19.8 mmHg) = 726 mmHg Hydrogen gas B. Use the combined gas law to calculate the “corrected” volume of hydrogen gas collected. (726 mmHg)*(.031 L) = (760 mmHg)*(V2) (295 K) (273 K) 6144.138 = 224200(X) V2= 0.027 L of H2 gas C. What is the theoretical number of moles of hydrogen that can be produced from 0.028 g of Mg? 0.028 g Mg x 1 mol Mg x1 mol H2 = .0012 mol H2 24.305 g 1 mol Mg D. Divide the corrected volume of hydrogen by the theoretical number of hydrogen to calculate the molar volume (L/mol) of hydrogen at STP. 0.027 L H2 0.0012 mol H2 = 22.5 L/mol H2 at STP II) Post Lab Questions 1. Calculate the mass of each piece of magnesium that you used if an analytical balance was unavailable? 0.012 g Mg and 0.014 g Mg 2. Calculate the number of moles of each piece of magnesium that you used. 0.012 g Mg x 1 mol Mg = 4.9 x 10.4 mol Mg for trial one. 24.305 g 0.014 g Mg x 1 mol Mg = 5.8 x 10-4 mol Mg for trial two. 24.305 3. Calculate the theoretical number of moles of hydrogen gas produced in each trial. 4.9 x 10-4 mol Mg x 1 mol H2 = 4.9 x 10-4 mol H2 for trial one. 1 mol Mg 5.8 x 10-4 mol Mg x 1 mol H2 = 5.8 x 10-4 mol H2 for trial two. 1 mol Mg

4. Use your results to calculate the molar amount of hydrogen gas that was produced in your reactions. Trial 1. (0.079-0.0277 atm)(0.146 L) = (n)(.0821)(296 K) 1 0.0082 = 24.3(n) n = 3.08x10-4 0.146L = V2 3.08x10-4mol 1 mol V2=473 L (0.0513 atm)(474 L) = (1 atm)(V2) 296 K 273K 296(V2)=6634.3 V2 = 22.4 L at STP for trial one Trial 2. (0.1-0.0277 atm)(0.1468 L) = (n)(.0821)(295 K) 1 0.0106 = 24.2(n) n = 4.38x10-4 0.1468L = V2 4.38x10-4mol 1 mol V2=335 L (0.0723 atm)(335 L) = (1 atm)(V2) 295 K 273K 295(V2)=66312.2 V2 = 22.4 L at STP for trial two 5. Compare your data calculated molar volume, at STP, with the accepted molar volume of an ideal gas at STP, 22.4 L/mol. Calculate the percent error for each trial. For both of the trials there was a 0% error. Calculations for % error= 22.4-22.4 = 0% for trials one and two. 22.4 6. A student fails to polish each sample of Mg ribbon before massing them. What effect does this error have on the calculated molar volume of hydrogen gas? Mathematically justify your answer. The mass of the Mg ribbon would be higher than it should which would lead to the moles of Mg and H2 (because they both have a 1:1 mole ratio) to be higher then it should. Then when you divide liters by moles in the molar volume calculation you will be dividing by a larger mole number, thus producing a smaller number for the molar volume. Which in turn would give lower liters per mole ratio and state that there is less H2 gas then there actually is. Molar Volume Calculation for distorted trial one =0.011 L H2 = 15.9m/L (4.9 x 10-4 + .0002 mol of distortion)

7. A different student fails to insert the stopper into the flask while determining the available volume of the 125 mL flask. What effect does this error has on the calculated molar volume? Mathematically justify your answer. Some of the H2 gas would be let out, thus in turn would lower the volume of the gas and giving it a lower molar volume. Molar Volume Calculation for distorted trial two: (0.013-0.002 L of gas let out) 5.8 x 10-4 = 19.0 m/L Summary: What was basically supposed to be done in this lab was we were to react a ribbon of magnesium and a HCl solution inside of a Erlenmeyer flask to form H2 gas. While we were doing this we were to collect all different types of data varibles such as temperature and pressure. During the performance of the lab we had to make sure that we set up the sensors, syringe and rubber stopper all correctly, otherwise errors such as gas escaping or an incorrect pressure reading could all lead to the molar volume calculation being distorted. Some things that could have been learned by doing this lab would be using gas equations and some lab data you can project a molar volume for literally any gas. Also I could have learned to operate many different lab equipment that I have never had any experience using before, such and the syringe and learning to correctly use a two-way rubber stopper. The data I was given because I wasn’t there to perform the lab gave me a zero percent error when I worked the numbers through the calculations, but as I previously stated there could be many sources of error including not polishing of the magnesium, adding to much of the HCL solution, not using the syringe correctly, and possibly letting out some of the H2 because someone might have accidentally left the two way valve open. Overall this looks like a lab were a tremendous amount of vital skills could have been learned, and I’m severely disappointed that I had to miss it.