6-1 Ideal Gas Law Constant Lab Kristen Brown

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Southern New Hampshire University *

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Chemistry

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Jan 9, 2024

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Determination of Ideal Gas Law Constant S tudent: Kristen Brown Date: 11/30/23 Activity 1 If your concentration, moles, or R calculations are incorrect or your calculation work does not actually result in the desired final units, or units are missing/incomplete, or cancelled incorrectly, you will lose points. Be sure that every number you include has its associated unit. Email me if you are unsure or need assistance. Data Table 1 Trial 1 (1 mL H 2 O 2 ) Trial 2 (1 mL H 2 O 2 ) Trial 3 (2 mL H 2 O 2 ) Trial 4 (2 mL H 2 O 2 ) Trial 5 (3 mL H 2 O 2 ) Trial 6 (3 mL H 2 O 2 ) Trial 7 (4 mL H 2 O 2 ) Trial 8 (4 mL H 2 O 2 ) Air temperature (°C) 25°C 25°C 25°C 25°C 25°C 25°C 25°C 25°C Volume H 2 O 2 liquid (mL) 1.0mL 1.0mL 2.0mL 2.0mL 3.0mL 3.0mL 4.0mL 4.0mL Initial Volume Gas (mL) 0.0mL 0.2mL 0.0mL 0.3mL 0.0mL 1.0mL 0.0mL 1.2mL Final Volume Gas (mL) 9.2mL 11.3mL 18.5mL 19.4mL 32.5mL 34.8mL 40.0mL 41.9mL ΔV (mL) 9.2mL- 0.0mL = 9.2mL 11.3mL- 0.2mL = 11.1mL 18.5mL - 0.0mL = 18.5mL 19.4mL - 0.0mL = 19.1mL 32.5mL - 0.0mL = 32.5mL 34.8mL - 1.0mL = 33.8mL 40.0mL - 0.0mL = 40.0mL 41.9mL - 1.2mL = 40.7mL ©2016 2 Carolina Biological Supply Company
2021 Activity 2 Data Table 2 Show work for determining Concentration of H 2 O 2 in the box below. Concentration should be mol/L. You need to begin your calculation using the provided 3%m/v H 2 O 2 concentration . Concentration H 2 O 2 (mol/L) 30.0g H 2 O 2 / 1000mL 30.0g H 2 O 2 x 1.0mol H 2 O 2 / 34.01g H 2 O 2 = 0.88mols H 2 O 2 / L Trial 1 1 mL H 2 O 2 Trial 2 1 mL H 2 O 2 Trial 3 2 mL H 2 O 2 Trial 4 2 mL H 2 O 2 Trial 5 3 mL H 2 O 2 Trial 6 3 mL H 2 O 2 Trial 7 4 mL H 2 O 2 Trial 8 4 mL H 2 O 2 Mole s H 2 O 2 0.0009mol H 2 O 2 0.0009mol H 2 O 2 0.0018mol H 2 O 2 0.0018mol H 2 O 2 0.0026mo l H 2 O 2 0.0026mo l H 2 O 2 0.0035mol H 2 O 2 0.0035mol H 2 O 2 Mole s O 2 * 0.00045mo l O 2 0.00045mol O 2 0.0009mol O 2 0.0009mol O 2 0.0013mo l O 2 0.0013mo l O 2 0.00175mo l O 2 0.00175mo l O 2 ΔV (L) 1L = 1000mL 9.2mL /1000 = 0.0092L 1L = 1000mL 11.1mL /1000 = 0.0111L 1L = 1000mL 18.5mL /1000 = 0.0185L 1L = 1000mL 19.1mL /1000 = 0.0191L 1L = 1000mL 32.5mL /1000 = 0.0325L 1L = 1000mL 33.8mL /1000 = 0.0338L 1L = 1000mL 40.0mL /1000 = 0.04L 1L = 1000mL 40.7mL /1000 = 0.0407L *Hint: Use reaction stoichiometry/molar ratio to solve for moles of O 2 . Show work for determining moles of H 2 O 2 for Trial 1 here: 1.0mL H 2 O 2 x 1L/1000mL x 0.88mols H 2 O 2 /1L = 0.00088mols H 2 O 2 Show work for determining moles of O 2 for Trial 1 here: 0.0009mols H 2 O 2 x 1.0mol O 2 /2.0mols H 2 O 2 = 0.00045mols O 2 Show work for determining moles of H 2 O 2 for Trial 3 here: 2.0mL H 2 O 2 x 1L/1000mL x 0.88mols H 2 O 2 /1L = 0.00176mols H 2 O 2 Show work for determining moles of O 2 for Trial 3 here: 0.0018mols H 2 O 2 x 1.0mols O 2 /2.0mol H 2 O 2 = 0.0009mols O 2 Show work for determining moles of H 2 O 2 for Trial 5 here: 3.0mL H 2 O 2 x 1L/1000mL x 0.88mols H 2 O 2 /1L = 0.00264mols H 2 O 2 Show work for determining moles of O 2 for Trial 5 here: 0.0026mols H 2 O 2 x 1.0mol O 2 /2.0mols H 2 O 2 = 0.0013mols O 2 © 2016 Carolina Biological Supply Company
2021 Show work for determining moles of H 2 O 2 for Trial 7 here: 4.0mL H 2 O 2 x 1L/1000mL x 0.88mols H 2 O 2 /1L = 0.00352mols H 2 O 2 Show work for determining moles of O 2 for Trial 7 here: 0.0035mols H 2 O 2 x 1.0mol O2/2.0mols H 2 O 2 = 0.00175mols O 2 Insert a copy of your graph. Your graph must include the equation of the line displayed, your name and date in the title. You cannot complete the rest of this lab without this graph and the equation of the line. See the Week 6 Announcement for an example. Don’t forget to set the y-intercept to 0. Graph: 0 0 0 0 0 0 0 0 0 0 0 0.01 0.01 0.02 0.02 0.03 0.03 0.04 0.04 0.05 f(x) = 23.45 x Volume of O2 vs Moles O2 Kristen Brown 12/2/2023 Moles O2 Vol. O2 (L) Show work, including all units , for the calculations/conversions for Air Temperature, Gas Constant R, and Percent Error in the boxes below. Calculations must include complete units to be correct. This includes the correct units for the slope. Air Temperature (K) Show work in the space provided. 25°C + 273.15 = 298.15K Air Pressure (atm) Show work in the space provided. 1 atm Equation of the Line Y = 23.451x © 2016 Carolina Biological Supply Company
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2021 Gas Constant R Show work in the space provided. 23.451 * l/mols * (1 atm) /298.15 k = 0.0787 L(atm)/mols(K) Percent Error Show work in the space provided. Use 0.0821 L*atm/mol*K in your calculation (|0.0821 L atm/mol K−0.0787 L atm/mols K| /0.0821 L atm/mol K) * 100 = 4.14% error Question 1: Would the volume of oxygen that is generated be affected if a smaller mass of yeast were used? Why or why not? Yes, if less yeast is used, there would be a smaller amount of catalase present to perform as the catalyst to the decomposition of H 2 O 2 . This would directly result in a smaller amount of O 2 being produced. Question 2: Identify at least two potential sources of error in the experiment. Are any assumptions made that would add to the experimental error? Some potential sources of error would be the scale having trouble reading such low numbers. Another would be how much gas is lost between the time of adding H 2 O 2 and putting the cap back on the bottle. Lastly, another potential error could be if any residual liquid is still in the container from the previous trials. © 2016 Carolina Biological Supply Company