experiment 6 : convert elements, Lab Reports of Chemistry

Download experiment 6 : convert elements and more Lab Reports Chemistry in PDF only on Docsity! Name: Tiffany Te-L-Peng Lab Instructor: Md Reazul Islam Post Lab Report Questions (submit to instructor when lab report is due) I. Show Calculations (include units) Part A ● Volume of NaOH(aq) used in first completed Trial: ○ 26.50mL - 0.1mL = 26.40 mL ● Molarity of NaOH(aq) calculated from first completed Trial: ○ 0.5067 g/KHP × 1 mole/ 204.22 g/KHP × 1 mole/ 1 mole of KHP × 1/0.026 L = 0.0939829 mol/L ~ 0.09398 mol/L ● Average molarity of NaOH(aq): ○ 0.149512 mol/L ~ 0.1495 mol/L Part B ● Volume of NaOH(aq) used in first completed Trial: ○ 16.50 mL - 0.00 mL = 16.50 mL ● Molarity of CH3COOH(vinegar) calculated from first completed Trial: ○ C1V1=C2V2 ○ 0.1495 × 16.5 = C × 5 = 0.49335 mol/L ~ 0.493 mol/L ● Average molarity of CH3COOH(vinegar): ○ 0.705640 mol/L ● Mass percent CH3COOH in the vinegar: ○ 5% II. Additional Questions 1. What is the percent error in your mass % CH3COOH compared to the label-reported value? ● The percent error would be 16.1%. 2. Does your calculated mass % CH3COOH support selling the vinegar as labeled? Why or why not? ● Yes because the mass % CH3COOH was 5% on the label. 3. A student, Pat, forgot to add phenolphthalein indicator before starting to add titrant to the diluted vinegar. After titrating a while Pat added three drops of indicator; the solution remained colorless. Pat asks you whether it’s okay to continue or whether to start over. What’s your advice? Why? ● I would advise Pat’s best option would be to restart her current trail from the beginning. There seems to be no possibility for the solution to have the color without the indication. She will wait here for a few moments and receive nothing. To identify the color, you’ll need the indicator. 4. If dilute sulfuric acid (H2SO4) had been used in Part B, instead of CH3COOH(vinegar), calculate the molarity of the H2SO4(aq). Assume 5.00 mL of sulfuric acid was used and your titration volumes were as in Part B. ● CH3COOH= NaOH=(1.21mol/L)*(5mL/1000mL)=0.00605 mol ● Average volume = (29.13mL + 29.39mL)/2 = 29.26mL = 29.26x10^-3 L ● NaOH average = (0.00605 mol/ 29.26x10^-3 L) = 0.2068 M ● H2SO4 average = (0.602M + 0.604 M)/2 = 0.603 M