Post-Lab Report (Use the In-lab observations to complete the laboratory report.
ID: 516371 • Letter: P
Question
Post-Lab Report (Use the In-lab observations to complete the laboratory report. Turn in to your Instructor when you have completed the report.) Sample Sample 2 2.5 (m ID of sample Buret reading of vol. of Hz collected l mL. vol. of H, in graduated portion of buret 22.0 m 8 volume of buret below 50-mL mark L mL mL Total volume of Ha collected 40.2. ml ml. Barometric pressure torr torr 114 mm 305 mm Difference in water levels Correction for difference in water levels 150.21 torr T40.57 130.38 ePressure of gas in buret 120.74 torr Temperature of water bath Water vapor pressure at water bath temp. torr la torr .Pressure of dry hydrogen 13 55 torr 700 torr Volume Ha corrected to STP mL. Moles H2 collected mol mol Mass Mg in sample Show your work for the determination of the volume of gas collected for Sample 1. Include calculatio of volume inside the graduated portion of the buret from buret reading and use of void volume. Experiment 13 Determination of the Mass of Magnesium Metal in a sample Terry on a n sayon IonExplanation / Answer
Total volume of H2 collected = Burette reading of volume of H2 collected ( or, Total Burette reading of 50 mL - Volume of H2 in graduated portion of burette)
Sample 1
Total volume of H2 collected = 17 mL
Sample -2
Total volume of H2 collected = 30.2 mL
Pressure difference in water levels =( Difference in water level x density of water ) / Density of mercury
= (Difference in water level x 1 gm/mL) / 13.5 gm/mL
Correction for difference in water levels = Barometric pressure - Pressure difference in water levels
Now for Sample - 1
Pressure difference in water levels = (174 x 1) /13.5 = 12.89 torr
Correction for difference in water levels = 763 - 12.89 = 750.11 torr
Now for Sample - 2
Pressure difference in water levels = (305 x 1) /13.5 = 22.59 torr
Correction for difference in water levels = 763 - 22.59 = 740.41 torr
Pressure of gas in burette = Correction for difference in water levels - vapour pressure of water
Sample-1
Pressure of gas in burette = 750.11 - 19.83 = 730.28 torr
Sample -2
Pressure of gas in burette = 740.41 - 19.83 = 720.58 torr
Pressure of dry hydrogen = Pressure of gas in burette - Pressure difference in water levels
Sample -1
Pressure of dry hydrogen = 730.28 - 12.89 = 717.39 torr
Sample - 2
Pressure of dry hydrogen = 720.58 - 22.59 = 697.99 torr
For STP correction we need to apply Combined gas law which is,
P1V1/T1 = P2V2/T2
Where P2 = 760 torr V2=? ,T2 = 273 K
Sample - 1
P1 = 717.39 torr
V1 = 17 mL
T1 = 297 K
V2 = (P1 X V1 X T2) / (T1 X P2) = ( 717.39 X 17 X 273) / ( 297 X 760) =14.75 mL
Sample - 2
P1 = 697.99 torr
V1 = 30.2 mL
T1 = 297 K
V2 = (P1 X V1 X T2) / (T1 X P2) = ( 697.99 X 30.2 X 273) / ( 297 X 760) = 25.49 mL
Moles of H2 collected (n) = PV / RT
Sample - 1
P in atm = P in torr / 760 = 717.39 / 760 = 0.94
V = 14.75 /1000 = 0.015 L
Moles of H2 collected (n) = (0.94 x 0.015) / (0.082 x 297) = 0.00058
Sample - 2
P in atm = P in torr / 760 = 697.99 / 760 = 0.92
V = 25.49/1000 = 0.0255 L
Moles of H2 collected (n) = (0.92 x 0.0255) / (0.082 x 297) = 0.00096
Mass of Mg in sample
Reaction is, Mg (s) + 2HCl (aq) –> MgCl2 (aq) + H2 (g)
i.e. 1 mole of H2 is generated by the reaction of 1 mole of Mg
Mass of Mg = Molar mass of Mg X no of Moles
Sample 1
Moles of Mg = 0.00058
Mass of Mg in the sample = 24.31 x 0.00058 = 0.014 gm = 14 mg
Sample 2
Moles of Mg = 0.00096
Mass of Mg in the sample = 24.31 x 0.00096 = 0.0233 gm = 23.3 mg