CCS Day 2 Summary PDF

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CCS Day 2 Lab Summary Michael Burand OSU...

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CCS Day 2 Summary

Day 1 Summary Questions: 1. Show how your group determined the values of a–d for the formula KaFeb(ox)c ·dH2O, including any relevant calculations. a: Trial 1: mass of K

+

g

1 mol H + 9.9 mL = 0.0391g = ( 1000 mL)(0.1009 M )( 1 mol K +)(39.10 mol)

% mass = ( 0.0391g ) 100 = 23.9290% 0.1634g * Trial 2: mass of K

+

9.6 mL H + )(39.10 g ) = ( 1000 )(0.1009 M )(( 11 mol = 0.0379g mL mol K + mol

% mass = ( 0.0379g ) 100 = 23.3231% 0.1625g * ave % mass = ( 23.9290 2+ 23.3231) * 100 = 23.6260% Ratio:

23.62% 39.10g /mol

= 0.6041

0.6041 0.2072

= 2.92 ≈ 3

b: g 1 mol N aOH 9.8 mL = 0.0184g Trial 1: mass of F e + = ( 1000 mL)(0.1009 M )( 3 mol F e+ )(55.85 mol ) 0.0184g

% mass = ( 0.1634g ) * 100 = 11.2607% g 1 mol N aOH mL = 0.0193g Trial 2: mass of F e + = ( 10.3 1000 mL)(0.1009 M )( 3 mol F e+ )(55.85 mol ) 0.0193g

% mass = ( 0.1625g ) * 100 = 11.8769% ave % mass = ( 11.2607 2+ 11.8769) * 100 = 11.5688% Ratio: c:

11.57% 55.85g /mol

= 0.2072

0.2072 0.2072

=1

mL − 1.55 mL)(0.009694 M )( 5 mol K MnO 4)(88.02 g ) Trial 1: mass of ox = ( 33.601000 = 0.06837g mL 2 mol oxalate mol

% mass = ( 0.06837g ) * 100 = 54.1330% 0.13g mL − 0.75 mL)(0.009694 M )( Trial 2: mass of ox = ( 31.601000 mL

5 mol K MnO 4 g )(88.02 mol) 2 mol oxalate

= 0.06581g

% mass = ( 0.06581g ) 100 = 52.2439% 0.1257g * ave % mass = ( 54.1330 2+ 52.2439) * 100 = 53.2439% Ratio:

53.24% 88.02g /mol

= 0.6049

0.6049 0.2072

= 2.92 ≈ 3

d: mass % water = 100 − (53.24 + 11.57 + 23.62) = 11.56% Ratio:

11.56% 18.016g /mol

= 0.6417

0.6417 0.2072

= 3.09 ≈ 3

2. Show the calculation of percent yield of your product. % Y ield = 3.2754 491.25

actual theoretical

* 1 00%

* 100 = 0.67%

Experimental Procedure: Part 1: Determination of the Oxalate by Titration Using an analytical balance, 0.1263 g of green crystals were weighed out and put into an Erlenmeyer flask. About 60 mL of deionized water was added to the flask along with 6.0 mL 6 M H2SO4 and 1.0 mL 85% (concentrated) H3PO4. The solution was heated to 83°C on a hotplate. While the solution was heating, a 50 mL burette was primed and filled with 0.009694 M KMnO4 solution. Once the solution reached 83°C, the titration began and a stir bar was added to the

flask. The solution was kept at a temperature above 60°C for the whole titration, until the solution turned light pink for at least 30 seconds. The final volume was recorded. Process was repeated, however the second trial had 0.1257 g of green crystals weighed out. The number of used moles of MnO4 -and the number of moles of C2 O 42− were calculated, then converted to grams. Lastly, the percent (by mass) of C2 O 42− and average values were calculated. Part 2: Determination of Potassium and Iron via Ion Exchange The ion exchange column was mounted to a ring stand before rinsing the column using 4.0 mL of DI water from a 10 mL graduated cylinder. The solution was collected in a 50 mL beaker and the solution tested acidic using pH paper. Once the liquid fell to the level of the resin another 4.0 mL of DI water was used to rinse the column. This was repeated 2 more times until the pH from the solution was around 7. 0.1634 g of green crystals were weighted out using an analytical balance and placed into a 50 mL beaker. 4.0 mL of DI water was added to the beaker using a 10 mL graduated cylinder. The beaker was swirled for a few minutes to dissolve the green crystals. A clean and dry 150 mL beaker was placed underneath the ion exchange column before the green solution in the 50 mL beaker was poured into the column. The liquid from the column was collected in the 150 mL beaker. The 50 mL beaker was rinsed twice more with 4.0 mL of DI water and the rinse was collected from the column in the 150 mL beaker. A Vernier system and pH probe were set up for titration. The probe was calculated using buffers with pH = 4 and pH = 7. A 50 mL burette was primed using 0.1009 M NaOH and then filled with the NaOH solution. A titration setup with stir bar and hotplate was used. The green solution was titrated with the NaOH solution until the pH reached 11. While the titration was occurring, the ion exchange column was recharged by running HCl through the column and stopping when the liquid tested

acidic using pH paper. Process was repeated, however 0.1625 g of green crystals were weighed out for the second trial. Results:

Data Before Total mass of Product

3.2754 g

KMnO4 Concentration

0.009694 M

NaOH Concentration

0.1009 M

Oxalation Data Table Trial

Mass of green crystals used

Initial Volume of KMnO4

Final Volume of KMnO4

Mass of Oxalate in Sample

1

0.13

1.55 mL

33.60 mL

0.06837

2

0.1257

0.75 mL

31.60 mL

0.06581

mL − 1.55 mL)(0.009694 M )( 5 mol K MnO 4 )(88.02 g ) M ass of oxalate = ( 33.60 1000 = 0.06837g mL 2 mol oxalate mol

K+ and Fe3+ Data Table Trial

Mass of green crystals used

Volume of NaOH for K+

Volume of NaOH for Fe+

Mass of K+

Mass of Fe+

1

0.1634

9.9 mL

9.8 mL

0.0391

0.0184

2

0.1625

9.6 mL

10.3 mL

0.0379

0.0193

9.9 mL H + )(39.10 g ) M ass of potassium = ( 1000 )(0.1009 M )( 11 mol = 0.0391g mL mol K + mol 9.8 mL N aOH)(55.85 g ) M ass of iron = ( 1000 )(0.1009 M )( 13mol = 0.0184g mL mol F e+ mol

Percent by Mass: ● Oxalate: 53.24% ● K+: 23.62% ● Fe3+: 11.57%

● H2O: 11.56% ● Ratio = 3:1:3:3 Conclusion: The oxalate was determined by titration and the potassium and iron by the ion exchange. 53.24% by mass of oxalate was titrated and calculated, then 23.62% by mass for potassium, 11.57% by mass for iron, and 11.56% by mass for water. In order to get the final ratios for the equation, each percent was divided by its molecular weight and then further divided by the smallest value (which was iron). The equation produced was K 3 F e(ox)3 · 3H 2O and the percent yield was 0.67%....