Yield Lab

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  • Topic: Sodium hydroxide, Mole, Kilogram
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  • Published : December 5, 2006
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Abstract: All the materials were measured and weighed. It was found in the experiment that the yield of copper hydroxide in 40%.

Introduction: The copper (II) sulphate is then placed in 100 mL of distilled water. Then 20 mL of CuSO4 is measured and placed 100 mL of distilled water. This can later be weighed to determine the mol of CuSO4 and the mol/L concentration. Then this was used to find out how many mL of 0.5 NaOH solution is needed to react completely with all the copper (II) sulphate in solution. Then titrate ¼ of the estimated amount number of mL of NaOH needed to react with the sulphate solution. Then add 5 mL at a time NaOH until Cu(OH)2 precipitate does not form. Record the correct number off mL which forms to NaOH solution. Fold the filter paper and stick inside the funnel, and pour the solution of copper (II) hydroxide through the filter. The purpose of this experiment is to determine the yield of copper (II) sulphate reaction.

Reagent preparation:

0.5 mol/L = mol/0.1L
0.5 (0.1L)=mol
mol = 0.05

0.05 x 40 (molar mass)= 2

There are 2 grams of solid sodium hydroxide.

Weigh 100 mL beaker
Transfer anhydrous copper (II) sulphate to that beaker
Heat beaker at level 6 for 3 minutes, stirring occasionally
Removing beaker from balance and weigh after cooling for 3 minutes Turn off heater (or leave for next group)
Measure 100 mL of distilled water into a graduated cylinder and place in 100 mL beaker with the copper (II) sulphate Heat and mix until solid material is all gone
Let cool for 5 minutes
Measure 20 mL of CuSO4 into 200 mL beaker, and add 100 mL of water Calculated mol of CuSO in solution and the mol/L concentration Predict how many mL of 0.5 NaOH solution is needed to react completely with all the copper(II) sulphate in solution Place 200 mL beaker under the titrating burnette

Record the number of mL in the burnette
Titrate quickly ¼ of the estimated number of mL of NaOH needed to react completely with the sulphate in the solution Carefully 5 mL at a time, add NaOH until Cu(OH) precipitate does not form anymore Record the final number of mL in the burnette

Weigh a piece of filter paper
Fold the paper and place in funnel. Place one of the discard solution beakers under the funnel Mix and pour (slowly) the solution of copper (II) hydroxide slurry through the filter When finished collecting the precipitate, use a paper towel to place it on the drying tray, and identify it with your name with tape. Clean your beakers. Pour your NaOH into a collection flask.

The results for the yield lab is:

7(actual mass)
12(estimated mass)
=0.5833333 x 100
=58.3 %

As illustrated on this paper it shows that the percentage we got on this lab is 58.3%. We got this by dividing the actual mass, and our estimated mass.

1.Weight of copper (II) sulphate: 2.13 grams

51.90 (beaker + copper (II) sulphate)
-49.77 (beaker)
2.13 (copper (II) sulphate)

2.Weight of anhydrous sulphate: 2.39 grams

63.11 (beaker + anhydrous sulphate)
- 60.72 (beaker)
2.39 (anhydrous sulphate)

3.Moles of anhydrous sulphate:0.01497 mol

2.39g (weight of anhydrous sulphate)
159.62 g/mol (molar mass copper sulphate)

= 0.01497 mol

4.Mol/L of anhydrous sulphate in 100 mL water: 0.0002994

=0.01497 x 0.01 (liters in water)
=0.000149 mols/L

5. Mol/L of sulphate when 3 is diluted to 100 mL: 0.006 mols

0.02L x 0.3=0.006 mols of NaOH

6. Moles of NaOH needed to react completely:0.06 mols

40 g/mol (molar mass)
= 0.059 mols

7. Number of mL of 0.5 mol/L NaOH: 30 mL
=0.03 x 1000
= 30 mL

8. Calculate the actual used:7 mL
38 (initial)
-31 (final)
= 7 mL

9. Number of moles of NaOH actually used: 0.175 mols
7 (actual)
40 (molar mass)
= 0.175

10. Percentage yield: 58.3 %

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