Lab Report 4 Dearing
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Name: Eric Dearing
College ID: 0550796
Thomas Edison State College
General Chemistry I with Labs (CHE-121)
Section no.:
Semester and year: May 2014
LABORATORY REPORT
Laboratory Assignment for Module 4: Gravimetric Analysis (eScience Lab 11)
I.
PURPOSE (10 POINTS)
Perform gravimetric analysis using precipitation, double displacement reaction CaCl
2
+ K
2
CO
3
→ 2KCl + CaCO
3
.
Utilize data to determine limiting reagent, theoretical yield, actual yield, and percentage of yield.
II.
TEST DATA (15 POINTS)
Trial 1
Mass of weigh boat: 2.2g
Mass of same weigh boat and CaCl
2
(
Note:
This sample will be set aside and exposed to air for 24 hours): 4.2g
Mass of same weigh boat and CaCl
2
after 24 hours: 5.0g Observations of CaCl
2
: CaCl
2
is wet, it has an accumulation of water over the 24hr period, total 0.7g in mass gained
Mass of 50 mL beaker: 6.5g
Mass of same beaker and K
2
CO
3
: 9g Amount of time beaker solution stirred: 4 min
Amount of time beaker solution set: 15 min
Mass of filter paper and watch glass: 36g
Mass of same filter paper, watch glass, and dried product: 37.7g, 1.7g of compound
Trial 2
1
Mass of 250 mL beaker: 17.2g
Mass of same beaker and CaCl
2
: 19.2g
Observations of CaCl
2
: Looks dry/normal
Mass of 50 mL beaker: 6.5g
Mass of same beaker and K
2
CO
3
: 9g
Amount of time beaker solution stirred: 4 min
Amount of time beaker solution set: 15 min
Mass of filter paper and watch glass: 36g
Mass of same filter paper, watch glass, and dried product: 37.7g, 1.7g of compound
III. CALCULATIONS (15 POINTS)
Trial 1
Mass of CaCl
2
: 2.7g
Mass of K
2
CO
3
: 2.5g
Mass of product: 1.7g
Mass of water absorbed with the CaCl
2
after 24 hours: 0.7g
Trial 2
Mass of CaCl
2
: 2.0g
Mass of K
2
CO
3
: 2.5g
Mass of product: 1.7g
IV.
RESULTS (20 POINTS)
Trial 1
Theoretical yield (CaCO
3
): 2.7g CaCl
2
/ 111amu = 0.024 mol x 1/1 = 0.024 mol x 100amu = 2.4g CaCO
3
Actual yield (CaCO
3
): 1.7g
Percent yield: 1.7g/2.4g x 100 =
71%
Moles of Ca
+2
present in original solution: 0.006 mol
2
Mass of CaCl
2
present in original solution: 2.7g
Percent water absorbed by CaCl
2
: 0.7g / 2.0g x 100 = 35% water absorbed
Trial 2
Theoretical yield (CaCO
3
): 2.0g CaCl
2
/ 111amu = 0.018 mol x 1/1 = 0.018 mol x 100amu = 1.8g CaCO
3
Actual yield (CaCO
3
):
1.7g
Percent yield: 1.7/1.8 x 100 = 94%
Moles of Ca
+2
present in original solution: 0.006 mol
Mass of CaCl
2
present in original solution: 2.0g
V.
CONCLUSION (10 POINTS)
CaCl
2
absorbed 0.7g the increase in water raises the theoretical yield. In actuality the yield is the same as CaCl
2
that was not left in the air for 24hrs.
VI.
QUESTIONS (30 POINTS)
1.
Write the balanced reaction equation for the precipitation of calcium carbonate from potassium carbonate and calcium chloride.
(6 points)
CaCl
2
+ K
2
CO
3
→ 2KCl + CaCO
3
2.
Using this balanced equation, determine the limiting reactant if 15 grams of calcium chloride reacted with 15 grams of potassium carbonate.
(7 points)
CaCl
2
,15g/111amu = .14 mol x 1/1 = .14 mol CaCO
3
K
2
CO
3
, 15g/138amu = .11 mol x 1/1 = .11 mol CaCo
3
K
2
CO
3
is limiting reactant
3.
Using your answer for Question 2, determine the mass of potassium carbonate needed to fully precipitate all the calcium from a 25 mL sample of 15% calcium chloride solution. (
Note:
this is the same concentration as the solution you used in Experiment 1.) (7 pts)
15% of 25ml = 3.75ml = 3.75g x 111amu = 0.034 moles CaCl
2 x 1/1 = 0.034 mol K
2
CO
3
x 138amu
= 4.7g K
2
CO
3 required
4.
Create a pie chart representing the percent by mass of calcium, chloride, and water in the 3
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original solution for each trial.
(10 points)
26%
27%
47%
Test 1
H20
Ca
Cl2
36%
64%
Test 2 Ca
Cl2
4
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