Archer G11
Partner: Judy
30-31 Aug 2011
Gravimetric Analysis of a Metal Carbonate
Purpose – The purpose of this lab is to identify the unknown carbonate. This can be done by finding the
mass of the product carbonate and using stoichiometry on that mass to find the molar mass of the
unknown carbonate. Once this is done, the molar mass of the unknown carbonate can be determined
The significance of this is that it allows chemists to identify unknown substances.
Hypothesis – The hypothesis is that the unknown carbonate can be identified. By finding the mass of the
unknown carbonate, displacing the unknown with calcium, finding the mass of the product carbonate,
and using stoichiometry the unknown metal in the carbonate can be identified.
Materials:
Materials
0.2 M Calcium chloride (CaCl2)
Distilled water
Unknown carbonate sample (M2CO3)
0.0001-g precision balance
600-mL beaker
Bunsen burner
30-mL crucible
Crucible tongs
Drying oven
Funnel
Filter paper
250-mL graduated cylinder
Wire gauze
Clay pipe stem triangle
Ring stand and clamp
Timer
Labeling tags
Watch glass
Micro spatula
Trial 1
Trial 2
375 mL
600 mL
1.7306 g
1.7078 g
1 balance
6 beakers
3 burners
3 crucibles
2 tongs
1 oven
3 funnels
0.6082 g
0.5976 g
3 cylinders
3 gauzes
3 triangles
3 stands and clamps
3 timers
3 tags
3 watch glasses
2 spatulas
Procedures:
1.)
2.)
3.)
4.)
5.)
6.)
7.)
8.)
9.)
Put a pipe stem triangle on a clamp of a ring stand
Put a crucible on the triangle
Gently heat the crucible for a minute using a Bunsen burner
Use a tong to transfer the crucible from the triangle to a wire gauze
Let the crucible cools down
Mass the crucible
Add 2 g of unknown carbonate into the crucible
Find the mass of the unknown carbonate plus the crucible
Move the crucible back to the triangle
Trial 3
1.7576 g
0.6006 g
Archer G11
Partner: Judy
30-31 Aug 2011
10.) Gently heat the crucible for 2-3 minutes
11.) Let the crucible cool down for about 30 seconds to a minute
12.) Mass the crucible with the unknown carbonate
13.) Repeat step 10-12 until the mass until the mass is about the same
14.) Pour the unknown carbonate from the crucible to a 600-mL beaker
15.) Add 200 mL of distilled water to the 600-mL beaker
16.) Swirl the beaker to dissolve the unknown carbonate
17.) Add 125 mL of 0.2 M CaCl2 to the beaker
18.) Wait until the precipitate settles
19.) Weigh a filter paper
20.) Fold the filter paper in a way that will maximize the area for filtering
21.) Put the filter paper in a funnel
22.) Put the funnel in a ring clamp on the ring stand
23.) Put another 600-mL beaker under the funnel
24.) Pour the solution from the first beaker through the filter paper
25.) When nearly all the solutions had been poured, swirl the beaker
26.) Pour the rest through the filter paper
27.) Use some distilled water to wash the beaker
28.) Put the filter paper on the watch glass using 2 micro spatulas
29.) Open the filter paper into a circle
30.) Put a label on the watch glass
31.) Put the watch glass in the drying oven
32.) Wait for a day
33.) Take out the watch glass
34.) Quickly mass the filter paper (using tongs to move it to the balance)
35.) Put it back in the drying oven
36.) Wait for 5 minutes
37.) Repeat step 33-36 for 2 more times
38.) Repeat step 1-37 for trial 2 and 3
Results: The unknown carbonate was a white powder. When it dissolved, it turned the water into a
milky color solution. When CaCl2 was added, the precipitate was not apparent because the precipitate
had the same color as the solution. When the filtrate dried, the filtrate formed a white crumb on the
filter paper. There were some black dots in the filtrate of unknown origin.
Archer G11
Partner: Judy
30-31 Aug 2011
M2CO3 + CaCl2  CaCO3 + 2MCl
CO32-(aq) + Ca2+(aq)  CaCO3(s)
Gravimetric Analysis Table
Trial 1
24.2250
Mass of crucible (g)
26.2175
Mass of crucible + M2CO3 (g)
25.8829
Mass of crucible + M2CO3 (dried) (1st weighing) (g)
25.9410
Mass of crucible + M2CO3 (dried) (2nd weighing) (g)
25.9500
Mass of crucible + M2CO3 (dried) (3rd weighing) (g)
25.9821
Mass of crucible + M2CO3 (dried) (4th weighing) (g)
25.9556
Mass of crucible + M2CO3 (dried) (5th weighing) (g)
1.7306
Calculated mass of M2CO3 (g)
0.6082
Mass of filter paper (g)
2.2464
Mass of filter paper + CaCO3 (1st weighing) (g)
2.2585
Mass of filter paper + CaCO3 (2nd weighing) (g)
2.2648
Mass of filter paper + CaCO3 (3rd weighing) (g)
1.6484
Calculated mass of CaCO3 (g)
0.01647
Calculated moles of CaCO3 (mole)
105.1
Calculated Molar mass of M2CO3 (g/mol)
Na2CO3
Identity of M2CO3
0.86
Calculated percent errors (%)
Trial 2
26.4086
28.4038
28.1780
28.1594
28.1164
1.7078
0.5976
2.1844
2.1941
2.1981
1.5946
0.01593
107.2
Na2CO3
1.14
Trial 3
26.7265
28.7263
28.4928
28.4886
28.4906
28.4841
1.7576
0.6006
2.2467
2.2469
2.2524
1.6481
0.01647
106.7
Na2CO3
0.71
Analysis:
Mass of M2CO3 = (Mass of crucible + M2CO3, last trial) – (Mass of crucible)
Mass of M2CO3 equals to mass of crucible and M2CO3 of the last trial subtract by the mass of crucible
Trial 1: 25.9556 – 24.2250 = 1.7306g M2CO3
Trial 2: 28.1164 – 26.4086 = 1.7078g M2CO3
Trial 3: 28.4841 – 26.7265 = 1.7576g M2CO3
Mass of CaCO3 = [(Mass of filter paper + CaCO3 (1st weighing) + Mass of filter paper + CaCO3 (2nd
weighing) + Mass of filter paper + CaCO3 (3rd weighing)) / 3] – (Mass of filter paper)
Mass of CaCO3 equals to the average mass of CaCO3 minus the mass of the filter paper
Trial 1: [(2.2464 + 2.2585 + 2.2648) / 3] – 0.6082 = 1.6484g CaCO3
Trial 2: [(2.1844 + 2.1941 + 2.1981) / 3] – 0.5976 = 1.5946g CaCO3
Archer G11
Partner: Judy
30-31 Aug 2011
Trial 3: [(2.2467 + 2.2469 + 2.2524) / 3] – 0.6006 = 1.6481g CaCO3
Moles of CaCO3 = (Mass of CaCO3) / (Molar mass of CaCO3)
Moles of CaCO3 equals the mass of CaCO3 divided by the molar mass of CaCO3
Trial 1: 1.6484 / 100.09 = 0.01647mole CaCO3
Trial 2: 1.5946 / 100.09 = 0.01593mole CaCO3
Trial 3: 1.6481 / 100.09 = 0.01647mole CaCO2
Molar mass of M2CO3 = (Mass of M2CO3) / (Moles of CaCO3)
Molar mass of M2CO3 equals to the mass of M2CO3 divide by the moles of CaCO3 (moles of CaCO3 =
moles of M2CO3 due to the mole ratio)
Trial 1: 1.7306 / 0.01647 = 105.1 g/mol
Trial 2: 1.7078 / 0.01593 = 107.2 g/mol
Trial 3: 1.7576 / 0.01647 = 106.7 g/mol
Identity of M = [(Molar mass of M2CO3) – (Molar mass of CO3)] / 2
Identity of M equaled the molar mass of M2CO3 minus the molar mass of CO3 then divided by 2
Trial 1: (105.1 – 60.01) / 2 = 22.50 = Na = Na2CO3
Trial 2: (107.2 – 60.01) / 2 = 23.60 = Na = Na2CO3
Trial 3: (106.7 – 60.01) / 2 = 23.35 = Na = Na2CO3
Percent Error = |(Molar mass of M2CO3) / (Molar mass of Na2CO3) – 100%|
Percent error is equal to the absolute value of the molar mass of M2CO3 divide by the molar mass of
Na2CO3 minus 100%
Trial 1: 105.1 / 105.99 – 100% = 0.86 %
Trial 2: 107.2 / 105.99 – 100% = 1.14 %
Trial 3: 106.7 / 105.99 – 100% = 0.71 %
The hypothesis has been proven to be true, according to the results above. The results show that the
unknown carbonate is actually sodium carbonate. Since the metal carbonate had been identified
correctly, the hypothesis on this lab is correct. The unknown carbonate could have been magnesium
carbonate. However, it is shown that the unknown metal in the unknown carbonate is an alkali metal, by
the subscript 2 after M. That is the reason that the unknown carbonate had been identified as Na2CO3.
Archer G11
Partner: Judy
30-31 Aug 2011
Conclusion: The results confirmed the hypothesis with an average of 99.1% accuracy. However, some
errors should be corrected in the future to increase the accuracy of the results. Some errors could have
occurred that caused the results to be slightly inaccurate. The alkali metal carbonates are hydroscopic so
it was heated in order to remove the water. However, the flame intensity used during this experiment
could have been too intense thus causing the substance to react with oxygen and form some new
compound. The inclusion of oxygen in the product’s mass could have caused an increase of molar mass
of the unknown carbonate. Not only that, the stirring rod was not used to direct the solution down to
the filter paper and so some of the solution might had pass through the side of the filter paper.
Therefore, the mass of the CaCO3 determined would be lower and so the molar mass of the unknown
carbonate would become higher. This is because the molar mass was determined by dividing the mass
of the unknown carbonate with the moles of carbonate, so if the mass of carbonate decreases, so does
the moles of carbonate which causes the mass of the unknown carbonate to be divide by a lower
number thus the molar mass increases. To prevent these errors from happening in the future, the flame
intensity should always be low when removing water and some equipments such as stirring rods should
be use to make sure that the solution is filtered properly.
Archer G11
Partner: Judy
30-31 Aug 2011