• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6

# Determining the mass of calcium carbonate obtained

Extracts from this document...

Introduction

Lab Experiment 2: Determining the mass of calcium carbonate obtained Purpose: The purpose of the experiment was to investigate the mass of calcium carbonate obtained from the reaction between calcium chloride and sodium carbonate. Apparatus: - Three beakers (250 - ml) - Spatula - Balance � 0.1g - Filtration setup - Filter paper - Stirring rod - Plastic wash bottle Materials: - Sodium Carbonate - Calcium chloride - Distilled water Procedure 1. Weigh out 4.0g of calcium chloride (111g/mol) and dissolve in enough distilled water. 2. Weigh out 6.0g of sodium carbonate (106g/mol) and dissolve in enough distilled water. 3. Pour the sodium carbonate solution into the beaker containing calcium chloride solution. 4. Stir the mixture. Set up the filtration apparatus. Weigh the filter paper and then filter the mixture. Rinse the beaker and empty the contents in the funnel. Wash the precipitate with distilled water several times. 5. Place the filter paper with the precipitate and leave it to dry out. After it is completely dry, then weigh the dry filter paper with the precipitate. ...read more.

Middle

Molar mass Having looked at the mole ratio, it is apparent that since calcium chloride has the lowest number of moles present, it is therefore the limiting reagent. The limiting reagent calcium chloride is therefore used to calculate the theoretical mass of calcium carbonate that can be obtained: Theoretical yield = number of moles of limiting reagent x mass of calcium . carbonate = 0.03604 x (40.08 + 12.01 + 16 + 16 + 16) = 0.03604 x 100.09 = 3.6g 3. Therefore theoretically the mass of the calcium carbonate that can be obtained is 3.6g. The theoretical yield assumes that everything reacts perfectly, and we are able to recover everything 100%. These ideal conditions are rarely present and so we would expect the actual yield to be less than the theoretical yield for this reason. To calculate the experimental mass, the following calculation is done: Experimental mass = Mass of filter paper with the precipitate - Mass of filter paper = 5.1g - 2g = 3.1g As expected the experimental mass is lower than the theoretical mass. ...read more.

Conclusion

In practise the theoretical yield based on the balanced chemical equation is never achieved owing to impurities in reagents, side reactions and other sources of experimental error. The possible sources of error in this experiment may include: - Material used may have been tampered with and so would affect the overall results. - Wrong measurements were taken. - Error arrising from human judgement. - The balance only recorded 2decimal points. - The filter paper may not have been left long enough to dry. A possible modification to this experiment would be to make the sodium carbonate the limiting reageant rather then the calcium chloride as it was in this case. This would be done so that we would have a smaller number of moles of sodium carbonate then calcium chloride. Although my experiment was successful, many improvement could have been made to both my experiment and too the experiment. This includes: - Repeating the measurements for more trials so that more accurate answers could be found. - Using an accurate method to measure the mass, so as to reduce the errors in the experiment. - Make sure that none of the compound is accidentally spilled out. - Use larger quantities so to reduce the error in their recording ?? ?? ?? ?? ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Chemistry essays

1. ## Can one determine the coefficients of a balanced chemical equation by having the mass ...

7. Record the properties of the copper(ii) chloride solution. This will allow one to identify the change in the solution in the reaction and to determine that there is a displacement going on within the reaction. Record this in "Data Table 1 - Qualitative Data Table" and record it under "Before Reaction".

2. ## How much calcium carbonate is in an eggshell

The amount of acid that reacted with the eggshell can be calculated by subtraction. 6.45 ml-4.07 ml=2.38 ml acid. 2.38 ml of 1.0 M HCl contains 2.38 x 10-3 moles HCl. Using the equation: 2HCl + CaCO3 --> CO2 + H2O + CaCl2, the amount of CaCO3 that will react with 2.38 x 10-3 moles of HCl can be determined.

1. ## The aim of the experiment is determining the percentage yield of the product (copper), ...

Coil the strip loosely to fit into the copper chloride solution in the beaker (strip is entirely immersed). 3. Mix the solution using the stirring rod for around five minutes, and notice that the bright blue color of the copper changes to gray, and that's a sign that the copper in the solution is being used.

2. ## Determining the Percent Yield of Calcium Carbonate

+ Na2CO3 (aq) ----> 2 NaCl(aq) + CaCO 3(s) Molar ratio: 1 : 1 : 2 : 1 Sample Calculation for determining number of moles: Number of moles of CaCl2= mass of CaCl2 � 1 mol Molar mass of CaCl2 = 2.09 � 0.02 g � 1 mol 110.98 g = 1.88 �

1. ## Percent Yield Lab. This experiment has proven that KI is the limiting reagent ...

Result This experiment has proven that KI is the limiting reagent in this chemical reaction. This was tested by adding more KI to a beaker full of KNO3 and adding more Pb(NO3)2 to a different beaker full of KNO3. With this simple experiment it was shown that when KI was added more precipitant formed where was Pb(NO3)2 no reaction occurred.

2. ## Aim: To estimate volumetrically the amount of Calcium carbonate present in the eggshell

involved in part A- C6H4COOH.COOK + NaOH → C6H4COOK.COONa Mole Ratio 1 1 The Strength of NaOH (C) = Wt. of Potassium hydrogen phthalate The Strength of NaOH (C) = 2(gm) = 0.93 (mol dm-3) 10.5/1000(dm3) X 204.23(gm) Uncertainty: 2/ (10.5/1000±0.1)

1. ## Measuring the fatty acid percentage of the reused sunflower oil after numerous times of ...

A triglyceride is formed from one molecule of glycerol (propan-1,2,3-triol, CH2(OH)CH(OH)CH2OH) and three fatty aliphatic acids (R-COOH). Structures are shown in the figure below: 1 1 Taken from < http://www.rawâmilkâfacts.com/images/GlycerolTrigly.gif> 6 ASLAN Özge Cemre D129077 Figure 1: Structures of glycerol and triglyceride- saturated.

2. ## Aim: To identify the limiting reagent in the reaction on the basis of practical ...

Results: Potassium iodide/cm3 Lead nitrate/cm3 Theoretical yield /gms 5 10 0.58 10 10 1.16 15 10 1.73 20 10 2.31 25 10 2.31 30 10 2.31 Potassium iodide/cm3 Lead nitrate/cm3 Mass of the filter paper/gms±0.01 Mass of the yield with filter paper/gms±0.01 Practical yield/gms±0.01 5 10 0.68 0.86 0.38 10

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work