An investigation to compare the % by mass of calcium carbonate in a variety of common rocks.

Preliminary tests:
In order to gather preliminary knowledge of the way the chemicals are likely to behave it is important to conduct some preliminary tests.

Test 1- this is to determine rate of thermal decomposition for both carbonates.

1. Weigh crucible
2. Add calcium carbonate and re-weigh
3. Subtract the mass of the crucible from the weight recorded in ‘2.’ to give the mass of calcium carbonate
4. Heat with a roaring blue flame, recording the weight every 5 minutes until there is no mass loss
5. Take the final weight from the initial weight to give the mass loss
6. Repeat as above for magnesium carbonate

Results:

                                CaCO3 (in g)                            MgCO3 (in g)

The calcium carbonate had not finished decomposing but it is clear to see that the magnesium had ceased to decompose after losing nearly 0.6g. This test re-enforces the evidence for a thermal decomposition being the best option as it clearly distinguishes between the two carbonates. So from this it can be acknowledged that magnesium carbonate will decompose before calcium carbonate.

Test 2- this test will demonstrate how easily calcium carbonate will decompose compared to magnesium carbonate.

1. Weigh crucible
2. Add calcium carbonate and re-weigh
3. Heat over a gentle blue flame for 5 minutes
4. Record mass loss
5. Repeat for magnesium carbonate

Results:

                                CaCO3 (in g)                            MgCO3 (in g)

From this experiment it is clear that magnesium carbonate decomposes far easier than calcium carbonate.

Safety factor-It is necessary to observe the safety protocols while conducting this experiment. Any equipment that is involved with the Bunsen burner will be hot and therefore must be handled with care. Long hair and loose clothing should be kept away from all apparatus. Care must be taken when handling the rock samples and safety goggles must be worn at all times.  

Main experiment:

For a thermal decomposition I will need the following apparatus:

1. Crucible- to contain the sample during heating while not changing mass itself
2. Forceps- to handle the crucible when hot
3. Tripod- to hold the crucible above the flame
4. Clay pipe triangle- to hold the crucible on the tripod
5. Asbestos mat- to protect the work surface
6. Balance- to weigh the samples
7. Rock crushing instruments- to break down the rock into useable samples  

Method:

1. Weigh the crucible and record the mass
2. Grind the rock up into tiny pieces
3. Add a sample of the crushed rock to the crucible and re-weigh
4. Subtract the mass of the crucible from the mass recorded in ‘3.’ to give the mass of the sample
5.  Heat for 5 minutes over a gentle blue flame and record the mass difference
6. Repeat ‘5.’ until there is negligible change in mass between two successive periods
7. Heat under a roaring blue flame for 5 minutes and record the mass
8. Repeat ‘7.’ until there is no mass loss
9. Subtract the mass recorded in ‘6.’ From that recorded in ‘8.’ to give the mass loss of calcium carbonate
10. Using information from previously stated sources that for every CO2 molecule given off, there remains one molecule of CaO and for every mole of CO2 given off, there is one mole of CaCO3.
11. Subtract the mass of CaO remaining from the mass of the other remaining substances and add it to the mass of CO2 lost. This will give the mass of CaCO3 that was in the rock
12. Work out the percentage of CaCO3 that was originally in the rock

Calculations:

In order to demonstrate the scientific calculation, I will use some ‘convenient’ numbers.

Example results:

Total mass of sample = 50g

Mass loss of CO2 = 5g

(Number of moles = mass / mr)

(Mass = number of moles x mr)

Example calculation:

Moles of CO2 = mass CO2 / mr CO2 

                 = 5 / (12+(16x2))

                 = 0.113636 mol dm-3  

Moles of CO2 = moles of CaCO3 

Therefore moles CaCO3 = 0.113636 mol dm-3  

Mass CaCO3 = moles CaCO3 x mr CaCO3 

                = 0.113636 x (40+12+(16x3))

                = 11.3636g

Therefore % by mass of CaCO3

= (Mass CaCO3 / total mass of rock sample) x 100

= (11.3636/50) x 100

= 22.7272 %

Bibliography:

Cambridge Advanced Sciences - OCR Advanced Chemistry 1

Advanced Chemistry for You

Practical Chemistry by Lambert and Muir

Chemistry Internet sites such as: