Determination of the Relative Atomic Mass of Calcium
In this experiment, I will determine the relative atomic mass of
calcium by two different methods.
v By measuring the volume of hydrogen produced.
v By titrating the lithium hydroxide produced.
Method 1
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[IMAGE]
v 0.10g of calcium used.
v 100cm3 of distilled water used.
Results
Method 1
Mass of Calcium (g)
Total volume of gas produced (cm3)
0.10
37
v Starting point of water in cylinder = 238cm3
v Ending point, after reaction complete, of water in cylinder= 201cm3
v (238 - 201 = 37) Deduction of 37cm3, therefore 37cm3 of hydrogen gas
produced.
1 mol of gas occupies 24000 cm3 at room temperature and pressure.
Ca (s) + 2H2O (l) à Ca(OH)2 (aq) + H2 (g)
v Number of moles of hydrogen = volume of hydrogen / 24000
= 37/24000 = 0.0015416 mol
= 0.001542 mol (4sf)
v Number of moles of calcium = 0.001542 (ratio 1:1 with hydrogen)
v Relative atomic mass of calcium = Mass of calcium / Moles
= 0.10 / 0.001542 = 64.85084
= 64.85 (4sf)
Method 2
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v 25.0cm3 of alkaline solution used.
v Amount of acid used:
Starting Amount of HCl used (cm3)
Amount of acid used till reaction complete- Test 1 (cm3)
Test 2 (cm3)
Test 3 (cm3)
Average (cm3)
100
9.00
8.20
8.20
8.47
Results
Ca(OH)2 (aq) + 2HCL (aq) à CaCl2 (aq) + 2H2O (l)
v Moles of hydrochloric acid used in titration = volume *
concentration
= 0.1 * (8.47/1000) = 0.0.000847
= 0.0008470 mol ( 4sf)
v Moles of Ca(OH)2 = the number of moles of HCl / 2 (as the ratio is 2
: 1) = 0.0008470 / 2 = 0.0004235 mol (4sf)
v Moles of Ca(OH)2 present in 100cm3 of alkaline solution = 0.
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Investigate how the concentration of hydrochloric acid effects the rate at which it reacts with calcium carbonate
NaOH(aq)Â Â Â Â Â Â Â Â +Â Â Â Â Â Â Â Â Â HCl(aq)Â Â Â Â Â Â Â Â Â Â Â Â Â Ã Â Â Â Â Â Â Â Â Â Â Â Â NaCl(aq) Â Â Â Â Â Â + Â Â Â H2O(l).
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Determining the Relative Atomic Mass of Lithium An experiment has been carried out to determine the relative atomic mass of Lithium by using two different types of methods The first method that was carried out was to determine the volume of Hydrogen produced. In this experiment a fixed amount of Lithium was used, in my case it was 0.11g. At the end of this experiment, the volume of Hydrogen gas I collected was 185cm³. Then using the solution of lithium hydroxide made from experiment one, I used it in the titrating experiment, to find out the total volume of Hydrochloric acid used to titrate the lithium hydroxide. RESULTS TABLE Experiment Initial Volume ( cm³) Final Volume ( cm³) Total volume Of HCl used ( cm³) Rough 0.2 30.3 30.1 1 6.3 35.8 29.5 2 2.7 32.0 29.3 Average 29.6 CONCLUSION Method 1 [IMAGE]2Li (s) + 2H20(l) LiOH(aq) + H2(g) Number of moles of Hydrogen. Volume of hydrogen gas was 185 cm³. Weight of Lithium was 0.11g. N = __V__ _185_ = 0.0077 MOLES 24000 24000 Number of moles of Lithium.