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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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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