Determination of the Relative Atomic Mass of Calcium

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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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