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.

  • By measuring the volume of hydrogen produced.
  • By titrating the lithium hydroxide produced.

Method 1

  • 0.10g of calcium used.
  • 100cm3 of distilled water used.

Results

  • Starting point of water in cylinder = 238cm3
  • Ending point, after reaction complete, of water in cylinder= 201cm3
  • (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) → Ca(OH)2 (aq) + H2 (g)

  • Number of moles of hydrogen = volume of hydrogen / 24000

 = 37/24000 = 0.0015416 mol

 = 0.001542 mol (4sf)

  • Number of moles of calcium = 0.001542 (ratio 1:1 with hydrogen)

  • Relative atomic mass of calcium  = Mass of calcium / Moles

      = 0.10 /  0.001542 = 64.85084

      = 64.85 (4sf)

Method 2

  • 25.0cm3 of alkaline solution used.
  • Amount of acid used:

Results

Ca(OH)2 (aq) + 2HCL (aq) → CaCl2 (aq) + 2H2O (l)

  • Moles of hydrochloric acid used in titration = volume * concentration

     = 0.1 * (8.47/1000) = 0.0.000847

     = 0.0008470 mol ( 4sf)

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  • 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)

  • Moles of Ca(OH)2 present in 100cm3 of alkaline solution = 0.0004235 / 25 = 0.0000169 * 100 = 0.001690 mol (4sf)

  • Relative atomic mass of calcium = mass / moles

                                               = 0.10 / 0.001690 = 59.171597

                                             = 59.17 (4sf)

Evaluation

The results gained by the two methods show that both of the ...

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