Calculating the relative atomic mass of lithium.

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Chemistry Assessed Practical Write-up

Calculating the relative atomic mass of lithium

1)

Amount of H2 collected = 129 ml

Mass of Lithium with oil = 1.2g

2Li + 2H20 → 2LiOH + H2

Vol H2 = 129

          24,000

         = 0.005375 moles → 1:2 → 0.01075

RAM = Mass

         Moles

        RFM Lithium = 0.12

                         0.01075

        RFM Lithium = 11.16g

The RFM of Li that I worked out is = 11.16g

RAM Li   = 6.9g

Therefore the mass of oil is: 11.16 – 6.9 = 4.26g

2)

Volume Lithium Hydroxide = 25.00 cm3 exactly (using a volumetric pipette)

Volume HCL (0.1 molar) to neutralize Lithium Hydroxide = See graph

LiOH + HCl → LiCl + H2O

Volume HCL = 27.1

Concentration of HCL = 0.l molar

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Moles         = Concentration * Volume

                   = 0.1 * 27.1

        = 2.71moles in ratio 1:1 = 2.71moles

Mass         = Moles * RFM

        = 2.71 * (6.9+16+1)

        = 64.77 g

Concentration         = Moles

                    Vol.

                = 2.71

                    25

                = 0.1084

Conc. = 0.1084 molar

Mass = 64.77g

Moles = 2.71 mol

Moles of Li in 100cm3 = 0.1084 * 100 = 10.84 moles

Mass of LiOH = 10.84 * 24 = 260.16 g

Mass of Li = (260.16/24) * 6.9 = 74.796 g

Calculated RAM of Li = 10.835 ...

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