DETERMINING THE MASS OF LITHIUM

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DETERMINING THE MASS OF LITHIUM

METHOD 1

RESULTS

Mass of lithium used                                 0.12g

Amount of hydrogen gas produced                182cm³

TREATMENT OR RESULTS

2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)

Moles of hydrogen gas collected

Moles of H2                volume / 24000(cm³)

                        82 / 24000

                        0.007583333 moles

Moles of H2                0.0076 moles

Lithium that reacted

H2        :        Li

1        :        2

0.0076        :        2 × 0.0076

0.0076        :        0.0152

Moles of Li                0.0152 moles

Relative atomic mass of lithium

R.A.M                        mass / moles

                        0.12 / 0.0152

                        7.894736842

R.A.M of Li                7.8947

METHOD 2

RESULTS

Titration of aqueous LiOH with 0.100 mol dm-3 HCl.

Average result

(34.80 + 34.90) / 2

= 69.7 / 2

= 34.85

TREATMENT OF RESULTS  

LiOH(aq) + HCl(aq) →  LiCl(aq) + H2O(l)

Moles of HCl used in titration

Moles of HCl                        concentration × (volume / 1000)

                                0.100 × (34.85 / 1000)

                                0.003485 moles

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Moles of HCl                        0.0035 moles

LiOH used in titration

HCl        :        LiOH

1        :        1

0.0035        :        0.0035

Moles of LiOH                0.0035 moles

Number of moles of LiOH present in 100cm³ of solution from method 1

25cm³ of LiOH is pipette each time.

There’s 100cm³ of LiOH, so it will be:

0.003485 × 4 = 0.01394

Moles present in 100cm³ of LiOH         0.0139 moles

Relative atomic mass of Lithium

R.A.M                        mass / moles

                        0.12 / 0.0139

                        8.633093525

R.A.M of Li                8.633

HAZARD OF CHEMICALS IN THIS EXPERIMENT

Care must be taking when handling chemicals in this experiment. Safety goggles ...

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