Chemistry determination of RAM of lithium

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Aminul Hoque

ANALYSIS:

Method 1: Measure the volume of hydrogen produced when a known mass of lithium reacts with water.

Set up apparatus. 250cm³ conical flask must contain 100cm³ of distilled water.

Weigh between 0.08g and 0.13g of lithium. Record the exact mass of lithium.

Remove the stopper, add lithium to the flask and quickly replace stopper.

Collect the gas and record final volume of hydrogen.

RESULTS:

I used 0.08g of lithium, this produced 168cm³ of hydrogen gas.

Treatment of results:

Assuming that 1 mole of gas occupies 24,000cm3 at room temperature and pressure

Calculating the number of moles of hydrogen:

168 / 24000 = 0.007 mol

Number of moles of lithium:

0.007 x 2 = 0.014 mol

Using above info and original mass of lithium, RAM of lithium can be calculated as:

1 / 0.014 x 0.08 = 5.7142

Method 2:

Titration of aqueous LiOH with 0.100 mol dm HCl.

Mean titre is: (22.7 + 22.9) / 2 = 22.8

On average, 25.0 cm of LiOH required 22.8cm of 0.100 mol of HCl

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Number of moles of HCl used:

22.8 / 1000 x 0.1 = 0.00228 mol

Number of moles of LiOH:

1:1 ratio so, 0.0228

Number of moles of LiOH present in 100cm of solution:

25cm = 0.00228 x 4 = 0.00912

RAM of Li is therefore :

1/ 0.00912 x 0.08 = 8.77

Hazard concepts:

Lithium reacts violently with water to give off flammable hydrogen gas and corrosive dust. Lithium hydroxide (LiOH) is a corrosive .Hydrochloric acid is corrosive to the eyes, skin, and mucous membranes.  Acute (short-term) inhalation exposure may cause eye, ...

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