Determination of the relative atomic mass of lithium

Introduction:

For this investigation I will determine the Relative atomic mass (Ar) by using two different methods. In the first method I will dissolve a piece of lithium of a known mass in water, I will then collect the hydrogen gas produced, which can be used to calculate the relative atomic mass of Lithium. In the second method I will titrate the resulting solution of lithium hydroxide with a known concentration of hydrochloric acid, this can also be used to calculate the relative atomic mass of lithium.

Readings:

*100 cm3 of water

*37.47 g of lithium plus the watch glass

*37.41 g of Watch glass

*122 cm3 of hydrogen gas collected

Calculations from method 1

Moles of hydrogen produced:

Assuming that 1 mole of gas occupies 24000 cm3 at pressure and room temperature!

I will now calculate the number of moles of the hydrogen gas produced.

I will do that by dividing the volume of gas produced by 24000cm3 :

Moles (n) = Volume (cm")

24000 cm"

So:

n = 122 cm" = 0.005083 mol of H2

24000 cm"

Equation:

2 Li(s) + 2 H2O (L)2LiOH(Ag) + H2 (g)

By looking at the Stoichiometry ratio in the reaction I can see that the ratio of Li: H2 is 2:1, therefore the number of moles of Li will be twice as the number of moles of H2 in the reaction.

So:

0.005083 x 2= 0.010166 mol of Li

Mass of Lithium =

Mass of the Lithium plus the watch glass - the mass of the watch glass

So:

37.47g - 37.41g = 0.06g

ÃÂ·0.06g of lithium was used

Relative atomic mass of lithium:

I will now calculate the relative atomic mass of the piece of lithium that I used by rearranging...

## Help

This stuff is foreign to me, but I can say that the language you use is very simple and easy to understand. The calculations part is where I lose it. Sorry, probably not much help to you.

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