RAM 7
Step 2 :
Mole ratio
Li : H2
2 : 1
0.0129 mol: 6.45 x 10-3 mol (0.0129 divided by 2)
Step 3 :
Volume of hydrogen, H2 = moles x 24000cm³
= 6.45 x 10-3 mol x 24000cm³
= 155cm³ of hydrogen produced
Step 4: calculating the RAM of Lithium
Moles of hydrogen gas = volume produced
24000cm³
155cm³
24000cm³
= 6.46 x 10-3 mol
Step 5:
Mole ratio
H2 : Li
1 : 2
6.46 x 10-3 mol : 0.013mol
Step 6 :
RAM of lithium = mass 0.09
= = 6.92
moles 0.013mol
Prediction – Method 2 (titration)
Step 1 :
Moles of lithium = 0.09
= 0.0129mol
7
Step 2 :
Li : LiOH
1 : 1
= 0.0129mol : 0.0129mol
Step 3 :
LiOH (aq) + HCL (aq) LiCl (aq) + H2O (l)
Mole ratio
LiOH : HCL
1 : 1
= 0.0129mol : 0.0129mol
Since I used 10cm³ of LiOH of the original 100cm³:
0.0129mol
= 1.29 x 10-3mol
10
Step 4 :
Volume of HCL (cm³) = moles (in 10cm3) x 1000
Concentration (M)
= 1.29 x 10-3 mol X 1000
= 12.9cm³
0.1M
Calculating the RAM of lithium
Step 5 :
In 100cm³ = 10 x (1.29 x 10-3 mol) = 0.0129mol
Step 6 :
RAM of lithium, Li = mass = 0.09g = 6.98
Moles 0.0129mol
Calculations: Method 1 – RAM of lithium
Volume of hydrogen gas = 188cm³
2Li (s) + 2H2O (l) 2LiOH (aq) + H2 (g)
Step 1 :
Moles of hydrogen gas = volume produced
24000cm³
188cm³
24000cm³
= 7.83 x 10-3 mol
Step 2 :
Mole ratio
H2 : Li
1 : 2
7.83 x 10-3 mol : 0.016mol
Step 3 :
RAM of lithium = mass 0.09
= = 5.63 (prediction 6.92)
moles 0.016mol
Calculations : Method 2 – RAM of lithium
Average titre = 12.6cm³ + 12.4cm³ + 12.5cm³
= 12.5cm³ (prediction 12.9cm3)
3
LiOH (aq) + HCL (aq) LiCl (aq) + H2O (l)
Number of moles of LiOH present in 100cm³ :
Step 1:
Moles of HCL = volume (cm³) x concentration (M)
1000
= 12.5cm³ x 0.1M
1000
= 1.25 x 10-3 mol
Step 2:
Mole ratio
HCL : LiOH
1 : 1
1.25 x 10-3 mol : 1.25 x 10-3 mol
Step 3 :
In 100cm³ = 10 x (1.25 x 10-3 mol) = 0.0125mol
I will use this result to calculate the RAM
Step 4 :
RAM of lithium, Li = mass
Moles
= 0.09g
= 7.20 (prediction 6.92)
0.0125mol
Skill E
Overall the accuracy of my experiment was fairly consistent, for example all readings were to 3 significant figures.
As you can see from the percentage errors (see below) method 2 was much more accurate. The actual results were also very close to the precitions. Even though I obtained the results that was expected, there were some limitations and errors within the procedure of my two methods, in which improvements could be made:
Limitation – The solid lithium (which was in fact stored in oil) is still able to react with the air and moisture in the air as it is a very reactive metal.
2Li (s) + O2 (g) 2LiO (s)
2Li (s) + 2H2O (l) 2LiOH (s) + H2 (g)
This will cause the lithium metal to tarnish and become dull. Therefore its reaction with water does not occur to its fullest, which affects the accuracy and reliability of my results.
Improvement – To prevent the lithium from reacting with anything else it can be stored in an inert atmosphere and also when using it, it can be handled in an inert box or area.
Limitation – Some hydrogen gas is lost when putting the bung back in conical flask.
Improvement – Use a piece of apparatus/ partition, for example a divided flask where the two reactants, lithium and distilled water is separate from each other, but already in the flask. So when ready to start, shake the flask in order for the reaction to begin.
Limitation – As this reaction between the lithium and distilled water is very exothermic, a lot of heat is given out and because of this the hydrogen gas expands. Due to this, the volume of hydrogen increases and therefore the readings would be less accurate. The room temperature varies from STP and due to the compressibility of gas the reaction vessel pressure varies.
Improvement – The solution to this is to wait for the temperature to return back to room temperature before measuring the volume of hydrogen gas. As all measurements in my experiment were taken at room temperature it is essential that the volume of hydrogen should be at room temperature as well. Another solution is to use a waterbath where the temperature is constant at all times (this can be adjusted by the thermostat) – this will surely increase the accuracy of the results.
Limitation – It was very unclear when measuring initial and final amount of water in the measuring cylinder and so this would produce inaccuracies in the final volume of hydrogen gas produced.
Improvement – Can use another piece of apparatus to improve the accuracy such as a gas syringe.
Limitation – As the lithium metal was stored in oil, it was very difficult to remove all the oil and so this could of affected the actual mass of the metal.
Improvement – Only solution to this is to measure the lithium metal many times. Or in an inert atmosphere, mix with hexane to remove oil and following this blow helium or nitrogen to remove the hexane.
Comparison of methods 1 and 2
Comparing the two methods, method 2 which was the titration experiment was much more accurate. By using a graduated burette, the accuracy of my obtained results improved greatly which can be seen in my results table. Both methods are linked, if there were inaccuracies in method 1 then this would affect the accuracy of method 2.
Relative error
Relative error = Difference in figures
Standard figure (in this case the standard figure is the
standard RAM of lithium, which is ‘7’)
METHOD 1 : 7 – 5.63
X 100% = 19.6%
7
METHOD 2 : 7.20 – 7
X 100% = 2.86%
7
