• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# The aim of our experiment was to find out the molar mass of the lithium sample and compare the molar mass to the actual molar mass of lithium from the periodic table.

Extracts from this document...

Introduction

Chemistry Write up # 4: Molar mass of lithium Table showing masses and measurements of chemicals in the experiment Description Value Unit Uncertainty Percentage Uncertainty Molar mass literature value 6.941 Mass of lithium # 1 0.035 g ?????? 2.86% Mass of Lithium #2 0.040 g ?????? 2.50% Air pressure 100.6 Kpa ????? 0.00994% Gas constant 8.31 Temperature of water #1 and #2 22.0 Celsius ??0.5?mL 2.27% Gas displaced # 1 40.0 Ml ??0.5?mL 1.25% Gas displaced # 2 50.0 Ml ??0.5?mL 1.00% Concentration of HCl 0.1 Mol dm-3 Observations: * There was a grey layer on the outside of the lithium which was the Lithium that had already reacted with oxygen to form lithium oxide. * Some gas escaped in the time it took to put the lid on after adding the lithium to the water. * The Lithium fizzed slightly when put in the 100ml of H2O. * While titrating the indicator change the lithium hydroxide blue and when titrated with the HCl the solution turned clear. * There was no smell throughout the experiment. Titration raw data Table showing the volume of (0.1) hydrochloric acid needed to neutralise the Lithium Hydroxide solution ml) ...read more.

Middle

% deviation = Experimental value - Literature value X 100 Literature value 1 % deviation = 9.8 - 6.941 X 100 6.941 1 % deviation = 41% Titration C = and N= N (HCl) = C * V N (HCl) = 0.1 * 0.0079 N (HCl) = 0.00079 1.26% + 0.00% =1.26% LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(aq) LiOH and HCl have a 1:1 ratio So N (LiOH) = 0.00079 N= Mr = m/N Mr = 0.035 / 0.00079 = 44.3 / 4 0.25ml used Mr = 11.1 1.26% 2.86% + 1.26% =4.12% � 0.5 % deviation = Experimental value - Literature value X 100 Literature value 1 % deviation = 11.1 - 6.941 X 100 6.941 1 % deviation = 60% C = and N= N (HCl) = C * V N (HCl) = 0.1 * 0.0090 N (HCl) = 0.00090 1.11% + 0.00% =1.11% LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(aq) LiOH and HCl have a 1:1 ratio So N (LiOH) = 0.00090 N= Mr = m/N Mr = 0.040 / 0.0090 = 44.4 / 4 0.25ml used Mr = 11.1 1.11% 2.50% + 1.11% =3.61% � 0.4 % deviation = Experimental value - Literature value X 100 Literature value 1 % deviation = 11.1 - 6.941 ...read more.

Conclusion

Even 10ml of gas escaped and not displaced would dramatically change our results. More displacement means more moles which in turn means a lower molar mass. The lower molar mass would be closer to the accurate value which shows that escaping gas was a major problem. Another problem with the experiment is that some of the Lithium had already oxidised. This Lithium oxide will not react with the water meaning that the actual mass of Lithium that reacts is smaller than the mass of Lithium that was weighed. A small mass would mean that the molar mass would be less. Because of the equation Mr = m / N . A smaller molar mass would mean that it would also be closer to the actual value of 6.941. Improvements: To solve the escaping gas problem an idea would be to suspend the Lithium on a piece of paper over the conical flask. You would then shove the bung into the paper with the lithium and into the conical flask. This would mean there would be no time for any gas to escape meaning the results would be more accurate. Another improvement is we could have used absolute pure non oxidised lithium because the lithium oxide or oxidized Lithium does not react with the water. ?? ?? ?? ?? Thomas Bishop IB13VF ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Chemistry essays

1. ## Butane Molar Mass Lab

(1 atm / 760 mmHg) = (20.45 mmHg x (1atm / 760 mmHg)) + Pressure of Butane Pressure of Butane = 0.976 atm n = PV/RT (Trial 1) n = (0.976 atm)(0.0134 L) / (0.08206 Latm/molK)(295.65) n = 5.39e-4 n= PV/RT (Trial 2)

2. ## Can one determine the coefficients of a balanced chemical equation by having the mass ...

When copper(ii) chloride is dissociated in an aqueous solution, the chlorine ions are strong enough to go through the porous oxide barrier and attract the aluminum ions. However this results in lone oxygen ions which will bond to copper. This would form copper(ii)

1. ## Molar Heat combustion chemistry - investigate the effect of molar mass on the molar ...

4 19.0 5 19.0 1-Butanol 1 23.5 2 20.5 3 24.0 4 22.0 5 23.0 1-Pentanol 1 21.0 2 21.0 3 20.0 4 22.0 5 23.0 Sample Calculations 3. Finding the heat of combustion To find heat of combustion, use the following equation: 1. Convert all masses to kg 2.

2. ## Enthalpy Change Design Lab (6/6)How does changing the initial temperature (19C, 25C, 35C, and ...

and KOH(aq) in a doubled polystyrene cup. Since molar enthalpy is determined through a few different values that need to be collected during the investigation, the steps taken to determine the molar enthalpy of the reaction of 1.00 mol dm-3 HCl(aq)

1. ## Biological oxygen demand (BOD) of water sample analysis.

Processed data Dissolved oxygen (mg of O2/dm3 or ppm) Sample A 11.8 Sample B 9.20 BOD 2.60 Results of others (already converted to mg of O2 per dm�) Name Source Sample A (mg/dm3) Sample B (mg/dm3) BOD (mg/dm3) Don Valley 11.6 9.00 2.60 Chong Pond 10.4 8.40 2.00 Keith Plant

2. ## The aim of this experiment is to examine the enthalpy of combustion of the ...

has been said that the volume of water is 40 cm ³. Therefore mass of water can be calcukated from here, if we assume that the density of water is ρ (H2O) = 1.0 g cm-3 . From here we can find the mass of water : m (H2O)

1. ## Biodiesel Investigation - How the concentration of Potassium Hydroxide solution would affect the yield ...

0.005 ÷ 33.119 = 0.02% (1sf) Uncertainty on mass of biodiesel 0.005 ÷ 5.95 % unc. 0.08% (1sf) Total % Uncertainty = 0.1% unc. 18% ± 0.018 Further Observation: Fig 4 Graph for % yield: Refer at end of document** (Page 11) Graph for mass obtained: Conclusion: Multiple inferences can be made from the data obtained.

2. ## Determining the relative atomic mass of Lithium

The Li was taken out of the container at the beginning of the experiment so it could be used by other students thus leaving time for the oxide to form. This would have increased the mass of lithium as LiO has a greater Mr than Li, thus would have increased our value for the relative atomic mass of lithium. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work 