• 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...


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.


% 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.


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

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

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

    Meaning, the temperature change is very sudden since the rate of reaction is quick fast, and so a lid is not necessary since that would only be truly influential in insulating heat for an investigation involving substances that react at slower rates.

  2. Aim: To find the molar mass of butane, by finding the number of moles ...

    * The Burette was filled completely with water. * We did not take the final volume reading when the level of water in the burette was equal level with the water in the trough. Instead, we measured the height of the water column above the level of water in the trough.

  1. Butane Molar Mass Lab

    the top end of the beaker because the beaker was turned upside-down PROCESSING RAW DATA - Change of Mass of Lighter (g +/-0.01g) Pressure of Butane (atm +/- 0.02 atm) Displaced water (mL +/- 0.1mL) Molar Mass of Butane Trial 1 0.030 0.976 13.4 ~55.65 g/mol Trial 2 0.040 0.976

  2. Aim. To find the identity of X(OH)2 (a group II metal hydroxide) by determining ...

    0.00196 0.0238 0.770 Ba(OH)2 171.36 0.00196 0.335 3.700 Uncertainties: The uncertainty in measurement: Uncertainty due to pipette of 25.000 cm3 : Volume of X(OH)2 = � 0.100 cm3 Percentage uncertainty = (0.1/25) X 100 = 0.400% Uncertainty due to Burrette of 50.000 cm3: Assumed due to measured volume of 19.675

  1. Determining the Molar Mass of Volatile Liquid

    The water bath containing the Erlenmeyer with volatile liquid was heated gently until the liquid was vaporized at around 90�C for 3 minutes. 6) The water bath temperature and the atmospheric pressure were recorded. 7) The vapor in the Erlenmeyer was condensed by immersing in an ice bath 8)

  2. The aim of our experiment was to find the mass of acetyl salicylic acid ...

    Our fairly small random error of 2.39% does not cover the 20% deviation so the rest of the error must be systematic error. Evaluation: A problem with this experiment was that when crushing the 4 aspirin tablets in the crucible although distilled water was used to wash most of the

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

    flame was different for each group and each alcohol * The coke cans used were previously used for the same experiment and as a result there was soot (carbon) on the bottom of the can. * The flame flickered due to movements around the apparatus Propanol * The distance between

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

    the sequence of exothermic chemical reactions between a fuel and an oxidant accompanied by the production of heat and conversion of chemical specie ) of a primary alcohol increases . This means that as the number of carbon- carbon bonds increase, with the increase in the carbon chain length, more

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work