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

Determination of the Relative Atomic Mass of Calcium

Extracts from this document...

Introduction

Determination of the Relative Atomic Mass of Calcium In this experiment, I will determine the relative atomic mass of calcium by two different methods. * By measuring the volume of hydrogen produced. * By titrating the lithium hydroxide produced. Method 1 * 0.10g of calcium used. * 100cm3 of distilled water used. Results Method 1 Mass of Calcium (g) Total volume of gas produced (cm3) 0.10 37 * Starting point of water in cylinder = 238cm3 * Ending point, after reaction complete, of water in cylinder= 201cm3 * (238 - 201 = 37) Deduction of 37cm3, therefore 37cm3 of hydrogen gas produced. 1 mol of gas occupies 24000 cm3 at room temperature and pressure. Ca (s) + 2H2O (l) --> Ca(OH)2 (aq) + H2 (g) * Number of moles of hydrogen = volume of hydrogen / 24000 = 37/24000 = 0.0015416 mol = 0.001542 mol (4sf) * Number of moles of calcium = 0.001542 (ratio 1:1 with hydrogen) * Relative atomic mass of calcium = Mass of calcium / Moles = 0.10 / 0.001542 = 64.85084 = 64.85 (4sf) Method 2 * 25.0cm3 of alkaline solution used. * Amount of acid used: Starting Amount of HCl used (cm3) Amount of acid used till reaction complete- Test 1 (cm3) ...read more.

Middle

If the gas syringe is used, it would give a percentage error of 1.35 % (0.5 / 37 * 100 = 1.3513). Where as the measuring cylinder gave a percentage error of 2.70 % (1 / 37 * 100 = 2.7027). This shows if the gas syringe is used it would reduce the gap for error. Another possible error is that the calcium could have been kept for so long that it might have oxidised when it was left out in the open. This would mean that less calcium would react and would have lowered the value of the gas produced. This would give a higher relative atomic mass when calculated. To prevent this, if the experiment is repeated, it can be made sure that the lid on the pot of calcium is always kept on and only taken off when it is needed. This would reduce the oxidising of the calcium. The reaction may have been stopped before it was completed. This would results in less gas being produced and this would make the relative atomic mass higher. Also there could have been a default in the equipment being used. There could have been leakage in the pipe, causing a certain amount of gas to escape into the atmosphere and not into the measuring cylinder. ...read more.

Conclusion

If too much hydrochloric acid had been let into the solution, the value recorded in the results would be higher therefore the average would come out higher and this would give a lower relative atomic mass. The weighing scale used measured to 2 decimal places. To make the results more accurate and reduce the percentage of error a weighing scale which measured to 3 decimal places could have been used. This aspect has already been discussed in the method 1 part of the evaluation. The experiment could also be carried out a higher number of times to get an more accurate average and more accurate relative atomic mass. The error with the oxidisation of the calcium could also apply to this method and has been discussed in the first part of the evaluation. In comparison method 2 is more accurate than method 1. This is because method 2 gave a relative atomic mass of 59.17 which is closer to the factual value of 40.1when compared to the value gained by method 1, 64.85. Also the accuracy can be linked to the percentage error as method one gave a higher percentage error than method two (method 1 = 61.72 %, method 2 = 49.56 %). Overall, both methods are very inaccurate (high percentage errors). But in comparison, method 2 is a better way of determining the relative atomic mass of calcium. DEV SEN 6L AS CHEMISTRY MODULAR COURSEWORK ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Aqueous 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 GCSE Aqueous Chemistry essays

  1. Determination of the relative atomic mass of magnesium.

    The lithium is kept in oil while in storage to prevent it from oxidising. When using the lithium the oil must be cleaned off to ensure that it reacts to its full potential. If the oil is not cleaned of the lithium will not produce as much hydrogen, also if

  2. Determination of the relative atomic mass of lithium.

    Lithium = 0.09g 1 mol of Li = 0.09 = 8.0g 0.0112 Thus the Relative Atomic Mass of Lithium is 8.0g mol-1 from experiment 2. Natural lithium exists as two isotopes: lithium-7 (92.5 percent) and lithium-6 (7.5 percent) so the relative atomic mass of lithium is; (92.5 x 7)

  1. The Determination of an Equilibrium Constant.

    dm-3 [CH3COOH] = 0.35 / 0.25 mole dm-3 [C2H5OH] = 1.638 / 0.25 mole dm-3 ?Kc4 = [CH3COOC2H5] * [H2O] / [CH3COOH] * [C2H5OH] = (0.434/0.25) * (5.206/0.25) / (0.35/0.25) * (1.638/0.25) =0.434 * 5.206 / 0.35 * 1.638 = 3.94 Kc = (Kc1 + Kc2 + Kc3 + Kc4)

  2. How much Iron (II) in 100 grams of Spinach Oleracea?

    x 7.17 cm3 1000 The volume of the average titre is divided by 1000 to change the units from cm3 to dm3. Moles = 0.00017925 mol dm-3 The ratio of Potassium Manganate (VII) (aq) to Iron (II) (aq) is 3:5 and therefore to work out the mols of Iron (II)

  1. Determination of the relative atomic mass of lithium.

    I need to use the triangle to find out how to re - arrange it. m n Mr Now using this triangle I can get the equation that I need to calculate my result (the relative atomic mass). If I want to figure out the Mr, then what I do is look at the other letters.

  2. Determine the relative atomic mass of lithium.

    The formula is as follows - LiOH(aq) + HCl(aq) LiCl(aq) + H2O(l) As you can see the mole ratio is 1:1 as I mole of lithium hydroxide is required to react with 1 mole of hydrochloric acid. So the number of moles of lithium hydroxide present is 0.00291 After that

  1. To determine the relative atomic mass of Lithium

    2 Li(s) + 2H2O(l) --> 2LiOH(aq) + H2(g) The stiochiometry ratio of LiOH to Li is 2:2, that is 1:1 Therefore number of moles of Li metal is 0.01462moles. The original mass of Lithium = 0.11g Number of moles of = Mass / Molar mass Molar mass = Relative atomic

  2. to determine the relative atomic mass of lithium. We will be doing this via ...

    LiOH in 100cm3 = 1.480 x 10-3 x 10 = 0.0148 moles Using this result we can now calculate the Relative Atomic Mass of Lithium. For this we must use the following formula: Relative Atomic Mass = Original Mass (g)

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