• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9

# The relative atomic mass of lithium

Extracts from this document...

Introduction

COURSEWORK Background information I'm going to produce a piece of coursework, which determine the relative atomic mass of lithium. Two methods are going to be used to approximate the relative atomic mass of lithium. 1) In the first method the RAM (Relative atomic mass) of lithium is determined by the Volume of hydrogen gas produced. 2) In the second method the RAM of lithium is determined by titrating the lithium hydroxide produced. To some extend, I think that the first method is the most practical and less complicated, because the less calculations has to be done. Hence it is more easer to measure the volume of hydrogen gas. Test-1 My mission was to find out the atomic mass of lithium. I took a sample and weighted it, removed parts of the sample so that it would weigh approximately 0.100g. The oil was removed from the sample by filter paper wiping it with,so no more oil can be seen. The scale was not a very sensitive one and we could weigh with a margin of error of 0.01g. Then I placed the sample in a beaker that contained 100 cm3 of distilled water. ...read more.

Middle

5) Then the average titre is recorded. Results I gathered a large enough range of results to make accurate observations from. The experiment was repeated until consistent results were obtained; Run 1 Run 2 Run 3 Initial volume (cm3) 1.3 2.7 1.4 Final Volume (cm3) 39.7 38.4 40 volume Titre used (cm3) 38.4 35.7 38.6 Mean titre (cm3) 56.35 Burette reagent HCl mol dm -1 Pipette solution LiOH mol dm-1 indicator Phenolphthalein mol dm -1 Analysis Calculation: 1st test- by measuring the volume of hydrogen produced A mass of 0.07 of lithium was used, which gave 116Cm3 of H2 * Calculate the number of moles of hydrogen 116/24000 = 0.0048 mol of H2 2Li (s) + 2H2O(l) ----->2LiOH(l) + H2(g) so, 2* 0.0048 = 0.0097mol of Li using the formula to find the RAM of Li --------> Mole = mass/Ar or Mr so, 0.07/0.0097 = 7.2g of Li The answer is very close to the standard RAM of Li Error calculations, 7.2*100/7=102.86 - which means that there is 2.86% error in the first method 2nd test - by titrating the lithium hydroxide with HCl. From the results table the mean titre is found by adding the runs altogether that divide the by 2 Mol =vol/1000*concentration = 38.4+35.7+38.6=56.35 /2 = 28.175cm3 = 28.175/1000*0.100=0.002818mol = ...read more.

Conclusion

if the methods had more details in it then this will inpruve the accuracy of the experiment. The method didn't give enough information of how to calculate the results, so i strugeled a bit, but in the end i worked out with the help of the book. To minimise the errors the method should have explained in detailed the performers that the experimenter has to carry out. To make the experiment the most accurate possible the equipment should be the more advanced ever, and variables should be pure with out any mixture. For further improvement a freshly cut pure lithium should be used, this could be done by putting the lithium in the vacuum before transferring it into the conical flask. A freshly cut pure lithium means having accurate results (out comes). Also after introducing the lithium into the conical flask the stopper should be replaced very quickly so a low minimum gas escapes. Then to measure this gas a more accurate apparatus should be used like(Eudiometer ).The mass balance despite taking quick readings, was not the most accurate possible, so to improve the accuracy of the experiment a more advanced electronic mass balance should be used. I think more time is also needed. If i was doing the experiment again i would like a bite more time to perform the experiment as this would improve the accuracy of the experiment ...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 Classifying Materials 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 GCSE Classifying Materials essays

1. ## The rates of reaction between CaCO3 and HCL

24.94 24.69 00.32 165 24.84 24.25 24.92 24.67 00.34 180 24.84 24.25 24.92 24.67 00.34 195 24.83 24.24 24.91 24.66 00.35 210 24.82 24.23 24.90 24.65 00.36 225 24.81 24.22 24.90 24.64 00.37 240 24.80 24.22 24.89 24.64 00.37 Analysis: Trends and patterns: The lines shown on the graph are

2. ## Periodic table

Hybrid - a plant with both alternatives such as tallness and shortness for characteristics Alleles - alternate forms for genes Monohybrid cross - if only one trait is to be passed on to the next generation then it is a monohybrid cross.

1. ## Electrolysis COursework

Take the mass after away from the original mass and record it in the table. Do this for all the results. Diagram Results Voltage (V) Original Mass (g) Mass After (g) Mass of Copper extracted (g) Volts Attempt Cathode Cathode Cathode 2 1 1.51 1.58 0.07 2 1.54 1.59

2. ## Affect of concentration on reaction

Conclusion I have decided the most appropriate mass of calcium carbonate to react with; highest concentration of 2 molar and lowest concentration of 0.4 molar hydrochloric acid, is 5g. The preliminary experiments have given me ideas to improve on the main experiment; calcium carbonate is ground to powder form, using

1. ## The role of mass customization and postponement in global logistics

Performance dimensions The relevant components where an enterprise demonstrates relative efficiency and effectiveness in satisfying its customers * Efficiency - economic utilisation of resources and * Effectiveness - extent to which requirements are met For Mass Customization, we can

2. ## Determination of the relative atomic mass of Lithium.

+ HCl (aq) LiCl (aq) + H O (l) � From the equation it can clearly be seen that the ratio of LiOH to HCl is 1:1.

1. ## Determination of the relative atomic mass of lithium

Method 1: procedure 1. Set up the apparatus as shown below. The 250cm3 conical flask must contain exactly 100 cm3 of distilled water. 2. Weigh between 0.08g and 0.13g of lithium. Record the exact mass of lithium using the measuring scale.

2. ## Determining the Relative Atomic Mass of Lithium.

very difficult to overcome as the phenolphthalene will still go clear, this shows that there is an error with the indicator a more sensitive indicator could have been used. But the other indicators available (methyl orange and Bromothynol blue) were not suitable for this experiment.

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