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# Magnesium Oxide

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Introduction

Chemistry Coursework Can Erdogan 11A Aim: What I was trying to find out is how magnesium and oxygen join up together so I would want to find out what formula the magnesium oxide would be in. Introduction: Magnesium is a metal which need get rid of two electrons; oxygen is a gas at room temperature and it needs two electrons and magnesium oxide is ionic compound because when they join up they are ions and both elements have an opposite charge which creates electrostatic attraction between them creating a very strong bond between the two ions, but they only become ions when oxygen takes electrons away from magnesium. 2Mg(s) + O2 (g) 2MgO(s). The conservation of mass is also involved in this experiment so the theory of conservation of mass is that if I use 48g of magnesium and 32g of oxygen I would get 80g of magnesium oxide, and with this reaction the ratio for mass is 24:16. What I would expect the formula to be is MgO because magnesium needs to lose 2 electrons and oxygen needs to gain 2 electrons so the ratio would be 1:1 so that in every 1million magnesium atoms there would be 1 million oxygen atoms. Preliminary Work: We would be using a total mass of 150cm, we initially have 200cm but encase of mistakes we would only be using 150cm. Since there would be two experiments I would need to divide the amount of magnesium I have in two so each experiment would be using 75g of magnesium. So this means that the lengths would need to go up in 5 (5cm, 10cm, 15cm, 20cm and 25cm). ...read more.

Middle

20cm 50.32 50.53 0.21 0.21 50.32 0.32 0.315 50.68 50.85 0.21 50.68 0.31 25cm 50.33 50.59 0.26 0.255 50.33 0.41 0.405 50.68 50.93 0.25 50.68 0.40 This table shows two results from each length showing the mass of magnesium and the mass of magnesium oxide produced. The two sets of results from each length were then turned to an average so that we can get an accurate figure. I then plotted a graph of these results so that I can see a clear trend. The reason why I took many readings from different lengths is because so that I could get different readings for mass so that I can plot them on the graph at different points so that it can give me a line of best fit. I can't have only 1 point connecting to zero to make a line of best fit because it would not be a very reliable line. Analysis The type of graph I have is as straight line in a positive correlation. The results are directly proportional which means as the mass of magnesium goes up, the mass of magnesium oxide goes up. The ratio is a constant ratio would always stay the same as long as the magnesium ribbon gets a constant supply of the same amount of oxygen every time throughout the whole experiment. This is because there are more magnesium atoms for oxygen to join with. The formula for working out the slope is: My equation was: This shows that the magnesium oxide which I have produces is Mg3O2 because 1.44 (gradient of Mg3O2) is the closest match to my gradient, 1.5. You can't use anything like half an atom so magnesium would lose 2 electrons and give them to oxygen and because they would become ions the structure of the molecule would be an ionic bond in a giant lattice. ...read more.

Conclusion

mainly small which indicates that my results are accurate, they're just not very reliable because I did not get the formula MgO. I am confident that my results are accurate but I'm not confident enough to believe that my results are reliable because when looking at the logical ratio it should be 24:16 but mine is 72:32 in atomic mass terms. Which does not really make much sense because naturally an atom would want a full shell but in this case only 4 out 5 atoms in the molecule are getting full shells, so I would believe that my results are very unreliable and when looking at what possible formulas there are, there really are loads which means that the formula which I have is very unlikely to be true, in fact many different formulas could have been made during just one test for one length. The method which I have used was quite accurate because of the equipment which we have used. One very accurate equipment which I used was the digital balance (0.01g) and that was very accurate because it was able to measure the small changes in mass. The Bunsen burner was also good to use because it gave a constant supply of heat at roughly the same temperature throughout the experiment. The pipe-clay triangle made your results accurate because it made sure that the crucible was placed securely on the tripod and it also made sure no heat was blocked for the crucible. A method such as waiting for there to be no change in mass was an accurate method because it was easy to see if there was no change in mass. Overall, my whole experiment is not very reliable my results are very accurate, according to my graph. ...read more.

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# Related GCSE Patterns of Behaviour essays

1. ## Formula of Magnesium Oxide

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moles of oxygen: 0.07 moles of magnesium: 0.9 16 24 moles of oxygen= 0.00438 moles = 0.0375 Empirical Formula: mass of magnesium: 0.9g mass of oxygen: 0.07g moles = mass moles = mass molar mass molar mass moles of magnesium: 0.9 moles of oxygen: 0.07 24 16 moles = 0.0375