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# This Is Lab Report About The Synthesis And Determination Of Empirical Formula For Magnesium Oxide.

517 words - 2 pages

The Synthesis and Determination of Empirical Formula for Magnesium OxideAim:To determine the empirical formula for magnesium oxide.Data Collection:Table 1 Mass of empty crucible, crucible + Mg (roughly 1cm pieces of a 15-cm piece of magnesium ribbon), and crucible +MgxOy. (mass measured with an analytical balance). The mass of crucible + MgxOy was obtained after a continuous process of heating, letting cool, heating with water enough to immerse the content and letting cool of the Mg.Mass of Empty crucible (g)Mass of Crucible + Mg (g)Mass of Crucible +MgxOy (g)16.04716.21416.294Uncertainty of analytical balance = + 0.002*Note that because we are uncertain if Mg has completely reacted with O2 in the first heating, if all the Mg have reacted with O2 instead of N and if the water added has completely vaporized in the second heating(with water), the data and the following calculations may not be as accurate.Data Presentation and Procession:1.mass of crucible + Mg - mass of empty crucible =Mass of Mg16.214 ±0.002 g- 16.047± 0.002 g0.167 ± 0.004 g2.mass of crucible + MgxOy -mass of crucible +Mg = mass of O16.294 ±0.002g- 16.214 ±0.002g0.0800±0.0043.mass of Mg =0.167±(0.004/0.167*100%)=0.167±2 % g4.mass of O = 0.0800±(0.004/0.0800*100%)=0.0800±5% g5.Mole O = (0.0800±5%g) /(15.999g/mol)=0.00500 ±5%mol6.Mole Mg = (0.167±2% g)/(24.305g/mol)=0.00687±2%mol7.Relative ratio of Mg = 0.00687/0.00500 ±(2 % +5%)mol=1.37±7%mol=1.37± (7*1.37/100)=1.37±0.18.Relative ratio of O = 0.00500/0.00500 ±(5%+5%)=1.00±10%=1.00±(10*1.00/100)=1.00±0.19.Empirical Formula= Mg1.37±0.1O1.00±0.1=MgOTable 2 Mass and relative ratio of...

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