Chemistry Lab Report: Determining the Percentage Composition of Magnesium Oxide Submitted by: Diwa Koirala Date:- 6th May 2015 Introduction: Law of Definite Proportions is the law stating that a specific compound always contains the same elements in definite proportions by mass. One has to experimentally determine the composition by mass of a substance and then convert the mass amounts to percentages in order to get the percentage composition. Magnesium is a silvery metal that burns with such a bright flame that it was once used in flashbulbs for photography. Magnesium oxide is a grey/white powder which is produced as a result of the synthesis reaction of combustion magnesium. Hypothesis: The percentage composition by mass of magnesium oxide will be 60.3% magnesium and 39.7% oxygen. Procedure/Material: Chemistry 11, F. Mustoe, MY. Jansen, T. Doram, McGraw-Hill Ryerson, Toronto, Canada, 2001, pg. 282- 283. Analyze and Interpret: 1. What mass of magnesium was used in the reaction? Solution 1: Mass of Mg = Mass of crucible, lid, and magnesium (g)-Mass of clean, empty crucible and lid(g) = 21.73g - 21.64g = 0.09g Solution 2: Mass of Mg = Mass of magnesium oxide produced (g)-Mass of oxygen that must have reacted(g) = 0.13g - 0.04g = 0.09g 1 64g = 0. Use the data to calculate the percentage composition of magnesium oxide.09g Mass of MgO = 0. and magnesium oxide(g) .13g = 30. lid. Calculate the mass of oxygen that reacted with the magnesium.04g Conclude and Communicate: 4.13g = 69.04g Mass of Mg = 0. Calculate the percentage error (PE) using the difference between the experimental mass percent (EP) of magnesium and the actual mass percent (AP) of magnesium.21.2.8% 5. Solution: Mass of O = 15. What mass of magnesium oxide was produced? Solution: Mass of MgO = Mass of crucible.Mass of clean. and magnesium oxide (g) .73 = 0.77g .09g X 100% 0. empty crucible and lid(g) = 21. lid.305g 2 . and magnesium (g) = 21. Solution: Mass of O = 0.999g Mass of Mg = 24. Solution: Mass of O = Mass of crucible.Mass of crucible.04g X 100% 0.13g Mass Percent of Mg = Mass of Mg X 100% Mass of MgO = 0.13g 3. lid.2% Mass Percent of O = Mass of O X 100% Mass of MgO = 0.21.77g . the amount of MgO smoke lost from 0.5 % 6. As a result.AP X 100% AP = (30.2 .7 % Percentage Error Calculation: PE(Mg) = ? EP(Mg) = 69. As a result.305 g/mol) + (15.Mass of MgO = (24. For example.04 g Mass of MgO = 0. Suppose that some magnesium oxide smoke escaped during the investigation.5 % = 22.8 % PE (O) = EP . 3 .7 = . so mass of Mg would be 0. the magnesium to oxygen ratio would remain unchanged.3 = 14.024g.13g of MgO is 0.13 g The loss of magnesium oxide would result in the loss of an equal amount of both magnesium and oxygen atoms.3 % Mass Percent of O = Mass of O X 100% Mass of MgO = 15.2 % AP(Mg) = 60. the 0.08g produced.22.999 g/mol) = 40.7 % Solution 1: Solution 2: PE (Mg) = EP .304 g = 60.3) X 100% 60. then the new mass of MgO would be 0.05 g.3 % PE(O) = ? EP(O) = 30.305 g X 100% 40.999 g X 100% 40.304 g = 39.8 .0.08g of MgO). decreased.08g of the reacting products must be in the equal ratio 9:4 to form the mass of 0.13g . or remained unchanged? Explain your answer using sample calculations. Mass of Mg = 0.39.056g and mass of O would be 0.8 % AP(O) = 39.7) X 100% 39.AP X 100% AP = (69.60.304 g/mol Mass Percent of Mg = Mass of Mg X 100% Mass of MgO = 24.08g (0.09 g Mass of O = 0. The loss of magnesium oxide would result in a loss of an equal amount of both magnesium and oxygen. Would the Mg:O ratio have increased.05g = 0. 8g (0.13g) Half Mg reacted.02g = 0. The apparent mass of MgO would be less than expected as you would have expected 0.02g) = (MgO: 0. Therefore.09g) + (O: 0.05g + 0.7. How would the value you calculated for the percentage composition of magnesium oxide have been affected if all the magnesium in the crucible had not reacted? Support your answer with sample calculations.05 g of MgO making the total mass appear to be 0.04g) = (MgO: 0. (Mg: 0.09g X 100% 0.05g = 60% 4 .13 g Mass Percent of Mg = Mass of Mg X 100% Mass of MgO = 0.04 g Mass of MgO = 0.03 g Mass of O = 0.05 g Mass Percent of Mg = Mass of Mg X 100% Mass of MgO = 0.09 g Mass of O = 0.13g = 69. in which makes the total mass lower than expected when under complete combustion. (Mg: 0.13g for the mass of MgO.03 g of unreacted Mg together with 0.02 g Mass of MgO = 0.5g) For instance. if only half of the Mg reacted you would have 0.03g X 100% 0. Current Percentage Composition: Mass of Mg = 0.03g) + (O: 0.2% Sample Percentage Composition: Mass of Mg = 0.03g). the mass is going to be less as no oxygen combines with the unreacted magnesium.03g + 0. 5 .