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# Determination of Formula (MgO)

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Please see attached PDF to review the experiment. I'm looking for calculations to complete the table, and answers to the proceeding questions.

https://brainmass.com/chemistry/stoichiometry/determination-formula-mgo-509895

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Calculations:
Trial 1 Trial 2
Wt of MgO 0.574 0.546
Wt of O 0.222 0.213
% MgO 61.32 60.99
Mean % 61.16 61.16
% error 1.69 1.14

The weight of MgO is the weight of MgO+crucible - wt of crucible
The weight of O is the wt of MgO - wt of Mg
The % Mg is the (wt of Mg / wt of MgO ) x 100
Mean % is the sum of both trials / 2
Error % = (Experimental value - Theoretical value) x 100 / Theoretical value

1. % Mg in MgO = (24.3 g / 40.3 g) x 100 = 60.30%

% O in MgO = (15.99 g / 40.3 g) = 39.70%

2. Amount of MgO = 2.033 g Mg x (1 mol Mg / 24.3 g Mg) x (2 mol MgO / 2 mol Mg) x (40.3 g MgO / 1 mol MgO) = 3.372 g MgO

3. a) % Ca = (40. 078g / 56.0774) x 100 = 71.47%

b) Amount of CaO = 1.358g Ca x (1 mol Ca / 40.078 g Ca) x (2 mol CaO / 2 mol Ca) x (56.1 g CaO / 1 mol CaO) = 1.900 g CaO

4. When crucibles are heated, they must be cooled properly to room temperature. Even a slight change in temperature can result in lower weights as moisture will not be absorbed.

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