1. Magnesium is the limiting reactant in this experiment. Calculate the theoretical yield of MgO for trial 1 (the same for trial 2)
Mass of Mg for each trial = Mass of Mg metal, crucible, and lid - Mass of empty crucible with lid = 26.931 g - 26.679 g = 0.252 g
Then, you need to write out the chemical reaction:
2Mg + O2 ==> 2MgO (balanced)
After this, based on the reaction above, 2 moles of Mg is needed to make 2 MgO.
To get the moles of Mg you need to take the grams you started with (0.252 g) and convert them to moles: 0.252 g/24.305 g/mol (molecular weight Mg) = 0.0100 moles.
2. Determine the percent yield of MgO for your experiment for trial 1
For the theoretical yield, we assume 100% of the Mg will be converted into MgO, so, based on the first equation, you would get 0.0100 moles of MgO (because all Mg are consumed to form MgO).
You multiply this by the molecular weight of MgO (40.30 g/mol) and the result is 0.403 g MgO.
The percent yield is what you actually got in the experiment, and you have to subtract the total mass from the crucible mass: 27.090 g - 26.679 g = 0.411 g of MgO obtained
Then % yield = Actual yield/ theoretical yield x 100
%yield = (0.403 g/0.411) g x 100 = 98.05 %
3. Determine the average percent yield of MgO for the two trials.
Average % yield = (% yield 1 + % yield 2) / 2
(remember to do the same procedure as 1 to get the results for trial number 2)