Report Sheet (Data must be recorded in pen) 1 Heating 2nd Heating Mass of the cr
ID: 550517 • Letter: R
Question
Report Sheet (Data must be recorded in pen) 1 Heating 2nd Heating Mass of the crucible (after heating followed by cooling to room temp.) Mass of the lid (after heating followed by cooling to room temp.) ooling to room temp) ll230%4 Combined mass of crucible and lid Mass of crucible and lid + magnesium Mass of magnesium 2 Mass of the crucible and lid + magnesium oxide (first weighing) Mass of the crucible and lid + magnesium oxide (second weighing) - Mass of magnesium oxide (actual yield) Theoretical Yield of magnesium oxid (For the mass of magnesium that you used in your experiment, what is the maximum amo expect to make? Show calculations below.) unt of magne Percent Yield of magnesium oxide. What was the percent yield of magnesium oxide in your experiment? Show calculations below.Explanation / Answer
Mass of magnesium oxide (actual yield) = (mass of crucible and lid + magnesium oxide) – (combined mass of crucible and lid) = (27.3326 g) – (27.2814 g) = 0.0512 g.
Theoretical yield of magnesium oxide
Write down the balanced chemical equation for the formation of magnesium oxide.
2 Mg + O2 -------> 2 MgO
As per the stoichiometric equation,
2 mole Mg = 2 mole MgO.
We will need the atomic and molar masses of Mg and MgO.
Atomic mass of Mg = 24.305 g/mol; molar mass of MgO = (1*24.305 + 1*15.9994) g/mol = 40.3044 g/mol.
Mole(s) of Mg corresponding to 0.0374 g Mg = (0.0374 g)/(24.305 g/mol) = 1.53878*10-3 mole.
Mole(s) of MgO expected = (1.53878*10-3 mole Mg)*(2 mole MgO/2 mole Mg) = 1.53878*10-3 mole MgO.
Mass of MgO expected = (1.53878*10-3 mole)*(40.3044 g/mol) = 0.0620 g (ans).
Percent yield of magnesium oxide
We expected 0.0620 g MgO, but we got 0.0512 g MgO instead. Therefore,
Percent yield = (actual yield)/(theoretical yield)*100 = (0.0512 g)/(0.0620 g)*100 = 82.5806% 82.581% (ans).