Simplest Formula for a Compound Experiment 9

Purpose –To find the simplest formula of a compound using experimental techniques.  Simplest or Empirical Formula – A chemical formula that shows the kind of atoms and their relative numbers in a substance. Theoretical yield – The quantity of product that is (predicted) calculated to form when the limiting reagent reacts completely. Experimental yield – The actual quantity of product formed during the experiment (Rarely = Theoretical yield). % Yield – Compares the experimental yield to the theoretical yield. Describes the efficiency of the reaction.

The limiting reagent controls how much product that can form. Givens Mg = 100grams O2 = 100 grams Mg + O2 Mg O2

How to determine limiting reactant in 3 steps We must balance the equation 2Mg+ O2 →2MgO

Step 2 Determine the theoretical yield of a product using both given reactants 100grams Mg 1 mol of Mg 1 mol of MgO 40g MgO 24.30g 1 mol of Mg 1 mol of MgO =1.6 grams MgO 100grams O2 1 mol of O2 1mol MgO 40 g MgO 32.00g 1mol O2 1 mol MgO =1.25 grams MgO

Step 3 The smaller theoretical yield is the one you can produce and the reactant that produces it is limiting 100grams Mg 1 mol of Mg 1 mol of MgO 40g MgO 24.30g 1 mol of Mg 1 mol of MgO =1.6 grams MgO 100grams O2 1 mol of O2 1mol MgO 40 g MgO 32.00g 1mol O2 1 mol MgO =1.25 grams MgO

1.00 grams MgO 1.02 grams MgO Determine the percent yield Experimental yield (actual) – this is determined by doing the lab. When I did the lab I produced 1.00g of MgO Determine the percent yield (actual yield)/(Theoretical yield) X 100 1.00 grams MgO 1.02 grams MgO X 100 = 98%

How to Determine Empirical Formula from data in 3 easy steps 1. Determine the # of moles present for each reactant based on the balanced formula 2Mg+ O2 →2MgO For my trial convert grams to moles .200 g O = 1.25 X10-2 mol .300 g Mg = 1.24X10-2mol

How to Determine Empirical Formula from data 2. Divide both by the smaller # of moles For my trial 1 .25 X10-2 mol O = 1.01 1.24X10-2 mol of Mg For my trial 1 .24 X10-2 mol O = 1.00 1.24X10-2 mol of Mg Decimals are not allowed so my final answer is MgO This is your answer for question # 1 pg 131

Magnesium reacts with air: 2Mg + O2  2MgO 3Mg + N2  Mg3N2 We only want MgO! That’s why you stop heating and add H2O. It removes the N2 group. Mg3N2 + 6H2O  3Mg(OH)2 + 2NH3 Mg(OH)2  MgO + H2O

Procedures: Obtain and wash a crucible and crucible cover. Make sure you use one of the newer Bunsen burners Dry crucible over a Bunsen burner for 10 minutes Once cool, determine the mass of the empty crucible. Obtain between 5.0 and 10.0 cm of Mg ribbon. Clean the Mg with steel wool. Once clean avoid touching the Mg Add the Mg to your crucible and determine the mass and the crucible.

Procedures continued Heat crucible for 10 minutes Turn off heat and allow to cool for 5 min. Add 30 drops of H2O Heat gently for 5 min. then strongly for 5 more minutes to remove the water. Allow crucible to cool approx. 10 minutes. Determine the mass of the crucible.

Due Next Week Complete data table on pg 131 Answer questions on pgs 131-133