Simplest Formula for a Compound Simplest Formula for a Compound Experiment 9

Empirical formula – one hot dog and 1 roll

Can not buy just one hot dog and roll!! 8 hot dogs 6 rolls

Which one would you run out of first !! 8 hot dogs 6 rolls Limiting reactant

When you buy just the right amount of each that is your balanced equation 4 Ro 6 + 3Hd 8 24 Hot Dogs Ro = hot dog rolls (6 rolls in a package) Hd = hot dogs (8 hot dogs in a package)

1.Simplest or Empirical Formula – A chemical formula that shows the smallest amount of atoms necessary to create the product. 2.Theoretical yield – The quantity of product that is (predicted) calculated to form when the limiting reagent reacts completely. 3.Experimental yield – The actual quantity of product formed during the experiment (Rarely = Theoretical yield). Purpose –To find the empirical formula of a compound using experimental techniques.

% Yield Compares the experimental yield to the theoretical yield. Describes the efficiency of the reaction. Formula (actual yield)/(Theoretical yield) X 100

Givens –Mg = 100grams – O 2 = 100 grams –Mg + O 2 The limiting reagent controls how much product that can form. Mg O

How to determine limiting reactant in 3 steps Step 1 We must balance the equation 2Mg+ O 2 → 2MgO

Step 2 Determine the theoretical yield of a product using both given reactants 2Mg+ O 2 → 2MgO 100grams Mg 1 mol of Mg 2mol of MgO 40g MgO 24.30g 2 mol of Mg 1 mol of MgO =82.3 grams MgO 100 grams O 2 1 mol of O 2 2mol MgO 40 g MgO 32.00g 1mol O 2 1 mol MgO =250 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 2 mol of MgO 40g MgO 24.30g 2 mol of Mg 1 mol of MgO =82.3 grams MgO 100grams O 2 1 mol of O 2 2mol MgO 40 g MgO 32.00g 1mol O 2 1 mol MgO =250 grams MgO

Experimental yield (actual) – this is determined by doing the lab. When I did the lab I produced 75.0g of MgO Determine the percent yield (actual yield)/(Theoretical yield) X grams MgO 82.3 grams MgO X 100 = 91.1%

Empirical Formula 1.A chemical formula that shows the smallest amount of atoms necessary to create the product.

How to Determine Empirical Formula from data in 3 easy steps 1. Determine the # of moles present for each element For my trial convert grams to moles.200 g Oxygen = 1.25 X10 -2 mol.300 g Mg = 1.24X10 -2 mol

How to Determine Empirical Formula from data 2. Divide both by the smaller # of moles Decimals are not allowed so my final answer is MgO For my trial 1.25 X10 -2 mol O = X10 -2 mol of Mg For my trial 1.24 X10 -2 mol O = X10 -2 mol of Mg

Procedures: 1.Obtain and wash a crucible and crucible cover. 2.Make sure you use one of the newer Bunsen burners 3.Dry crucible over a Bunsen burner for 10 minutes 4.Once cool, determine the mass of the empty crucible. 1.Obtain between 5.0 and 10.0 cm of Mg ribbon. 2.Clean the Mg with steel wool. 3.Once clean avoid touching the Mg 4.Add the Mg to your crucible and determine the mass and the crucible.

1.Heat crucible for 10 minutes 2.Turn off heat and allow to cool for 5 min. 3.Add 30 drops of H 2 O 4.Heat gently for 5 min. then strongly for 5 more minutes to remove the water. 5.Allow crucible to cool approx. 10 minutes. 6.Determine the mass of the crucible. Procedures continued

2Mg + O 2  2MgO 3Mg + N 2  Mg 3 N 2 We only want MgO! That’s why you stop heating and add H 2 O. It removes the N 2 group. Mg 3 N 2 + 6H 2 O  3Mg(OH) 2 + 2NH 3  Mg(OH) 2  MgO + H 2 O  Nitrogen reacts with water:

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