Simplest Formula for a Compound
Experiment 9
Empirical formula – one hot dog and 1 roll
Can not buy just one hot dog and
roll!!
6 rolls
8 hot dogs
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 Ro6 + 3Hd8
24 Hot Dogs
Ro = hot dog rolls (6 rolls in a package)
Hd = hot dogs (8 hot dogs in a package)
Purpose –To find the empirical formula of a compound using
experimental techniques.
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).
% Yield
Compares the experimental yield to the
theoretical yield. Describes the
efficiency of the reaction.
Formula
(actual yield)/(Theoretical yield) X 100
The limiting reagent controls
how much product that can form.
• Givens
– Mg = 100grams
– O2 = 100 grams
– Mg + O2
Mg O
How to determine limiting reactant
in 3 steps
Step 1
We must balance the equation
2Mg+ O2 →2MgO
Step 2
Determine the theoretical yield of a product using
both given reactants
2Mg+ O2 →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 O2 1 mol of O2 2mol MgO 40 g MgO
32.00g 1mol O2
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 O2 1 mol of O2 2mol MgO 40 g MgO
32.00g 1mol O2
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 100
75.0 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 = 1.24X10-2mol
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
Procedures:
1.
2.
3.
4.
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.
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.
Procedures continued
1.
2.
3.
4.
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.
5. Allow crucible to cool approx. 10 minutes.
6. Determine the mass of the crucible.
Nitrogen reacts with water:
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
Due Next Week
• Complete data table on pg 131
• Answer questions on pgs 131-133