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% Composition, Empirical and
Molecular Formulas
What makes up a compound?
PERCENT COMPOSITION
Mass of element
x 100
Total mass of compound
What is the % composition of Hydrogen
and Oxygen in H2O?
H = 1 amu x 2 atoms = 2
O = 16 amu x 1 atom = 16
total
18 g/mol
% H is 2 x 100 = 11%
18
% O is 16 x 100 = 89%
18
Another example:
What is the percent composition of CaSO4?
1. Ca: 40
2.
S: 32 O: 4x16 = 64
40 + 32 + 64 = 136
% Ca
40 x 100 = 29.4%
136
%S
32 x 100 = 23.6%
136
%O
64 x 100 = 47.1%
136
29.4 + 23.6 + 47.1 = 100%
We can also calculate the mass of an element
in a given amount of a compound:
What is the mass of Oxygen in 5 g of H2O?
x g O = 16 g O
5 g H 2O
18x = 80
18 g H2O
x = 4.44 g O
Empirical and Molecular formulas
ARRIVING AT CHEMICAL
FORMULAS
We can use math to show the relationship
between the empirical and molecular formulas.
Molar mass
= 294.3 g/mol = 1
Empirical formula
294 g/mol
So the empirical formula is the molecular formula.
Determining Empirical Formula
1. If given as a %, remove the % and change to grams
(assume you have 100 g of the compound)
2. Convert g to moles
3. Select the lowest number of moles and divide each
number of moles by this number
4. If the number divides out evenly, these are the
subscripts for the elements in the compound
5. If any of the numbers have a .5, MULTIPLY them ALL
by TWO & then place these numbers as the
subscripts.
Example:
What is the empirical formula of a substance
that is 40% carbon, 6.7% hydrogen and
53.3% oxygen by mass?
Step 1: change % to g (assume 100 g): mc = 40 g, mH = 6.7 g, mO = 53.3 g
Step 2: convert mass (g) to moles: nc = 40 g/12 g/mol = 3.33 mol
nH = 6.7 g/1.01g/mol = 6.63 mol; nO = 53.3 g/16 g/mol = 3.33 mol
Step 3: divide each element by the lowest quantity:
3.33 mol C = 1
3.33
6.63 mol H = 1.99
3.33
3.33 mol O = 1
3.33
Step 4: Since the numbers divide evenly, these are the subscripts
for the compound.
The empirical formula for this compound is CH2O
Adipic acid contains 49.32% C, 43.84% O
Another Example: and 6.85% H by mass. Determine its
empirical formula.
Treat % as mass
and convert grams
to moles:
Divide each value of
moles by the smallest of
the values:
Because we don’t have whole number ratios, we
have to determine the lowest number that will
give you whole number ratios (or step 5: because
we have a .5. multiply everything by 2)
Carbon
Hydrogen
Oxygen
1.5 x 2 = 3
(rounded)
2.5 x 2 = 5
1.5 x 2 = 3
These are the whole number ratios that become the
subscripts for the empirical formula. Therefore:
Adipic acid’s empirical formula is C3H5O2
Calculating the molecular formula for a
compound:
What is the molecular formula for a
compound with empirical formula of CH4 and
a molecular mass of 48 g/mol?
1. Given the empirical formula: CH4
2. Calculate the molar mass of the empirical
formula: 16 g/mol
3. Divide the molecular mass by the molar
mass: molecular mass of compound
molar mass empirical formula
What is the molecular formula for a compound with empirical formula of
CH4 and a molecular mass of 48 g/mol?
3. Divide the molecular mass by the molar mass:
molecular mass = 48 g/mol = 3
molar mass
16 g/mol
4. This is the ratio (relationship) between the
empirical and molecular formulas! Use this
number to multiply the subscripts of the
empirical formal to get the molecular
formula.
CH4 x 3 = C3H12
Example:
P2O5
A white powder is analyzed and found to
have an empirical formula of P2O5. The
compound has a molecular mass of 283.88
g/mol. What is its molecular formula?
Step 1: calculate the molar mass of the empirical formula:
P = 2(30.97) = 61.94 g
O = 5(16) = 80.00 g
= 141.94 g
Step 2: divide molecular mass
empirical formula molar mass
283.88 = 2
141.94
Step 3: multiply empirical formula by this ratio
(P2O5)2 = P4O10
Another example:
You found Adipic acid’s empirical formula to be
C3H5O2. It’s molecular mass is 146 g/mol. What is its
molecular formula?
Molecular mass (compound)
Molar mass (empirical formula)
Multiply the empirical formula by the ratio number.
STOICHIOMETRY
How much of a product
can you expect?
Stoichiometry
Reaction stoichiometry involves the mass relationships
(mole ratios*) between reactants and products in a
chemical reaction.
For example: in the reaction 4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l)
• Start with a balanced equation
• Compare the mole ratios: 4 mol NH3; 6 mol NO; 4 mol NH3; 6 mol NO
5 mol N2
5 mol N2
6 mol 6H2O 6 mol 6H2O
• Use these ratios as conversion factors to solve problems.
*mole ratio: a conversion factor relating the amounts, in moles, of any 2
substances in a chemical reaction.
Example (ch. 9, pg 305)
In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by
reacting it with lithium hydroxide, LiOH. Based on the following reaction,
how many moles of LiOH are required to react with 20 mol of CO2 (the ave.
amt exhaled by a person each day)?
Step 1: write & balance the equation
CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O(l)
Step 2: choose a conversion factor. In this problem we are asked
to compare LiOH and CO2 . 2 mol LiOH
1 mol CO2
Step 3: Set up and solve – we want our answer in mol LiOH
20 mol CO2 x 2 mol LiOH = 40 mol LiOH
1 mol CO2
Another example - converting mol to mass (ch. 9, pg 306)
In photosynthesis, plants use energy from the sun to produce glucose,
C6H12O6, and oxygen from the reaction of carbon dioxide and water. What
mass, in grams, of glucose is produced when 3 mol of water react with
carbon dioxide?
Start with the balanced equation: 6CO2(g) + 6H2O(l) → C6H12O6 (s) + 6O2(g)
Use the mole ratio between water and glucose.
3 mol H2O x 1 mol C6H12O6 = ½ mol C6H12O6
6 mol H2O
Use the molar mas of C6H12O6 and convert mol to mass.
C6H12O6 = 180.18 g/mol
180.18 g x .5 mol C6H12O6 = 90.1 g
mol
What if you don’t have enough of something?
LIMITING REAGENTS
We want to bake cookies! Here’s the recipe:
2 cups of flour,
1 cup chocolate chips,
2 eggs, and water
Yields: 12 cookies.
How many cookies can we
make if we’re given 100
cups of flour, 100 cups of
chocolate chips, an
unlimited amount of water,
and 4 eggs.
But we can only bake 24 cookies. Why?
We run out of eggs after making the 24 cookies. So
we say that the eggs are our limiting ingredient.
In chemistry, we use the word reagent instead
of ingredient, but it means the same thing. Eggs
are our limiting reagent.
What about cheese sandwiches?
We have 6 slices of whole-grain bread* and 2 slices of
thickly sliced Swiss cheese. Our recipe calls for 1
slice/sandwich. How many sandwiches can we make?
Just 2. Even
though we have
more bread left
over, cheese is
our limiting
reagent.
*Don’t even think about eating that bleached, refined white bread nonsense.
How about a more sophisticated sandwich?
How about a more sophisticated sandwich?
Limiting & Excess Reactants
The limiting reactant is the reactant that limits the
amount of the other reactant(s) that can combine and
therefore, the amount of products that can form in a
chemical reaction.
The excess reactant
is the compound
that is not used up
completely in the
chemical reaction.
Limiting
reactant
Now you try it – before we do the
chemistry….
Smore recipe:
1 marshmallow
2 squares of chocolate
2 halves of graham crackers
1. How many smores can you make?
2. What is the limiting ingredient?
A chemical example:
10 g of Aluminum reacts with 35 g of chlorine gas to produce
aluminum chloride. Which reactant is limiting? Which is in
excess? And how much product is produced?
2 Al + 3 Cl2 → 2 AlCl3
Al
10 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3 = 49.4 g AlCl3
27 g Al 2 mol Al
1 mol AlCl3
Cl2
35 g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3 = 43.9 g AlCl3
71 g Cl2 3 mol Cl2 1 mol AlCl3
Cl2 is the limiting reagent, Al is in excess
One more for good measure:
Ca(s) +2H2O(l) → Ca(OH)2(aq) + H2(g)
If we have 2 g of each reactant, identify the limiting reagent
and determine the mass of Ca(OH)2 produced.
Ca
2 g Ca 1 mol Ca 1 mol Ca(OH)2 74.09 g Ca(OH)2 = 3.70 g Ca(OH)2
40.08 g Ca 1 mol Ca
1 mol Ca(OH)2
H2O
2 g H2O 1 mol H2O 1 mol Ca(OH)2 74.09 g Ca(OH)2 = 4.11 g Ca(OH)2
18 g H2O
2 mol H2O
1 mol Ca(OH)2
Calculating % yield
• Stoichiometry tells us what should happen for
a reaction. This is the theoretical yield.
• We predict using stoichiometry conversions
• The amount that a reaction actual produces is
known as the actual yield.
The ratio of the two gives the % yield
New vocabulary:
yield: the amount of product produced in a
chemical reaction
theoretical yield: the amount of product predicted
by the balanced equation, when all of the limiting
reactant has reacted (what you expect from math
calculations)
actual yield : the amount of product actually
obtained in a reaction (what you measure in the lab)
percentage yield: the ratio of the actual yield
to the percentage yield
What Percent Yield Means
• Percent yield is often lower than 100%
because of error
• Every experiment has sources of error:
• Not all product is recovered (e.g. spattering)
• Reactant impurities (e.g. weigh out 100 g of
chemical which has 20 g of junk)
• A side reaction occurs (e.g. MgO vs. Mg3N2)
• The reaction does not go to completion
• Accurate measurements are not taken.
Calculating the Percent Yield
• Percent yield can be calculated by the
equation:
actual yield
´100% = Percent yield
theoretical yield
• Calculate the theoretical yield using
stoichiometry.
• Often, you experiment to obtain the actual
yield.
When 5.00 g of KClO3 is heated it decomposes
according to the equation:
2KClO3 → 2KCl + 3O2
Calculate the theoretical yield of oxygen.
5g KClO3 mol KClO3 3 mol O2
32 g O2 = 1.96 g O2
122.55 g 2 mol KClO3 1 mol O2
Give the % yield if 1.78 g of O2 is produced.
% yield =
actual x 100%
theoretical
=
1.78 g O2 x 100% = 90.9%
1.958 g O2
How much O2 would be produced if the percentage
yield was 78.5%?
3. How much O2 would be produced if the percentage yield
was 78.5%?
2KClO3 → 2KCl + 3O2
actual
x 100%
% yield =
theoretical
78.5% =
x g O2
1.958 g O2
x 100%
x g O2 = .785 x 1.958 g O2 = 1.537 g O2
Try another problem:
What is the % yield of H2O if 138 g H2O is
produced from 16 g H2 and excess O2?
Step 1: write the balanced chemical equation
Step 2: determine actual and theoretical yield.
(Actual is given, theoretical is calculated)
Step 3: Calculate % yield
What is the % yield of H2O if 138 g H2O is
produced from 16 g H2 and excess O2?
Step 1: write the balanced chemical equation
2 H2 + O2 → 2 H2O
By this equation we see the compounds react
in a ratio of:
2 moles H2: 1 mole O2: 2 moles H2O
What is the % yield of H2O if 138 g H2O is produced
from 16 g H2 and excess O2?
Step 2: determine actual and theoretical yield.
Actual is given, theoretical is calculated:
2H2 + O2 → 2H2O
16 g H2 x1 mol H2 x2 mol H2O x18.02 g H2O= 143 g
2.02 g H2 2 mol H2
1 mol
H2O
theoretical yield
What is the % yield of H2O if 138 g H2O is produced
from 16 g H2 and excess O2?
actual yield = 138 g
theoretical yield = 143 g
H2O
Step 3: Calculate % yield
% yield =
actual
x 100%
theoretical
138 g H2O
x 100% = 96.7%
143 g H2O
Try another problem
Q - What is the % yield of NH3 if 40.5 g NH3 is
produced from 20.0 mol H2 and excess N2?
Step 1: write the balanced chemical equation
Step 2: determine actual and theoretical yield.
Actual is given, theoretical is calculated:
Step 3: Calculate % yield
What is the % yield of NH3 if 40.5 g NH3 is
produced from 20.0 mol H2 and excess N2?
Step 1: write the balanced chemical equation
N2 + 3 H2 → 2 NH3
By this equation we see the compounds react
in a ratio of:
1 moles N2: 3 mole H2: 2 moles NH3
What is the % yield of NH3 if 40.5 g NH3 is
produced from 20.0 mol H2 and excess N2?
N2 + 3H2 → 2NH3
Step 2: determine actual and theoretical
yield. Actual is given, theoretical is
calculated:
# g NH3= 20.0 mol H2 x 2 mol NH3 x 17.04 g NH3 = 227 g
3 mol H2
1 mol NH3
theoretical yield
What is the % yield of NH3 if 40.5 g NH3 is
produced from 20.0 mol H2 and excess N2?
actual yield = 40.5 g
theoretical yield = 227 g
g NH3
Step 3: Calculate % yield
% yield =
actual
x 100%
theoretical
40.5 g H2O
x 100% = 17.8%
227 g H2O
Let’s Try it
for reals!