Empirical Formula - Waterford Schools

Using the Mole to Calculate
% Composition, Empirical Formulas
and Molecular Formulas
Illustrating Compound
Composition
• So far, we have seen the composition of many chemical compounds
in the form of chemical formulas
• Up until now, you believed a chemical formula was just the ratio of
number of atoms in a compound
• Chemical formulas are also the ratio of MOLES of atoms in a
compound!
• For example:
In 1 MOLE of CO2, there is 1 MOLE of C and 2 MOLES of O
Law of Definite Proportions
• In chemistry, a chemical compound always contains
the same elements in certain definite proportions
• Defined by a scientist named Proust in 1799
• Called the Law of Definite Proportions
• Because the atoms of the elements in a compound
are combined in constant proportion, they are also
combined in a definite proportion by mass
• Examples:
• Sodium chloride is always 39.3% sodium and 60.7%
chlorine by mass, no matter what its source
• Water is always 11.2% hydrogen and 88.8% oxygen by
mass
More on the Law of Definite
Proportions
• In a compound, the ratios by mass of the elements in
that compound are fixed independent of the origins
or preparation of that compound
• A compound is unique because of the specific
arrangement and weights of the elements which
make up the compound
• In other words, elements combine in specific ratios to
form compounds
• That is, elements combine in whole numbers!
• It is not possible to have a compound with a portion of an
atom
A drop of water, a glass of water, and a lake of water all contain
hydrogen and oxygen in the same percent by mass.
Example – Atom View
Example – Mass View
How is the Composition of a
Compound Measured?
• A chemist could measure composition of a compound in two
different ways:
• Count numbers of constituent atoms
• Atoms are too small to see so this method is doesn’t work!
• Percentages (by mass) of its elements
• Molar masses are used because these are measureable!
• So to calculate percent composition of a covalent molecule
or an ionic formula unit, you must assume you have one
mole of the substance:
Molar Mass of Element
% Composition =
× 100
Molar Mass of Molecule or Formula Unit
Steps to Calculate %
Composition
1. Write correct formula of compound with subscripts (if
necessary)
2. Calculate mass of each element in 1 mole of the compound
•
In other words, get the molar mass of each element
3. Calculate the molar mass of compound
4. Find the fraction of the total mass contributed by each
element and convert it to a percentage
Molar Mass of Element
× 100
Molar Mass of Molecule or Formula Unit
5. Sum the individual mass percent values to make sure they
add up to 100%!
Practice!
• Carvone is a substance that occurs in two forms having different arrangements of the
atoms but the same molecular formula of C10H14O. One type of carvone gives caraway
seeds their characteristic smell, and the other type is responsible for the smell of
spearmint oil. Compute the mass percent of each element in carvone.
• Step 1: Write correct formula of compound with subscripts
C10H14O
• Step 2: Find the masses of each element in 1mole of carvone:
12.01 g
Mass of C in 1 mole = 10 mol ×
= 120.1 g
1 mol
Mass of H in 1 mole = 14 mol ×
Mass of O in 1 mole = 1 mol ×
1.008 g
= 14.11 g
1 mol
16.00 g
= 16.00 g
1 mol
• Step 3: Get molar mass of compound
Mass of 1 mol C10 H14 O = 120.1 + 14.11 + 16.00 = 150.2 g
Practice!
• Step 4: Find the fraction of the total mass contributed by each
element and convert it to a percentage:
120.1 g C
Mass percent of C =
× 100% = 79.96%
150.2 gC10 H14 O
14.11 g H
Mass percent of H =
× 100% = 9.394%
150.2 gC10 H14 O
16.00 g O
Mass percent of O =
× 100% = 10.65%
150.2 gC10 H14 O
• Step 5: Sum the individual mass percent values to make sure they
add up to 100%!
Practice!
• What is the percent composition of potassium permanganate?
• Formula of potassium permanganate = KMnO4
• Molar mass of KMnO4:
g
39.1 + 54.9 + 4 ∙ 15.99 = 157.69 mol
• %K=
39.1 g K
157.69 g KMnO4
• % Mn =
• %O=
× 100 = 24.7%
54.9 g Mn
157.69 g KMnO4
63.96 g O
157.69 g KMnO4
× 100 = 34.8%
× 100 = 40.5%
Practice!
• What is the percent composition of sodium oxalate?
• What is the mass of bromine in 50.0 grams of potassium
bromide? HINT – First, calculate the mass % of bromine then
multiply by the mass of sample!
Introducing a New Type of Compound…The
Hydrated Salt
• A hydrated salt is an ionic compound that
has water molecules trapped within the
crystal lattice
• Examples:
CuSO4·5 H2O
Copper (II) sulfate pentahydrate
FeCl3·6 H2O
Iron (III) chloride hexahydrate
• An anhydrous salt is an ionic compound
without water molecules
• Examples:
CaCl2
CuCl2
• The molar mass of hydrated compounds are
larger due to the waters
Using % Composition to Calculate
Water in a Hydrated Salt
• When a hydrated crystal is heated to very high temperatures, the
crystal can lose water and become dehydrated (anhydrous)
• For example, the chemical reaction for the dehydration of
magnesium sulfate is:
MgSO4 ∙ n H2 O s → MgSO4 s + n H2 O (g)
• The mass of water lost and the mass of the hydrated salt can be
used to calculate the percent water in the hydrated salt
Mass of H2 O lost
% H2 O =
× 100
Total mass of hydrate
• Can use this information to calculate the empirical formula of the
salt
• More on this later!
TIME FOR A DEHYDRATION LAB!
What is an Empirical Formula?
• As mentioned, percent composition allows you to calculate the
empirical formula of a compound
• An empirical formula is the simplest whole number ratio of
elements in a compound
• How is an empirical formula different than what we’ve seen
before?
• Up until now, you’ve been working with molecular formulas!
• A molecular formula is the actual ratio of elements in a compound
Differentiating between
Molecular Formulas
Empirical
and
• C2H4 and C3H6 are both molecular formulas
• To find the empirical formulas, find the lowest whole number
ratio between atoms
• So, both C2H4 and C3H6 have empirical formulas of CH2
• Note - Empirical formulas and molecular formulas can be the
same!
• Example is H2O
• Example:
• Molecular: C6H12O6
• Empirical: CH2O
• Example:
• Molecular: CH4N
• Empirical: CH4N
Using % Composition to Determine Empirical
Formula of a Compound
1. Pretend that you have a 100 gram sample of the
compound
•
In other words, change the % to grams
2. Convert the grams to moles for each element
•
Use Mole Road Map
3. Write the number of each element as a subscript
in a chemical formula
•
Keep each number as a decimal at this point!
4. Divide each subscript by the smallest number
5. Multiply the result by some integer to get rid of any
fractions
•
May not be necessary
Practice!
• Calculate the empirical formula of a compound composed of 38.67
% C, 16.22 % H, and 45.11 % N
• Step 1: Pretend that you have a 100 gram sample of the compound
• Step 2: Convert the grams to moles for each element
Example (continued)
• Step 3: Write the number of each element as a subscript in a
chemical formula
C3.22H16.09N3.22
• Step 4: If we divide all of these by the smallest subscript, it will
give us the empirical formula
CH5N
Practice!
• Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What
is its empirical formula? What is the empirical molar mass?
Using an Empirical Formula to Determine the
Molecular Formula
• Since the empirical formula is the lowest ratio, the actual
molecule would have a bigger mass
• Molecular formula can always be obtained by multiplying the
empirical formula by some whole number
• To do so, follow the steps below:
1. Calculate the empirical molar mass from the empirical formula
2. Divide the actual molecular molar mass (usually given in the
problem) by empirical molar mass
• Gives a whole number
3. Multiply empirical formula by the whole number to get the
molecular formula
Practice!
• Look back at the previous problem. Caffeine has a molar mass
of 194 g/mol. What is its molecular formula?
• A compound is known to be composed of 71.65 % Cl, 24.27%
C, and 4.07% H. Its molecular molar mass is known to be
98.96 g/mol. What is its empirical formula? What is its
molecular formula?
A REAL-WORLD DETERMINATION OF
EMPIRICAL FORMULA
Real-world determination of
Empirical Formula
• Combustion analysis is
one of the most common
methods for determining
empirical formulas
• A weighed compound is
burned in oxygen and its
products are analyzed by
a gas chromatogram
• It is particularly useful
for analysis of
hydrocarbons!
• Products are CO2 and H2O
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More on Combustion Analysis
• Combustion Analysis
• The technique of finding the mass composition of an unknown sample (X)
by examining the products of its combustion
X + O2 → CO2 + H2O
• Example Data
• 0.250 g of compound X produces 0.686 g CO2 and 0.562 g H2O
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Determining empirical formulas from
Combustion Analysis
X + O2 → CO2 + H2O
•
Step 1: Find the mass of C & H that must have been present in X
•
Multiply masses of products by percent composition (decimal form) of the
products
C: 0.686 g x (12.01 g/44.01 g) = 0.187 g C
H: 0.562 g x [(2 x 1.008 g)/18.02 g]= 0.063 g H
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Combustion Analysis
X + O2 → CO2 + H2O
•
Step 2: Add masses of products and compare to mass of original
compound
0.187 g C + 0.063 g H = 0.250 g total
•
•
Based on original compound mass of 2.90 g, compound X must contain only
C and H!!
Step 3: Find the number of moles of C and H
C: 0.187 g x mole/12.01 g = 0.0156 moles C
H: 0.063 g x mole/1.008 g = 0.063 moles H
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Combustion Analysis
X + O2 → CO2 + H2O
•
Step 4: Write the formula using the mole numbers as subscripts
C.0156H.063
•
Step 5: Divide by smallest number of moles
C: 0.0156/0.0156 = 1
H: 0.063/0.0156 = 4
•
If these numbers are fractions, multiply each by the same whole
number
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Combustion Analysis
X + O2 → CO2 + H2O
•
Step 6: Rewrite formula with new mole whole numbers
•
You now have the empirical formula!
CH4
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