Unit 10: Stoichiometry 1
Calculations with Chemical
Formulas
Stoichiometry
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Mass in
Elements and
Compounds
Stoichiometry
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Atomic Mass
Atoms are so small, it is difficult to discuss
how much they weigh in grams.
• Use atomic mass units or amus
– 1 amu is 1/12 the mass of a carbon-12 atom.
This gives us a basis for comparison.
• The decimal numbers on the periodic table
are atomic masses in amu.
Stoichiometry
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Gram Atomic Mass
• Atomic mass in grams instead of amu’s.
– Represents an amount that we can actually
measure in lab.
• Also known as a mole, or molar mass,
which we will come back to later…
Stoichiometry
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Gram Formula Mass
• The gram formula mass is the sum of
the atomic masses for the atoms in a
chemical formula.
• So, the gram formula mass of calcium
chloride, CaCl2, would be
Ca: 1 x 40.1 = 40.1
+ Cl: 2 x 35.5 = 71.0
111.1 amu
• Gram formula mass is generally used for
either molecular or ionic compounds.
Stoichiometry
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Gram Formula Mass
• For the molecule, ethane (C2H6), the
formula mass would be: C: 2 x 12.0 = 24.0
+ H: 6 x 1.0 = 6.0
30.0 amu
• For the ionic compound, (NH4)2CO3
N: 2 x 14.0 amu = 28.0
H: 8 x 1.0 amu = 8.0
C: 1 x 12.0 amu = 12.0
+O: 3 x 16.0 amu = 48.0
Stoichiometry
= 96.0 amu
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Percent Composition
You can find the percentage (%) of the mass of
a compound that comes from each of the
elements in the compound by using these steps:
1. Calculate the formula mass of the
compound.
2. Divide the mass of each element by
the formula mass and multiply that
fraction x 100.
Stoichiometry
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Percent Composition
So the percentage of carbon in
ethane,C2H6, is…
C is 2 x 12.0 = 24.0
H is 6 x 1.0 = 6.0
30.0
% Carbon =
24.0 amu
30.0 amu
= 80.0%
x 100
amu
Stoichiometry
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(stop)
Stoichiometry
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Moles
Stoichiometry
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Avogadro’s Number
• 6.02 x 1023
• 1 mole of 12C has a
mass of 12 g.
Stoichiometry
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Molar Mass
• By definition, a molar mass is the mass
of 1 mol of a substance (i.e., g/mol).
– The molar mass of an element is the mass
number for the element that we find on the
periodic table.
– The formula weight (in amu’s) will be the
same number as the molar mass (in
g/mol).
Stoichiometry
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Using Moles
Moles provide a bridge from the molecular
scale to the real-world scale.
Stoichiometry
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Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles.
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or
ions of each element in the compound.
Stoichiometry
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Finding
Empirical
Formulas
Stoichiometry
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Calculating Empirical Formulas
• Empirical formula: smallest, wholenumber ratio of atoms of elements in a
compound.
• You can calculate the empirical formula from
the percent composition.
Stoichiometry
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Steps for Empirical Formula
1. For each element, convert mass to
moles.
– If you have percents, use the percent as the
number of grams/100 grams of compound.
Example: 67% would be 67 grams.
2. Find the mole ratio: Divide all the
numbers by the smallest number of moles
3. Use the smallest, whole number ratio as
the subscripts for the formula.
Stoichiometry
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Calculating Empirical Formulas
Example:
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of:
carbon (61.31%)
hydrogen (5.14%)
nitrogen (10.21%)
oxygen (23.33%)
Find the empirical formula of PABA.
Stoichiometry
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Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C:
Sunscreen PABA
carbon (61.31%)
hydrogen (5.14%)
nitrogen (10.21%)
oxygen (23.33%)
H:
1 mol
61.31 g x
= 5.105 mol C
12.01 g
5.14 g x
1 mol
= 5.09 mol H
1.01 g
N:
1 mol
10.21 g x 14.01 g = 0.7288 mol N
O:
1 mol
23.33 g x 16.00 g = 1.456 mol O
Stoichiometry
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Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
C:
5.105 mol
0.7288 mol
= 7.005 = 7
H:
5.09 mol
0.7288 mol
= 6.984 = 7
N:
0.7288 mol
0.7288 mol
= 1.000 = 1
O:
1.458 mol
0.7288 mol
= 2.001 = 2
Stoichiometry
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Calculating Empirical Formulas
These numbers are the subscripts for the empirical
formula:
C7H7NO2
The molecule is shown here.
Stoichiometry
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Let’s work some examples
in your notes packet…
Stoichiometry
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Molecular Formula
Empirical Formula is the smallest whole number ratio of
atoms of elements in a compound.
Molecular Formula is the real formula of a compound. It is
a multiple of the Empirical Formula. They may be the same!
1. Calculate the mass of the empirical formula.
2. Divide the molecular mass given in the problem
by the empirical formula mass.
3. Multiply the subscripts in the empirical formula
by the number you get to make new subscripts.
Molecular formula is a multiple
of the empirical formula!
Stoichiometry
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Molecular Formula Problem
• Analysis of a chemical used in
photographic developing fluid indicates
a chemical composition of:
65.4% C
5.45% H
29.09% O
• The molar mass is found to be 110.0
g/mol. Determine the empirical and
Stoichiometry
molecular formulas.
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Find Empirical Formula First!
65.4g C 1 mole C = 5.45 ÷ 1.82 = 3
12g C
5.45g H 1 mole H = 5.45 ÷1.82 = 3
1g H
29.09g O 1 mole O = 1.82 ÷ 1.82 = 1
16g O
C3H3O = Empirical Formula
Stoichiometry
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Molecular Formula
1. Find molar mass C3H3O
C – 3 x 12 = 36
H–3x1=
3
O – 1 x 16 = +16
55 g/mole is Empirical Formula Mass
2. Divide the given Molecular Formula Mass
(110g/mole) by the Empirical Formula Mass
110/55 = 2, so C3x2H3x2O1x2
3. Molecular Formula = C6H6O2
Stoichiometry
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Molecular formula is some multiple of
the empirical formula. It may be the
same as the empirical formula!
Two samples of a compound must have
the same percent composition to be the
same compound or they would have
Stoichiometry
different empirical formulas.
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Hydrates
A hydrate is an ionic compound
that has a specific number of
water molecules bound to the
atoms in its crystals.
(This is NOT the same as being dissolved in water)
Many compounds are found in nature as hydrates,
such as protein crystals
Stoichiometry
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Naming Hydrates
To name a hydrate, give the name of the
compound, and add the prefix for the
number of waters + hydrate:
CuSO4·5H2O
copper(II) sulfate pentahydrate
Stoichiometry
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Formula of a Hydrate
1. Weigh the hydrate, then heat it in a partly
covered crucible to drive off the water.
2. Weigh it again to determine the amount of
water lost from the anhydrous (no water)
compound. Repeat until mass stops
changing.
3. Calculate the empirical formula with the
anhydrous (no water) compound as one unit
and the water as the second.
Stoichiometry
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Hydrate Problem
Problem:
Molar mass of CoCl2
If 11.75 g of cobalt(II)
Co – 1 x 59 = 59
chloride is heated, 9.25 g
Cl – 2 x 35.5 = +71
of anhydrous cobalt
chloride, CoCl2, remains.
130 g/mol
What is the formula and
name for this hydrate?
9.25 g CoCl2 1 mole = .0712 ÷.0712 = 1
130 g
11.75 – 9.25 = 2.5 g water
removed
9.25 g anhydrous CoCl2
2.5 g H20
1 mole = .139 ÷ .0712 = 2
18 g
CoCl2·2H2O is cobalt(II) chloride dihydrate
Stoichiometry
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Hydrate Problem
A hydrate is found to have the following percent
composition: 48.8% MgSO4 and 51.22% H2O.
What is the formula and name of this hydrate?
Molar mass of MgSO4 = 24.3 + 32 + 64 =120.3g/mol
Molar mass of H2O = 2 + 16 = 18 g/mol
= .406 ÷ .406 =
1
51.22 g 1 mole = 2.85 ÷ .406 =
7
48.8g 1 mole
120.3 g
18 g
Stoichiometry
Magnesium sulfate heptahydrate: MgSO4· 7H
2O
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Hydrates
Sometimes hydration can have nifty color changes!
Stoichiometry
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(end)
Stoichiometry
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Empirical Formula
• A blue solid is
found to contain
36.84% nitrogen
and 63.16%
oxygen. What is
the empirical
formula for this
solid?
• Determine the
empirical formula
for a compound
that contains
35.98% aluminum
and 64.02%
sulfur.
Stoichiometry
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