Stoichiometry: Calculations with
Chemical Formulas and Equations
Sponsored by
Lavoisier & Avogadro
Chapter 3
September 7th, 2004
In this chapter
• The mole concept.
• Relationships between chemical formulas,
atomic masses and moles.
• Empirical and molecular formulas
• Writing chemical equations.
• Using moles to find the formulas of
compounds.
• Limiting reagent problems.
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The Mole
• Atoms and molecules are extremely small
making it very difficult to measure their
masses individually.
– It is easier to weigh a large collection of these.
• A mole (abbreviated as mol) is defined as the
number of C atoms in exactly 12 grams of
pure C-12.
A mole of anything = 6.022 x 1023 things.
• This number is called Avogadro’s number.
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Fun facts about the Mole
• The American Chemical
Society celebrates the Mole
Day annually.
• If 10,000 people started to
count Avogadro’s # at the
rate of 100/minute every
day, it would take them 1
trillion years.
• 1 mole of dollars will provide each person of the
Earth’s population (approx. 6 billion) an income of
$5000/s for about 100 years.
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Mole of atoms: The Molar Mass
• The mass in grams of one mole of atoms of
any element is known as the molar mass of
the element. The unit used is g/mol (gmol-1).
• Molar mass of sodium (Na) = mass of exactly
one mole of Na atoms = 22.99 g/mol = mass
of 6.022 x 1023 atoms of Na.
• Molar mass of Uranium (U) = 238.03 g/mol.
• The atomic mass number of an element in a
periodic table = molar mass for the element
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Uses of the Mole
• Two important conversions that we must know:
• Grams to moles:
Grams ×
1 mole
= moles
grams
1/Molar mass
• The evil twin; moles to grams:
Moles ×
grams
= grams
1 mole
Molar mass
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Grams to moles (look at Sample ex 3.7-3.12)
• Calculate the number of moles in 225.0 g of
Na.
• Need to look up the molar mass of Na from the
periodic table and then calculate the # of
moles.
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Moles to grams
• Calculate the number of grams of Na in 145.95
moles of Na.
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Moles of Compounds
• 1 mole of any compound = 6.022 x 1023 units
of the compound.
• Molar masses of compounds are calculated by
adding the molar masses of the atoms present
in the compound.
• Calculate the molar mass of Ca3(PO4)2
• Need to look the molar masses of each of the
element present and multiply by the number of
the atoms present.
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Molar mass of Ca3(PO4)2
3 Ca
2P
8O
3 moles Ca x
40.078 gmol-1 =
2 moles P x
30.974 gmol-1 =
8 moles O x
15.999 gmol-1 =
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Working with Molar Masses
• Calculate the # of moles of Ca3(PO4)2 in 77.5 g
of Ca3(PO4)2.
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Another example
• Calculate the # of moles of O atoms in 77.5 g
of Ca3(PO4)2.
• The chemical formula gives us a clue (a
conversion factor):
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Percent composition of compounds
• Composition of compounds can be described
in 2 ways:
– By the number of its constituent atoms.
– By the mass (%) of each element present.
• The mass percents of elements are obtained
by comparing the mass of each element
present in 1 mole of the compound to the total
mass of the compound.
• Useful today with unknown compounds.
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Using % composition
• Calculate the % by weight of each element in
Ca3(PO4)2.
– Find the molar mass of Ca3(PO4)2 = (310.174 gmol-1).
– Find the mass of each element in 1 mole of Ca3(PO4)2,
divide by the molar mass of Ca3(PO4)2 and multiply by
100%
3 Ca
3 moles Ca x
40.078 gmol-1 =
120.234
(No rounding)
2P
2 moles P x
30.974 gmol-1 =
61.948
8O
8 moles O x
15.999 gmol-1 =
127.992
1 mole Ca3(PO4)2 =
310.174 gmol-1
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Using % composition
• Now divide the mass of each element by the
molar mass and multiply by 100.
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Empirical Formulas
• Molecular formulas tell us 2 things:
– the relative number of atoms (atom ratio) of each
element in a compound.
– the total number of atoms in a molecule
• Empirical formulas simplify the chemical
formula so that the molecular formula is
always a whole-number multiple of the
subscripts in the empirical formula.
• Molecular formula = (Empirical formula) x a
whole number
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Empirical formulas
• Following are possible combinations between
Nitrogen and Hydrogen:
N2 H4
N3 H6
N4 H 8
All of these are whole number
multiples of the simplest possible ratio
between N and H which is NH2
The empirical formula for all of these
compounds = NH2
• Sometimes the empirical and molecular formula can
be the same; H2O, CO, CO2.
• To find the molecular formula of a compound the
molar mass must be known.
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Mass % to empirical formula
• Determine the empirical and molecular formula
of a compound that has the following mass %:
71.65% Cl, 24.27% C and 4.07% H.
The molar mass of this compound is known to
be = 98.96 g/mol.
– Assume we have 100.00 g of the compound.
– Convert the mass % to mass in grams.
Thus in 100.00 grams of this compound there are
71.65 g of Cl, 24.27 g of C and 4.07 g H.
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– Convert these grams to moles.
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– Divide the moles by the smallest # of moles to
obtain the empirical formula.
The smallest # of moles is 2.021 and dividing
throughout gives us the empirical formula as
ClCH2, (empirical formula mass = 49.48 g mol-1).
– Obtain the molecular formula by dividing the molar
mass by the empirical formula mass.
Molar mass
98.96 g mol −1
=
=2
Empirical formula mass 49.48 g mol −1
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