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Objectives
Additional Stoichiometry
problems
• Use stoichiometry to determine the empirical
and molecular formula of a compound
Review
• In the previous lesson we learned about using
the balanced equation to convert from the
moles of one substance to the moles of another
substance (stoichiometry)
• In our previous unit we learned how to go
from the percent by mass to the empirical
formula of a compound.
• Now we are going to combine both of these
concepts.
• If you are given the name of a compound that
you are unfamiliar with or it says an unknown
compound, you can use the information about
the products to determine the empirical
formula of the compound.
– Example – 8.410 g of propane gas combusts in
excess oxygen gas to form 25.23 g of carbon
dioxide and 13.76 g of water.
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• Example 2 – 1.97 g of an unknown compound
completely combusts and produces 3.94 g of
carbon dioxide and 1.61 g of water.
– Determine the empirical formula of the compound.
– The molecular formula has a molar mass of 88
g/mol. What is the molecular formula of this
compound?
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