Molar mass

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Chapter 3
Stoichiometry:
Ratios of Combination
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Mass % = [(mass of component)/(total mass)] x 100
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111.70
 100 = 69.9% Fe
159.70
What is the % oxygen in this sample? (hint :100%)
% by mass =
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•
Equations can represent physical changes:
KClO3(s)  KClO3(l)
or chemical changes
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Note the symbol for heat above the arrow:
2 KClO3(s)


2 KCl(s) + 3 O2(g)
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Exercises:
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(Molecular mass)/(Empirical mass) = n
and
Molecular formula = (Empirical formula)n
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• The data given allows an empirical formula
determination with just a few more steps.
• The mass of products (carbon dioxide and water)
will be known so we work our way back.
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Eercise
Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)
Theoretical yield = 2.14 g
ZnCl2
Actual yield
ZnCl2
=
1.58 g
The % yield is:
1.58g
% yield =
 100  73.8%
2.14g
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Reaction Types
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Combination: one product is formed
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Decomposition: one reactant produces more
than one product
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Combustion: a hydrocarbon reacts with oxygen
to produce carbon dioxide and water
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