Download Chapter 7 - Chemical Quantities

Ch7-p1
Chapter 7 - Chemical Quantities
The Mole
In everyday life, we use counting units to deal with large
quantities. For example, we buy eggs by the dozen (12),
paper by the ream (500), and pop by the case (24). In
chemistry, we count molecules, atoms, and ions by the
mole. A mole contains 6.02 X 1023 items. This very large
number is called Avogadro’s number, after the Italian
physicist Amedeo Avogadro.
One mole of any element, molecule, or ionic compound
contains Avogadro’s number of atoms.
Therefore:
1mol of C contains 6.02 X 1023 carbon atoms.
1mol of Al contains 6.02 X 1023 aluminum atoms.
1mol of Br contains 6.02 X 1023 bromine atoms.
1mol of H2O contains 6.02 X 1023 H2O molecules.
1mol of NaCl contains 6.02 X 1023 NaCl formula
units.
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The subscripts in a chemical formula also represent
moles. For example, the chemical formula for aspirin is
C9H8O4. Therefore, 1 mole of aspirin contains 9 moles of
carbon atoms, 8 moles of hydrogen atoms, and 4 moles of
oxygen atoms.
How many C atoms are in 0.350 mol of C6H12O6?
How many Cl atoms in 2.0 mole of PCl3?
How many moles of C2H6O contain 5.0 x 10 24 atoms of
H?
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Molar Mass
The mass in grams of 1 mole of substance is called its
molar mass. The molar mass (in grams) of any substance
is always equal to its formula weight (in amu).
Determine the molar masses for the following substances:
a)
b)
c)
d)
e)
C
Ag
H2O
NaCl
NO3-
What is the mass of 1 mole of glucose, C6H12O6?
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How many moles of glucose are in 5.38 g?
How many grams of glucose are in 2.74 moles?
The concept of the mole provides a bridge between
masses and numbers of particles. Using molar mass and
Avogadro’s Number as conversion factors, we can
convert grams
moles
atoms.
For example, calculate the number of Cu atoms in a 3g
copper penny (assuming the penny is 100% Cu).
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How many glucose molecules are in 5.23 g of C6H12O6?
Percent Composition and Empirical Formula
Recall that the subscripts in a molecular formula represent
a definite proportion of the elements within a compound.
Therefore, a mole of any compound contains a definite
proportion by mass of its elements. By using the molar
mass of a compound, we can determine its percent
composition. For example, determine the percent
composition of K2CO3.
The molecular formula of a compound represents the
true formula of a compound. A formula that gives the
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lowest whole-number ratio of atoms in the compound is
called the empirical formula. For example, hydrazine
with molecular formula of N2H4, has an empirical formula
of NH2. The mole concept can be used to determine
empirical formulas.
e.g. A sample of a compound contains 3.24 g Na, 2.26 g
S, and 4.51 g O. What is its empirical formula?
e.g. Calculate the empirical formula of a compound that
has percent composition 36% Al and 64% S.
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Stoichiometry
Consider the following reaction:
2H2 + O2
2H2O
The coefficients from the balanced equation tell us that
we need two molecules of hydrogen to react with one
molecule of oxygen to give two molecules of water.
From the mole concept, we can interpret the equation in
terms of moles. Thus, 2 moles of hydrogen react with 1
mole of oxygen to give 2 moles of water. The
coefficients in the above equation are called
stoichiometrically equivalent quantities. The
relationship between these quantities can be expressed as
follows:
2 mol H2 K 1 mol O2 K 2 mol H2O
These stoichiometric relations can be used to give
conversion factors for relating quantities of reactants and
products in a chemical reaction.
e.g. Calculate the number of moles of H2O produced
from 1.57 moles of O2.
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e.g. Using the equation below, calculate the mass of CO2
that is produced in burning 1.00 g of butane, C4H10.
2C4H10 + 13O2
8CO2 + 10H2O
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e.g. Nitrogen gas reacts with hydrogen gas to produce
ammonia by the following equation:
N2 + 3H2
2NH3
a) If you have 1.8 moles of H2, how many grams of NH3
can be produced?
b) How many grams of H2 are needed to react with 2.80 g
of N2?
c) How many grams of NH3 can be produced from 12 g
of H2?
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Percent Yield.
All of the equations we’ve seen so far have assumed that
100% of product was formed. In reality, this is rarely the
case. Side reactions or the equilibrium can prevent
reactions from going to completion. When chemists
perform reactions, there are two things they want to
know; a) what is the theoretical yield of the reaction and
b) what is the actual yield of the reaction.
• The theoretical yield is the calculated amount of
product if the reaction proceeds to completion.
• The actual yield is the amount of product obtained
after the reaction is complete.
If we know the values of the theoretical yield and actual
yield, then we can express these values as a percentage,
called the percent yield.
e.g. What is the percent yield if 40.0 g of CO are
produced from the reaction of 30.0 g O2?
2C + O2
2CO
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e.g. From the following equation, calculate the mass of
CO2 that can be produced if the reaction of 45.0 g of
propane (C3H8) and sufficient oxygen has a 60% yield.
C3H8 + 5O2
3CO2 + 4H2O
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Equilibrium Constants.
We’ve learned that when a reversible reaction has reached
equilibrium, the forward and reverse reaction occur at the
same rate. We’ve also learned that at equilibrium, the
amounts of reactants and products remain constant.
Because of this, we can express the ratio of products to
reactants. This relationship between reactants and
products is called an equilibrium constant.
equilibrium constant = Keq =
products
reactants
Consider the general reaction:
aA + bB
cC + dD
In this equation, the capital letters A and B are reactants
and C and D are products. The lower case letters
represent their coefficients in the balanced equations. In
the equilibrium constant, the products are multiplied in
the numerator and divided by the reactants in the
denominator.
[C]c[D]d
Keq =
[A]a[B]b
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The square brackets indicate that the concentrations
(moles per 1 litre of solution) of the products and
reactants are used. Also note that the concentrations of
reactants and products are raised to the power of their
respective coefficients.
e.g. Write the expression for the equilibrium constant Keq
for the following reactions:
2HBr
CO + 2H2
CH4 + 2H2S
H2 + Br2
CH3OH
CS2 + 4H2
Write the balanced chemical equation that would give the
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following equilibrium constant expression.
[NO2]2
Keq =
[NO]2[O2]
When Keq > 1, the equilibrium favours the products.
When Keq < 1, the equilibrium favours the reactants.
When Keq = 1, the concentration of reactants and products
are equal.