Transcript Chapter 7 Chemical Quantities

Chapter 7
Chemical Quantities
Charles Page High School
Dr. Stephen L. Cotton
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Section 7.1
The Mole: A Measurement of
Matter
 OBJECTIVES:
–Describe how Avogadro’s
number is related to a mole of
any substance.
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Section 7.1
The Mole: A Measurement of
Matter
 OBJECTIVES:
–Calculate the mass of a mole of
any substance.
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What is a Mole?
 You
can measure mass,
 or volume,
 or you can count pieces.
 We measure mass in grams.
 We measure volume in liters.
 We
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count pieces in MOLES.
Moles (abbreviated: mol)
 Defined
as the number of carbon
atoms in exactly 12 grams of
carbon-12.
 1 mole is 6.02 x 1023 particles.
 Treat it like a very large dozen
 6.02
x 1023 is called Avogadro’s
number.
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Representative particles
 The
smallest pieces of a substance.
–For a molecular compound: it is
the molecule.
–For an ionic compound: it is the
formula unit (ions).
–For an element: it is the atom.
»Remember the 7 diatomic
elements (made of molecules)
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Types of questions
 How
many oxygen atoms in the
following?
–CaCO3
–Al2(SO4)3
 How many ions in the following?
–CaCl2
–NaOH
–Al2(SO4)3
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Types of questions
 How
many molecules of CO2 are
there in 4.56 moles of CO2 ?
 How many moles of water is 5.87 x
1022 molecules?
 How many atoms of carbon are
there in 1.23 moles of C6H12O6 ?
 How many moles is 7.78 x 1024
formula units of MgCl2?
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Measuring Moles
 Remember
relative atomic mass?
 The amu was one twelfth the mass
of a carbon-12 atom.
 Since the mole is the number of
atoms in 12 grams of carbon-12,
 the decimal number on the periodic
table is also the mass of 1 mole of
those atoms in grams.
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Gram Atomic Mass (gam)
 Equals
the mass of 1 mole of an
element in grams
 12.01 grams of C has the same
number of pieces as 1.008 grams of
H and 55.85 grams of iron.
 We can write this as
12.01 g C = 1 mole C
 We can count things by weighing
10 them.
Examples
 How
much would 2.34 moles of
carbon weigh?
 How many moles of magnesium is
24.31 g of Mg?
 How many atoms of lithium is 1.00
g of Li?
 How much would 3.45 x 1022 atoms
of U weigh?
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What about compounds?
in 1 mole of H2O molecules there are
two moles of H atoms and 1 mole of O
atoms
 To find the mass of one mole of a
compound
– determine the moles of the elements
they have
– Find out how much they would weigh
– add them up
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
What about compounds?
What is the mass of one mole of CH4?
1 mole of C = 12.01 g
4 mole of H x 1.01 g = 4.04g
1 mole CH4 = 12.01 + 4.04 = 16.05g
 The Gram Molecular Mass (gmm) of
CH4 is 16.05g
– this is the mass of one mole of a
molecular compound.

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Gram Formula Mass (gfm)
The mass of one mole of an ionic
compound.
 Calculated the same way as gmm.
 What is the GFM of Fe2O3?
2 moles of Fe x 55.85 g = 111.70 g
3 moles of O x 16.00 g = 48.00 g
The GFM = 111.70 g + 48.00 g =
159.70 g

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Section 7.2
Mole-Mass and Mole-Volume
Relationships
 OBJECTIVES:
–Use the molar mass to convert
between mass and moles of a
substance.
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Section 7.2
Mole-Mass and Mole-Volume
Relationships
 OBJECTIVES:
–Use the mole to convert among
measurements of mass,
volume, and number of
particles.
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Molar Mass
 Molar
mass is the generic term for
the mass of one mole of any
substance (in grams)
 The same as: 1) gram molecular
mass, 2) gram formula mass, and 3)
gram atomic mass- just a much
broader term.
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Examples
Calculate the molar mass of the
following and tell what type it is:
 Na2S
 N2O4
C
 Ca(NO3)2
 C6H12O6
 (NH4)3PO4

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Molar Mass
 The
number of grams of 1 mole
of atoms, ions, or molecules.
 We can make conversion factors
from these.
–To change grams of a
compound to moles of a
compound.
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For example

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How many moles is 5.69 g of NaOH?
For example

How many moles is 5.69 g of NaOH?

5.69 g

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


For example

How many moles is 5.69 g of NaOH?
mole 

5.69 g


g 
 need to change grams to moles
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For example

How many moles is 5.69 g of NaOH?
mole 

5.69 g


g 
 need to change grams to moles
 for NaOH
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For example

How many moles is 5.69 g of NaOH?
mole 

5.69 g


g 
 need to change grams to moles
 for NaOH
 1mole Na = 22.99g 1 mol O = 16.00 g
1 mole of H = 1.01 g
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For example

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How many moles is 5.69 g of NaOH?
mole 

5.69 g


g 
 need to change grams to moles
 for NaOH
 1mole Na = 22.99g 1 mol O = 16.00 g
1 mole of H = 1.01 g
 1 mole NaOH = 40.00 g
For example

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How many moles is 5.69 g of NaOH?
1 mole 

5.69 g


40.00 g 
 need to change grams to moles
 for NaOH
 1mole Na = 22.99g 1 mol O = 16.00 g
1 mole of H = 1.01 g
 1 mole NaOH = 40.00 g
For example

How many moles is 5.69 g of NaOH?

5.69 g

need to change grams to moles
 for NaOH
 1mole Na = 22.99g 1 mol O = 16.00 g
1 mole of H = 1.01 g
 1 mole NaOH = 40.00 g

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1 mole 
 = 0.142 mol NaOH
40.00 g 
Examples
 How
many moles is 4.56 g of
CO2?
 How many grams is 9.87 moles of
H2O?
 How many molecules is 6.8 g of
CH4?
 49 molecules of C6H12O6 weighs
how much?
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Gases
 Many
of the chemicals we deal with
are gases.
–They are difficult to weigh.
 Need to know how many moles of gas
we have.
 Two things effect the volume of a gas
–Temperature and pressure
 We need to compare them at the same
temperature and pressure.
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Standard Temperature and
Pressure
 0ºC
and 1 atm pressure
 abbreviated STP
 At STP 1 mole of gas occupies
22.4 L
 Called the molar volume
 1 mole = 22.4 L of any gas at STP
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Examples
 What
is the volume of 4.59 mole
of CO2 gas at STP?
 How many moles is 5.67 L of O2
at STP?
 What is the volume of 8.8 g of
CH4 gas at STP?
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Density of a gas
D=m/V
–for a gas the units will be g / L
 We can determine the density of any
gas at STP if we know its formula.
 To find the density we need the mass
and the volume.
 If you assume you have 1 mole, then
the mass is the molar mass (from PT)
32 At STP the volume is 22.4 L.

Examples
 Find
the density of CO2 at STP.
 Find the density of CH4 at STP.
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The other way
 Given
the density, we can find the
molar mass of the gas.
 Again, pretend you have 1 mole at
STP, so V = 22.4 L.
m = D x V
 m is the mass of 1 mole, since you
have 22.4 L of the stuff.
 What is the molar mass of a gas with a
density of 1.964 g/L?
34  2.86 g/L?
Summary
 These
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four items are all equal:
a) 1 mole
b) molar mass (in grams)
c) 6.02 x 1023 representative
particles
d) 22.4 L at STP
Thus, we can make conversion
factors from them.
Section 7.3
Percent Composition and
Chemical Formulas
 OBJECTIVES:
–Calculate the percent
composition of a substance
from its chemical formula or
experimental data.
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Section 7.3
Percent Composition and
Chemical Formulas
 OBJECTIVES:
–Derive the empirical formula
and the molecular formula of a
compound from experimental
data.
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Calculating Percent Composition of
a Compound
 Like
all percent problems:
Part
x 100 %
whole
 Find the mass of each
component,
 then divide by the total mass.
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Example
 Calculate
the percent
composition of a compound that
is 29.0 g of Ag with 4.30 g of S.
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Getting it from the formula
 If
we know the formula, assume
you have 1 mole.
 Then you know the mass of the
pieces and the whole.
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Examples
 Calculate
the percent
composittion of C2H4?
 How about Aluminum carbonate?
–Sample Problem 7-11, p.191
 We can also use the percent as a
conversion factor
–Sample Problem 7-12, p.191
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The Empirical Formula
 The
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lowest whole number ratio of
elements in a compound.
 The molecular formula = the actual
ratio of elements in a compound.
 The two can be the same.
 CH2 is an empirical formula
 C2H4 is a molecular formula
 C3H6 is a molecular formula
 H2O is both empirical & molecular
Calculating Empirical
 Just
find the lowest whole number
ratio
 C6H12O6
 CH4N
 It is not just the ratio of atoms, it is
also the ratio of moles of atoms.
 In 1 mole of CO2 there is 1 mole of
carbon and 2 moles of oxygen.
 In one molecule of CO2 there is 1
43 atom of C and 2 atoms of O.
Calculating Empirical
 We
can get a ratio from the
percent composition.
 Assume you have a 100 g.
 The percentages become grams.
 Convert grams to moles.
 Find lowest whole number ratio
by dividing by the smallest.
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Example
 Calculate
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the empirical formula of a
compound composed of 38.67 % C,
16.22 % H, and 45.11 %N.
 Assume 100 g so
 38.67 g C x 1mol C
= 3.220 mole C
12.01 gC
 16.22 g H x 1mol H
= 16.09 mole H
1.01 gH
 45.11 g N x 1mol N = 3.219 mole N
14.01 gN
Example
 The
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ratio is 3.220 mol C = 1 mol C
3.219 molN
1 mol N
 The ratio is 16.09 mol H = 5 mol H
3.219 molN
1 mol N
 = C1 H 5 N 1
 A compound is 43.64 % P and 56.36
% O. What is the empirical formula?
 Caffeine is 49.48% C, 5.15% H,
28.87% N and 16.49% O. What is its
empirical formula?
Empirical to molecular
 Since
the empirical formula is the
lowest ratio, the actual molecule
would weigh more.
 By a whole number multiple.
 Divide the actual molar mass by the
empirical formula mass.
 Caffeine has a molar mass of 194 g.
what is its molecular formula?
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Example
A
compound is known to be
composed of 71.65 % Cl, 24.27%
C and 4.07% H. Its molar mass is
known (from gas density) to be
98.96 g. What is its molecular
formula?
 Sample Problem 7-14, p.194
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