Transcript Chapter 10 Chemical Quantities - CNG Chemistry | Resources

Chapter 10
Chemical Quantities
Get ready for some
serious finger exercise!
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10.1 The Mole: Measurement of Matter
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OBJECTIVES:
– Describe methods of measuring the
amount of something
– Define Avogadro’s number as it
relates to a mole of a substance
– Distinguish between the atomic mass
of an element and its molar mass
– Describe how the mass of a mole of a
compound is calculated
How do we measure items?
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You can measure mass,
or volume,
or you can count pieces.
We measure mass in grams.
We measure volume in liters (or cm3).
We count matter in MOLES.
What is the mole?
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We’re not talking about this
kind of mole!
Mole (abbreviated mol)
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A mole is simply an amount (like a dozen)
It is defined as the number of carbon
atoms in exactly 12 g of carbon-12 (12C).
That amount (1 mole) = 6.022 x 1023 of
the representative particles
6.022 x 1023 = Avogadro’s number.
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Similar Words for Amounts
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Pair: 1 pair of shoes
= 2 shoes
Dozen: 1 dozen roses
= 12 roses
Ream: 1 ream of paper
= 500 sheets of paper
Representative Particles?
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Representative particles are the smallest
“units” of a substance that define it
1) For a molecular compounds: it is the
molecule.
2) For an ionic compounds: it is the
formula unit (lowest whole-number
ratio of the ions).
3) For an element: it is the atom.
» Remember the 7 diatomic elements?
(made of molecules)
» Br2 I2 N2 Cl2 H2 O2 F2
Types of questions
How many oxygen atoms in the
following?
3 atoms of oxygen
CaCO3
Al2(SO4)3 12 (3 x 4) atoms of oxygen
 How many ions in the following?
3 total ions (1 Ca ion and 2 Cl ions)
CaCl2
2 total ions (1 Na ion and 1 OH ion)
NaOH
Al2(SO4)3 5 total ions (2 Al + 3 SO ions)
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2+
1-
1+
1-
3+
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2-
Parts of a Whole Mole
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Molecules are made of groups of atoms
Ionic compounds are made of ions
The subscripts in the formulas of compounds
give the numbers of each type of element
– CO2 has 2 O and 1 C atom per molecule
» So one mole of CO2 has one mole of C atoms and 2 moles of O atoms
– CaCl2 has 2 Cl− and 1 Ca+ ion per formula unit
» So one mole of CaCl2 has 2 moles of Cl− and 1 mole Ca+ ions
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Tricycle analogy
– One tricycle has 1 seat, 2 pedals and 3 wheels
– So the “formula” would be StPe2Wh3
9 – How many of each “element” in a dozen tricycles? How
many in one mole of tricycles?
Practice problems
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How many molecules in 4.56 moles of CO2? How
many atoms?
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How many moles of water is 5.87 x 1022 molecules?
How many atoms are in 1.23 moles of C6H12O6?
 How many moles is 7.78 x 1024 formula units of
10MgCl2?
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Measuring Moles
 Remember the mole is based on
the number of C atoms in 12
grams of carbon-12.
 1 AMU in atoms = 1 g in moles
 This proportion is true for all
elements in the periodic table.
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Gram Atomic Mass
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Equals the mass of 1 mole of an element in
grams (from periodic table)
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
(this is the molar mass)

(
12.01 g C
1 mole C
=
1 mole C
12.01 g C
We can count things by weighing them!
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)
Practice Problems
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What is the mass in grams of 2.34 moles of carbon?
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How many moles of magnesium is 24.31 g of Mg?
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How many atoms of lithium is 1.00 g of Li?
13 How much would 3.45 x 1022 atoms of U weigh?
What about compounds?
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in 1 mole of H2O molecules there are two
moles of H atoms and 1 mole of O atoms
(think of a compound as a molar ratio)
To find the mass of one mole of a compound
– determine the number of moles of the
elements present
– Multiply the number times their mass (from
the periodic table)
– add them up for the total mass
Calculating Formula Mass
Calculate the formula mass of
magnesium carbonate, MgCO3.
24.3 g
+
12.0 g
+ 3 x (16.0 g) =
84.3 g
the formula mass (or molar mass) for MgCO3
is 84.3 g/mol
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More Practice
1. What is the molar mass of sucrose (C12H22O11)?
 2. What is the molar mass of each of the following
compounds?
a) phosphorus pentachloride (PCl5)
b) uranium hexafluoride (UF6)
 3. Calculate the molar mass of each of the following
ionic compounds:
a) KMnO4
b) Ca3(PO4)2
 4. How many moles is 3.52 x 1024 molecules of water?
 5. How many atoms of zinc are in 0.60 mol of zinc?
 6. What is the mass of 1.00 mol of oxygen (O2)?
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
10.2 Mole-Mass and Mole-Vol
Relationships
 OBJECTIVES:
– Describe how to convert the mass of a
substance to the number of moles of a
substance, and moles to mass
– Identify the volume of a molar quantity of
gas at STP
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Molar Mass
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Molar mass is the generic term for the mass
of one mole of any substance (expressed in
grams/mol)
The same as:
1) Gram Molecular Mass (for molecules)
2) Gram Formula Mass (ionic compounds)
3) Gram Atomic Mass (for elements)
– molar mass is a broad term that encompasses
all these other specific masses
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Examples
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5.
6.
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Calculate the molar mass of the following
and tell what the units are for each:
Na2S
N2O4
C
Ca(NO3)2
C6H12O6
(NH4)3PO4
Molar Mass = Conversion Factor
 Molar mass is the number of grams in
1.0 mole of atoms, ions, or molecules
 We can make conversion factors from
molar masses to change:
- from grams of a substance to moles
- OR, from moles to grams
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For Example
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How many moles is 5.69 g of Ca(OH)2?
– Want to convert grams to moles, so need a
conversion factor with grams in denominator (to
cancel grams) and moles in numerator to change
the units to moles.
5.69 g Ca(OH)2
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? mole Ca(OH)2
? g Ca(OH)2
For Example
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Molar masses are equivalent values of
moles ↔ grams of a substance
1mole Ca
 1 mol O
 1 mole H
1 mole Ca(OH)2

=
=
=
=
40.1 g x 1 = 40.1 g
16.0 g x 2 = 32.0 g
1.0 g x 2 = 2.0 g
74.1 g
1 mol Ca(OH)2
74.1 g Ca(OH)2
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=
2 O and 2 H
atoms per formula
unit of Ca(OH)2
74.1 g Ca(OH)2
1 mol Ca(OH)2
For Example

How many moles is 5.69 g of Ca(OH)2?
– Looking to convert grams to moles, so need a conversion
factor with grams in denominator (to cancel grams) and moles
in numerator to change the units to moles.
5.69 g Ca(OH)2
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1 mol Ca(OH)2
74.1 g Ca(OH)2
For Example

How many moles is 5.69 g of Ca(OH)2?
– Looking to convert grams to moles, so need a conversion
factor with grams in denominator (to cancel grams) and moles
in numerator to change the units to moles.
5.69 g Ca(OH)2
1 mol Ca(OH)2
74.1 g Ca(OH)2
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= 0.077 mol Ca(OH)2
The Mole-Volume Relationship
 Many chemicals we’ll deal with are gases
- difficult to measure their masses
 But, we will still need to know how many moles of
gas we have
 we must know how many particles (atoms, ion,
molecules) are taking part in chemical reactions
 Easy to measure the volume of a gas, but 2 things
effect volumes of molar amounts of gases:
a) Temperature and b) Pressure
 If we compare all gases at the same T and P they all
occupy the approximately same volume
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Standard Temperature & Pressure
(STP)
 = 0ºC (273 K) and 1 atm pressure
 At STP, 1 mole of any gas occupies a
volume of 22.4 L
- Called the molar volume
 This equality can also be a conversion factor
 1 mole of any gas at STP = 22.4 L
1 mole gas
22.4 L
22.4 L
1 mole gas
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Practice Examples
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What is the volume of 4.59 mole of CO2
gas at STP?
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How many moles is 5.67 L of O2 at STP?
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What is the volume of 8.8 g of CH4 gas at
STP?
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Density of a gas
 Density = mass / volume
 for a gas the units are usually g / L
 We can determine the density of any gas at STP
if we know its formula.
 Use formula to calculate molar mass (= grams
in 1 mole of gas)
 Divide molar mass by molar volume (all gases
are 22.4 L / mole)
Mass of 1 mole in grams = grams in 1 L = density!
Volume of 1 mole in L
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Practice Examples (D=m/V)
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Find the density of CO2 at STP
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Find the density of CH4 at STP
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Another way:
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If given the density, we can find molar
mass of a gas.
Again, “pretend” you have 1 mole at
STP, so V = 22.4 L.
m=DxV
D = m/V so…
“m” = mass of 1 mole
Using Density to Find Molar Mass
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What is the molar mass of a gas with a density of
1.964 g/L at STP?
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How about a density of 2.86 g/L at STP?
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Summary
These four items are all equal
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a) 1 mole
b) molar mass (in grams)
c) 6.02 x 1023 representative particles (atoms,
molecules, or formula units)
d) 22.4 L of gas at STP
Thus, we can make conversion factors from
these 4 values!
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Practice Problems
1. What is the molar mass of each of the following compounds?
a. C6H12O6
b. NaHCO3
c. C7H12
d. KNH4SO4
2. Calculate the mass in grams of each of the following:
a. 8.0 mol lead oxide (PbO)
b. 0.75 mol hydrogen sulfide (H2S)
c. 1.50 x 10-2 mol oxygen (O2)
d. 2.30 mol ethylene glycol (C2H6O2)
3. How many grams are in 1.73 mol of dinitrogen pentoxide (N2O5)?
4. How many grams are in 0.66 mol of calcium phosphate [Ca3(PO4)2]?
5. Calculate the number of moles in each of the following:
a. 0.50 g sodium bromide (NaBr)
b. 13.5 g magnesium nitrate [Mg(NO3)2 ]
c. 0.0010 g chloromethane (CH3Cl)
d. 1.02 g MgCl2
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More Practice Problems
6. A chemist plans to use 435.0 grams of ammonium nitrate
(NH4NO3) in a reaction. How many moles of the compound is
this?
7. A solution is to be prepared in a laboratory. The solution
requires 0.0465 mol of quinine (C20H24N2O2). What mass, in
grams, should the laboratory technician obtain in order to
make the solution?
8. What is the volume at STP of 2.66 mol of methane (CH4) gas?
9. How many moles is 135 L of ammonia (NH3) gas at STP?
10. What is the density of carbon dioxide (CO2) gas at STP?
11. What is the molar mass of ethene (C2H4) if its density at STP
is 1.25 g/L?
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10.3: % Composition, Chemical Formulas
 OBJECTIVES:
–Describe how to calculate the
percent by mass of an element in a
compound
–Interpret an empirical formula
–Distinguish between empirical and
molecular formulas
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Calculating Percent Composition
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 Like all percent problems:
part
x 100 % = percent
whole
1) Find the mass of each of the
components (the elements),
2) Next, divide by the total mass of
the compound; then x 100
Example
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Calculate the percent composition
of a compound that is made of
29.0 grams of Ag with 4.30 grams
of S.
29.0 g Ag
X
100
=
87.1
%
Ag
33.3 g total
4.30 g S
X 100 = 12.9 % S
33.3 g total
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Total = 100 %
Using Formulae to Calculate % Mass
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assume you have 1 mole of the
compound…
If we know the formula we know the mass
of the elements and the whole compound
(from the periodic table!).
Practice Problems
Calculate the percent composition of
C2H4
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Aluminum carbonate
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Sample Problem 10.10, p.307
We can also use the percent composition
as a conversion factor (see p. 308)
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Read the
explanation at
left.
Use that
information to
help you
complete the
practice problems
below.
Empirical and Molecular Formulas
Empirical formula: the lowest whole
number ratio of atoms/ions in a compound
Molecular formula: the true number of
atoms of each element in the molecules of a
molecular compound.
• Example: molecular formula for benzene is
C6H6 (note that everything is divisible by 6)
• Therefore, the empirical formula = CH (the
lowest whole number ratio)
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Formulas (continued)
Formulas for ionic compounds are
ALWAYS empirical (the lowest whole
number ratio = cannot be reduced).
Examples:
NaCl
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MgCl2
Al2(SO4)3
K2CO3
Formulas (continued)
Formulas for molecular compounds
MIGHT be empirical (lowest whole
number ratio).
Water
Glucose
Octane
C4H9
Molecular:
H2O
C6H12O6
Empirical:
H2O
CH2O
(Lowest whole
number ratio)
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C8H18
Determining Empirical Formulas
 Just find the lowest whole number ratio
C6H12O6
= CH2O
CH2Cl2 = already the lowest ratio
 A formula is not just the ratio of atoms, it is
also the ratio of moles.
 In 1 mole of CO2 there is 1 mole of carbon
and 2 moles of oxygen.
 In one molecule of CO2 there is 1 atom of
C and 2 atoms of O.
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Calculating Empirical Formulas
Can be calculated from the percent composition
because the sum of the parts always = 100%
1) Assume you have a 100 g sample
- the percentage become grams (75.1% =
75.1 grams)
2) Convert grams to moles.
3) Find lowest whole number ratio by dividing
each number of moles by the smallest value.
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Example
Calculate the empirical formula of a
compound composed of 38.67 % C, 16.22
% H, and 45.11 %N.
1) Assume 100 g sample, so
 38.67 g C x 1mol C
= 3.22 mole C
12.0 g C
 16.22 g H x 1mol H
= 16.22 mole H
1.0 g H
 45.11 g N x 1mol N
= 3.22 mole N
14.0 g N
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Example
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The ratio is 3.22 mol C = 1 mol C
3.22 mol N
1 mol N
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The ratio is 16.22 mol H = 5 mol H
3.22 mol N
1 mol N
= C1 H5 N1
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= CH5N
Practice Problem
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Caffeine is 49.48% C, 5.15% H, 28.87% N
and 16.49% O. What is its empirical formula?
Empirical to Molecular Formula
 An empirical formula is the lowest ratio of atoms in
the compound, the actual molecule may have more
atoms.
 By a whole number multiple.
 For this type of problem, you are usually given the
data to determine an empirical formula, then given a
molar mass of the compound.
 Divide the compound’s actual molar mass by the
empirical formula mass
 yielding a whole number to increase each
coefficient in the empirical formula
 Caffeine has a molar mass of 194 g. what is its
molecular formula?
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Empirical to Molecular Formula
 Caffeine has a molar mass of 194 g. what is its molecular
formula?
 Recall the empirical formula for caffeine from the previous
example problem:
C4H5N2O
Find empirical formula mass
C = 4 x 12 g
= 48 g/mol
H = 5 x 1 g/mol
= 5 g/mol
N = 2 x 14 g/mol
= 28 g/mol
O = 1 x 16 g/mol
= 16 g/mol
Molar mass C4H5N2O = 97 g/mol
Molecular mass = 194 g
EmpForm mass = 97 g
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=2
Video Homework
Remember, every time you watch a video for
chemistry, you are expected to take notes and
complete all practice problems pointed out in the
lecture.
You are also expected to THINK.
From now on, you are also expected to bring to
class one intelligent question to ask me about the
lecture.
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