Introducing
Quantitative Chemistry
Chapter 3
Quantities in Chemistry
3.1 Formula Mass
3.2 Counting by Weighing
3.3 Working with Moles
3.4 Percentages by Mass
3.5 Finding the Formula
3.6 Chemical Equation
3.7 Working with Equations
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Quantities in Chemistry
Epinephrine (Adrenaline)
2x10-8 g/L in blood
Aspirin
Pain killer with 600 mg
Fatal death with 60g
Under extreme stress, it ramps to
a thousand-fold.
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3.1. Formula Masses
• Formula mass is often called formula
weight or molecular mass.
• It is the sum of the atomic masses of
the elements in a molecule or
compound.
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Sample Problem
Calculate the formula mass
of sucrose, C12H22O11
12 carbon atoms @ 12.01amu = 144.12
22 hydrogen atoms @ 1.01amu= 22.22
11 oxygen atoms @ 16.00amu= 176.00
Formula mass of sucrose
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= 342.34amu
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3.2. Counting by Weighing
• The mass in grams of an element
having the same numerical value as
the formula mass in amu’s will contain
6.022 × 1023 atoms of the element.
• The same relationship applies to
compounds.
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Sample Problem -1
How many atoms are there in 1.50g of iron?
1 amu

 1 atom Fe 
=
(1.50g Fe) 

-24 
 1.6605 ×10  55.85 amu Fe 
1.62 ×1022 atoms of Fe
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Sample Problem -2
How many molecules of carbon dioxide
are in a 3.61g sample?
 6.022 × 1023 molecules CO2 
( 3.61 g CO2 ) 
=
44.01 g CO2


4.94 × 1022 molecules CO2
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3.3. Working with Moles
• 1 mole of any substance
contains 6.022×1023
particles of that substance.
• Avogadro’s number
(NA) = 6.022×1023
• The molar mass of a
compound is the formula
mass expressed in grams
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Sample Problem
How many moles of gold are there
in 8.62×1020 atoms of gold?


1mol Au
8.62 × 10 atoms Au 
=
23
 6.022 × 10 atoms Au 
−3
1.43 × 10 mol Au
(
20
)
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Sample Problem
How many moles of HCl are in 102g?
 1 mol HCl 
(102 g HCl) 
 = 2.80 mol HCl
g HCl 
 36.46mol
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Sample Problem
What is the mass of 5.22 mol of urea,
CH4N2O?
 60.07 g urea 
= 314 g
( 5.22 mol urea ) 

 1 mol urea 
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3.4. Percentages by Mass
total mass component
mass percent =
× 100%
total mass whole substance
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Sample Problem
Calculate the mass percent of nitrogen in
urea, CH4N2O.
 2 × 14.01 
 60.07  × 100% = 46.65%


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3.5. Finding the Formula
• The empirical formula of a compound is the
simplest whole number ratio of the atoms.
• The molecular formula is the actual number of
each type of element in the compound.
• The molecular formula equals the empirical
formula or is a whole number multiple of the
empirical formula.
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Sample Problem
A compound contains 3.46g of sulfur and
5.18g of oxygen. What is the empirical
formula? -- get the simple ratio of elements
 1 mol S 
( 3.46 g S ) 
 = 0.108 molS
 32.02 g S 
 1 mol O 
= 0.324 molO
( 5.18 g O ) 

 16.00 mol O 
Dividing both mol values by the smallest:
0.108
0.324
= 1S
= 3 O or SO3
0.108
0.324
0.108
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Sample Problem
The empirical formula of a carbohydrate is
CH2O. If the molar mass is 180 g/mol, what
is the molecular formula?
The empirical formula mass is:
12.0 + ( 2 × 1.0 ) + 16.0 = 30.0 g/mol
Dividing the empirical formula mass by
the molar mass:
n=
180 g/mol
=6
30.0 g/mol
( CH2O )n = C6H12O6
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3.6. Chemical equations
Fundamental process of change in chemistry is the chemical reaction
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Mass balance
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3.6. Chemical Equations
A chemical equation uses chemical symbols to
show what happens during a reaction.
2H2(g) + O2(g) → 2H2O(g)
Reactants
→
Products
2 molecule of hydrogen + 1 molecule of oxygen → 2 molecule of water
2 moles of hydrogen
+ 1 mole of oxygen
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→
2 moles of water
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Chemical Equations
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Chemical Equations
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Balancing Equations
1. Write the formulas for the reactants on the left
side and the formulas for the products on the
right side of the equation.
Propane reacts with oxygen to form carbon dioxide and water
C3H8 + O2 → CO2 + H2O
2. Adjust the numbers in front of the compounds
until the number of atoms of each element is the
same on both sides of the equation.
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Balancing Equations
C3H8 + O2 → CO2 + H2O
3C
8H
2O→ 1C
2O
2H
1O
1) Balance C:
C3H8 + O2 → 3CO2 + H2O
2) Balance H:
C3H8 + O2 → 3CO2 + 4H2O
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Balancing Equations
3) Balance O:
C3H8 + 5O2 → 3CO2 + 4H2O
4) Check if equation is balanced:
3C
8H
10O → 3C
6O
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8H
4O
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More information from the equations
Phase : another term for the state of matter.
(s)= the solid phase
(l) = the liquid phase
(g) = the gaseous phase
(aq) = aqueous phase solutions
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l)
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3.7. Working with Equations
How many grams of water will be produced by the
combustion of 122g of propane with excess oxygen?
C3H8(g) + 2O2(g) → 3CO2(g) + 4H2O(l)
1 mol
4 mol
122 g
?? g
X mol
4X mol
 1 mol C3H8  4 mol H2O   18.0g H2O 
(122g C3H8 ) 


 = 199g H2O
 44.1g C3H8  1 mol C3H8   1 mol H2O 
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Limiting Reactant
v
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Limiting Reactant
If a sandwich is made of a slice a cheese and two
slices of bread, how many sandwiches can be
made?
Since there is less cheese than bread, the cheese
is the limiting reactant.
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Limiting Reactant
• The limiting reactant is
the reactant used up first
in a reaction.
• The excess reactant is
the reactant present is
greater amount than is
necessary for the reaction.
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Sample Problem
If 12.0g of NO are reactant with 10.0g of O2,
how much NO2 will be formed? Which is the
limiting reactant?
2NO(g) + O2(g) → 2NO2(g)
 1 mol NO   2 mol NO2   46.0g 

 = 18.4g NO2


 30.0g   2 mol NO   1 mol NO2 
(12.0g NO ) 
 1 mol O2   2 mol NO2   46.0g 
10.0g
O
(

 = 28.8g NO2

2 )
 32.0g   1 mol O2   1 mol NO2 
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Sample Problem
• Theoretical Yield is the amount of
product that would result if all the limiting
reagent reacted.
• Actual Yield is the amount of product
actually obtained from a reaction.
% Yield =
Actual Yield
Theoretical Yield
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x 100
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Sample Problem
The theoretical yield of NO2 based on NO was
18.4g. If 13.7g were obtained from the
reaction, what is the percentage yield?
actual yield
× 100%
%yield =
theoretical yield
13.7g
=
× 100%
18.4g
= 74.5%
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