# Chapter 3 Stoichiometry

Topics: Stoichiometry, Concentration, Mole Pages: 23 (2090 words) Published: March 10, 2015
Chapter 4 — Intro—1

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CHAPTER 3

Topic Scopes:

Stoichiometry and
Solution Concentration

• Molarity, molality, parts per million &
percentage (w/w, w/v and v/v)
• Stoichiometry calculation
• Limiting reactant
• Theoretical yield, actual yield and
percentage yield
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Mole Concept
No. of Moles =

Molarity (M)
• Molarity (molar concentration) is the
number of moles of a solute that is
contained in 1 liter of solution

Mass (g)
molar mass (g/mol)

No. of Moles = Molarity (mol/L)  volume (L)
Molarity (M) = Amount of solute (Mol)
Volume of solution (L)
• 1 mole contains 1 Avogadro’s number
(6.022 x 1023)

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Example: Saline Water
Concentration
Typical seawaters contain sodium chloride,
NaCl, as much as 2.7 g per 100 mL.
(Molar mass of Na = 22.99 g/mol; Cl = 35.45
g/mol; Mg = 24.30 g/mol)
(a) What is the molarity of NaCl in the saline
water?
(b) The MgCl2 content of the saline water is
0.054 M. Determine the weight (grams)
of MgCl2 in 50 mL of the saline water? 5

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Solution:
(a) Molar mass of NaCl = 22.99 +35.45
= 58.44 g mol-1
Moles of NaCl in 100 mL of saline water
= 2.7g /(58.44 g mol-1) = 0.046 mol
 Molarity of saline water = Mol/L
= 0.046 mol /(100/1000)L = 0.46 M
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Chapter 4 — Intro—1

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Solution:
(b) Molar mass of MgCl2
= 24.30 + 2(35.45) = 95.20 g mol-1

Molality (m)
• Molality is the number of moles of solute
per kilogram (1000 g) of solvent

Moles of MgCl2 =
Molarity (M) x volume of solution (V)
Weight of MgCl2 in 50 mL of saline water
= (M x V) x MW
= 0.054 mol L-1 x (50/1000)L x 95.20 g mol-1
= 0.26 g

Molality (m) = Amount of solute (Mol)
Mass of solvent (kg)

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Percent Composition
(Concentration In Percentage)

Example:
A solution contains 118.5 g KI per liter of
solution. Calculate the concentration in (a)
% w/v & (b) % w/w. Given the density of the
solution at 25C is 1.078 g mL-1

(g)
(g)

(ml)
(ml)

Solution:
(a) % w/v = 118.5 g x 100%
1000 mL
= 11.85 % w/v

(g)
(ml)

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Units of Low Concentration
Solution:
(b) % w/w = 118.5 g x 1 mL x 100%
1000 mL 1.078 g
= 10.99 % w/w

• Parts per million, (ppm) is grams of solute
per million grams of total solution/ mixture
• ppm = mass of solute (g)
x 106
mass of sample(g)

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Unit of ppm: w/w = µg/g or mg/kg
w/v = µg/mL or mg/L
v/v = nL/mL or µL/L

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Chapter 4 — Intro—1

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Conversion of ppm to molarity

Units of Low Concentration

An aqueous solution contains 56 ppm SO2. Calculate
the molarity of the solution. (Molar mass of S = 32.06
g/mol; O = 16.00 g/mol)

• Parts per billion, (ppb) is grams of solute
per billion grams of total solution/ mixture
• ppb = mass of solute (g)
x 109
mass of sample(g)

Solution:
Molar mass of SO2= 32.06 + 2(16.00) = 64.06 g/mol
56 ppm  56 mg/L

Unit of ppb: w/w = ng/g or µg/kg
w/v = ng/mL or µg/L
v/v = nL/L

M
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56 mg
1g
1 mol

1L
1000 mg 64.06 g

 8.74 10 4 M

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Reaction of Phosphorus with Cl2

Stoichiometry

Cl2

• The relationship
between the
quantities of
chemical reactants
and products
• Depend on the
principle of the
conservation of
matter

PCl3

P4

Notice the stoichiometric coefficients and the
physical states of the reactants and products
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Chemical Equations
• Depict the kind of reactants and products and
their relative amounts in a reaction
reactants

products

4 Al(s) + 3 O2(g)  2 Al2O3(s)
stoichiometric coefficients
• (s),(g),(l) – physical states of compounds
• (s) – solid, (g) – gas, (l) – liquid
(aq) – aqueous solution

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Law of The
Conservation of Matter
• States that matter can be
neither created nor
destroyed
• An equation must be
balanced
• It must have the same
number of atoms of the
same kind on both sides
of the equation

Lavoisier, 1788
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Chapter 4 — Intro—1

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Balanced Chemical Equation

Law of The
Conservation of Matter
12 Cl atoms...