Chapter 3 Stoichiometry

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