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Chemistry 20H
Chapter 3
Stoichiometry: Calculations with Chemical Formulas and Equations
Objectives
1. Identify changes which indicate that a chemical reaction has taken place.
2. Define chemical and physical reactions.
3. Understand why a chemical reaction occurs.
4. Understand that both atoms and mass are conserved in chemical reactions: the law of conservation of mass.
5. Be able to write, understand and read chemical equations. (3.1)
6. Be able to balance chemical equations and understand the principles behind balancing equations. (3.1)
7. Write and balance chemical equation from a word equation, experimental evidence, or a description of a chemical
reaction taking place. (3.1)
8. Know and recognize the six most common types of chemical reactions. (3.3)
9. Use your knowledge of chemical reactions to predict the products of a chemical reaction.
10. Use and understand the energy terms in a chemical reaction.
11. Calculate molecular weights and formula weights. (3.3)
12. Calculate the percent composition of elements in molecules. (3.3)
13. Understand what is meant by Avogadro’s number. (3.4)
14. Convert from number of atoms to moles to mass and back again. (3.4)
15. Calculate empirical and molecular formulas from lab data. (3.5)
16. Use the principles of gravimetric stoichiometry to combine data from balanced chemical equations, moles, and mass.
(3.6)
17. Be able to perform calculations around limiting reactants. (3.7)
18. Perform calculations involving theoretic yield. (3.7)
Vocabulary
reactant
chemical change
combination reaction
double replacement
molecular weight
mole
theoretical yield
product
precipitate
decomposition
neutralization
percent composition
empirical formula
percent yield
stoichiometry
aqueous
combustion
water forming
avogadro’s number
molecular formula
physical change
synthesis
single replacement
formula weight
molar mass
limiting reactant
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Notes
There are two types of reactions:
A physical change
Examples:
A chemical change
For instance:
A chemical reaction can be identified by the following:
a)
b)
c)
d)
Stoichiometry
Chemical equations
o
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Subscripts vs Coefficients
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Balancing Equations
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Matter cannot be lost in any chemical reaction.
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Balance the following equation:
Na(s) + H2O(l)  H2 (g) + NaOH(aq)
CH4 (g) + O2 (g)  CO2 (g) + H2O (g)
Pb(NO3)2 (aq) + NaI
Remember:
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(aq)
 PbI2 (aq) + NaNO3 (aq)
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Indicating the States of Reactants and Products
•
The physical state of each reactant and product may be added to the equation:
(s) (l) (g) (aq) (ppt) -
•
Reaction conditions occasionally appear above or below the reaction arrow (e.g., "Δ" is often used to indicate the
addition of heat).
Complete questions 3.11 to 3.14
Reaction Types
There are several basic types of chemical reactions. The ones covered in this chapter are:
1) Synthesis (Combination) A + B  AB
another way to look at it is:
element + element  compound
Examples:
2) Decomposition AB  A + B
another way to look at it is:
Examples:
compound  element + element
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3) Combustion -
CxHy + O2  CO2 + H2O
Examples:
Other reaction types will be covered in chapter 4.
Complete questions 3.15 to 3.20
3.3 Formula Weights
•
Formula weight (FW)
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Example: CaCl2
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Molecular weight (MW) is the sum of the atomic weights of the atoms in a molecule as shown in the molecular
formula.
• Example: MW (C6H12O6)
• = 6(12.01 amu) + 12 (1.01 amu) + 6 (16.00 amu)
• = 180.18 amu.
•
Formula weight of the repeating unit ( formula unit) is used for ionic substances.
• Example: FW (NaCl)
• = 22.99 amu + 35.45 amu
• = 58.44 amu.
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Complete questions 3.21 and 3.22
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Percentage Composition
•
Percentage composition is obtained by dividing the mass contributed by each element (number of atoms times AW)
by the formula weight of the compound and multiplying by 100.
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example:
% composition of C2H6
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example:
What is the percent composition of each element in H 2SO4
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the sum of the percent compositions will always equal 100 %
Complete questions 3.24 and 3.26. Calculate the percentage of every element in each compound
3.4
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Avogadro’s Number and The Mole
The mole (abbreviated "mol") is a convenient measure of chemical quantities.
o 1 mole of something = 6.0221421 x 1023 of that thing.
o This number is called Avogadro’s number.
Thus, 1 mole of carbon atoms = 6.0221421 x 10 23 carbon atoms.
Experimentally, 1 mole of 12C has a mass of 12 g. (an exact number)
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Molar Mass
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Interconverting Masses and Moles
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Look at units:
• Mass: g
• Moles: mol
• Molar mass: g/mol
To convert between grams and moles, we use the molar mass.
record the examples in class.
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Interconverting Masses and Number of Particles
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Units:
• Number of particles: 6.022 x 1023 mol–1 (Avogadro’s number).
• Note: g/mol x mol = g (i.e. molar mass x moles = mass), and
• mol x mol–1 = a number (i.e. moles x Avogadro’s number = molecules).
To convert between moles and molecules we use Avogadro’s number.
record the examples in class
Complete questions 3.32 to 3.42, even
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3.5 Empirical Formulas from Analyses
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Finding empirical formula from mass percent data:
1.
2.
3.
4.
5.
6.
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Assume we start with 100 g of sample.
The mass percent then translates as the number of grams of each element in 100 g of sample.
From these masses, calculate the number of moles (use atomic weight from Periodic Table)
The lowest number of moles becomes the divisor for the others. (gives a mole ratio greater than 1)
Adjust mole ratios so all numbers are whole (1, 2, etc)
The lowest whole-number ratio of moles is the empirical formula.
Example:
The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical
formula of PABA.
Assumptions:
1. These are % by mass; in 100g of the compound, the percentages represent masses in grams.
Mass divided by molar mass gives moles of each atom.
Complete questions 3.43 to 3.46
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Molecular Formula from Empirical Formula
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The empirical formula (relative ratio of elements in the molecule) may not be the molecular formula (actual ratio of
elements in the molecule).
o Example: ascorbic acid (vitamin C) has the empirical formula C 3H4O3.
o The molecular formula is C6H8O6.
To get the molecular formula from the empirical formula, we need to know the molecular weight, MW.
The ratio of molecular weight (MW) to formula weight (FW) of the empirical formula must be a whole number.
Complete questions 3.47 to 3.50
Combustion Analysis
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Empirical formulas are routinely determined by combustion analysis.
A sample containing C, H, and O is combusted in excess oxygen to produce CO2 and H2O.
o The amount of CO2 gives the amount of C originally present in the sample.
o The amount of H2O gives the amount of H originally present in the sample.
o Watch the stoichiometry: 1 mol H2O contains 2 mol H.
o The amount of O originally present in the sample is given by the difference between the amount of sample and
the amount of C and H accounted for.
More complicated methods can be used to quantify the amounts of other elements present, but they rely on
analogous methods.
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Complete questions 3.51 and 3.52
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3.6
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Quantitative Information from Balanced Equations
The coefficients in a balanced chemical equation give the relative numbers of molecules (or formula units) involved in
the reaction.
The stoichiometric coefficients in the balanced equation may be interpreted as:
o the relative numbers of molecules or formula units involved in the reaction or
o the relative numbers of moles involved in the reaction.
The molar quantities indicated by the coefficients in a balanced equation are called stoichiometrically equivalent
quantities.
Stoichiometric relations or ratios may be used to convert between quantities of reactants and products in a reaction.
It is important to realize that the stoichiometric ratios are the ideal proportions in which reactants are needed to form
products.
The number of grams of reactant cannot be directly related to the number of grams of product.
To get grams of product from grams of reactant:
 convert grams of reactant to moles of reactant (use molar mass),
 convert moles of one reactant to moles of other reactants and products (use the stoichiometric ratio from the
balanced chemical equation), and then
 convert moles back into grams for desired product (use molar mass).
Complete questions 3.57 to 3.66
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3.7
Limiting Reactants
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It is not necessary to have all reactants present in stoichiometric amounts.
Often, one or more reactants is present in excess.
Therefore, at the end of reaction those reactants present in excess will still be in the reaction mixture.
The one or more reactants that are completely consumed are called the limiting reactants or limiting reagents.
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Reactants present in excess are called excess reactants or excess reagents.
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Consider 10 H2 molecules mixed with 7 O2 molecules to form water.
o The balanced chemical equation tells us that the stoichiometric ratio of H 2 to O2 is 2 to 1:
2H2(g) + O2(g)  2H2O(l)
o
o
o
o
This means that our 10 H2 molecules require 5 O2 molecules (2:1).
Since we have 7 O2 molecules, our reaction is limited by the amount of H2 we have (the O2 is present in
excess).
So, all 10 H2 molecules can (and do) react with 5 of the O2 molecules producing 10 H2O molecules.
At the end of the reaction, 2 O2 molecules remain unreacted.
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Complete questions 3.69 to 3.76
Theoretical Yields
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The amount of product predicted from stoichiometry, taking into account limiting reagents, is called the theoretical
yield.
This is often different from the actual yield – the amount of product actually obtained in the reaction.
The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:
Percent yield =
Complete questions 3.77 to 3.80
actual yield
100
theoretica l yield