Chapter 3
6/2015
Chemical Equations and Stoichiometry
Stoichiometry: the study of quantities of substances used and produced
in a chemical equation. Stoichiometry is based on…
 chemical equations represent chemical reactions
 Law of Conservation of Matter: Matter (mass) cannot be
created or destroyed in a chemical reaction. The atoms are
only re-arranged. (Antoine Lavoisier)
Total mass reactants = total mass of products
Chemical reaction: a process where one or more substances are
converted into new substances with different chemical
properties.
Substances react so atoms can have a full octet and attain a more
favorable (lower) energy state.
reactants: the starting substances in a chemical reactions
written on the left
2HCl + Zn  ZnCl2 + H2
products: substances produced in a chemical reaction written on the
right
Chemical equation: a “statement” using words or chemical symbols and
formulas to describe a chemical reaction.
- reactants on left , products on right
- + means “ with”
-  means yields or produces, and shows
the direction of the reaction
1. Skeleton equation – just shows reactants and products
Ex:
H2 + O2  H2O it is not balanced
2. Balanced chemical equation – includes coefficients to indicate the
number of each substance needed to make the equation
follow the Law of Conservation of Mass, meaning that the total
number and type of atoms on the reactant side of the equation
are the same as on the product side of the equation. (Atoms
cannot be created or destroyed)
Coefficients: indicate the number of molecules
(or atoms in the case of pure elements)
* written as whole numbers in front of a
formula or symbol to balance an equation
Coefficients show 2 H2 molecules & 2 H2O molecules
Ex: 2H2 + O2  2H2O
Subscripts tell you how many of that type of
atom is in one unit /molecule of the
compound.
Subscripts are NEVER altered when balancing an equation.
* Ex:
Ca + O2  CaO skeleton, unbalanced
2Ca + O2  2 CaO balanced
RULES FOR BALANCING EQUATIONS:
1. Write a formula equation with correct symbols and formulas
(remember ionic compounds are neutral)
Na + Cl2  NaCl
2. Count the number of atoms of each element on each side of the
arrow.
3. Balance atoms by adding coefficients in FRONT of a formula!
2Na + Cl2  2NaCl
4. Check your work by counting atoms of each element , they should be
the same on both sides of the equation.
5. If you cannot balance an equation, you have probably written a
formula incorrectly, Check and try again.
Hints for balancing and writing chemical equations:
* Remember the seven elements that exist as
diatomic molecules when they are by themselves and
not in a compound. They are all gases:
N2 O2 F2 Cl2 Br2 I2 H2
* Remember to write the correct formulas for ionic products of
A reaction. (Total positive charges must equal total neg
charges)
Complete balanced equations include the states of all reactants and
products written as subscripts, in parentheses, to lower
right of the substance Ex: H2O (g)
(s) solid
(l) liquid
pure substance in the liquid state, usually
only H2O and NH3
(g) gas
N2 O2 F2 Cl2 Br2 I2 H2, sometimes H2O
(aq) aqueous solution – anything mixed in water and all acids
( ppt) or ( ), precipitate: a solid product of a reaction, forms in
small particles , makes solution cloudy, the solid will
eventually settle to the bottom
∆
heat added
Balancing Equations Practice
1. Ca + O2  CaO
2.
Mg + N2  Mg3N2
3. CH4 + O2  CO2 + H2O
4. Cl2 + LiBr  LiCl + Br2
5. FeCl2 + Na3PO4  NaCl + Fe3(PO4)2
www.sciencegeek.net/Chemistry/taters/EquationBalancing.htm
http://phet.colorado.edu/en/simulation/balancing-chemicalequationshttp://www.files.chem.vt.edu/RVGS/ACT/notes/scripts/bal_eq1.html
http://education.jlab.org/elementbalancing/question.php?85239516
Types of chemical equations:
1. Synthesis reactions (aka direct combination)
* 2 or more reactants come together to form a single product
General Form: A + B  C
Example: 2Ca + O2  2CaO
2. Decomposition Reactions
* a single compound (reactant) breaks down into 2 or more
smaller compounds or elements
General form: AB  A + B
Example : CaCO3  CaO + CO2
3. Combustion Reactions:
General form: X + O2  products…


combustion of a hydrocarbon (H-C compound) will usually
form H2O and CO2.
If the oxygen is limiting, the product is CO instead of CO2.
Relating chemical equations to quantities we can accurately measure
in the lab…..
Atomic mass: mass of one atom
* unit is amu; we cannot easily measure this
* Ex: mass of one atom of Fe is 55.8 amu
Formula mass: mass of all atoms in one piece of a substance
* unit is amu
* calculated by adding up atomic masses of all
atoms in the formula
* Formula mass of H2O =
(2 x mass of hydrogen ) + (mass of oxygen)
(2 x 1.01) + 15.99
18.01 amu
Molar mass
 The sum of all atomic masses of the atoms in one unit or
molecule of the substance, in grams, is molar mass of a
substance.
 We can easily weigh out grams
Ex: What is the molar mass of NaCl?
23 + 35.4 = 58.4 grams
Ex: What is the molar mass of CO2?
12 + (2 x 16) = 44g
Group practice:
What is the molar mass of MgCl2?
What is the molar mass of KOH?
Calculating % Composition of a Compound
Percent composition of a compound is the percentage of each
element in the compound by mass.
 You need to find the % for each element in the compound
using the formula:
% element = total grams of element in formula x 100
Total formula or molecular mass
Sample Problem 1: 20.0 g sucrose is analyzed and found to contain
8.44 g C, 1.30g H and 10.26 g O. Calculate the % composition of
sucrose.
Sample Problem 2: Calculate the percent composition of H2O.
(hint – you must look up the atomic masses of the elements in the
periodic table).
Sample Problem 3: How much oxygen is contained
in 120g H2O?
Calculating percent (a math refresher)
 you must change % to a decimal before multiplying it to
get a quantity
 do this by moving the decimal place left two places!
20% becomes .20
6.5% becomes .065

So 20 % of 125 is .20 x 125 = 25
http://science.widener.edu/svb/tutorial/percentcompcsn7.html
MOLE: (mol) metric unit for quantity of small particles
* also called Avagadro’s number
1 mole = 6.02 x 1023 particles (smallest piece)
Three words used to describe a particle of a substance
Atom: smallest piece of a pure element
Molecule: smallest piece of a covalent substance
(2 or more nonmetals)
Ex: CO2
Formula Unit: smallest piece of an ionic compound
( metal + nonmetal or polyatomic ions)
Ex: NaHCO3
A mole can also be used to describe a mass or a volume of gas:
1 mole = molar mass of a substance in grams
Molar mass : the mass of all atoms in one unit / molecule of a
substance
 same as the formula mass or molecular mass, except
the unit is grams
 Ex: molar mass of H2O = 18.01 grams
1 mole = 22.4 L volume of any gas at standard temp. and
pressure (STP)
* STP = 0oC, 1 atm
SUMMARY:
6.02 X 1023 particles
1 MOLE
Molar mass, grams
22.4 L volume of any gas at STP
Because 1 mole is equal to these three things , we can use this
relationship to convert from one quantity to another!
CONVERSIONS WITH MOLES
Equivalent quantities:
1 mole
6.02 x 1023 particles
Molar mass, grams
22.4 L gas at STP
Because these values are all equivalent to each other, we can use any
two to make a conversion factor.
Converting:
Conversion factor
Moles – Particles
6.02 x 1023 particles
1 mole
Moles – Grams
molar mass *, grams
1 mole
* you need to calculate molar
mass for each substance
Moles – Volume, L
22.4 L
1 mole
Particles – Grams
molar mass, grams
6.02 x 1023 particles
Particles – Volume
22.4 L
6.02 x 1023 particles
Grams – Volume
22.4 L
Molar mass, grams
Steps for solving problems:
1. Read problem
2. Write given value as a fraction over 1
3. Write out conversion factor. (If problem involves grams,
you must figure out the molar mass at this point)
Remember to write the conversion factor so the given
unit is on the bottom of the conversion factor and cancels out,
and the unit you want is on the top.
4. Multiply across the top and bottom of the fractions, then divide the
top value by the bottom value.
5. Write your answer , remembering the units.
EXAMPLES
A. How many moles of NaCl are in 11.2 g NaCl?
* you will need to use 1 mole = molar mass of NaCl
* calculate molar mass of NaCl = 23 + 35.4 = 58.4 g
11.2 g NaCl x 1 mole NaCl = 0.197 mol NaCl
1
58.4 g NaCl
B. How many grams are in 2.5 mol NaCl?
C. How many molecules are in 2 moles of H2O?
D. How many moles are in 8.74 x 1023 formula units of CaCO3?
E. A 1 L flask is filled with CO2 gas at standard temp. and pressure.
How many moles of gas are in the flask?
F. Six moles of CO2 gas at STP will occupy what volume?
G. 14.5 L of H2 gas contains how many molecules?
H. 56 grams of O2 will occupy what volume?
More fun calculations involving moles…
Empirical and molecular formulas are written to show the ratio or actual
number of atoms in a compound. These ratios or numbers can also be
thought of as a ratio of MOLES….
So…H2O can be thought of as 2H atoms : 1 O atom
Or 2 moles of H atoms : 1 mole O atoms.
Solving problems for empirical formulas:
1. If you are given grams of elements…
Convert grams of each element in the compound to moles.
2. if you are given % values instead of grams, just change the %
sign to a grams, and convert to moles.
3. Divide all mole values by the smallest mole value to get a ratio
(remember empirical formulas are lowest whole number ratios
of atoms in a compound)
4. If you do not get whole numbers, then multiply all values by 2, 3 ,
4, or 5 to get whole numbers.
5. These whole numbers become the subscripts for the empirical
formula.
Sample problems:
A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O and
0.675 g H. What is the empirical formula of this compound?
First Convert g to mol for each
Then divide all mole values by the smallest value to get a ratio
13.5 g Ca x 1 mol Ca = 0.337 mol Ca / .337 = 1 Ca
1
40 g Ca
10.8 g O x 1 mol O = 0.675 mol O / .337 = 2 O
1
16 g O
0.675g H x 1 mol H = 0.675 mol H / .337 = 2 H
1
1gH
So…you write the formula as Ca(OH)2
Group practice:
Determine the empirical formula of a compound containing 7.30 g Na,
5.08 g S and 7.62 g O.
Solving for molecular formulas:
A molecular formula is a whole number MULTIPLE of an empirical
formula.
Ex The empirical formula of glucose is CH2O
Multiply all subscripts x 6 = molecular formula
C6H12O6
To solve these problems you need
Empirical formula (often have to calculate this)
Empirical formula mass (add masses of atoms)
Molecular formula mass (aka molar mass), given in the problem.
Calculate the number to multiply the empirical formula subscripts by to
get the molecular formula using the equation:
Multiple = molar mass (mol. Form. Mass)
Empirical formula mass
Sample problems.
Ribose has a molar mass of 150 g/mol. It is 40% C, 6.67% H and 53.3%
O. What is the molecular formula of ribose?
Find the molecular formula of a compound that contains 4.9g N and
11.2 g O The molar mass is 92.0 g/mol.
Stoichiometry: the quantitative (measurable) relationships between
chemicals in a chemical reaction.
* Coefficients in a balanced chemical equation tell
us the ratio of individual particles OR moles
Thus, given the balanced chemical equation:
2H2 + O2  2H2O
* there is a 2 mole H2 : 1 mole O2 ratio
* we call these ratios mole: mole ratios
We can use mole: mole ratios from a balanced chemical equation to
convert from moles of one substance to moles of another substance.
Solving Mole to Mole Problems – where you are given moles of one
substance and asked to solve for moles of another substance in that
reaction.
STEPS:
1. Write the balanced chemical equation.
2. Write the given quantity as a fraction over 1.
3. Find the mole to mole ratio for the substance you were given and the
requested substance (from the coefficients in front of these substances
in the balanced chemical equation)
4. Multiply by your mole to mole conversion ratio, with the moles given
substance on the bottom so they will cross out, and the moles requested
substance on top.
EXAMPLE:
NH4NO3 decomposes to form water and N2O gas. How many moles of
H2O are produced from the decomposition of 2.25 mol NH4NO3 ?
1. NH4NO3  N2O + 2H2O
2. 2.25 mol NH4NO3
1
3.
Mole to mole ratio is 1 mole NH4NO3 : 2 moles H2O
4. 2.25 mol NH4NO3 x 2 mol H2O
1
1 mol NH4NO3
= 4.5 mol H2O
MORE STOICHIOMETRY:
Remember that we can convert from moles of a substance to particles,
grams, or volume of a gas at STP?
6.02 X 1023 particles
1 MOLE
Molar mass, grams
22.4 L volume of any gas at STP
…and remember we can convert from one substance to another using
the mole:mole ratio from the coefficients of a balanced chemical
equation….
x moles A
y moles B
…and we can convert moles of substance B to particles, or volume in
liters or mass in grams…..
We can use these relationships to convert any quantity of substance A to
any quantity of substance B using the following steps….
1. Write the balanced chemical equation.
2. Write the given quantity as a fraction over 1.
3. Convert the given quantity to moles using the
conversion factor:
1 mole
6.02 x 1023 particles
1mole
molar mass, g
appropriate
1 mole_
22.4 L
4. Multiply by the mole : mole ratio (from the coefficients of the
balanced equation) to convert from moles of the given substance to
moles of the requested substance.
x moles requested substance
moles given substance
5. Convert from moles to the requested unit by using the appropriate
mole conversion factor (see step 3)
6. Do the math….
Multiply all numbers on the top together.
Multiply all numbers on the bottom together.
Divide the top answer by the bottom answer
EXAMPLE:
Glucose, C6H12O6 , reacts with oxygen gas to produce water and carbon
dioxide. How many grams of water are produced from 1.5 g glucose?
1. C6H12O6 + 6O2 6H2O + 6CO2
2. Convert grams glucose to moles glucose
– need to calculate molar mass of glucose….
(12 x 6 ) + (1 x 12) + (16 x 6) = 180 g
1.5 g glu x 1 mol glu
1
180 g glu
3. Now convert moles glu to moles water using the mole to mole ratio
which is 1 mole glucose : 6 moles H2O
1.5 g glu x 1 mol glu x 6 mole H2O
1
180 g glu
1 mol glu
4. And the last step is to convert moles H2O to grams using the
molar mass of water, 18 g = 1 mole
1.5 g glu x 1 mol glu x 6 mole H2O x 18 g H2O = 0.899 g H2O
1
180 g glu
1 mol glu
1 mol H2O
Group practice: In a thermite reaction, powdered aluminum reacts with
Iron (III) oxide to produce aluminum oxide and molten iron. What mass
of aluminum oxide is produced when 2.3g of aluminum reacts with
excess iron (III) oxide? 2Al + Fe2O3  2Fe + Al2O3
Limiting Reactant Problems:
1. You are given amounts of reactants and asked to
predict the quantity of products formed
2. The limiting reactant is the reactant that is used up
first and determines the total amount of product
that can be made.
Steps to solving limiting reactant problems:
a. Write balanced equation
b. Calculate the reactant mass  product mass for
EACH reactant
( hint – pick the simplest product to make your
work easier)
c. The reactant that yields the least amount of
product is your limiting reactant
Example: Identify the limiting reactant when 1.7 g sodium reacts with
2.6L of chlorine gas at STP to produce sodium chloride.
Percent Yield: the percent of the expected product that you actually
obtain in a chemical reaction. It is a measure of the efficiency of a
reaction.
1. % yield =
actual yield
x 100
theoretical yield
2. theoretical yield: the amount of product you expect to
get according to your calculations
3. actual yield: the amount of product actually obtained
in the lab. You are usually given this in a
problem.
* The actual yield is usually less than expected
because:
i. some reactant might not react
ii. a side reaction may be using up some of
your reactant
iii. some product may be lost when transferring
or isolating it
Solving % Yield Problems:
1. Write the balanced equation
2. Calculate the expected mass of the product from
amount of reactant given.
Sometimes you need to identify the limiting
reactant, then the theoretical value would be the
quantity of product produced from the limiting
reactant
3. You are usually given the actual yield.
4. Calculate % yield using equation above
Example: A piece of copper with a mass of 5 g is placed in a solution of
silver nitrate. Assume there is excess silver nitrate. The silver metal
produced has a mass of 5.2 g. What is the percent yield of this reaction?
Example: Determine the percent yield for the reaction between 15.0g N2
and 15.0 g H2 if 10.5 g NH3 is produced?