Chapter Three:
Stoichiometry
Nick Tang
2nd Period
Ms. Ricks
3.1 Chemical Equations
• + means reacts with,  means produces
• Chemicals to the left represent reactants, Chemicals to the
right represent products
• The numbers in front of the formulas are coefficients while
the small number on the bottom are called subscripts
• The coefficients lets you know how much there is of that
substance, while the subscripts indicate its identity
Balancing Equations
• An equation is balanced once each side of the equation has
the same amount of atoms of each element
• You can change the coefficients but never the subscripts
• It’s usually best to try to first balance the elements that occurs
the least in the equation
CH4 + O2  CO2 + H2O
• We have 1 C on both sides, 4 H’s on the left while 2 H’s on the
right, and 2 O’s on left and 3 O’s on the right
• Balance the H first on the left by putting a 2 in front of H2O
and then balance the O’s by putting a 2 in front of the O2
Final equation:
CH4 + 2O2  CO2 + 2H2O
Check Up
• Balance the following
__Fe + __O2  __Fe2O3
___C2H4 + __O2  __CO2 + __ H2O
___Al + __HCL __ALCL3 + __H2
•
•
•
4,3,2
1,3,2,2
2,6,2,3
Indicating the States of Reactants
and Products
• The 4 states are solid (s), liquid (l), gas (g), and aqueous (aq)
• We use this to indicate the physical state of each substance
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
3.2 Some Simple Patterns of
Chemical Reactivity
• Combination reactions: two or more substances react to form
one product. Also when a metal combines with a nonmetal
the product is an ionic solid.
Ex: 2Mg + O2  2MgO
• Decomposition reactions: one substance undergoes a
reaction to produce two or more other substances.
Ex: CaCO3  CaO +CO2
3.2 Some Simple Patterns of
Chemical Reactivity Cont.
• Combustion in Air. Combustion reactions are rapid reactions
that produce a flame.
• Most of these reactions involve O2 as a reactant
• When hydrocarbons are combusted they react with O2 to
make CO2 and H2O
C3H8 + 5O2  3CO2 + 4H2O
Check Up
• Write equation for:
• Solid mercury(II) sulfide decomposes into its
component elements when heated
• Write the balanced equation for the reaction
that occurs when ethanol, C2H5OH is burned
in air
• HgS  Hg + S
• C2H5OH + 3O2  2CO2 + 3H2O
3.3 Formula Weights
• The formula weight of a substance is the sum of the atomic
weights of each atom in its chemical formula
• Use the atomic masses from the periodic table and multiply with
the subscripts and coefficients to find the weight
• Ex: FW of H2SO4 = 2(AW of H) + (AW of S) + 4(AW of O) = 98.1amu
• Percent Composition: How much of a element contributes to the
mass of the entire substance. Use this equation
Check Up
• Calculate the formula weight of sucrose
C12H22O11
• Calculate the percentage of nitrogen, by mass,
in Ca(NO3)2
• 342.0 amu
• 17.1%
3.4 Avogadro’s Number and the
Mole
• Mole(abbreviated mol) is the unit for dealing with atoms,
ions, or molecules
• Avogadro’s number is 6.02E23
• Molar Mass the mass in grams of one mole of a substance
• Make sure you know how to convert moles to atoms, grams
to moles, moles to grams, and number of molecules and
number of atoms from mass
Converting Moles to Number of
Atoms
Calculate the number of H atoms in 0.350 mol of C6H12O6
First put how many moles you have and convert it to molecules.
We know we have 6.02E23 molecules for every mole so multiply
by 6.02E23 and then multiply by how many H atoms there are
which in this case is 12
There are 2.53E24 H atoms
Converting grams to moles
Calculate the number of moles of glucose (C6H12O6) in 5.380 g of C6H12O6
First put how much you have of the substance and then divide by it’s
molar mass and multiply by 1 mol
Put that in the calculator and you should get .02989 mol C6H12O6
Converting moles to grams
Calculate the mass, in grams, of 0.433 mol of calcium nitrate
First put down how many moles of calcium nitrate you have and
we know that one mol is equivalent to the molar mass of the
compound so multiply .433mols of Ca(NO3)2 by its molar mass,
164.1g Ca(NO3)2
Answer: 71.1g Ca(NO3)2
Check Up
Super Problem: How many glucose molecules are in 5.23g of
C6H12O6? How many oxygen atoms are in this sample?
• 1.75E22 molecules C6H12O6
• 1.05E23 atoms O
3.5 Empirical Formulas from
Analyses
• The empirical formula tells us the simplest whole number
ratio of moles of each element in a compound
• To find the empirical formula, first take the given masses of
each element and divide by their atomic masses
• Afterwards determine the simplest whole number ratio of
moles by dividing each number of moles by the smallest
number of moles
• If you get decimals either round them if they are close(.9, .1)
but if it’s too far from being a whole number such as 1.5
multiple it(and everything else) by a ratio to make it even
3.5 Empirical Formulas from
Analyses Cont.
What is the empirical formula of vitamin C which contains
40.92% C, 4.58% H, and 54.50% O by mass
3.5 Empirical Formulas from
Analyses Cont.
• Molecular formula indicates the actual number of atoms of
each element in one molecule of a substance
• To find the molecular formula you must have the molecular
weight and the empirical formula weight
• You divide the molecular weight by the empirical formula
weight to find your whole number multiple, which you use to
multiply the subscripts in the empirical formula
molecular weight
Whole-number multiple= -------------------------------empirical formula weight
3.5 Empirical Formulas from
Analyses Cont.
Determine the molecular formula of C3H4 which has a molecular
weight of 121 amu
3.5 Empirical Formulas from
Analyses Cont.
• Combustion analysis is commonly used for compounds
containing carbon and hydrogen as their component elements
• CO2 and H2O are produced and from the masses of CO2 and
H2O we can calculate the empirical formula
• If there is a third element we can find it’s mass by subtracting
the total mass by the mass of C and H
Combustion Analysis
Combustion of 0.255g of isopropyl alcohol produces 0.561g of
CO2 and 0.306g of H2O. Determine the empirical formula
Combustion Analysis Cont.
3.6 Quantitative Information
from Balanced Equations
2H2 + O2  2H2O
• The quantities 2 mol H2, 1 mol O2, and 2 mol H2O are called
stoichiometric ally equivalent quantities
• This shows the stoichiometric relations can can be used to
convert between quantities of reactants and products in a
chemical reaction
3.6 Quantitative Information
from Balanced Equations
How many grams of water are produced in the oxidation of
1.00g of glucose, C6H12O6?
C6H12O6 + 6O2  6CO2 + 6H2O
3.7 Limiting Reactants
• The limiting reactant is the reactant that is completely
consumed and determines the amount of product formed
• The other reactant is called the excess reactant which is what
is left over
N2 + 3H2  2NH3
How many moles of NH3 can be formed from 3.0 mol of N2 and
6.0 mol of H2
3.7 Limiting Reactants Conti.
• We don’t have 9 moles of H2 therefore H2 is the limiting
reactant
• Now go from moles of H2 to moles of NH3
3.7 Theoretical Yields
• Theoretical yield is the quantity of product that is calculated
to form when all of the limiting reactant reacts
• The amount of product actually obtained in a reaction is the
actual yield
• The actual yield is almost always less than the theoretical due
to some parts of the reactants not reacting or some other
outside force but never will it be higher than the theoretical
• Percent yield relates the actual to the theoretical using this
equation
Actual Yield
Percent Yield= ---------------------------- X 100%
Theoretical Yield
3.7 Theoretical Yields Cont.
2C6H12 + 5O2  2H2C6H8O4 + 2H2O
Assume that you carry out this reaction starting with 25g of
C6H12 and that C6H12 is your limiting reactant. What’s the
theoretical yield of 2H2C6H8O4? If you obtain 33.5g of 2H2C6H8O4
what is the percent yield for 2H2C6H8O4?