Transcript Stoichiometry

Topic 4
Reactants chemically change into
products
 must be balanced
 use symbols to indicate states
 on AP exam, all equations are written as
net ionic (have to know solubility rules)

CO2(g) + 2NH3(g) + H2O(l)  2NH4+(aq) +
CO32-(aq)
Balance the following reaction:
H2SO4(aq) +
NaHCO3(s)  CO2(g) +
Na2SO4(aq)
H2O(l) +
Combination/synthesis – multiple
reactants come together to form one
product
2Na(s) + Cl2(g)  2NaCl(s)
 Decomposition – one reactant “falls
apart” into multiple products
MgCO3(s)  MgO(s) + CO2(g)
 Combustion – compound reacts with
oxygen (usu. from the air)
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)

We don’t use “single replacement” any
more (more on that later)
 We also don’t use “double
replacement”, they are called
precipitation reactions (more on that
later, too)

Avogadro’s # = 6.02 x 1023
 mole = amount of a substance that
contains 6.02x1023 “things”
 molar mass = mass of 1 mole of a
substance (numerically the same as
atomic mass, but measured in g/mol)

percentage by mass of each element in
a compound
 Divide the mass of an element in a
compound by the molar mass of the
compound and multiply by 100.

What is the %comp of each element in
Na2CO3?

simplest ratio of elements in a
compound
› for ionic compounds this IS the actual
formula
› for molecular compounds it could be the
formula or it may need to by multiplied by
some number to equal the actual formula
start with %comp, convert to grams, then
to moles, then to simplest ratio
 this is what your lab will be based on…

What is the empirical formula of a
compound containing 68.4% chromium
and the rest oxygen?
Molecular formulas are multiples of the
empirical formula
 if the molar mass of the compound is
known, the MF can be determined from
the EF…

A compound containing only carbon,
hydrogen, and oxygen is 63.16% C and
8.77% H. It has a molar mass of 114
g/mol. What is its empirical formula and
molecular formula?
When a compound containing C, H, & O
is completely combusted, all of the C
becomes CO2 and all of the H becomes
H2O.
 EF can be calculated from the amounts
of products…

A 3.489 g sample of a compound
containing C, H & O yields 7.832 g of CO2
and 1.922 g of H2O upon combustion.
What is the simplest formula of the
compound?
Ionic compounds form hydrates with
water molecules (ex. sodium thiosulfate,
decahydrate = Na2S2O310H2O)
 When hydrates are heated, the extra water
molecules are evaporated off.
 Can calculate how many water molecules are in a
hydrate by comparing the masses before and
after heating (this will be another lab)

When 21.91 g of a hydrate of copper (II)
sulfate is heated to drive off the water,
14.00 g of anhydrous copper (II) sulfate
remain. What is the formula for the
hydrate?
Must have a balanced reaction
 Pay attention to sig figs
 Keep it organized!!!

How many moles of octane will burn in the
presence of 37.0 moles of oxygen?
How many grams of carbon dioxide are
obtained when 695 g of octane are
burned in atmospheric oxygen?
In a chemical reaction with multiple
reactants, one will be completely used
up (the limiting reactant) and the others
will be left over (the excess reactants).
 Involves lots of steps and the process is
key to getting these right.

255 g of octane and 1510 g of oxygen gas are present
are the beginning of a reaction that goes to
completion and forms carbon dioxide and water
according to the following equation.
2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(g)
a) What is the limiting reactant
b) How many grams of water are formed when all of the
limiting reactant is consumed?
c) How many grams of excess reactant is consumed?
d) How many grams of excess reactant is left
unreacted?
Theoretical yield = calculated amount of
product that should be formed
 Actual yield = amount of product that is
actually formed in the lab, almost always
less than the theoretical yield
 Percent yield = comparison of
theoretical to actual yield
(AY/TY) x 100%
