Stoichiometry
The relationship between
masses of (neutral)
compounds, as well as, the
mass relationship of
products and reactants in
balanced equations
(molar ratios)
Compounds:
 We have worked with the relationship between
cations and anions in neutral compounds to make
mass relationships between the different
elements involved adding appropriate subscripts
(We have balanced at least 300 compounds!).
1. Write the formula for copper(II)nitrate.
Cu(NO3)2
2. Write the formula for mercury(I)chloride.
Hg2Cl2
3. Write the formula for lead(II)hydroxide.
Pb(OH)2
Molar ratios
relate masses according to the
Law of conservation of Mass:
 By balancing each of the four types of
chemical reactions (synthesis,
decomposition, single displacement, double
displacement) we can relate mass of all
products and reactants through the molar
ratios of the equation. The ratios from the
coefficients are used to calculate the
amounts of all products and reactants used.
 #1: Write the balanced equation between
lead(II)nitrate and potassium chromate
that results in the yellow paintish ppt.,
lead(II)chromate from one of our last lab
in class.
Pb(NO3)2 + K2CrO4  PbCrO4 + 2KNO3
 The ratios from
the coefficients
are used to
calculate the
amounts of all
products and
reactants used.
Suppose that only
.25 moles of
potassium chromate
are available, how
much of each of the
other substances is
involved in the
balanced equation?
.25
.25
.25
.50
Pb(NO3)2 + K2CrO4  PbCrO4 + 2KNO3
 Note that the mole ratio for the
balanced equation is:
 1 :
1
:
1
: 2
 So the ratio for the .25 moles is:
 .25 : .25
:
.25 : .50
How many grams is .25 moles of the
potassium chromate?
.25 mo.
Pb(NO3)2 + K2CrO4  PbCrO4 + 2KNO3
The molar mass
of K2CrO4 is:
K = 39 x 2 = 78
Cr = 52
O = 16 x 4 = 64
Total = 194g
.25mo (194g) =
(1 mo)
g
How many grams of
precipitate are made?
.25 mo.
Pb(NO3)2 + K2CrO4  PbCrO4 + 2KNO3
molar mass of PbCrO4 is:
Pb = 207
Cr = 52
O = 16 x 4 = 64
Total = 323g
.25mo (323g) = g
(1 mo)
Mass-Mass mole problems
The following slides will
contain a set of 4 more word
equations that need to be
written in balanced chemical
form on the separate sheet of
paper provided.
Have them checked over prior
to completing the
stoichiometry calculations
The 4 balanced equations and
stoich for #2 and #3 are due
Monday:
 You have had plenty
 You need to write
of time to work
the 4 balanced
through these… if you
equations for the
need more help, come
stoich problems first
in to the science
and get them
resource during your
checked off before
free time.
you leave… then you
can proceed through
the slides and use
the hints to work on
the actual stoich!
2. 15.7g of aluminum and excess nitrogen
gas form how many grams of aluminum
nitride?
3. The dissociation of
3.2 moles of
ammonium nitrate
yield how many
grams of dinitrogen
monoxide (N2O) and
how many grams of
water in a
dissociation reaction?
4. 27.0g of silver nitrate and
excess zinc metal will form how
many grams of silver metal?
(what else is formed?)
5. Excess aluminum oxide and 15.0g of
magnesium form magnesium oxide
and how many grams of aluminum?
The next slides are a guide to help
with #2 and #3.
 All of these should
be done before you
leave…
 Tomorrow we will
go over #5-8 on
the salmon
bellwork and
 The pink stoich
problems will be
collected and
corrected.
That means March 5th,
Not April 15th!
2. 15.7g of aluminum and excess
nitrogen gas form how many grams
of aluminum nitride?
 mo:
.58mo ----.29------ .58

 gram:
2Al
+
N2
15.7g (1 mo)
(27g)
= .58mo.

2AlN
?g
This is only a mole-mole, mole
gram problem…
 Have you noticed that some problems
use all three steps, gram-mole, molemole and mole-gram?
 3.2mo. -----------6.4 mo.
 NH4NO3  N2O + 2HOH

6.4mo (18 g)


(1 mo.)
=?g