Download Stoichiometry and the Mole

Stoichiometry and the Mole
Chapter 8
Stoichiometry-What is it?
• The study of the numerical relationship between
chemical quantities in a chemical reaction is called
reaction stoichiometry
• The amount of every substance used and made in
a chemical reaction is related to the amounts of all
the other substances in the reaction
– Law of Conservation of Mass
– balancing equations by balancing atoms
Stoichiometry-What is it?
• the number of pancakes you can make depends
on the amount of the ingredients you use
1 cup Flour + 2 Eggs + ½ tsp Baking Powder  5 Pancakes
• this relationship can be expressed mathematically
1 cu flour  2 eggs  ½ tsp baking powder  5 pancakes
Stoichiometry-What is it?
• if you want to make more or less than 5 pancakes
you can use the number of eggs you have to
determine the number of pancakes you can make
– assuming you have enough flour and baking powder
5 pancakes
8 eggs 
 20 pancakes
2 eggs
How you measure how
much?
•
•
•
•
•
You can measure mass,
or volume,
or you can count pieces.
We measure mass in grams.
We measure volume in liters.
• We count pieces in MOLES.
The Mole
• The mole is a number.
• A very large number, but still, just a number.
• 6.022 x 1023 of anything is a mole
– A large dozen.
– The number of atoms in exactly 12 grams of
carbon-12.
The Mole
The mole is Avogadro’s Number of items (6.02 x 1023).
A mole of atoms weighs the same number of grams as
the atomic weight. 1 mole of hydrogen weighs 1.0079g.
1 mole of carbon atoms weighs 12.011g. The atomic
weight is not only the number of protons and neutrons
but is the grams of 1 mole of atoms.
Using the mole and the atomic weight at grams/mole is
stoichiometry.
The Mole
• the balanced equation is the “recipe” for a chemical
reaction
• the equation 3 H2(g) + N2(g)  2 NH3(g) tells us
that 3 molecules of H2 react with exactly 1 molecule
of N2 and make exactly 2 molecules of NH3 or
3 molecules H2  1 molecule N2  2 molecules NH3
– in this reaction
• and since we count molecules by moles
3 moles H2  1 mole N2  2 moles NH3
Mole-to-Mole Conversions
How many moles of NaCl result from the complete reaction
of 3.4 mol of Cl2 in the reaction below?
2 Na(s) + Cl2(g)  2 NaCl(s)
How many moles of sodium oxide result from the complete
combination of 8.3 mol of O2 with sodium
How many moles of water are formed when 3.6 moles of
phosphoric acid react with barium hydroxide
More Practice
• 2C2H2 + 5 O2  4CO2 + 2 H2O
• If 3.84 moles of C2H2 are burned, how many moles
of O2 are needed?
• How many moles of C2H2 are needed to produce
8.95 mole of H2O?
• If 2.47 moles of C2H2 are burned, how many moles
of CO2 are formed?
Representative particles
• The smallest pieces of a substance.
• For a molecular compound it is a molecule.
• For an ionic compound it is a formula unit.
• For an element it is an atom.
Types of questions
• How many molecules of CO2 are the in 4.56
moles of CO2 ?
• How many moles of water is 5.87 x 1022
molecules?
• How many atoms of carbon are there in 1.23
moles of C6H12O6 ?
• How many moles is 7.78 x 1024 formula units
of MgCl2?
Measuring Moles
• The amu was one twelfth the mass of a
carbon 12 atom.
• Since the mole is the number of atoms in 12
grams of carbon-12,
• the decimal number on the periodic table is
also the mass of 1 mole of those atoms in
grams.
Atomic Mass
• The mass of 1 mole of an element in grams.
• 12.01 grams of carbon has the same
number of pieces as 1.008 grams of
hydrogen and 55.85 grams of iron.
• We can right this as
12.01 g C = 1 mole
Examples
• How much would 2.34 moles of carbon
weigh?
• How many moles of magnesium in 24.31 g
of Mg?
• How many atoms of lithium in 1.00 g of Li?
• How much would 3.45 x 1022 atoms of U
weigh?
What about compounds?
• in 1 mole of H2O molecules there are two moles
of H atoms and 1 mole of O atoms
• To find the mass of one mole of a compound
– determine the moles of the elements they have
– Find out how much they would weigh
– add them up
What about compounds?
•
•
•
•
•
•
What is the mass of one mole of CH4?
1 mole of C = 12.01 g
4 mole of H x 1.01 g = 4.04g
1 mole CH4 = 12.01 + 4.04 = 16.05g
The Gram Molecular mass of CH4 is 16.05g
The mass of one mole of a molecular
compound.
Examples
• Calculate the molar mass of the
following.
• Na2S
• N2O4
• C
• Ca(NO3)2
• C6H12O6
• (NH4)3PO4
Molar Mass
• The number of grams of 1 mole of
atoms, ions, or molecules.
• We can make conversion factors from
these.
• To change grams of a compound to
moles of a compound.
For example
• How many moles is 5.69 g of NaOH?
Examples
• How many moles is 4.56 g of CO2 ?
• How many grams is 9.87 moles of H2O?
• How many molecules in 6.8 g of CH4?
• 49 molecules of C6H12O6 weighs how
much?
Mole to Mole conversions
• How many moles of O2 are produced
when 3.34 moles of Al2O3 decompose?
• 2 Al2O3 Al + 3O2
3.34 moles
3 mole O2
= 5.01 moles O2
Al2O3 2 moles Al O
2 3
Your Turn
• 2C2H2 + 5 O2  4CO2 + 2 H2O
• If 3.84 moles of C2H2 are burned, how
many moles of O2 are needed?
• How many moles of C2H2 are needed
to produce 8.95 mole of H2O?
• If 2.47 moles of C2H2 are burned, how
many moles of CO2 are formed?
How do you get good at this?
Mass in Chemical Reactions
How much do you make?
How much do you need?
The Steps in a Stoichiometric
Calculation
Mass of substance A
Use molar mass of A
Moles of substance A
Use coefficients of A & B in
balanced eqn
Moles of substance B
Use molar mass of B
Mass of substance B
The equation is :
2Al(s) + Fe2O3(s)
Al2O3(s) + 2Fe(l)
A certain welding operation, requires that at least 86.0 g
of Fe be produced. What is the minimum mass in grams
of Fe2O3 that must be used for the operation?
Calculate also how many grams of aluminium are
needed.
Strategy:
2Al(s) + Fe2O3(s)
mass of Fe
mol of Fe
mol of Fe
mol of Fe2O3
mol of Fe2O3
mass of Fe2O3
Al2O3(s) + 2Fe(l)
For example...
• If 10.1 g of Fe are added to a solution of
Copper (II) Sulfate, how much solid
copper would form?
• Fe + CuSO4  Fe2(SO4)3 + Cu
• 2Fe + 3CuSO4  Fe2(SO4)3 + Cu
10.1 g Fe
1 mol Fe
= 0.181 mol Fe
55.85 g Fe
2Fe + 3CuSO4  Fe2(SO4)3 + 3Cu
3 mol Cu
0.181 mol Fe
= 0.272 mol Cu
2 mol Fe
63.55 g Cu
0.272 mol Cu
= 17.3 g Cu
1 mol Cu
Could have done it
10.1 g Fe 1 mol Fe 3 mol Cu 63.55 g Cu
55.85 g Fe 2 mol Fe 1 mol Cu
= 17.3 g Cu
More Examples
• To make silicon for computer chips they
use this reaction
• SiCl4 + 2Mg  2MgCl2 + Si
• How many grams of Mg are needed to
make 9.3 g of Si?
• How many grams of SiCl4 are needed
to make 9.3 g of Si?
• How many grams of MgCl2 are
produced along with 9.3 g of silicon?
For Example
• The U. S. Space Shuttle boosters use
this reaction
• 3 Al(s) + 3 NH4ClO4 
Al2O3 + AlCl3 + 3 NO + 6H2O
• How much Al must be used to react with
652 g of NH4ClO4 ?
• How much water is produced?
• How much AlCl3?
Gas and Moles
Gases
• Many of the chemicals we deal with are
gases.
• They are difficult to weigh.
• Need to know how many moles of gas we
have.
• Two things effect the volume of a gas
– Temperature and pressure
Standard Temperature and
Pressure (STP)
• 0ºC and 1 atm pressure
• At STP 1 mole of gas occupies 22.4 L
• Called the molar volume
• Avogadro's Hypothesis - at the same
temperature and pressure equal volumes
of gas have the same number of particles.
For Example
• If 6.45 grams of water are decomposed,
how many liters of oxygen will be
produced at STP?
• 2H2O  2H2 + O2
6.45 g H2O 1 mol H2O
1 mol O2 22.4 L O2
18.02 g H2O 2 mol H2O 1 mol O2
Example
• How many liters of CH4 at STP are
required to completely react with 17.5 L
of O2 ?
• CH4 + 2O2  CO2 + 2H2O
1 mol O2 1 mol CH4 22.4 L CH4
17.5 L O2
22.4 L O2 2 mol O2 1 mol CH4
= 8.75 L CH4
Examples
• What is the volume of 4.59 mole of CO2
gas at STP?
• How many moles is 5.67 L of O2 at
STP?
• What is the volume of 8.8g of CH4 gas
at STP?
Example
• How many liters of CO2 at STP will be
produced from the complete combustion
of 23.2 g C4H10 ?
• What volume of oxygen will be
required?
Density of a gas
• D = m /V
• for a gas the units will be g / L
• We can determine the density of any gas
at STP if we know its formula.
• To find the density we need the mass
and the volume.
• If you assume you have 1 mole than the
mass is the molar mass (PT)
• At STP the volume is 22.4 L.
Examples
• Find the density of CO2 at STP.
• Find the density of CH4 at STP.
The other way
• Given the density, we can find the molar
mass of the gas.
• Again, pretend you have a mole at STP,
so V = 22.4 L.
• m=DxV
• m is the mass of 1 mole, since you have
22.4 L of the stuff.
• What is the molar mass of a gas with a
density of 1.964 g/L?
• 2.86 g/L?
Limiting Reagent
Limiting Reactants
The limiting reactant (or limiting reagent) is
the reactant that is entirely consumed when a
reaction goes to completion.
The moles of product are always determined
by the starting moles of the limiting reactant.
The Cheese Sandwich Analogy
Which is the limiting reactant?
Strategy:
• Use the relationships from the balanced
chemical equation
• You take each reactant in turn and ask how
much product would be obtain, if each were
totally consumed.
• The reactant that gives the smaller amount
of product is the limiting reactant.
Example
• Copper reacts with sulfur to form copper
( I ) sulfide. If 10.6 g of copper reacts
with 3.83 g S how much product will be
formed?
• If 10.6 g of copper reacts with 3.83 g S.
How many grams of product will be
formed?
Cu is
• 2Cu + S  Cu2S
Limiting
1 mol
Cu2S 159.16 g Cu2S
1
mol
Cu
Reagent
10.6 g Cu
63.55g Cu 2 mol Cu
1 mol Cu2S
= 13.3 g Cu2S
1
mol
S
3.83 g S
32.06g S
1 mol Cu2S 159.16 g Cu2S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S
Limiting
Limiting
Reactant
Reactant: Example
• 10.0g of aluminum reacts with 35.0 grams of
chlorine gas to produce aluminum chloride. Which
reactant is limiting, which is in excess, and how
much product is produced?
2 Al + 3 Cl2  2 AlCl3
• Start with Al:
10.0 g Al
1 mol Al
27.0 g Al
2 mol AlCl3 133.5 g AlCl3
2 mol Al
1 mol AlCl3
= 49.4g AlCl3
• Now Cl2:
35.0g Cl2
1 mol Cl2
71.0 g Cl2
2 mol AlCl3 133.5 g AlCl3
3 mol Cl2
1 mol AlCl3
= 43.9g AlCl3
LR Example Continued
• We get 49.4g of aluminum chloride from the given
amount of aluminum, but only 43.9g of aluminum
chloride from the given amount of chlorine.
Therefore, chlorine is the limiting reactant. Once
the 35.0g of chlorine is used up, the reaction
comes to a complete
.
Limiting Reactant Practice
• 15.0 g of potassium reacts with 15.0 g of
iodine. Calculate which reactant is limiting
and how much product is made.
Finding the Amount of Excess
• By calculating the amount of the excess
reactant needed to completely react with
the limiting reactant, we can subtract that
amount from the given amount to find the
amount of excess.
• Can we find the amount of excess
potassium in the previous problem?
Finding Excess Practice
• 15.0 g of potassium reacts with 15.0 g of iodine.
2 K + I2  2 KI
• We found that Iodine is the limiting reactant, and
19.6 g of potassium iodide are produced.
15.0 g I2
1 mol I2
2 mol K
39.1 g K
254 g I2
1 mol I2
1 mol K
= 4.62 g K
USED!
15.0 g K – 4.62 g K = 10.38 g K EXCESS
Given amount
of excess
reactant
Amount of
excess
reactant
actually
used
Note that we started with
the limiting reactant! Once
you determine the LR, you
should only start with it!
Limiting Reactant: Recap
1. You can recognize a limiting reactant
problem because there is MORE THAN
ONE GIVEN AMOUNT.
2. Convert ALL of the reactants to the SAME
product
3. The lowest answer is the correct answer.
Limiting Reactant: Recap
1. The reactant that gave you the lowest answer is
the LIMITING REACTANT.
2. The other reactant (s) are in EXCESS.
3. To find the amount of excess, subtract the
amount used from the given amount.
4. If you have to find more than one product, be
sure to start with the limiting reactant. You don’t
have to determine which is the LR over and over
again!
Your turn
• If 10.1 g of magnesium and 2.87 L of
HCl gas are reacted, how many liters of
gas will be produced?
• How many grams of solid?
• How much excess reagent is left?
Your Turn II
• If 10.3 g of aluminum are reacted with
51.7 g of CuSO4 how much copper will
be produced?
• How much excess reagent will remain?
Yield
• The amount of product made in a
chemical reaction.
• There are three types
• Actual yield- what you get in the lab
when the chemicals are mixed
• Theoretical yield- what the balanced
equation tells you you should make.
• Percent yield = Actual
x 100 %
Theoretical
Example
• 6.78 g of copper is produced when 3.92
g of Al are reacted with excess copper
(II) sulfate.
• 2Al + 3 CuSO4  Al2(SO4)3 + 3Cu
• What is the actual yield?
• What is the theoretical yield?
• What is the percent yield?
Empirical Formula
Percent Composition
Tells the relative mass each
element contributes to the mass
of the whole compound.
• Formula:
Mass of one element
Mass of compound
X 100%
Practice
• Find the percent composition of the idicated
element in the following:
•
•
•
•
•
•
Al in aluminum sulfate
O in Tin (IV) phosphate
Iron in Iron (II) nitride
Nitrogen in ammonium phosphite
Hydrogen in perchloric acid
Carbon in glucose
The Empirical Formula
• The lowest whole number ratio of elements
in a compound.
• The molecular formula the actual ratio of
elements in a compound.
• The two can be the same.
– CH2 empirical formula
– C2H4 molecular formula
– C3H6 molecular formula
– H2O both
Calculating Empirical
• Just find the lowest whole number ratio
• C6H12O6
• CH4N
• It is not just the ratio of atoms, it is also the ratio of moles
of atoms.
• In 1 mole of CO2 there is 1 mole of carbon and 2 moles
of oxygen.
• In one molecule of CO2 there is 1 atom of C and 2 atoms
of O.
Calculating Empirical
• We can get ratio from percent composition.
• Assume you have a 100 g.
• The percentages become grams.
• Can turn grams to moles.
• Find lowest whole number ratio by dividing by the
smallest.
• Be careful! Do not round off numbers prematurely
Example
• Calculate the empirical formula of a
compound composed of 38.67 % C, 16.22
% H, and 45.11 %N.
• Assume 100 g so
• 38.67 g C x 1mol C
= 3.220 mole C
12.01 gC
• 16.22 g H x 1mol H
= 16.09 mole H
1.01 gH
• 45.11 g N x 1mol N = 3.219 mole N
14.01 gN
Example
• The ratio is 3.220 mol C = 1 mol C
3.219 mol N
1 mol N
• The ratio is 16.09 mol H = 5 mol H
3.219 mol N 1 mol N
• C1H5N1
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
 Convert the grams to moles
1 mol C
47g C 
 3.9 mol C
12.01g
1 mol H
6.0 g H 
 6.0 mol H
1.008g
1 mol O
47 g O 
 2.9 mol O
16.00g
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
 Divide each by the smallest number of
moles
3.9 mol C  2.9  1.3
6.0 mol H  2.9  2
2.9 mol O  2.9  1
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen

If any of the ratios are not a whole number, multiply all
the ratios by a factor to make it a whole number
– If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then multiply by 3;
if ?.25 or ?.75 then multiply by 4
Multiply all the
Ratios by 3
Because C is 1.3
3.9 mol C  2.9  1.3 x 3  4
6.0 mol H  2.9  2 x 3  6
2.9 mol O  2.9  1 x 3  3
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
° Use the ratios as the subscripts in the
empirical formula
3.9 mol C  2.9  1.3 x 3  4
6.0 mol H  2.9  2 x 3  6
2.9 mol O  2.9  1 x 3  3
C4H6O3
Empirical formula from Composition
Consider the following flow-diagram:
Percent composition
Mass Composition
Number of moles of
each element
Divide by smallest number of
moles to find the molar ratios
Multiply by appropriate number to
get whole number subscripts
Practice
• A compound is 43.64 % P and 56.36 % O.
What is the empirical formula?
• Caffeine is 49.48% C, 5.15% H, 28.87% N
and 16.49% O. What is its empirical
formula?
Molecular Formulas
• The molecular formula is a multiple of the
empirical formula
• To determine the molecular formula you
need to know the empirical formula and
the molar mass of the compound
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
 Determine the empirical formula
• May need to calculate it as previous
C5H3
 Determine the molar mass of the
empirical formula
5 C = 60.05 g, 3 H = 3.024 g
C5H3 = 63.07 g
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
 Divide the given molar mass of the
compound by the molar mass of the
empirical formula
–Round to the nearest whole number
252 g
4
63.07 g
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
 Multiply the empirical formula by the
calculated factor to give the molecular
formula
(C5H3)4 = C20H12
Example
• A compound is known to be composed
of 71.65 % Cl, 24.27% C and 4.07% H.
Its molar mass is known (from gas
density) is known to be 98.96 g. What is
its molecular formula?