Transcript Stoichiometry: Calculations with Chemical

Stoichiometry: Calculations
with Chemical Formulas and
Equations
Chapter 3
An equation
• Describes a reaction
• Must be balanced to follow Law of
Conservation of Energy
• Can only be balanced by changing the
coefficients.
• Has special symbols to indicate state, and if
catalyst or energy is required.
Symbols used in equations
• the arrow separates the reactants from the
products.
• (s) after the formula –solid
• (g) after the formula –gas
• (l) after the formula –liquid
• (aq) after the formula-aqueous (Dissolved in
water, and aqueous solution)
Symbols used in equations
•
indicates a reversible reaction

heat
• 
 ,   shows that heat is
supplied to the reaction
• Pt is used to indicate a catalyst
used supplied, in this case, platinum.
What is a catalyst?
• A substance that speeds up a reaction
without being changed by the reaction.
– ex.. enzymes-biological catalysis
protein catalysts
Convert these to equations
• Solid iron (II) sulfide reacts with
gaseous hydrogen chloride to form iron
(II) chloride and hydrogen sulfide gas.
• Nitric acid dissolved in water reacts with
solid sodium carbonate to form liquid
water and carbon dioxide gas and
sodium nitrate dissolved in water.
Balancing Chemical
Equations
Balanced Equation
• Atoms can’t be created or destroyed
• A balanced equation has the same
number of each element on both sides of
the equation.
Rules for balancing
Equations
 Write the correct formulas for all the reactants
and products
 Count the number of atoms of each type
appearing on both sides
 Balance the elements one at a time by adding
coefficients (the numbers in front)
 Check to make sure it is balanced.
Never
• Change a subscript to balance an equation
• If you change the formula you are describing a
different reaction
• H2O is a different compound than H2O2
• Never put a coefficient in the middle of a formula
• 2 NaCl is okay, Na2Cl is not.
Balance the following
equation
• AgNO3 + Cu → Cu(NO3)2 + Ag
• Mg + N2 → Mg3N2
• P + O2 → P4O10
• Na + H2O → H2 + NaOH
• CH4 + O2 → CO2 + H2O
Types of Reactions
Predicting The products
Types of Reactions
• There are millions of reactions.
• Fall into several categories.
•
•
•
•
•
Combination
Decomposition
Single Replacement
Double Displacement
Combustion
Combination Reactions
2 elements, or compounds combine to
make one compound.
• Ca +O2 CaO
• SO3 + H2O  H2SO4
• We can predict the products if they are
two elements.
• Mg + N2 
Write and balance
• Ca + Cl2 →
• Fe + O2 →iron (II) oxide
• Al + O2 →
• Remember that the first step is to write the
formula
• Then balance
Decomposition Reactions
• decompose = fall apart
• one reactant falls apart into two or more
elements or compounds.
electricity
• NaCl   
 Na + Cl2

• CaCO3   CaO + CO2
Double Replacement
• Two things replace each other.
• Reactants must be two ionic compounds or acids.
• Usually in aqueous solution
• NaOH + FeCl3 
• The positive ions change place.
• NaOH + FeCl3 Fe+3 OH- + Na+1Cl-1
• NaOH + FeCl3 Fe(OH)3 + NaCl
Double Replacement
• Will only happen if one of the products
– doesn’t dissolve in water and forms a solid
– or is a gas that bubbles out.
– or is a covalent compound usually
water.
Complete and balance
• assume all of the reactions take place.
• CaCl2 + NaOH 
• CuCl2 + K2S 
• KOH + Fe(NO3)3 
• (NH4)2SO4 + BaF2 
How to recognize which type
• Look at the reactants
• E+E
Combination
• C
Decomposition
• C+C
Double replacement
Combustion
• A compound composed of only C H and
maybe O is reacted with oxygen
• If the combustion is complete, the
products will be CO2 and H2O.
• If the combustion is incomplete, the
products will be CO and H2O.
Reactions
• Come in 5 types. (We learned 4)
• Can tell what type they are by the
reactants.
• Double Replacement happens if the
product is a solid, water, or a gas.
The Process
• Determine the type by looking at the
reactants.
• Put the pieces next to each other
• Use charges to write the formulas
• Use coefficients to balance the equation.
Chemical Quantities
Moles
• Defined as the number of carbon atoms in
exactly 12 grams of carbon-12.
• 1 mole is 6.02 x 1023 particles/molecules.
• Treat it like a very large dozen
• 6.02 x 1023 is called Avogadro's number.
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 oxygen atoms in the
following?
– CaCO3
– Al2(SO4)3
• How many ions in the following?
– CaCl2
– NaOH
– Al2(SO4)3
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.
Gram 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.
Atomic Mass
• The mass of one mole of an ionic
compound.
• Calculated the same way.
• What is the atomic mass of Fe2O3?
• 2 moles of Fe x 55.85 g = 111.70 g
• 3 moles of O x 16.00 g = 48.00 g
• The atomic mass = 111.70 g + 48.00 g =
159.70g
Molar Mass
• The generic term for the mass of one
mole.
• The same as gram molecular mass,
gram formula mass, and gram atomic
mass.
Examples
• Calculate the molar mass of the
following.
• Na2S
• N2O4
• C
• Ca(NO3)2
• C6H12O6
• (NH4)3PO4
Using Molar Mass
Finding moles of compounds
Counting pieces by weighing
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?
For example
• How many moles is 5.69 g of NaOH?

5.69 g




For example
• How many moles is 5.69 g of NaOH?

5.69 g

1 mole 
 = 0.142 mol NaOH
40.00 g 
need to change grams to moles
 for NaOH
 1mole Na = 22.99g 1 mol O = 16.00 g
1 mole of H = 1.01 g
 1 mole NaOH = 40.00 g

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?
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
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?
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
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 Reagent
• If you are given one dozen loaves of bread, a
gallon of mustard and three pieces of salami,
how many salami sandwiches can you make
(rhetorical question).
• The limiting reagent is the reactant you run
out of first.
• The excess reagent is the one you have left
over.
• The limiting reagent determines how much
product you can make
How do you find out?
• Do two stoichiometry problems.
• The one that makes the least product is
the limiting reagent.
• For 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
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 remains?
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?
Details
• Percent yield tells us how “efficient” a
reaction is.
• Percent yield can not be bigger than
100 %.
Empirical Formula
From percentage to formula
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.
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
• 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?
Empirical to molecular
• Since the empirical formula is the lowest ratio
the actual molecule would weigh more.
• By a whole number multiple.
• Divide the actual molar mass by the the mass
of one mole of the empirical formula.
• Caffeine has a molar mass of 194 g. what is its
molecular mass?
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?