Transcript Stoichiometry1

Topic 1:Stoichiometry
(12.5 hours)
1.1 The mole concept and Avogardo’s number
1.2 Formulas
1.3 Chemical equations
1.4 Mass and gas vol. relationships in chemical reactions
1.1 The mole concept and
Avogadro’s constant
1.2 Formulas
1.1.1 Apply the mole concept to substances
1.1.2 Determine the number of particles and the amount of
substance (in moles)
1.2.1 Define the terms relative atomic mass (Ar) and
relative molecular mass (Mr)
1.2.2 Calculate the mass of one of mole of species from its
formula
1.2.3 Solve problems involving the relationships between
the amount of substance in moles, mass, and molar mass
Quick Review:
 Atoms are the smallest unit
 The same atoms make up elements (Na or
Cl2)
 Different whole combination of atoms make
up compounds (NaCl) or molecules (CH4)
Atomic mass
 Atoms contain different numbers of protons
and neutrons, so they have different masses
(isotopes)
 Carbon-12 (6 p+, 6 no)has a mass 12 times
greater than hydrogen-1(1 p+, 0 no)
 It’s impossible to measure the mass of
individual atoms on a balance, because it is
so small
 Chemists developed their own unit, called the
mole
The Mole
 Similar to a dozen, except instead of 12,
it’s 602,000,000,000,000,000,000,000
 6.02 X 1023 mol-1(in scientific notation)
 It applies to all kinds of particles: atoms,
particles, molecules, ions, electrons,
formula units depending the way the
question is asked, so be careful.
 This number is named in honor of
Amedeo Avogadro (1776 – 1856)
Just how big is Avogadro's constant?
 The earth is estimated to be 4.54 billion years
old, that is approximately 1.43 x 1017 s. Still
much smaller than a mole.
 If you were to count every grain of sand in the
Sahara desert, assuming its 4000 km by
1500 km, and its depth is 10 m, and
assuming that 1 cm3 contains 1000 grains of
sand. It would contain 6 x 1022 grains of
sand. Only one tenth the size of a mole!!!
Relative atomic mass (Ar)
 As the weighted mean mass of all the
naturally occurring isotopes of an element
relative to one twelfth of the mass of a
carbon-12 atom.
 Not whole numbers
 No mass unit
 Also applies to relative molecular mass (Mr)
Molar Mass (or atomic mass)
 The Mass of 1 mole (in grams)
 Equal to the numerical value of the average atomic
mass (get from periodic table), or add the atoms
together for a molecule
1 mole of C atoms
=
12.0 g
1 mole of Mg atoms
=
24.3 g
1 mole of O2 molecules =
32.0 g
Molar Mass of Compounds
 The molar mass (MM) of a compound is determined
the same way, except now you add up all the atomic
masses for the molecule (or compound)




Ex. Molar mass of CaCl2
Avg. Atomic mass of Calcium = 40.08g
Avg. Atomic mass of Chlorine = 35.45g
Molar Mass of calcium chloride =
40.08 g/mol Ca + (2 X 35.45) g/mol Cl
 110.98 g/mol CaCl2
20
Ca
40.08
17
Cl
35.45
Atoms or
Molecules
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Moles
Divide by
atomic/molar mass
from periodic table
Multiply by
atomic/molar mass
from periodic table
Mass
(grams)
Practice
 Calculate the Molar Mass of calcium
phosphate


Formula =
Ca3(PO4)2
Masses elements:
 Ca: 3 Ca’s X 40.1 =
 P: 2 P’s X 31.0 =
 O: 8 O’s X 16.0 =
120.3 g
62.0 g
128.0 g

Molar Mass =
120.3g
310.3
+ 62.0g
g/mol
+128.0g
Calculations
molar mass
Grams
Avogadro’s number
Moles
particles
Everything must go through
Moles!!!
Atoms/Molecules and Grams
How many moles of Cu are present in 35.4 g of Cu?
How many atoms?
35.4 g x 1 mol = 0.557 mol Cu
63.55 g
0.557 mol Cu x 6.02x1023 atoms = 3.35x1023 atoms
1 mol
On your own!
How many moles of Fe are present in 102.4 g of
Fe? How many atoms?
102.4 g x 1 mol = 1.83 mol Fe
55.85 g
1.83 mol Fe x 6.02x1023 atoms = 1.10x1024 atoms
1 mol
Work backwards!
What is the mass (in grams) of 1.20x1024 molecules
of glucose (C6H12O6)?
1.20x1024 molec. X 1 mol = 1.99 mol
6.02x1023 molec.
1.99 mol x 180.12 g = 358.4 g
1 mol
From periodic table
This is as tricky as it gets!
 How many atoms of carbon are found in 2.6g
of glucose (C6H12O6)?
2.6 g x 1 mole= 0.014 mol C6H12O6
180.12 g
0.014 mol x 6.02 x 1023 molecules = 8.4 x 1021 molecules
mol
8.4 x 1021 molecules x 6 atoms of C = 5.06 x 1022 atoms
1 molecule C6H12O6
1.3 Chemical Equations
1.3.1 Deduce chemical equations when all reactants and
products are given
1.3.2 Identify the mole ratio of any two species in a
chemical equation
1.3.3 Apply the state symbols (s), (l), (g) and (aq)
Chemical equations (review)
 To deduce the products of an equation we
1.
2.
3.
4.
5.
must know the type of reaction occuring.
Single displacement
Double displacement
Combustion
Synthesis
Decomposition
Single displacement:
 either the metal changes position with the
metal ion…
AgNO3 (aq) + Cu(s)  CuNO3(aq) + Ag(s)
 or the non-metal changes position with the
non-metal ion.
F2(g) + NaCl(aq)  NaF(aq)+ Cl2(g)
Double displacement
 The cations (positively charged ions) change
position with each other.
NaOH + HCl  HOH + NaCl
Fe(OH)2 + 2KBr  2KOH + FeBr2
Balancing an equation means that both sides must have
the same number and type of atoms present.
Coefficients are red (indicating total number of
compounds), subscripts are blue (indicating number
of atoms).
Organic Combustion
 A hydrocarbon reacting with oxygen to produce
carbon dioxide and water
CH4 (g) + 2O2 (g) 2H2O (g) + CO2 (g)
2C2H6 (g) + 7O2 (g) 6H2O (g) + 4CO2 (g)
The coefficient is also related the mole concept. For the
1st reaction, we would say that one mole of methane
reacts with two moles of oxygen to produce two
moles of water and one mole of carbon dioxide gas.
Synthesis
 The combination of two or more particles to
produce one compound.
N2 (g) + H2 (g)  NH3 (l)
Use (g) for gas, (l) for liquid, (s) for solids and
(aq) if it dissolves in water/aqueous.
Decomposition
 A single large compound will be broken down
into more simple parts.
H2O (l)  H2 (g) + O2 (g)
1.4 Mass relationships in
chemical equations
1.4.1 Calculate theoretical yields from chemical equations
1.4.2 Determine the limiting reactant the reactant in excess
when quantities of reacting substances are given.
1.4.3 Solve problems involving theoretical, experimental
and percentage yield.
Chemistry Recipes
 Looking at a reaction tells us how much of
something you need to react with something
else to get a product
 Be sure you have a balanced reaction before
you start!



Example: 2 Na + Cl2  2 NaCl
This reaction tells us that by mixing 2 moles of
sodium with 1 mole of chlorine we will get 2 moles of
sodium chloride
What if we wanted 4 moles of NaCl? 10 moles?
50 moles?
Practice
 Write the balanced reaction for hydrogen gas reacting with
oxygen gas.


2 H2 + O2  2 H2O
2 mol H2
How many moles of reactants are needed?
1 mol O2
What if we wanted 4 moles of water?
4 mol H2
2 mol O2

What if we had 3 moles of oxygen, how much hydrogen would
we need to react and how much water would we get?
6 mol H2, 6 mol H2O

What if we had 50 moles of hydrogen, how much oxygen
would we need and how much water produced?
25 mol O2, 50 mol H2O
Mole Ratios
 These mole ratios can be used to calculate the
moles of one chemical from the given amount of a
different chemical
 Example: How many moles of chlorine is needed to
react with 5 moles of sodium (without any sodium left
over)?
2 Na + Cl2  2 NaCl
5 moles Na 1 mol Cl2
2 mol Na
= 2.5 moles Cl2
Mole-Mole Conversions
 How many moles of sodium chloride will be
produced if you react 2.6 moles of chlorine gas
with an excess (more than you need) of sodium
metal?
2 Na + Cl2  2 NaCl
2.6 moles Cl2 2 mol NaCl
1 mol Cl2
= 5.2 moles NaCl
You practice
 Aluminum reacts with oxygen to produce
aluminum oxide. If I have 2.6 mol of Al, how
much Al2O3 would I produce?
4Al + 3O2  2Al2O3
2.6 mol Al x 2 mol Al2O3 = 1.3 mol Al2O3
4 mol Al
Mole-Mass Conversions
 Most of the time in chemistry, the amounts are given in
grams instead of moles
 We still go through moles and use the mole ratio, but now
we also use molar mass to get to grams
 Example: How many grams of chlorine are
required to react completely with 5.00 moles of
sodium to produce sodium chloride?
2 Na + Cl2  2 NaCl
5.00 moles Na 1 mol Cl2
2 mol Na
70.90g Cl2
1 mol Cl2
= 177g Cl2
You Practice
 Calculate the mass in grams of Iodine
required to react completely with 0.50 moles
of aluminum.
2 Al + 3 I2  2 AlI3
Mass-Mole
 We can also start with mass and convert to moles of
product or another reactant
 We use molar mass and the mole ratio to get to
moles of the compound of interest


Calculate the number of moles of ethane (C2H6)
needed to produce 10.0 g of water
2 C2H6 + 7 O2  4 CO2 + 6 H20
10.0 g H2O 1 mol H2O
2 mol C2H6 = 0.185
18.0 g H2O 6 mol H20
mol C2H6
Mass-Mass Conversions
 Most often we are given a starting mass and
want to find out the mass of a product we will get
(called theoretical yield) or how much of another
reactant we need to completely react with it (no
leftover ingredients!)
 Now we must go from grams to moles, mole
ratio, and back to grams of compound we are
interested in
Mass-Mass Conversion
 Ex. Calculate how many grams of ammonia
are produced when you react 2.00g of
nitrogen with excess hydrogen.
 N2 + 3 H2  2 NH3
2.00g N2
1 mol N2
2 mol NH3
28.02g N2 1 mol N2
= 2.4 g NH3
17.06g NH3
1 mol NH3
Practice
 How many grams of calcium nitride are
produced when 2.00 g of calcium reacts with
an excess of nitrogen?
 __Ca + __N2  __Ca3N2
Grilled Cheese
Sandwich
Bread +
2B
+
100 bread
Cheese
C
30 slices
 ‘Cheese Melt’

B2C
? sandwiches
Limiting Reactants
aluminum
+
chlorine gas

Al(s)
+
Cl2(g)

AlCl3
2 Al(s)
+
3 Cl2(g)

2 AlCl3
100 g
A. 200 g
aluminum chloride
100 g
B. 125 g
?g
C. 667 g
D. ???
Limiting Reactant
 Most of the time in chemistry we have more of one
reactant than we need to completely use up other
reactant.
 That reactant is said to be in excess (there is too
much).
 The other reactant limits how much product we get.
Once it runs out, the reaction
s. This
is called the limiting reactant.
Limiting Reactant
 To find the correct answer, we have to try all of the
reactants. We have to calculate how much of a product
we can get from each of the reactants to determine
which reactant is the limiting one.
 The lower amount of a product is the correct answer.
 The reactant that makes the least amount of product is
the limiting reactant. Once you determine the limiting
reactant, you should ALWAYS start with it!
 Be sure to pick a product! You can’t compare to see
which is greater and which is lower unless the product is
the same!
Limiting
Limiting
Reactant:
Example
Reactant
 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
 Now Cl2:
35.0g Cl2
1 mol Cl2
71.0 g Cl2
2 mol AlCl3 133.5 g AlCl3
2 mol Al
1 mol AlCl3
2 mol AlCl3 133.5 g AlCl3
3 mol Cl2
1 mol AlCl3
= 49.4g 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.
2.
3.
4.
5.
6.
7.
You can recognize a limiting reactant problem because there is
MORE THAN ONE GIVEN AMOUNT.
Convert ALL of the reactants to the SAME product (pick any
product you choose.)
The lowest answer is the correct answer.
The reactant that gave you the lowest answer is the LIMITING
REACTANT.
The other reactant(s) are in EXCESS.
To find the amount of excess, subtract the amount used from
the given amount.
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!
Percent Yields
 Theoretical yield: max amount of a product
that is formed in a reaction.
 Actual yield: amount of product that is actually
obtained in a reaction

Usually less than theoretical. Why?
Why?
 Theoretical has assumed that all of limiting
reagent has completely reacted.






Many reactions do not go to completion
Unexpected competing side reactions limit the
formation of products.
Some reactants are lost during the separation
process (remember in the lab, pouring off
water, leaving silver behind!)
Impure reactants
Faulty measuring
Poor experimental design or technique
How to calculate?
Percent Yield = Actual Yield
x 100%
Theoretical yield
Practice together:
20g of HBrO3 is reacted with excess HBr.
1. What is the theoretical yield of Br2?
2. What is the percent yield, if 47.3g is
produced?
HBrO3 + 5HBr  3H2O + 3 Br2
Practice alone:
When 35g of Ba(NO3)2 is reacted with
excess Na2SO4, 29.8g of BaSO4 is
recovered.
Ba(NO3)2 + Na2SO4  BaSO4 + 2NaNO3
1.
2.
Calculate the theoretical yield of BaSO4
Calculate the percent yield of BaSO4