Download Stoichiometry, % Comp, Empirical & Molecular Formula

STOICHIOMETRY
(The Mathematics of Chemical
Equations!)
Rice
Atomic Mass
 Atoms are so small, it is difficult to discuss
how much they weigh in grams.
So……. We use atomic mass units.
 an atomic mass unit (amu) is 1/12 the mass
of a carbon-12 atom.
 This gives us a basis for comparison.
 The decimal numbers on the periodic table
are average atomic masses in amu.
They are not whole numbers
Why????
 Because they are based on averages of
atoms and of isotopes.
 We figure out the average atomic mass of
each element from the mass of the
isotopes and their relative abundance.
Massaverage = Σ (% abundance x Massisotope)
Examples
 There are two isotopes of carbon C-12
with a mass of 12.00000 amu (98.892%),
and C-13 with a mass of 13.00335 amu
(1.108%). What is the average atomic
mass.
 C-12 = 0.98892 x 12.00000 = 11.86704
 C-13 = 0.01108 x 13.00335 = 0.14408
C = 12.01
Wow! it matches the periodic table
Your Turn
There are two isotopes of
nitrogen, one with an atomic
mass of 14.0031 amu and one
with a mass of 15.0001 amu.
What is the percent abundance of
each isotope?
Solution
14.0067 • 100 = Σ [(14.0034 • x) + (15.0001 • (100-x)]
99.668907 = x
So % abundance of N-14 = 99.668907%
% abundance of N-15 = 0.3310926%
CHECK:
N-14 = 14.0031 x 0.9968907 = 13.95956
N-15 = 15.0001 x 0.003310926 = + 0.04966422
close enough 14.0092 amu
The Mole
The Mole
The mole is a
number.
Amounts in
chemistry are
measured in
moles.
The Mole
The average mass of each element
on the periodic table is = 1 mole.
How we measure how much?
We can measure mass, or volume,
or we can count pieces.
 We measure mass in grams.
 We measure volume in liters.
 We count pieces in MOLES.
1 mole is equal to:
1. The GMW of any compound
2. 22.4 L volume
3. 6.02 x 1023 molecules, particles, or
atoms (6.02 x 1023 is called Avagadro’s
number) Treat it like a very large dozen!
Molar Mass
Some Confusing Vocabulary
 We call the smallest pieces of a
substance … particles.
 We refer to molecular compounds as
…molecules.
 We refer to ionic compound as …
formula units.
 We call an element …an atom.
Molar Mass: The mass of one
mole of anything.
Also called:
1.
2.
3.
4.
5.
6.
7.
gram molecular mass (MM)
gram formula mass (GFM)
gram atomic mass (GAM)
molecular weight (MW)
formula weight (FW)
atomic mass (AM)
formula mass (FM)
Molar mass
The molar mass of an element is
the mass written on the periodic
table.
molar mass of C is 12.011g/mol
FOR DIATOMCS:
molar mass of O2 is 32.00g/mol
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
1. Determine the number of moles of
each element in the compound
2. Find out the mass of 1 mole then
multiply it by the # of moles
3. Add them all up
Example
What is the mass of one mole of CH4?
 1 mole of C = 12.01 g
 4 moles of H so (4 x 1.01g) = 4.04g
Add them together = 16.05g
So the GMW of CH4 is 16.05g
Note: We call this the Gram Molecular mass of
CH4 since it is the mass of a molecular
compound.
Gram Formula Mass
 The mass of one mole of an ionic
compound.
 Calculated the same way.
What is the GFM of Fe2O3?
 2 moles of Fe (2 x 55.85g) = 111.70 g
 3 moles of O (3 x 16.00g) = 48.00 g
The GFM = 111.70 g + 48.00 g = 159.70g
Your Turn
Find the molar mass of:
1) CH4
2) Mg3P2
3) Ca(NO3)3
4) Al2(Cr2O7)3
5) CaSO4 · 2H2O
Solution
1) CH4 = 16.04 g/mole
2) Mg3P2 = 134.85 g/mole
3) Ca(NO3)3 = 226.10 g/mole
4) Al2(Cr2O7)3 = 701.96 g/mole
5) CaSO4 · 2H2O = 172.14 g/mole
More Examples
Calculate the molar mass of the following:
1. Na2S
2. N2O4
3. C
4. Ca(NO3)2
5. C6H12O6
6. (NH4)3PO4
All the things we
can change
Mass
Moles
MW
Moles
Mass
Volume
MW
Moles
Mass
Volume 22.4 L
Moles
MW
Mass
Volume 22.4 L
MW
Moles
Representative
Particles
Mass
Volume 22.4 L
MW
Moles
6.02 x
23
10
Representative
Particles
Mass
Volume 22.4 L
MW
Moles
6.02 x
23
10
Representative
Particles
Atoms
Mass
Volume 22.4 L
MW
Mass
Moles
6.02 x
23
10
Representative
Particles
Atoms
Ions
Can You answer these Questions?
 How many oxygen atoms in the
following?
CaCO3
 Al2(SO4)3

 How many ions in the following?
CaCl2
 NaOH


Al2(SO4)3
Using Molar Mass
Conversions using Molar
Mass
We make conversion factors
using molar mass.
How many moles in 5.69 g of NaOH?
Need to change grams to moles for NaOH:
 1 mole Na = 22.99g
 1 mol O = 16.00 g
 1 mole of H = 1.01 g
So 1 mole NaOH = 40.00 g
Use Factor-Label:
5.69g NaOH
1
1 mol NaOH
40g NaOH
= 0.142 mol NaOH
Your Turn





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 has what mass?
How many molecules of CO2 are the in 7.61 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?
Other 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
Gases and the Mole
+
THE
Gases
Many of the chemicals we deal with are gases.
Two things effect the volume of a gas:
1) Temperature
2) pressure
 Always compare gases at the same temperature and
pressure.
Standard Temperature and
Pressure (abbreviated STP)
 Standard Temperature = 0ºC
 Standard Pressure = 1 atm pressure
 At STP 1 mole of gas occupies 22.4 L
 22.4L/mole is called the molar volume
 Avagadro’s Hypothesis - at the same
temperature and pressure equal
volumes of gas have the same number
of particles.
Example
1. What is the volume of 4.59 mole of CO2
gas at STP?
4.59 mol CO2
1
22.4L CO2 = 102.82 L CO2
1 mol CO2
Your Turn
1) What is the volume of 4.59 mole of
CO2 gas at STP?
2) How many moles is 5.67 L of O2 at
STP?
3) What is the volume of 8.8g of CH4 gas
at STP?
More Problems
4) What is the mass of 2.34 moles of
carbon?
5) How many moles of magnesium in
24.31g of Mg?
6) How many atoms of lithium in 1.00g of
Li?
7) What is the mass of 3.45 x 1022 atoms
of U?
MASS
DENSITY
VOLUME
D=m/V
3
(g/cm
Density of a Solid
or
g/ml) is easy, but how do you
get the density of a gas?
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 chemical
formula (CO2, H2, Cl2)
 To find the density we need the mass
and the volume.
Examples
1. Find the density of CO2 at STP.
1 mole = 22.4L and 44.011g
D = 44.011g = 2.0g/L
22.4L
Your Turn
1. Find the density of CH4 at STP.
The other way
Given the density, we can find the molar mass of any gas.
Ex: What is the molar mass of a gas with a
density of 1.964 g/L?
1. Pretend you have a mole @ STP, V= 22.4L
2. D = m / V
3. 1.964g/L = x / 22.4L
x = 43.994g
Try This!!! What is the molar mass of a gas
with a density of 2.86 g/L?
We have learned how to
 change moles to grams
 moles to atoms
 moles to formula units
 moles to molecules
 moles to liters
 molecules to atoms
 formula units to atoms
 formula units to ions
Percent Composition
Percent Composition
Percent of each element (by mass) in a
compound:
Part x 100 = %
Whole
(must equal 100%)
Example:
Find the % composition of CH4
C = 12.011 x 100 = 74.8722 = 74.87%
16.042
H = (1.0079 x 4) x 100 = 25.131 = 25.13%
16.042
100%
Your Turn
Find the % composition of each element in:
1) Al2(Cr2O7)3
2) CaSO4 · 2H2O
Solution
1) Al2(Cr2O7)3
Al = 7.68 %
Cr = 44.45 %
O = 47.87 %
2) CaSO4 · 2H2O
Ca = 23.28 %
S = 18.62 %
O = 55.76 %
H = 2.34 %
Empirical & Molecular
Formulas
Empirical Formula
The lowest ratio of atoms in a molecule.
 From percent composition, you can
determine the empirical formula.
 Empirical Formulas are based on mole
ratios.
 Steps:
1) Find the moles
2) Divide by the smallest # of
moles
3) Use answer from #2 as
subscripts of each element
Example
A sample is 52.11% C, 13.14% H and 34.75%
O. What is its empirical formula?
C = 52.11g C 1 mole
= 4.339 moles = 2
12.011g C
2.17
H = 13.14g H
O = 34.75g O
1 mole
= 13.037 moles = 6
1.0079g H
2.17
1 mole = 2.17 moles = 1
16.00g O
2.17
So the empirical formula is: C2H6O
Your Turn
A 1.334 gram sample of a
compound contains 0.365g Na,
0.758g O and Nitrogen. What is
the empirical formula of this
compound?
Answer
NaNO3
Molecular Formula –
The True Formula
Empirical To Molecular
Formulas




Empirical is lowest ratio.
Molecular is actual formula of the molecule.
Need Molar mass of the empirical molecule.
Ratio of empirical to molar mass will tell you
the molecular formula.
 Must be a whole number because
subscripts are never decimals.
Empirical to molecular
 Since the empirical formula is the
lowest ratio, the actual molecule could
have a greater mass….
 By a whole number multiple or the
empirical and the molecular formula
could be the same.
To Find Molecular Formula from
the Empirical Formula
 Divide the actual molar mass of the
molecule given by the mass of one mole of
the empirical formula molecule.
Molecular Mass of Molecular Formula
Molar Mass of the Empirical Formula
= # to multiply the subscripts
of the empirical formula by to
give you the molecular formula.
Example
Determine the molecular formula of the
compound with an empirical formula of
C4H5N2O. Formula mass of the compound is
194.19 g.
Formula mass of Molecular Formula = 194.19g =
Molar mass of Empirical Formula
= 97.09g
2
Multiply each subscript by 2 to get molecular formula:
Molecular Formula = C8H10N4O2
HEY…
THAT’S CAFFIENE !!
Example
 A compound is made of only sulfur and
oxygen. It is 60.9% S by mass. Its molar
mass is 192.24 g/mol. What is its
molecular formula?
S = 60.9g 1mol = 1.9 mol = 1
32.06g 1.9
O = 30.4g 1mol = 1.9 mol = 1
16.00g
1.9
SO = 48.06g
so 192.24 = 4
48.06
so Molecular Formula is: S4O4
Your Turn
The molar mass of a compound
is 92g/mol. Analysis of the
compound indicates that it
contains 0.606gN and 1.390g O.
Find its molecular formula.
Solution
N = 0.606g N 1mol
= 0.043 = 1
14.007g
0.043
O = 1.390g O 1mol
= 0.087 = 2
16.00g
0.043
NO2 = 46.007g
so
92
=2
46.007
so Molecular Formula is: N2O4
Your Turn Again
A compound is known to be
composed of 71.65 % Cl, 24.27% C
and 4.07% H. Its molar mass is
known to be 98.96g/mol.
What is its molecular formula?
Chemical Equations
Types of Chemical Reactions
Key: A, B = cations X, Y = anions
 Combination or Synthesis: A + X  AX
 Decomposition: AX  A + X
 Single Replacement / Displacement:
AX + B  BX + A or AX + Y  AY + X
4) Double Replacement / Displacement:
AX + BY  AY + BX
5) Combustion: ____ + O2  ____ +
____
Balancing Equations
Balancing equations
CH4 + O2 → CO2 + H2O
Products
Reactants
1
C
1
4
H
2
2
O
3
Balancing equations
CH4 + O2 → CO2 + 2 H2O
Reactants
Products
1 C 1
4 H 2 4
2 O 3 4
Balancing equations
CH4 + 2O2 → CO2 + 2 H2O
Reactants
Products
1 C 1
4 H 2 4
4 2 O 3 4
And We’re Done!!!
Abbreviations
 (s) ↓ = solid
 (l) = liquid
 (g) ↑ = gas
 (aq) = aqueous
 Δ or heat or
 Catalyst Ex:

~
MnO2
= yields
= reversible
rxn
↔ = resonance structure
20 C
= heat
= catalyst
= electricity
Example
Solid iron (III) sulfide reacts with
gaseous hydrogen chloride to form
solid iron (III) chloride and
hydrogen sulfide gas.
Fe2S3(s) + 6HCl(aq)
2FeCl3(s) + 3H2S(g)
Your Turn
Write the word equations:
 KClO3 (s)
Cl2 (g) + O2 (g)
 Fe2O3 (s) + Al (s)
Fe (s) + Al2O3 (s)
What does a balanced equation
mean?
1) A balanced equation can be used to
describe a reaction in molecules, atoms
and mole ratios.

NOT grams.
2) Chemical reactions happen molecules
at a time.


or dozens of molecules at a time.
or moles of molecules.
LET’S DO
STOICHIOMETRY
(The Mathematics of Chemical
Equations!)
Stoichiometry
 Given an amount of either a reactant or
product STOICHIOMETRY allows us to
determine the other quantities.
 We use conversion factors from:
 molar
mass (g/mole, L/mole, atoms/mole)
 mole ratios in the balanced equation
Examples
 One way of producing O2(g) involves
the decomposition of potassium
chlorate into potassium chloride and
oxygen gas. A 25.5 g sample of
Potassium chlorate is decomposed.
1) How many moles of O2(g) are
produced?
2) How many grams of potassium
chloride are produced?
3) How many grams of oxygen?
Solution
25.5g
? mol
 K+1ClO3-1(s)  K+1Cl-1(aq) + O2(g)
 KClO3(s)  KCl(aq) + O2(g)
 Balance: 2KClO3(s)  2KCl(aq) + 3O2(g)
Answers to questions:
1)
2)
3)
25.5 g KClO3
1 mol KClO3
3 O2
= 0.3 molO2
1
122.553g KClO3
2KClO3
0.312 mol O2 2KCl
74.553g KCl = 15.5 g KCl
1
3O2
1 mol KCl
0.312 mol O2
1
32.00g O2
1 mol O2
= 9.98 g O2  10. g O2
Your Turn
 A piece of aluminum foil 5.11in x 3.23 in
x 0.0381in is dissolved in excess
HCl(aq). How many grams of H2(g) are
produced? (Hint: the density of Al is 2.7g/cm3)
 How many grams of each reactant are
needed to produce 15g of iron from the
following reaction?
Fe2O3 (s) + Al (s)  Fe (s) + Al2O3 (s)
Examples
Given:
K2PtCl4 (aq) + NH3 (aq)  Pt(NH3)2Cl2 (s) + KCl (aq)
Find:
1. The mass of Pt(NH3)2Cl2 that can be
produced from 65g of K2PtCl4 ?
2.
How much KCl will be produced?
3.
How much from 65 grams of NH3?
Limiting Reactant
Limiting Reagent
 Reactant that determines the amount of
product formed.
 The one you run out of first.
 Makes the least product.
 You can use a ratio method or…
Limiting reagent
 To determine the limiting reagent
requires that you do two stoichiometry
problems.
 Figure out how much product each
reactant makes.
 The one that makes the least is the
limiting reagent.
Excess Reactant
 The reactant you don’t run out of.
 The amount of stuff you make is the
yield.
 The theoretical yield is the amount you
would make if everything went perfect.
 The actual yield is what you make in
the lab.
Example
Ammonia is produced by the following
reaction:
N2 + H2  NH3
1. What mass of ammonia can be produced
from a mixture of 100. g N2 and 500. g H2 ?
2. How much unreacted material remains?
3.57mol
247.52mol
100.g
500.g
N2 + 3H2
100. g N2
1
 2NH3
1 mol N2 2NH3 17.03g NH3 = 121.58g
28.014g N2 N2
1 mol NH3
NH3
500. G H2 1 mol H2 2NH3 17.03g NH3 = 2810.23g
1
2.02g H2 3H2 1 mol NH3
NH3
Can only produce 121.58g NH3 !
Moles
3.57mol
247.52mol
Given:
100.g
500.g
N2 + 3H2
LIMITING
 2NH3
EXCESS
To determine the unreacted material start with
limiting:
3.57 mol N2
1
HAVE
So
3H2
N2
= 10.71 mol H2 (needed)
NEED
247.52 mol – 10.71 mol = 236.81mol H2 (excess)
Your Turn
Hg + Br2  HgBr2
1) If 10.0g of Hg and 9.00g of Br2 are reacted,
how much HgBr2 will be produced?
2) If the reaction did go to completion, how
much excess reagent would be left?
%
Percent Yield
How much you get from a chemical reaction!
Percent Yield
 % yield =
Actual (Experimental)
Theoretical (calculated using stoichiometry)
 % yield =
Experiment results
Calculated results
x 100
x 100
% Yield vs % Error
% error = (experimental – accepted)
accepted
% yield =
Actual
Theoretical
x 100
x 100
What are the three types of
yield?
Answer
1. % Yield
2. Actual Yield
3. Theoretical Yield
Your Turn
 Aluminum burns in bromine producing
aluminum bromide. In a laboratory 6.0g of
aluminum reacts with excess bromine and
50.3g of aluminum bromide are produced.
What is the % yield?