Transcript Stoichiometry: Calculations with Chemical Formulas and

CHM 1045: General Chemistry and
Qualitative Analysis
Unit # 3
Stoichiometry:
Calculations with Chemical
Formulas and Equations
Dr. Jorge L. Alonso
Miami-Dade College –
Kendall Campus
Miami, FL
Textbook Reference:
•Module #3 & 4 Stoichiometry
Chemical
Reaction
H2O(g)
CO2 (g)
flame
The actual
phenomenon that
occurs when chemical
interact with each other.
Methane gas is mixed with air and then
it is light-up by a spark.
O2 (g)
What is happening here?
CH
4 (g)
Stoichiometry
Chemical Equations
Symbolic representations of chemical reactions
Reactants
CH4 (g) + 2 O2 (g)
States
Subscripts
Products
CO2 (g) + 2 H2O (g)
Coefficients
Balanced Chemical Equations represent events that occur at the
Stoichiometry
atomic level which we cannot perceive; but they explain the
mass
ratio of substances involved in a chemical equation (stoichiometry)
Predicting Products: Types of Reactions
What happens when substances react?
(1) Decomposition: AB  A + B
(2) Combination (Synthesis): A + B  AB
(3) Double Displacement (Replacement) or Metathesis,
Exchange
AB + CD  AD + CB
where A & C are Metals, B & D Nonmetals
(4) Single Displacement (Replacement)
MN +
M
MN + M
or
N
MN + N
(5) Combustion : reactions of oxygen with an organic compounds
(hydrocarbons, alcohols) that produce CO2 + H2O and a flame. Stoichiometry
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Chemical Equations
What happens when you mix (cause a reaction of) the following?
• Sodium + Chlorine  Sodium Chloride
• Dihydrogen Monoxide  Hydrogen + oxygen
• Magnesium + Hydrochloric Acid 
Magnesium Chloride + Hydrogen
• Hydrochloric Acid + Calcium Hydroxide 
Hydrogen hydroxide + Calcium chloride
• Combustion (burning with oxygen) of:
Sucrose (C12H22O11)
+ Oxygen  Carbon dioxide + water
Octane + Oxygen  Carbon dioxide + water
Write balanced chemical equations for each.
Stoichiometry
*
Predicting Products, writing Formulas
and Balancing Equations
•
Sodium + Chlorine 
2 Na + Cl2
•
Dihydrogen Monoxide 
•
2 NaCl
Hydrogen + oxygen

2 H2 + O2
Mg + 2 HCl 
MgCl2 + H2
2 HCl + Ca(OH)2  2 HOH + CaCl2
C12H22O11 +12 O2  12CO2 +11H2O
2 C8H18 +12.5
8 CO2 +18
9 H2O
25 O2  16
2 H2O
•

Sodium Chloride
Alonso’s Rules for BE:
(1) Easy element 1st hard
elements last.
(2) One element at a time.
(3) Use fractions when
necessary.
Magnesium + Hydrochloric Acid 
Magnesium Chloride + Hydrogen
Hydrochloric Acid + Calcium Hydroxide 
Hydrogen hydroxide + Calcium chloride
Combustion (burning with oxygen) of:
Sucrose (C12H22O11)
Octane
Carbon dioxide + water
Carbon dioxide + water
Stoichiometry
Decomposition Reactions
Simple: Binary compounds break down into their
constituent elements
2H2O
electrolysis
2NaCl(l)
2NaN3(s)
2H2 + O2
electrolysis
heat
2Na (l) + Cl2(g)
2Na(s) + 3N2(g)
{AirBags Movie*}
sodium azide (N31-)
∆
Important Exception:
2H2O2
Catalyst
{Peroxide Movie}
2H2O + O2
Stoichiometry
Decomposition Reactions
Complex Compounds decompose into simpler compounds
All carbonates break down to metal oxides and carbon dioxide
CaCO3 (s)  CaO (s) + CO2 (g)
Chlorates break down to metal chlorides and oxygen
2 KClO3 (s)  2 KCl (s) + 3O2 (g)
Acids break down to nonmetal oxides and∆water
2 H3PO4 (aq)  P2O5(g) + 3H2O (l)
2HNO3 (aq)  N2O5(g) + H2O (l)
Bases break down to metal oxides and water
2 NaOH (aq)  Na2O (s) + H2O (l)
Stoichiometry
 Na2O + CO2 + H2O
Ammonium carbonate powder is heated strongly
Stoichiometry
Stoichiometry
Combination (Synthesis) Reactions
Simple:
• Two or more elements react to form one compound
A + B  AB
Now let’s balance equations
2 Mg (s) + O2 (g)  2 MgO (s)
Zn (s) + S (s)  ZnS (s)
2 H2 (g) + O2 (g)  2 H2O (l)
2 Al (s) + 3 Br2 (l)  2 AlBr3 (s)
{Mg Movie}
{ZnS Movie*}
{H2O Movie*}
{AlBr3 Movie*}
Stoichiometry
Bromine liquid is poured over aluminum metal
Hydrogen chloride and ammonia gas are mixed together.
Sulfur dioxide gas is bubbled into water.
Stoichiometry
Stoichiometry
Metathesis (Double Displacement)
AB + CD  AD + CB
where A & C are Metals, B & D Nonmetals
• Involve two Compounds
• Elements (or polyatiomic
groups) in the two
compounds exchange
partners
Example: what quantity of Baking Soda will react with 100mL of vinegar?
NaHCO3 (s) + HC2H3O2 (l)  NaC2H3O2 (aq) + HHCO3 (aq)
H2CO3 (aq)  H2O (l) + CO2 (g)
(2nd Rx decomposition) Stoichiometry
{Movie: Bicarb + Vineg with Stoichio&LimitReag *}
Metathesis (Double Displacement):
Acid-Base Neutralization Reaction
Acid: compound containing hydrogen and a non metal (HN)
Bases: a metal hydroxide (MOH)
HN + MOH  MN + HOH
Acid
+
Base
 Salt
+ Water
Examples:
HCl (aq) + NaOH (aq)  NaCl (aq) + HOH (l)
2 HCl (aq) + Ca(OH)2(aq)  CaCl2 (aq) + 2 HOH (l)
HCl (aq) + NH4OH (aq)  NH4Cl (aq) + 2 HOH (l)
Stoichiometry
{Movie: A-B RxNo Ind}
{Movie: A-B Rx Ind+pH meter*}
Single Displacement Reactions
(Single Replacement Rx.)
A more active element displacing a less active elements
from a compound.
element
compound
MN +
M
MN + M
or
N
MN + N
Activity (Electromotive) Series:
Metals: Li > K > Ba > Sr > Ca > Na > Mg > Al > Zn > Cr > Fe >
Cd > Co > Ni > Sn > Pb > (H) > Sb > Bi > Cu > Hg > Ag > Pd >
Pt > Au
Stoichiometry
Halogens: F >Cl > Br > I
Single Displacement Reactions
A more active element displacing a less active
elements from a compound.
{Movie:Cu+AgNO3}
• Examples:
Cu (s) + 2 AgNO3 (aq)  2 Ag + Cu(NO3)2 (aq)
Cu (s) + Zn(NO3)2 (aq) 
No Reaction
Cl2 (g) + 2 NaBr (aq)  2 NaCl (aq) + Br2 (aq)
Activity (Electromotive) Series:
Metals: Li > K > Ba > Sr > Ca > Na > Mg > Al > Zn > Cr > Fe >
Cd > Co > Ni > Sn > Pb > (H) > Sb > Bi > Cu > Hg > Ag > Pd >
Pt > Au
Halogens: F >Cl > Br > I
Activity series can also be found in
form of Reduction Potential table.
Stoichiometry
Most
Active
Nonmetal
Most
Active
Metal
Stoichiometry
H+OH-
Stoichiometry
Reactions with Oxygen
What is the difference?
Oxidation Rx.
(1) Oxidation Reactions: are combination
reactions involving oxygen.
{Movie: Mg, Fe, P, S + conc. O2 
{Metal Oxides*}
Combustion Rx
(2) Combustion Reaction: Rapid reactions of
oxygen with an organic compounds
(hydrocarbons, alcohols) that produce CO2 + H2O
and a flame.
Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Stoichiometry
{Movie: CH3OH + O2*}
When oxygen is scarce….
*
Stoichiometry
gram-Molar Mass (g-MM)
= Atomic Weigh, Formula
Weigh or Molecular Weight
Mole = 6.022 x 1023 particles
Mass : Weight (in grams)
Stoichiometry
gram-Molar Mass (g-MM): AW, FW, MW
*
the mass in grams of 1 mole of
a substance (units= g/mol)
For an element we find it on the
periodic table.
For compounds the same as the formula & molecular weight
(but in g/mol)
Example: the g-MM of Al2(SO4)3, would be
2 Al: 2x(26.98 amu) = 53.96
+ 3 S: 3x(32.06 amu) = 96.18
+3x4 O: 12x(16.00 amu) =192.00
Stoichiometry
342.14 amu (g/mol)
Formula Weight (FW)
•
•
Sum of the atomic weights for the atoms in a chemical formula unit
(ionic compound)
So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1x(40.1 amu)
2 x Cl: 2x(35.5 amu)
111.1 amu
Molecular Weight (MW)
•
•
Sum of the atomic weights of the atoms in a molecule (covalent
compound)
For the molecule ethane, C2H6, the molecular weight would be
2 x C: 2x(12.0 amu)
6 x H: 6x(1.0 amu)
30.0 amu
Stoichiometry
(Mass) Percent Composition
Percentage mass of a element (Na) in compound (NaCl):
23g
mass Na
x 100  40%
% Na 
x 100 
58g
mass NaCl
part
%
x 100
whole
g-MM = 2,301g/
Nagyagite Gold Ore: Pb5Au(TeSb)4S5
Problem: (1) calculate mass % Au in Nagyagite.
(2) If you buy 1 kg of the ore, how much gold
does it have?
% element =
(# atoms of element) (atomic weight of Au)
x 100
(MW of Pb5Au(TeSb)4S5)
% Au =
(1) ( 197)
( 2,301)
x 100
= 8.56 %
 8.56g Au 
  85.6g Stoichiometry
Au
? g of Au  1000g Ore 
 100g Ore 
Using mass % to determine mass of
one particular element in a sample
What is the mass of carbon in a 25g sample of carbon dioxide?
There are two parts to this problem:
(1) What is the percentage mass of carbon in carbon dioxide?
part
mass of C
12 g
 100 
100 
 100  27 %
%C=
whole
mass of CO 2
44 g
(2) What is the mass of carbon in a 25 g sample of
carbon dioxide?
x 27%
 27 g C 
  6.75g
? g C = 25 g CO2  
 100 g CO 2 
Stoichiometry
The Mole Concept
Dermatological
Chemical
Biological
Avogadro's Number:
6.022,141,410,704,090,840,990,72 x 1023
602,214,141,070,409,084,099,072 .
sextillion
pentillion
quadrillion
trillion
billion
602 sextillion
million
thousand
Stoichiometry
Using Equivalences as Mole Ratios:
g-MM 1 Mole ()
=
NaCl = 58g/η
=
6.022 x 10 23
particles
(Atoms or molecules)
From Equivalences we obtain useful Ratios or Conversion factors:
 g - MM 


 1 mole 
or

1 mole
g-MM

 

  6.022x 1023 particles
23

 6.022 x 10 particles  
or
 1 mole   6.022 x 10 23 particles

 
g-MM
 g - MM  
or
  6.022x 1023 particles


Stoichiometry

1 mole


Mole Calculations:
g-MM
Moles
g-MM
Moles:
? g = 3.20 mol of NaCl
# of Particles
Which ratios will you need?  1 mol NaCl

58.443g

 187 g
? mol = 3.20 g of NaCl
Moles
*
# of Particles:
? f.u. = 3.2 mol NaCl
? mol = 3.2 x 10 52 f.u. NaCl
 0.0548 mol



or
 58.443g 


 1 mol NaCl 
 1 mol NaCl 


23
23
 6.023 x 10 f.u.
= 1.9 x 10 24 f.u. or
23
23

6
.
023
x
10
f.u. 
28

6
.
023
x
10
= 5.3 x 10 Stoichiometry
molf.u.
 1 mol NaCl 
 1 mol NaCl 
Mole Calculations:
g-MM
Moles
g-MM
# of Particles
# of Particles:
? g = 4.2 x 10
34
f. u. of NaCl
58.443g




23
 6.022x 10 f.u. 
4.1 x 1012 g
1 mol

 58.443g 
12 g





4.1
x
10
23
 6.022x 10 f.u.  1 mol 
? f. u. of NaCl = 3.2 g of NaCl
 6.022x 1023 f.u 

  3.3 x 1022 f.u.
 58.443g

 1 mole  6.022 x 10 23 f.u. 
  3.3 x 1022 f.u.


1 mole
 58.443g 

Stoichiometry
Mole Calculations:
g-MM
Moles
*
# of Particles
(atoms or molecules)
How many molecules of H2O in 29g of water? How many atoms?
 1 mole   6.0221023 m olecules
? molecules 29gH2O 
 2.2 1023 molecules
 


1 m ole
 18 gH 2O  

? atoms 2.2 1023 moleculesof H 2O


3 atoms


 1 molecule H 2O 
 6.6 x 1023 at oms
23

6.022
x
10
f.u. 
? atoms = 0.50 mole Fe(CO)3(PH3)2 

= 3.0 X 1023 f.u.


1
mole


1+3+3+ 2+ 6 = 15


15 atoms


 1 f.u.of Fe (C O 3) (PH3 ) 2 
Stoichiometry
= 4.5 x 1024 atoms
The Determination of
Empirical Formulas of
Compounds by
Elemental Analysis
CxHy
Combustion
furnace
Stoichiometry
Types of Formulas
• Structural formulas (skeletal or spacefilling) show the order in which atoms are
bonded and their three-dimensional shape.
• Molecular formulas give the exact
number of atoms of each element in a
compound.
• Empirical formulas give the lowest
whole-number ratio of atoms of each
element in a compound.
HO
CH
H2O
Stoichiometry
Why are empirical formulas needed?
Benzene, C6H6
Elemental Analyses
How do we determine the formula of a compound?
Compounds are broken down
and the masses of their constituent
elements are measured. From
these masses the empirical
formulas can be determined.
HxOy
moles
Mole Ratio
Expt. Data: (68g)
 1mole
  4 H
4g H 
 1 H
 1g H 
4
 1mole
  4 O
64g O
 1 O
 16g O 
4
EmpF
Stoichiometry
HO
Calculating Empirical Formulas
*
Problem:
The compound para-aminobenzoic acid (you may have seen it
listed as PABA on your bottle of sunscreen) is composed of
carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and
oxygen (23.33%). Find the empirical formula of PABA.
carbon (61.31%),
hydrogen (5.14%),
nitrogen (10.21%),
oxygen (23.33%).
Percent means out of 100, so assume a
100g sample of the compound, then….
carbon (61.31g),
hydrogen (5.14g),
nitrogen (10.21g),
oxygen (23.33g)
Calculate the empirical formula (mole ratio) from
the percent composition (% mass).
Stoichiometry
Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
1 mol
12.01 g
? mol H = 5.14 g x 1 mol
1.01 g
1 mol
? mol N = 10.21 g x
14.01 g
? mol O = 23.33 g x 1 mol
16.00 g
61% C ? mol C = 61.31 g x
= 5.105 mol C
5% H
= 5.09 mol H
10% N
23% O
= 0.7288 mol N
= 1.456 mol O
5.105 mol
0.7288 mol = 7.005  7
5.09 mol
= 6.984  7
0.7288 mol
0.7288 mol
= 1.000
0.7288 mol
1.458 mol
= 2.001  2
0.7288 mol
What is the smallest mole ratio of the elements in this compound?
Calculate the mole ratio by dividing by the smallest number of moles.
These are the subscripts for the empirical formula: C7H7NO2
Stoichiometry
Combustion Analysis: is a method of
experimentally determining empirical formulas
CxHy +
O2

mass
How do you calculate the mass of C
in CO2 and that of H in H2O?
mass of CO2
CxHy
{Movie}
CO2
+
H2O
mass of CO2
mass of H2O
mass of C?
mass of H?
 12.011 g C

 44.011 g CO 2



Magnesium
perchlorate
 1.0 g H 

18.0
g
H
O
2


mass of H2O 
Sodium
hydroxide
Combustion
furnace
•
Compounds containing C, H and O are routinely analyzed through combustion in a
chamber like this
– C is determined from the mass of CO2 produced
– H is determined from the mass of H2O produced
Stoichiometry
– O is determined by difference after the C and H have been determined
Calculating Empirical Formulas
(12g:32g=44g)
CxHy (g) + O2 (g)  CO2 (g) + H2O (g)
14.6 g
5.00 g
A 5.00 g sample of an unknown hydrocarbon was burned and produced
14.6 g of CO2. What is the empirical formula of the unknown
compound?
? g C = 14.6 g CO2
 12.011 g C

 44.011 g CO 2

 = 3.98 g C

g H = 5.00 CxHy – 3.98 g C = 1.02 g H
Empirical
Formula
CH3
 1 moleC 
 = 0.332 mol C/ 0.332 = 1.00
? mol C = 3.98 g C 
 12.011g 
 1 moleH 
Stoichiometry
 = 1.01 mol H / 0.332 = 3.04
? mol H = 1.02 g H 
 1.011g 
2006 A
?g C =
?g N =
Stoichiometry
Stoichiometry
2003 B
Stoichiometry
Stoichiometry
Stoichiometry
(mass relationships within chemical equations)
Stoichiometry
Stoichiometric Calculations
2
2
The coefficients in the balanced equation can also be
interpreted as mole ratios of reactants and products
Mole Ratios from Balanced Equation:
 2 mol H 2 


 1 mol O 2 
 2 mol H 2 


 2 mol H 2 O 
 1 mol O 2 


 2 mol H 2 O 
Stoichiometry
Stoichiometric Calculations
How many grams of O2 are
required to form 2.35 g of MgO?
*
2 Mg (s) + O2 (g)  2 MgO (s)
? g O2  2.35g MgO
grams
No direct
calculation
grams
Change:
1.
2.
3.
grams of MgO  mol MgO
mol of MgO  mol O2
mol of O2  grams of O2
2 Mg (s) + O2 (g)  2 MgO (s)
 1 mol MgO  1 molO2  32.000g O 2 


? g O2  2.35g MgO
  0.933g O2

2
mol
MgO
 1 molO 2 
 40.304g MgO 
Stoichiometry
mole ratio from balanced equation
Stoichiometric Calculations
Balanced Eq. uses MOLE language
C6H12O6 + 6 O2  6 CO2 + 6 H2O
? g H 2O
Grams H2O
Grams C6H12O6
(3)
(1)
Moles C6H12O6
(2)
Moles H2O
(1) Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
(2) use the coefficients to find the moles of H2O…
Stoichiometry
(3) and then turn the moles of water to grams
Stoichiometry: Limiting Reactants
(or, too much of one reactant and not enough of the other)
Make cookies until you run out of
one of the ingredients
In this example the sugar would be the limiting reactant,
because it will limit the amount of cookies you can make.
Stoichiometry
{ MovieLimitingReactants: Zn + 2 HCl  ZnCl2 + H2 }
Limiting & Excess Reactants
• The limiting reactant is the reactant present in the
smallest stoichiometric amount
– In other words, it’s the reactant you’ll run out of first
Stoichiometry
Which is Limiting which is Excess?
Limiting H2; Excess O2
Limiting & Excess Reactants
Problem
2 Mg (s) + O2 (g)  2 MgO (s)
If 5.0g of both Mg and O2 are used:
(1) Which is the limiting and the excess
reactants?
(2) How much of the excess will be left
unreacted ?
(3) How much MgO will be produced ?
Stoichiometry
Limiting & Excess Reactants
Problem
5g
5g
2 Mg (s) + O2 (g)  2 MgO (s)

 1 molO 2  32.0g 
? g O2 = 5.0 g Mg  1 mol Mg 
 = 3.29 g O2

 24.3g  2 mol Mg  1 molO 2 
 1 molO 2  2 mol Mg  24.3g


? g Mg = 5.0 g O2 
 32.0g  1 molO 2  1 mol Mg

 = 7.59 g Mg

Mg is limiting reactant and O2 is excess reactant
Which one do I use to determine the MgO produced by rx?
Stoichiometry






1
mol
Mg
2
mol
MgO
40.3
g
?g MgO = 5.0 g Mg 
 24.3g  2 mol Mg  1 mol MgO   8.29g MgO




Limiting Reactant Experiments
H2
Zn+HCl
 ZnCl2
H2O
{Movie:LimitReac}
Experiments:
Leveling
Bulb: to
maintain
pressure of
H2 same as
that of
atmosphere
Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
(1)
0.0025 η + 0.0050 η 
61.0 mL
(2)
0.0012 η + 0.0050 η 
30.5 mL
Stoichiometry
(3)
0.0031 η + 0.0050 η 
61.0 mL
Which are the limiting and excess reactants in each expt.?
Limiting Reactant Experiments
Experiments: Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
0.0025 g + 0.0050 g 
 1 moleZn   2 m ole HCl   66.40g HCl 
? gHCl  0.0025gZn 

 

 35.453gZn   1 m oleZn   1 m oleHCl 
HCl is Limiting Reactant   0.0093gHCl
 1 m oleHCl   1 m oleZn   35.453g Zn 
 


? gZn  0.0050gHCl 
 66.40g HCl   2 m ole HCl   1 m oleZn 
 0.0013gZn
Zn is Excess Reactant 
How many grams of ZnCl2 are produced in this reaction?
 1 m oleHCl  1 ZnCl2   166.233g ZnCl2 
  0.0063gZnCl2
 

? gZnCl2  0.0050gHCl 
Stoichiometry
 66.40g HCl  2  HCl   1 m oleZnCl2 
How many grams of ZnCl2 are produced in this reaction?
Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
 1 m oleHCl  1 ZnCl2   166.233g ZnCl2 
  0.0063gZnCl2

 
? gZnCl2  0.0050gHCl 
 66.40g HCl  2  HCl   1 m oleZnCl2 
Theoretical Yield
the amount of product that can be made as
calculated by stoichiometry.
Actual Yield
Using Balance 0.0045gZnCl2
the amount reaction actually produces (less)
Percent Yield
Percent Yield =
Actual Yield
Theoretical Yield
 .0045

 100  71.4%
 .0063
x 100
Stoichiometry
*
The following reaction has a 95% yield:
GeH4 + 3GeF4  4GeF3H
g-MM: 76.622
148.5756
130.58
Problem:
How many grams of the product are formed, when 23.4 g of GeH4 are
reacted with excess GeF4?
 1  GeH 4
 76.622 g
?g GeF3H = 23.4 g GeH4 
 4  GeF3 H  130.58 g 

  160.g GeF3H

 1  GeH 4  1  GeF3 H 
Since % yield is only 95%, then actual yield is:
(160.g GeF3 H)  0.95  152 g of GeF3 H
What would the yield be if reaction Actually produced 130g only?
 130g 
 100  81 %
%Yield  
 160g 
 120g 
75%  
 100 
 whole
 120g 
whole  
 100  160g
 75% 
Stoichiometry
What would be the Theoretical yield if 120 g was a 75% yield ?