Download APch3-2009

Stoichiometry
Chapter 3 -MW
What Is Stoich?
Stoichiometry - The study of quantities of materials
consumed and produced in chemical reactions.
Stoichiometry is the study of reactions:




Why do reactions occur?
How fast do they proceed?
What intermediary products if any are used?
How much of the reactants react?
Mass Spectrometer
Compares mass of atoms
Atomic mass is defined by Carbon 12 = 12amu
Parts of a mass spectrometer:
Vaporizer
Electron
beams
Ionaccelerating
electric field
Magnetic field
Detector Plate

Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
4
Mass Spectrometer Process
1. A heater vaporizes a sample
2. A beam of high speed electrons knocks electrons
off test atoms/molecules.
3. An electric field accelerates the sample ions
4. The accelerating ions have a magnetic field. They
interacts with an applied magnetic field deflecting
their path. The ions separate.
5. A detector plate measures the deflections
-comparison of deflections gives ions’ masses
-less massive particles deflect more
Atomic Masses
Elements occur in nature as mixtures of
isotopes
Carbon =
98.89% 12C
1.11% 13C
<0.01% 14C
Carbon atomic mass = 12.01 amu
.
6
Change % abundance into decimals &
multiply by respective isotopic weights.
Add together.
If want % abundance; use “x” & “1 – x” to
represent abundance.
7
Examples
1) There are two isotopes of carbon 12C with a
mass of 12.00000 amu(98.892%), and 13C
with a mass of 13.00335 amu (1.108%).
2) 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 if the weighted average is
14.01amu?
Answers
1) 0.98892(12.00000amu) + 0.01108(13.00335amu)
= 11.86704amu + .1440771amu = 12.011117amu
Matches P.table
14.0031(x) + 15.0001(1-x) = 14.01
14.0031x + 15.0001amu - 15.001x = 14.01 amu
-0.997x = -0.9901
X = .9930192 or 99.3%, 1-x = .692%
9
The Mole
The number equal to the number of carbon
atoms in exactly 12 grams of pure 12C.
1 mole of anything = 6.022 x 1023 units of
that thing
Avogadro’s number equals 6.022 x 1023
units
10
Molar Mass
A substance’s molar mass (molecular
weight) is the mass in grams of one mole of
the compound.
CO2 = 44.01 grams per mole
11
Find the molar mass of
CH4
Mg3P2
Ca(NO3)3
Al2(Cr2O7)3
CaSO4 · 2H2O
CH4
=12.0 + (4) 1.01 = 16.0g/mol
Mg3P2
= (3) 24.3 + (2) 30.974 = 135g/mol
Ca(NO3)3 = 40.1 + (3) 14.0 + (9) 16.0 = 226g/mol
Al2(Cr2O7)3 = (2) 27.0 + (6) 52.0 + (21) 16.0 =
= 702g/mol
CaSO4 · 2H2O = 40.1 + 32.1 + (4) 16.0 + (2) 18.0
= 156g/mol
13
Percent Composition
Mass percent of an element:
mass of element in compound
mass % 
 100%
mass of compound
For iron in iron (III) oxide, (Fe2O3)
111.69
mass % Fe 
 100%  69.94%
159.69
Working backwards
From percent composition, you can determine the
empirical formula.
Empirical Formula the lowest ratio of atoms in a
molecule.
Based on mole ratios.
Empirical Formula Determination
1. Base calculation on 100 grams of
compound.
2. Determine moles of each element in 100
grams of compound.
3. Divide each value of moles by the smallest
of the values.
4. Multiply each number by an integer to
obtain all whole numbers.
16
A sample is 59.53% C, 5.38% H, 10.68% N,
and 24.40% O. What is its empirical
formula?
C
59.53
12.0
H
5.38
1.01
N
10.68
14.0
O
24.40
16.0
4.96
.7628
5.33
.7628
.7628
.7628
1.525
.7628
6.5
7
1
2
C13H14N2O4
(mult. By 2)
17
A 0.2000 gram sample of a compound
(vitamin C) composed of only C, H, and O
is burned completely with excess O2 .
0.2998 g of CO2 and 0.0819 g of H2O are
produced. What is the empirical formula?
Get C from CO2, H from H2O and O from
subtracting C + O from original amount.
C 0.2998g x 12g = 0.0817636g C
CO2
44g
H 0.0819g x 2.02g = 0.00919g H
H 20
18.0g 0.0909546g
0.2000g - 0.0909546g = 0.1090454g O
19
0.0817636g C
12.0
1
(x 3)
0.00919g H
1.01
1.33
0.1090454g O
16.0
1
C3 H4 O3
20
Molecular Formula
Molar mass = (empirical formula)n
[n = integer]
empirical formula = CH, Molar mass = 78.0g
(CH)x = 78.0g, (12 + 1.01)x= 78.0g, x = 6
molecular formula = (CH)6 = C6H6
21
Example
A compound is made of only sulfur and
nitrogen. It is 69.6% S by mass. Its molar
mass is 184 g/mol. What is its formula?
69.6/32.1 = 2.168
30.4/14.0 = 2.171
46.1x = 184,
x = 3.99
S4N4
Chemical Equations
Chemical change involves a
reorganization of the atoms in one or
more substances.
23
Chemical Equation
A representation of a chemical reaction:
C2H5OH + 3O2 -> 2CO2 + 3H2O
reactants
products
24
Chemical Equations
Are sentences.
Describe what happens in a chemical reaction.
Reactants -> Products
Equations should be balanced.
Have the same number of each kind of atoms
on both sides because ...
Abbreviations
(s) 
(g)
(aq)
heat
D
catalyst
Chemical Equation
C2H5OH + 3O2 -> 2CO2 + 3H2O
The equation is balanced.
1 mole of ethanol reacts with 3 moles of
oxygen
to produce
2 moles of carbon dioxide and 3 moles of
water
27
Practice
Ca(OH)2 + H3PO4 -> H2O + Ca3(PO4)2
Cr + S8 -> Cr2S3
KClO3(s) -> Cl2(g) + O2(g)
Solid iron(III) sulfide reacts with gaseous
hydrogen chloride to form solid iron(III)
chloride and hydrogen sulfide gas.
Fe2O3(s) + Al(s) -> Fe(s) + Al2O3(s)
3Ca(OH)2 + 2H3PO4 -> 6H2O + Ca3(PO4)2
16Cr + 3S8 -> 8Cr2S3
2KClO3(s) -> 2KCl(s) + 3O2(g)
Fe2S3(s) + 6HCl(g) -> 2FeCl3(s) + 3H2S(g)
Fe2O3(s) + 2Al(s) -> 2Fe(s) + Al2O3(s)
29
All chemical reactions can be placed into
one of six categories. Here they are, in no
particular order:
1) Combustion: A combustion reaction is
when oxygen combines with another
compound to form water and carbon
dioxide. These reactions are exothermic,
meaning they produce heat. An example of
this kind of reaction is the burning of
napthalene:
C10H8 + 12 O2 ---> 10 CO2 + 4 H2O
.
30
2) Synthesis: A synthesis reaction is when
two or more simple compounds combine to
form a more complicated one. These
reactions come in the general form of:
A + B ---> AB
One example of a synthesis reaction is the
combination of iron and sulfur to form iron
(II) sulfide:
8 Fe + S8 ---> 8 FeS
31
3) Decomposition: A decomposition reaction
is the opposite of a synthesis reaction - a
complex molecule breaks down to make
simpler ones. These reactions come in the
general form:
AB ---> A + B
One example of a decomposition reaction is
the electrolysis of water to make oxygen
and hydrogen gas:
2 H2O ---> 2 H2 + O2
32
4) Single displacement: This is when one
element trades places with another element
in a compound. These reactions come in the
general form of:
A + BC ---> AC + B
One example of a single displacement
reaction is when magnesium replaces
hydrogen in water to make magnesium
hydroxide and hydrogen gas:
Mg + 2 H 2O ---> Mg(OH)2 + H2
33
5) Double displacement: This is when the
anions and cations of two different
molecules switch places, forming two
entirely different compounds. These
reactions are in the general form:
AB + CD ---> AD + CB
One example of a double displacement
reaction is the reaction of lead (II) nitrate
with potassium iodide to form lead (II)
iodide and potassium nitrate:
Pb(NO3) 2 + 2 KI ---> PbI 2 + 2 KNO3
34
6) Acid-base: This is a special kind of double
displacement reaction that takes place when
an acid and base react with each other. The
H+ ion in the acid reacts with the OH- ion in
the base, causing the formation of water.
Generally, the product of this reaction is
some ionic salt and water:
HA + BOH -> H2O + BA
One example of an acid-base reaction is the
reaction of hydrobromic acid (HBr) with
sodium hydroxide:
35
HBr + NaOH -> NaBr + H2O
Calculating Masses of Reactants
and Products
1.
2.
3.
4.
Balance the equation.
Convert mass to moles.
Set up mole ratios.
Use mole ratios to calculate moles of
desired substituent.
5. Convert moles to grams, if necessary.
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
36
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?
3) How many grams of oxygen?
2KClO3(s) -> 2KCl(s) + 3O2(g)
25.5g KClO3 x 1 mol x 3 mol = 0.3109756 mol O2
123g 2 mol
25.5g KClO3 x 1 mol x 2 mol x 74.6g = 15.5g KCl
123g 2 mol 1 mol
25.5g KClO3 x 1 mol x 3 mol x 32.0g = 9.95g O2
123g 2 mol 1 mol
38
Examples
1) A piece of aluminum foil 5.11 in x 3.23 in x
0.0381 in is dissolved in excess HCl(aq). How
many grams of H2(g) are produced?
2) How many grams of each reactant are needed
to produce 15 grams of iron from the
following reaction?
Fe2O3(s) + Al(s) -> Fe(s) + Al2O3(s)
5.11in=12.98, 3.23in=8.20cm, 0.0381in=0.0968cm
vol = 10.3cm3
D = 2.7g/cm3
27.8g Al x 1mol x 3 mol H2 x 27.0g = 3.12g H2
27g 2 mol Al 1 mol
-----------------------------------------------------------------15g Fe x 1 mol x 2 mol Al x 27.0g = 7.25g Al
55.85g 2 mol Fe 1 mol
15g Fe x 1 mol x 1 mol Fe2O3 x 159.7g = 21.5g Al
55.85g 2 mol Fe
1 mol
40
Examples
K2PtCl4(aq) + NH3(aq) ->
Pt(NH3)2Cl2 (s)+ KCl(aq)
What mass of Pt(NH3)2Cl2 can be produced
from 65 g of K2PtCl4 ?
How much KCl will be produced?
How much from 65 grams of NH3?
K2PtCl4(aq) + 2NH3(aq) -> Pt(NH3)2Cl2 (s)+ 2KCl(aq)
65 g K2PtCl4 x 1mol x 1mol x 300g = 47g Pt(NH3)2Cl2
415g 1mol 1mol
How much KCl will be produced?
65 g K2PtCl4 x 1mol x 2mol x 74.6g = 23g KCl
415g 1mol 1mol
How much from 65 grams of NH3?
65g NH3 x 1mol x 1mol x 300g = 574g Pt(NH3)2Cl2
17.0g 2mol 1mol
42
65g NH3 x 1mol x 2mol x 74.6g = 285g KCl
Limiting Reagent
Reactant that determines the amount of
product formed.
The one you run out of first.
Makes the least product.
Book shows you a ratio method.
It works.
So does mine
Example
Ammonia is produced by the following reaction
N2 +
H2 -> NH3
What mass of ammonia can be produced from a
mixture of 100. g N2 and 500. g H2 ?
How much unreacted material remains?
100g N2 x 1mol x 2mol x 17.0g = 121g NH3
28.0g 1mol 1mol
500g H2 x 1mol x 2mol x 17.0g = 2805g NH3
2.02g 3mol 1mol
121g NH3 x 1mol x 3mol x 2.02g = 21.6g H2
17.0g 2mol 1mol
500g – 21.6g = 478g H2 unreacted
45
Excess Reagent
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.
Percent Yield
% yield = Actual
Theoretical
% yield =
x 100%
what you got
x 100%
what you could have got
Examples
Aluminum burns in bromine producing
aluminum bromide. In a laboratory 6.0 g of
aluminum reacts with excess bromine. 50.3
g of aluminum bromide are produced. What
are the three types of yield.
2Al + 3Br2 -> 2AlBr3
6.0g Al x 1mol x 2mol x 267g = 59.3g AlBr3
27.0g 2mol 1mol
50.3 x 100 = 84.8%
59.3
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
49
Examples
Years of experience have proven that the percent
yield for the following reaction is 74.3%
Hg + Br2 > HgBr2
If 10.0 g of Hg
and 9.00 g of Br2 are reacted, how much HgBr2
will be produced?
If the reaction did go to completion, how much
excess reagent would be left?
Hg + Br2 -> HgBr2
If 10.0 g of Hg and 9.00 g of Br2 are reacted, how much
HgBr2 will be produced?
10.0 g Hg x 1mol x 1mol x 361g = 17.96g
201g 1mol 1mol
9.00 g Br2 x 1mol x 1mol x 361g = 20.31g
160g 1mol 1mol
how much excess reagent would be left?
.743 x 17.96g = 13.34 g HgBr2
13.34 g HgBr2 x 1mol x 1mol x 160g = 5.91gBr2
361g 1mol 1mol
9.00-5.91=3.09g excess
Examples
Commercial brass is an alloy of Cu and Zn. It
reacts with HCl by the following reaction
Zn(s) + 2HCl (aq) -> ZnCl2 (aq) + H2(g)
Cu does not react.
When 0.5065 g of brass is reacted with excess
HCl, 0.0985 g of ZnCl2 are eventually
isolated. What is the composition of the
brass?
Zn(s) + 2HCl(aq) -> ZnCl2 (aq) + H2(g)
0.5065 g of brass is reacted with excess HCl,
0.0985g ZnCl2 x 1mol x 1mol x 65.39g =
.0473597g
136g 1mol 1mol
0.5065 - 0.0473597g = 0.4591 x 100 = 90.6% Cu
9.35% zn
53