stoichiometry chapter 3 -mw. what is stoich? stoichiometry is the study of reactions: why do...
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Stoichiometry
Chapter 3 -MW
What Is Stoich?
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?
Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.
Mass SpectrometerCompares mass of atomsAtomic mass is defined by Carbon 12 = 12amuParts of a mass spectrometer:
VaporizerElectron beamsIon-accelerating electric fieldMagnetic fieldDetector Plate
Copyright©2000 by Houghton Mifflin Company. All rights
reserved.
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Mass Spectrometer Process
1. A heater vaporizes a sample2. A beam of high speed electrons knocks electrons
off test atoms/molecules.3. An electric field accelerates the sample ions4. 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
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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
Change % abundance into decimals & multiply by respective isotopic weights.
Add together.
If want % abundance; use “x” & “1 – x” to represent abundance.
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Examples1) 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%
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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
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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
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/mol13
Percent Composition
Mass percent of an element:
For iron in iron (III) oxide, (Fe2O3)
mass Fe%..
. 111 69159 69
100% 69 94%
massmass of element in compound
mass of compound% 100%
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.
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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.
A sample is 59.53% C, 5.38% H, 10.68% N, and 24.40% O. What is its empirical formula?
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C H N O59.53 5.38 10.68 24.4012.0 1.01 14.0 16.0
4.96 5.33 .7628 1.525.7628 .7628 .7628 .7628
6.5 7 1 2 (mult. By 2)
C13H14N2O4
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
H0.0819g x 2.02g = 0.00919g H
H20 18.0g 0.0909546g
0.2000g - 0.0909546g = 0.1090454g O19
0.0817636g C 0.00919g H 0.1090454g O
12.0 1.01 16.0
1 1.33 1(x 3)
C3 H4 O3
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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
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
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Chemical Equations
Chemical change involves a reorganization of the atoms in one or
more substances.
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Chemical Equation
A representation of a chemical reaction:
C2H5OH + 3O2 2CO2 + 3H2O
reactants products
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
catalyst
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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
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) Cl(s) + 3O2(g)
Fe2S3(s) + 6HCl(g) FeCl3(s) + 3H2S(g)
Fe2O3(s) + 2Al(s) 2Fe(s) + Al2O3(s)
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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
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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
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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
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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
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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
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:
HBr + NaOH ---> NaBr + H2O
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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:
HBr + NaOH -> NaBr + H2O35
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reserved.
36
Calculating Masses of Reactants and Products
1. Balance the equation.
2. Convert mass to moles.
3. Set up mole ratios.
4. Use mole ratios to calculate moles of desired substituent.
5. Convert moles to grams, if necessary.
ExamplesOne 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) Cl(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 Cl
123g 2 mol 1 mol
25.5g KClO3 x 1 mol x 3 mol x 32.0g = 9.95g O2
123g 2 mol 1 mol38
Examples1) 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
ExamplesK2PtCl4(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 K2PtCl4x 1mol x 1mol x 300g = 47g Pt(NH3)2Cl2
415g 1mol 1mol
How much KCl will be produced?
65 g K2PtCl4x 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
65g NH3x 1mol x 2mol x 74.6g = 285g KCl
17.0g 2mol 1mol 42
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 x 100% Theoretical
% yield = 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
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