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Transcript of Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chapter 3 Stoichiometric Atomic...
Copyright©2000 by Houghton Mifflin Company. All rights reserved.
1
Chapter 3Stoichiometric
Atomic Masses, Mole concept, and Molar Mass (Average atomic mass).
Number of atoms per amount of element.Percent composition and Empirical formula
of molecules.Chemical equations, Balancing equations,
and Stoichiometric calculations including limiting reagents.
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2
By definition: 1 atom 12C “weighs” 12 amu
On this scale
1H = 1.008 amu
16O = 16.00 amu
Atomic mass is the mass of an atom in atomic mass units (amu)
Micro Worldatoms & molecules
Macro Worldgrams
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3
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
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4
Measuring Atomic Mass
Figure 3.1: (left) A scientist injecting a sample into a mass spectrometer. (right) Schematic diagram of a mass spectrometer.
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6
Atomic mass of NeonAtomic mass of NeonConsider 100 atoms of neonConsider 100 atoms of neon
18.20100/
2018)
82.8(
)257.0
()92.90(
22
2120
amux
atom
xatoms
xatoms
atomamu
atomamu
atomamu
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7
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
7.42 x 6.015 + 92.58 x 7.016100
= 6.941 amu
Average atomic mass of lithium:
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8
The Mole
The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.
1 mole of anything = 6.022 1023 units of that
thing
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9
Avogadro’s number equals
6.022 1023 units
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10
Molar Mass
A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.
C=12 O=16
CO2 = 44.01 grams per mole
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11
Molar mass is the mass of 1 mole of in gramseggsshoes
marblesatoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
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12
1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu
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13
EX. 3.2: Calculate the mass of EX. 3.2: Calculate the mass of 6 atoms of americium in grams6 atoms of americium in grams• From periodic table, Am has a mass of 243
amu.• 6.022x1023 atoms weigh 243 g.
gxatomsAmAmatomsx
g x21
23
1042.2610022.6
243
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14
Molar Mass
A chemical compound is a collection of atoms.
What is the mass of 1 mol of CH4?
Mass of 1 mol C = 12.01 g
Mass of 4 mol H = 4x1.008 g
Mass of 1 mol CH4 = 16.04 g
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15
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol H = 6.022 x 1023 atoms H
5.82 x 1024 atoms H
1 mol C3H8O molecules = 8 mol H atoms
72.5 g C3H8O1 mol C3H8O
60 g C3H8Ox
8 mol H atoms
1 mol C3H8Ox
6.022 x 1023 H atoms
1 mol H atomsx =
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16
S.Ex. 3.7Calculate Molar mass of CaCO3
4.86 moles of CaCO3 → g CaCO3 → g CO32-
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17
Percent composition of an element in a compound =
n x molar mass of elementmolar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the compound
C2H6O
%C =2 x (12.01 g)
46.07 gx 100% = 52.14%
%H =6 x (1.008 g)
46.07 gx 100% = 13.13%
%O =1 x (16.00 g)
46.07 gx 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
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18
Figure 3.5: A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen.
Determining Elemental Composition(Formula)
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19
The masses obtained (mostly CO2 and H2O and sometimes N2)) will be used to determine:
1. % composition in compound
2. Empirical formula
3. Chemical or molecular formula if the Molar mass of the compound is known or given.
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20
Example of Combustion
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
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21
1mol CO244gCO2 1mol C 12g C
22gCO2 mc
mc =
Collect 22.0 g CO2 and 13.5 g H2O
12 16 x 2
44gCO2
22gCO2 x 12gC
% C in CO2 collected
= 6g C
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22
1mol C 12g Cnmol C 6g
6g x 1molCnc =
12 gC
= 0.5 mol
Convert g to mole:
Repeat the same for H from H2O
1mol H2O 18g H2O 2 mol H 2g H
13.5g H2O nmol H
2x13.5nmol H = = 1.5 mol H
18 mol H Faster H but still need O
mH
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23
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
2x13.5mH = = 1.5 g H
18mO = 11.5g – mC – mH
= 11.5 – 6 – 1.5 = 4g
m 4nO = = = 0.25 mol O
MM 16
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24
OR for C
mC 12.01
1. Fraction of C in CO2 = = =
MMCO2 44.01
2. Mass of C in compound = mass of CO2 x Fraction of C
mC
3. % C in compound = x 100
msample
4. If N then % N = 100 - % C - % H
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25
ThenEmpirical Formula
Using the previously calculated % in compound:
% in gram
a. Number of mole of C =
Atomic mass of C
% in gram
b. Number of mole of H =
Atomic mass of H
a b cThen divide by the smallest number:
smallest smallest smallest
: :
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26
NoteIf results are : 0.99 : 2.01 : 1.00Then you have to convert to whole numbers:
1 : 2 : 1
CH2N
If results are : 1.49 : 3.01 : 0.99Then you have to multiply by 2:
3 : 6 : 2
C3H6N2
Hence, empirical formula is the simplest formula of a compound
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27
S. E.3.1171.65% Cl; 24.27% C ; 4.07 % H
2.021 mol Cl 2.021 mol C 4.04 mol H1 : 1 : 2
Empirical formula is ClCH2
Molar mass = 98.96 g/molMolar mass/ empirical formula mass =
98.96/49.48 = 3
Molecular formula – (ClCH2)2 = Cl2C2H4
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28
Formulasmolecular formula = (empirical formula)n
[n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH
Then
Molecular Mass
= n
Empirical Mass
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29
Figure 3.6: Examples of substances whose empirical and molecular formulas differ. Notice
that molecular formula = (empirical formula)n, where n is a integer.
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30
Exercise 3.12
A compound analyzed and found to contain only P and O.
Mass % of P = 43.64%
Molar mass = 283.88 g/mol.
What are the empirical and molecular formulas?
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31
Figure 3.8 The Structure of P4O10.
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32
Chemical Equations
Chemical change involves a reorganization of the atoms in one or
more substances.
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33
Chemical Equation
A representation of a chemical reaction:
C2H5OH + O2 CO2 + H2O
reactants products
Unbalanced !
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34
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
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35
Methane Reacts with Oxygen to Produce Flame
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37
S.Ex. 3.14
Balance the equation
(NH4)2Cr2O7 (s) → Cr2O3(s) + N2(g) + H2O(g)
N and Cr are balanced; 4 for H2O balances H
It also balances O.
(NH4)2Cr2O7 (s) → Cr2O3(s) + N2(g)+4H2O(g)
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38
Decomposition of Ammonium Dichromate
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39
Decomposition of Ammonium Dichromate
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40
S. Ex.3.15Balance the equation
NH3(g) + O2(g)→ NO (g) + H2O (g)
All molecules are of equal complexity.
Balance H first; 2 for NH3 and 3 for H2O
N can be balanced with a coeff. of 2 for NO
O can be balanced by 5/2 for O2
2NH3(g) + 5/2 O2(g)→ 2NO (g) + 3H2O (g)
x2
4NH3(g) + 5O2(g)→ 4NO (g) + 6H2O (g)
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41
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.
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42
Methanol burns in air according to the equation
2CH3OH + 3O2 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion, what mass of water is produced?
grams CH3OH moles CH3OH moles H2O grams H2O
molar massCH3OH
coefficientschemical equation
molar massH2O
209 g CH3OH1 mol CH3OH
32.0 g CH3OHx
4 mol H2O
2 mol CH3OHx
18.0 g H2O
1 mol H2Ox =
235 g H2O
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43
2CH3OH + 3O2 2CO2 + 4H2O
2 mol 4 mol
2x(12+4+16) g 4x(2+16) g
209 g m
209 x 4(2+16)m =
2(12+4+16)
OR
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44
S. Ex 3.16
What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?
Unbalanced eq.
LiOH(s) + CO2 (g) → Li2CO3 (s) + H2O(l)
Balanced equation
2 LiOH(s) + CO2 (g) → Li2CO3 (s) + H2O(l)
23.95 g/mol 44.0 g/mol
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45
Limiting Reactant
The limiting reactant is the reactant that is consumed first, limiting the
amounts of products formed.
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46
6 green used up6 red left over
Limiting Reagents
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48
Figure 3.12 Hydrogen and Nitrogen React to Form Ammonia
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49
Solving a Stoichiometry Problem
1. Balance the equation.
2. Convert masses to moles.
3. Determine which reactant is limiting.
4. Use moles of limiting reactant and mole ratios to find moles of desired product.
5. Convert from moles to grams.
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50
Limiting Reactant
N2 (g) + 3 H2 (g) → 2 NH3 (g)
25.0kg 5.00 kg ??
8.93x102 2.48x103
8.93x102 mol N2 x 3 mol H2 = 2.68x103 mol H2
1 mol N2
H2 is limiting; 2.48x103 mol H2 x 2 mol NH3 = 1.65x103
3 mol H2
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51
Limiting Reactant Calculations
What mss of molten iron is produced by 1 kg each of the reactants?
Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)1 mol 2 mol6.26mol 37.04 ??
If all Fe2O3(s) reacts 2x6.26 = 12.5 mol Fe -Limiting
If all Al reacts 37.04 mol Fe -- ExcessThe 6.26 mol Fe2O3 will Disappear first
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52
S. Ex 3.182NH3(g) +3CuO(s) → N2(g) +3Cu(s) +3H2O(g)1.06 mol 1.14 mol ??g
If CuO is limiting; 1.14 mol CuO x 1 mol N2
3 mol CuO
= 0.380 mol N2
If NH3 is limiting; 1.06 mol NH3 x 1 mol N2
2 mol NH3
= 0.530 mol N2
Least amount of product obtained with CuO; Cuo is the limiting reactant
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53
Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted. Its amount isCalculated using the balanced equation.
Actual Yield is the amount of product actually obtainedfrom a reaction. It is usually given.
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54
Percent Yield
Actual yield = quantity of product actually obtained
Theoretical yield = quantity of product predicted by stoichiometry
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55
Percent Yield Example
14.4 g
0.104 mol
excess
Actual yield = 6.26 g
1 mol S.A. → 1 mol aspirin0.104 mol S.A. → 0.104 mol aspirin = 18.7 g aspirin
Percent yield = 6.26 g x 100 = 33.5 % 18.7 g
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56
Solving a Stoichiometry Problem Invovling Masses of Reactants and Products
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57
Sample exercise 3.19 - % yield calculation
Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5Kg CO(g) is reacted with 8.60Kg H 2(g). Calculate the theoretical yield of methanol. If 3.57x104g CH3OH is actually produced, what is the percent yield of methanol?
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58
2 H2 (g) + CO(g) → CH3OH (l)
8.60 kg 68.5 kg 35.7 kg(actual yield) 4.27x103mol 2.44x103 mol
If all H2 are converted to methanol,
4.27x103mol H2 x 1 mol methanol = 2.14x103 mol methanol
2 mol H2
If all CO are converted to methanol,2.44x103 mol CO x 1 mol methanol = 2.44x103 mol methanol
1 mol CO H2 is the limiting reactant; theoretical yield = 2.14x103 mol methanol
= 68.6 kg methanol
% yield = actual yield x 100 = 35.7 kg x100 = 52.0%
theoretical yield 68.6 kg
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59
Sample Exercise
Titanium tetrachloride, TiCl4, can be made by combining titanium-containing ore (which is often impure TiO2) with carbon and chlorine -
TiO2(s) + 2 Cl2(g) + C(s) TiCl4(l) + CO2(g)
If one begins with 125 g each of Cl2 and C, but plenty of titanium-containing ore, which is the limiting reagent in the reaction? What quantity of TiCl4 can be produced?
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61
Practice Example 1
A compound contains C, H, N. Combustion of 35.0mg of the compound produces 33.5mg CO2 and 41.1mg H2O. What is the empirical formula of the compound?
Solution:1. Determine C and H, the rest from 33.5mg is N.2. Determine moles from masses.3. Divide by smallest number of moles.
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62
Practice Example 2Caffeine contains 49.48% C, 5.15% H, 28.87% N and 16.49% O by mass and has a molar mass of 194.2 g/mol. Determine the molecular formula.
Solution:1. Convert mass to moles.2. Determine empirical formula.3. Determine actual formula.
C8H10N4O2
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63
Practice Example 3
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of N2 will be formed.
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64
Practice Example 5
SnO2(s) + 2 H2(g) Sn(s) + 2 H2O(l)
a) the mass of tin produced from 0.211 moles of hydrogen gas.
b) the number of moles of H2O produced from 339 grams of SnO2.
c) the mass of SnO2 required to produce 39.4 grams of tin.d) the number of atoms of tin produced in the reaction of 3.00
grams of H2.
e) the mass of SnO2 required to produce 1.20 x 1021 molecules of water.