stoichiometry and the mole chapter 8 what is stoichiometry? quantitative aspects of chemistry ...
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Stoichiometry and the Stoichiometry and the molemole
Chapter 8Chapter 8
What is stoichiometry?What is stoichiometry?
Quantitative aspects of chemistryQuantitative aspects of chemistry Stoicheon Greek root (element)Stoicheon Greek root (element) Metron Greek root( to measure)Metron Greek root( to measure) Calculate how much of a reactant Calculate how much of a reactant
needed to produce a product or how needed to produce a product or how much product could be expectedmuch product could be expected
StoichiometryStoichiometry
How much do I want?How much do I want? How much do I have?How much do I have? How much will I get?How much will I get?
The Mole•Defined as the number of carbon atoms in exactly 12 grams of carbon-12. •1 mole is 6.02 x 1023 particles. •Treat it like a very large dozen •6.02 x 1023 is called Avogadro's number.
The MoleThe Mole
One mole = 6.022 x 10One mole = 6.022 x 1023 23 (Avogadro’s (Avogadro’s number)number)
Avogadro’s hypothesisAvogadro’s hypothesis
When all of the following are the same:When all of the following are the same: Volumes of containersVolumes of containers Temperature of the gasesTemperature of the gases The pressure exerted by and on each The pressure exerted by and on each
gasgas
Then, the number of molecules in each Then, the number of molecules in each container will be the same, too. container will be the same, too. However, the masses are not equal.However, the masses are not equal.
NN2(g)2(g) + 3H + 3H2(g)2(g)2NH2NH3(g)3(g)
11 33 22
1 dozen1 dozen
(12)(12)3 dozen3 dozen
(36)(36)2 dozen2 dozen
(24)(24)
1 gross1 gross
(144)(144)3 gross3 gross
(432)(432)2 gross2 gross
(288)(288)
1mole1mole(6.02x10(6.02x102323))
3 moles3 moles(18.06x10(18.06x102323))
2 moles2 moles(12.04 x10(12.04 x102323))
Representative particles
•The smallest pieces of a substance.
•For an element it is an atom. –Unless it is diatomic
•For a molecular compound it is a molecule.
•For an ionic compound it is a formula unit.
Molar MassMolar Mass
A substance’s A substance’s molar mass molar mass (molecular weight) is the mass in (molecular weight) is the mass in grams of one mole of the compound. grams of one mole of the compound.
COCO22 = 44.01 grams per mole = 44.01 grams per mole
HH22O = 18.02 grams per moleO = 18.02 grams per mole
Ca(OH)Ca(OH)22 = 74.10 grams per mole = 74.10 grams per mole
Chemical EquationsChemical Equations
Chemical change involves a Chemical change involves a reorganization of reorganization of
the atoms in one or more substances.the atoms in one or more substances.CC22HH55OH + 3OOH + 3O22 2CO2CO22 + 3H + 3H22OO
reactantsreactants productsproducts
11 mole of ethanol mole of ethanol reacts with reacts with 33 moles of moles of oxygen oxygen
to produce to produce 22 moles of carbon dioxide moles of carbon dioxide and and 33 moles of watermoles of water
When the equation is balanced it has When the equation is balanced it has quantitative significance:quantitative significance:
Mole Mole RelationsRelations
Calculating Masses of Reactants Calculating Masses of Reactants and Productsand Products
1.1. Balance the equation. Balance the equation.
2.2. Convert mass to moles. Convert mass to moles.
3.3. Set up mole ratios. Set up mole ratios.
4.4. Use mole ratios to calculate moles Use mole ratios to calculate moles of desired substance. of desired substance.
5.5. Convert moles to grams, if Convert moles to grams, if necessary.necessary.
Working a Stoichiometry Working a Stoichiometry ProblemProblem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.
1. Identify reactants and products and write the balanced equation.
Al + O2 Al2O3
b. What are the reactants?
a. Every reaction needs a yield sign!
c. What are the products?
d. What are the balanced coefficients?
4 3 2
How to solve stoichiometry How to solve stoichiometry problems?problems?
Write the balanced chemical equationWrite the balanced chemical equation Convert to moles information on Convert to moles information on
reactants and productsreactants and products Use mole ratios from equation to Use mole ratios from equation to
determine number of moles of determine number of moles of unknownunknown
Convert moles to unit desiredConvert moles to unit desired g xg xmoles xmoles xmoles ymoles yg yg y
Working a Stoichiometry Working a Stoichiometry ProblemProblem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?
4 Al + 3 O2 2Al2O3
=6.50 g Al
? g Al2O3
1 mol Al
26.98 g Al 4 mol Al
2 mol Al2O3
1 mol Al2O3
101.96 g Al2O3
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =
12.3 g Al2O3
Percent yieldPercent yield
The % yield is calculated from:The % yield is calculated from:
(actual yield / theoretical yield) x 100(actual yield / theoretical yield) x 100 The theoretical yield is how much The theoretical yield is how much
product is predicted from balanced product is predicted from balanced chemical equation.chemical equation.
The actual yield is how much is The actual yield is how much is recovered when actual experiment is recovered when actual experiment is conducted.conducted.
Limiting Limiting ReactantReactant
The The limiting reactantlimiting reactant is the reactant is the reactant
that is that is consumed firstconsumed first,, limiting the amounts of limiting the amounts of products formed.products formed.
Limiting reactantLimiting reactant
Reactant present in short supplyReactant present in short supply
Excess reactant•Reactant in excess relative to limiting reactant
FormulasFormulas
molecular formula = (empirical molecular formula = (empirical formula)formula)nn [ [nn = integer] = integer]
molecular formula = Cmolecular formula = C66HH66 = = (CH)(CH)6 6
empirical formula = CHempirical formula = CH
Empirical formula: the lowest whole number ratio of atoms in a compound.
Molecular formula: the true number of atoms of each element in the formula of a compound.
FormulasFormulas (continued)(continued)
Formulas for Formulas for ionic compoundsionic compounds are are ALWAYSALWAYS empirical (lowest whole empirical (lowest whole number ratio).number ratio).Examples:Examples:
NaCl MgCl2 Al2(SO4)3 K2CO3
FormulasFormulas (continued)(continued)
Formulas for Formulas for molecular compoundsmolecular compounds MIGHTMIGHT be empirical (lowest whole be empirical (lowest whole number ratio).number ratio).
Molecular:Molecular:
H2O
C6H12O6 C12H22O11
Empirical:
H2O
CH2O C12H22O11
Empirical Formula Empirical Formula DeterminationDetermination
1.1. Base calculation on 100 grams of Base calculation on 100 grams of compound. compound.
2.2. Determine moles of each element in 100 Determine moles of each element in 100 grams of compound. grams of compound.
3.3. Divide each value of moles by the Divide each value of moles by the smallest of the values. smallest of the values.
4.4. Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.
Empirical Formula Empirical Formula DeterminationDetermination
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?
49.32 14.107
12.01
g C mol Cmol C
g C
6.85 16.78
1.01
g H mol Hmol H
g H
43.84 12.74
16.00
g O mol Omol O
g O
Empirical Formula Empirical Formula DeterminationDetermination
(part 2)(part 2)
4.1071.50
2.74
mol C
mol O
6.782.47
2.74
mol H
mol O
2.741.00
2.74
mol O
mol O
Divide each value of moles by the smallest Divide each value of moles by the smallest of the values.of the values.
Carbon:Carbon:
Hydrogen:Hydrogen:
Oxygen:Oxygen:
Empirical Formula Empirical Formula DeterminationDetermination
(part 3)(part 3)Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.
Carbon: 1.50Carbon: 1.50 Hydrogen: 2.50Hydrogen: 2.50 Oxygen: 1.00Oxygen: 1.00x 2 x 2 x 2
33 55 22
Empirical formula:C3H5O
2
Finding the Molecular Finding the Molecular FormulaFormula
The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?
1. Find the formula mass of 1. Find the formula mass of CC33HH55OO22
3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g
Finding the Molecular Finding the Molecular FormulaFormula
The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?
3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g
2. Divide the molecular mass by 2. Divide the molecular mass by the mass given by the emipirical the mass given by the emipirical formula.formula.
1462
73
Finding the Molecular Finding the Molecular FormulaFormula
The empirical formula for adipic acid The empirical formula for adipic acid is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?
3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g146
273
3. Multiply the empirical formula by 3. Multiply the empirical formula by this number to get the molecular this number to get the molecular formula.formula.
(C(C33HH55OO22) x 2 ) x 2 ==
CC66HH1010OO44
Combustion AnalysisCombustion Analysis Technique that requires burning of an unknown substance and trap Technique that requires burning of an unknown substance and trap
the gases from burning.the gases from burning.
Used to determine molecular Used to determine molecular formula of an unknownformula of an unknown
Combustion= burningCombustion= burning Using oxygen Using oxygen
Points about combustionPoints about combustion Element that makes unknown almost always contain carbon Element that makes unknown almost always contain carbon
and hydrogen. Oxygen is often involved and nitrogen is and hydrogen. Oxygen is often involved and nitrogen is involved sometimes.involved sometimes.
Must know mass of the unknown substance before burning Must know mass of the unknown substance before burning itit
Unknown will be burnt in pure oxygen, present in excessUnknown will be burnt in pure oxygen, present in excess Carbon dioxide and water are the productsCarbon dioxide and water are the products All the carbon winds up as carbon dioxide and all the All the carbon winds up as carbon dioxide and all the
hydrogen winds up as waterhydrogen winds up as water The end result will be to determine the empirical formula of The end result will be to determine the empirical formula of
the substancethe substance