Upload
leslie-nelson
View
217
Download
2
Tags:
Embed Size (px)
Citation preview
StoichiometryStoichiometry
Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet
“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”
The MoleThe Mole
1 dozen =1 gross =
1 ream =
1 mole =
12
144
500
6.022 x 1023
There are exactly 12 grams of carbon-12 in one mole of carbon-12.
Avogadro’s NumberAvogadro’s Number6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855).
Amadeo Avogadro
I didn’t discover it. Its just named after
me!
Calculations with Moles:Calculations with Moles:Converting moles to gramsConverting moles to grams
How many grams of lithium are in 3.50 moles of lithium?
3.50 mol Li= g Li
1 mol Li
6.94 g Li45.1
Calculations with Moles:Calculations with Moles:Converting grams to molesConverting grams to moles
How many moles of lithium are in 18.2 grams of lithium?
18.2 g Li= mol Li
6.94 g Li
1 mol Li2.62
Calculations with Moles:Calculations with Moles:Converting grams to molesConverting grams to moles
How many moles of lithium are in 18.2 grams of lithium?
18.2 g Li= mol Li
6.94 g Li
1 mol Li2.62
Calculations with Moles:Calculations with Moles:Using Avogadro’s NumberUsing Avogadro’s Number
How many atoms of lithium are in 3.50 moles of lithium?
3.50 mol Li = atoms Li
1 mol Li
6.022 x 1023 atoms Li 2.11 x 1024
Calculations with Moles:Calculations with Moles:Using Avogadro’s NumberUsing Avogadro’s Number
How many atoms of lithium are in 18.2 g of lithium?
18.2 g Li
= atoms Li
1 mol Li 6.022 x 1023 atoms Li
1.58 x 1024
6.94 g Li 1 mol Li
(18.2)(6.022 x 1023)/6.94
Calculating Formula MassCalculating Formula Mass
Calculate the formula mass of magnesium Calculate the formula mass of magnesium carbonate, MgCOcarbonate, MgCO33..
24.31 g + 12.01 g + 3(16.00 g) 24.31 g + 12.01 g + 3(16.00 g) ==
84.32 g84.32 g
Calculating Percentage Calculating Percentage CompositionComposition
Calculate the percentage composition of Calculate the percentage composition of magnesium carbonate, MgCOmagnesium carbonate, MgCO33..
From previous slide:From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g84.32 g 24.31
100 28.83%84.32
Mg 12.01
100 14.24%84.32
C 48.00
100 56.93%84.32
O
100.00
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 forFormulas 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 forFormulas 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 valuesof 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 (C(C33HH55OO22) x 2 =) x 2 = CC66HH1010OO44
ReviewReview: Chemical : Chemical EquationsEquations
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 withreacts with 33 moles of moles of oxygen oxygen
to produceto produce 22 moles of carbon dioxide moles of carbon dioxide andand 33 moles of watermoles of water
When the equation is balanced it has When the equation is balanced it has quantitative significance:quantitative significance:
Solving a Stoichiometry Solving a Stoichiometry ProblemProblem
1.1. Balance the equation. Balance the equation.
2.2. Convert masses to moles. Convert masses to moles.
3.3. Determine which reactant is limiting. Determine which reactant is limiting.
4.4. Use moles of limiting reactant and Use moles of limiting reactant and mole ratios to find moles of desired mole ratios to find moles of desired product. product.
5.5. Convert from moles to grams.Convert from moles to grams.
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
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
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.
More of limiting reagents
% yield = Measured value x100
Calculated value• A student reacts 6.50 grams of Aluminum
with excess oxygen and measures 11.7 g of Al2O3 is produced. What is the students % yield of aluminum oxide?
11.7 g x 100 = 95.1 % yield12.3 g
From previous calculation calculated value is 12.3 g of Al2O3
6.50 grams of aluminum reacts with 2.50 g of Oxygen gas. How many grams of aluminum oxide are formed?
Limiting Reagent Example Problem
4 Al + 3 O2 2Al2O3
6.50 g Al 1 mol Al
26.98 g Al
= 0.241 mol Al
2.50 g O2 1 mol O2
31.99 g O2
= 0.0781 mol O2
Need Have
4Al = 1.33 0.241 = 3.083 O2 1 0.078 1
This shows we have excess Al and Oxygen is the limiting reagent. We use the moles of oxygen (limiting reagent) to find the mass of aluminum oxide that can be formed.
= ? g Al2O3
3 mol O2
2 mol Al2O3
1 mol Al2O3
101.96 g Al2O3
4 Al + 3 O2 2Al2O3
0.078 mol O2
Aluminum is the excess reagent. How would you find the amount of excess aluminum present ?.
5.31