Transcript 2.50 g O 2 1 mol O 2 31.99 g O 2

Stoichiometry
“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.”
Sherlock Holmes, in Sir Arthur Conan
Doyle’s A Study in Scarlet
The Mole
1 dozen = 12
1 gross = 144
1 ream = 500
1 mole = 6.022 x 1023
There are exactly 12 grams of
carbon-12 in one mole of carbon-12.
Avogadro’s Number
6.022 x 1023 is called “Avogadro’s Number” in
honor of the Italian chemist Amadeo Avogadro
(1776-1855).
I didn’t discover it. Its
just named after me!
Amadeo Avogadro
Calculations with Moles:
Converting moles to grams
How many grams of lithium are in 3.50 moles of
lithium?
3.50 mol Li
6.94 g Li
1 mol Li
=
45.1
g Li
Calculations with Moles:
Converting grams to moles
How many moles of lithium are in 18.2 grams of
lithium?
18.2 g Li
1 mol Li
6.94 g Li
=
2.62
mol Li
Calculations with Moles:
Converting grams to moles
How many moles of lithium are in 18.2 grams of
lithium?
18.2 g Li
1 mol Li
6.94 g Li
=
2.62
mol Li
Calculations with Moles:
Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of
lithium?
3.50 mol Li 6.022 x 1023 atoms Li
1 mol Li
= 2.11 x 1024 atoms Li
Calculations with Moles:
Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li 1 mol Li
6.94 g Li
6.022 x 1023 atoms Li
1 mol Li
(18.2)(6.022 x 1023)/6.94
= 1.58 x 1024 atoms Li
Calculating Formula Mass
Calculate the formula mass of magnesium carbonate,
MgCO3.
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
From previous slide:
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
 24.31 
Mg  
  100  28.83%
 84.32 
 12.01 
C 
  100  14.24%
 84.32 
 48.00 
O
  100  56.93%
 84.32 
100.00
Formulas
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.
 molecular formula = (empirical
formula)n [n = integer]
 molecular formula = C6H6 = (CH)6
 empirical formula = CH
Formulas
(continued)
Formulas for ionic compounds are ALWAYS
empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas
(continued)
Formulas for molecular compounds MIGHT
be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
Empirical:
H2O
CH2O
C12H22O11
C12H22O11
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.
Empirical Formula Determination
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 g C 1 mol C   4.107 mol C
12.01 g C 
 6.85g H 1 mol H   6.78 mol H
1.01 g H 
 43.84 g O 1 mol O   2.74 mol O
16.00 g O 
Empirical Formula Determination
(part 2)
Divide each value of moles by the smallest of the
values.
4.107
mol
C
Carbon:
 1.50
2.74 mol O
6.78 mol H
Hydrogen:
 2.47
2.74 mol O
2.74 mol O
Oxygen:
 1.00
2.74 mol O
Empirical Formula Determination
(part 3)
Multiply each number by an integer to obtain all
whole numbers.
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Oxygen: 1.00
x 2
2
Empirical formula: C3H5O2
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular
formula of adipic acid?
2. Divide the molecular mass by the
mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
146
 2 (C3H5O2) x 2 = C6H10O4
73
Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2  2CO2 + 3H2O
reactants
products
When the equation is balanced it has quantitative
significance:
1 mole of ethanol reacts with 3 moles of oxygen
to produce 2 moles of carbon dioxide and 3 moles
of water
Solving a Stoichiometry Problem
1.
2.
3.
4.
Balance the equation.
Convert masses to moles.
Determine which reactant is limiting.
Use moles of limiting reactant and mole
ratios to find moles of desired product.
5. Convert from moles to grams.
Working a Stoichiometry Problem
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.
4 Al
+ 3 O2
2 Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?
Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4 Al
6.50 g Al
+
3 O2  2Al2O3
1 mol Al
2 mol Al2O3 101.96 g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
= ? g Al2O3
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al2O3
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the amounts of
products formed.
More of limiting reagents
% yield = Measured value
Calculated value
x100
• 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?
From previous calculation calculated value is 12.3 g of Al2O3
11.7 g
12.3 g
x 100
= 95.1 % yield
Limiting Reagent Example Problem
6.50 grams of aluminum reacts with 2.50 g of
Oxygen gas. How many grams of aluminum oxide are formed?
4 Al + 3 O2  2Al2O3
6.50 g Al
1 mol Al
Need
= 0.241 mol Al
26.98 g Al
2.50 g O2
1 mol O2
31.99 g O2
Have
4Al = 1.33
3 O2
1
= 0.0781
mol O2
0.241 = 3.08
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.
4 Al + 3 O2  2Al2O3
0.078 mol O2
2 mol Al2O3 101.96 g Al2O3
3 mol O2
1 mol Al2O3
= 5.31 ? g Al2O3
Aluminum is the excess reagent.
How would you find the amount of excess aluminum present ?.