Transcript AP Chem Cpt 3

Stoichiometry
Calculations with Chemical
Formulas and Equations
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
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.
The Mole Highway
Form Mass 1 mol
22.4 L
x
x
x
1 mol
Form Mass
1 mol
Mass (g)
x
1 mol
22.4 L
Vol (L)
6.02x1023
Rep Part
1 mol
x
x
6.02x1023
1 mol
Rep Part
Rep Part
Calculating Formula Mass
Calculate the formula mass of magnesium carbonate,
MgCO3.
24.3 g + 12.0 g + 3(16.0 g) =
84.3g
Calculations with Moles:
Converting moles to grams
How many grams of lithium are in 3.50 moles of
lithium?
3.50 mol Li
6.9 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
6.9 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:
Two-Step Problem
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li 1 mol Li
6.02 x 1023 atoms Li
6.9 g Li
(18.2)(6.02 x 1023)/6.9
1 mol Li
= 1.58 x 1024 atoms Li
Calculating Percentage Composition
Calculate the percentage composition of magnesium
carbonate, MgCO3.
From previous slide:
24.3 g + 12.0 g + 3(16.0 g) = 84.3 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
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Empirical Formula Determination
Rules
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
Example
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(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g
Finding the Molecular Formula
(continued)
2. Divide the molecular mass by the
mass given by the emipirical formula.
3(12.0 g) + 5(1.0) + 2(16.0) = 73.0 g
146
2
73
3. Multiply the empirical formula by this
number to get the molecular formula.
(C3H5O2) x 2 = C6H10O4
Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2  2CO2 + 3H2O
reactants
products
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. Write and balance the equation.
2. Convert masses to moles.
3. Determine which reactant,if any, is
limiting.
4. Use moles of limiting reactant and mole ratios
to find moles of desired product.
5. Convert from moles to desired units. It must
be grams if percent yield is to be calculated.
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
(continued)
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  2 Al2O3
1 mol Al
2 mol Al2O3 102.0 g Al2O3
27.0 g Al
4 mol Al
1 mol Al2O3
(6.50 x 2 x 102.0) ÷ (27.0 ÷ 4) =
= ? g Al2O3
12.3 g Al2O3
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the amounts of products
formed.