Download Chapter 3 - Stoichiometry

Ch 3 Stoichiometry
Atomic Masses
 Atomic masses are assigned relative to
12C
which has exactly 12 amu
 Mass spectrophotometry is used to
accurately determine the relative mass of
atoms (see p. 80 for description)
Average Atomic Mass
 Elements occur in nature as a mixture of
isotopes.
 Average Atomic Mass – weighted average
of the masses of the naturally occurring
isotopes of an element
Atomic Mass Example
 In nature carbon is composed of 98.89%
12C
atoms and 1.11% 13C atoms. 12C has
a mass of 12 amu and 13C has a mass of
13.0034 amu. What is the average atomic
mass of carbon?
Mass C = (0.9889)(12 amu) + (0.0111)*(13.0034 amu)
Mass C = 12.01 amu
The Mole
 Mole – number of carbon atoms in exactly
12 g of pure 12C
 1 mole = 6.022 x 1023 units of a substance
 Avogadro’s Number = 6.022 x 1023
 1 mole = atomic mass of an atom in grams
Molar Mass
 Molar Mass – mass in grams of 1 mole of
the compound
EXAMPLE:
Calculate the molar mass of H2O.
H – 1.008 g/mol
O – 16.00 g/mol
Molar mass of H2O = (2(1.008) + 16.00) g/mol
Molar mass of H2O = 18.02 g/mol
Percent Composition
 Mass Percent =
(mass of 1 mole of element in compound/
mass of 1 mole of compound) * 100%
 Mass percent of each component of a
compound should add up to 100%
Percent Composition Example
 What is the mass percent of carbon in
glucose, C6H12O6.
Mass Percent = (mass of 1 mole of element in compound)/
(mass of 1 mole of compound) * 100%
Molar mass C = 12.01 g/mol
Molar mass C6H12O6 = (6(12.01) + 12(1.008) + 6(16.00))
g/mol
Molar mass C6H12O6 = 180.16 g/mol
Mass percent C = ((6*12.01 g/mol)/(180.16 g/mol))*100%
Mass percent C = 40.00 %
Determining the Formula of a
Compound
 Empirical Formula – a formula with the
lowest whole-number ratio of elements in
a compound
 Molecular Formula = (empirical formula)n
where n is an integer
Determining the Formula of a
Compound Example
 A white powder is found to contain 43.64% phosphorus
and 56.36% oxygen by mass. The compound has a
molar mass of 283.88 g/mol. What are the compound’s
empirical and molecular formulas?
In a 100.00 g sample:
43.64 g P (1 mol P/30.97 g P) = 1.409 mol P
56.36 g O (1 mol O/16.00 g O) = 3.523 mol O
Divide both by smaller value to get
1.409/1.409 = 1 P and 3.523/1.409 = 2.5 O
Multiply both by 2 to get whole numbers to get an empirical formula of
2 P and 5 O or P2O5 with a molar mass of 141.94 g/mol
Since molar mass of compound was 283.88 g/mol, which is twice as
much as 141.94, the molecular formula is (P2O5)2 or P4O10
Chemical Equations
 Chemical equation – an expression
representing a chemical reaction; the
formulas of the reactants (on the left) are
connected by an arrow with the products
(on the right); states of matter are also
given
Example:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Reactants
Yield
Products
Balancing Chemical Equations
 Chemical equations must adhere to the
Law of Conservation of Mass
 Whole number coefficients are added in
front of chemical formulas to balance
atoms on each side
 Chemical formulas are never changed!
Balancing Chemical Equations
Example
Balance the following chemical equation:
C2H5OH(l) + O2(g) → CO2(g) + H2O(g)
Reactants – 2 C, 6 H, 3 O
Products – 1 C, 2 H, 3 O
Add coefficients:
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
Reactants – 2 C, 6 H, 7 O
Products – 2C, 6 H, 7 O
Mass is conserved and the equation is balanced.
Stoichiometric Calculations
 Stoichiometry – portion of chemistry
dealing with numerical relationships in
chemical reactions; the calculation of
quantities of a substances involved in
chemical equations
 A means to determine quantities of
reactants or products based on the
amount of other species present in a
reaction
Mole Ratio
 Mole ratio – factor used in stoichiometry to
convert moles of one substance to moles
of another substance using the coefficients
from the balanced chemical equation
Steps Used in Stoichiometry
Step 1: balance the equation for the reaction
Step 2: convert known mass of reactants or
products into moles of that substance
Step 3: use the balanced chemical equation
to set up the appropriate mole ratios
Step 4: use the appropriate mole ratio to
calculate the number of moles of the desired
reactant or product
Step 5: convert moles to proper units required by
problem
Stoichiometry Example
5.00 grams of solid magnesium reacts with
hydrochloric acid to produce hydrogen gas
and magnesium chloride. How many grams
of hydrochloric acid must be present to
ensure the complete reaction of magnesium
in this reaction?
Solution
Write balanced chemical equation:
Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)
Convert known value to moles:
5.00 g Mg (1 mol Mg/24.31 g) = 0.206 mol Mg
Set-up appropriate mole-ratio and find desired moles:
0.206 mol Mg (2 mol HCl/1 mol Mg) = 0.412 mol HCl
Covert to proper units required by problem:
0.412 mol HCl (36.46 g HCl/1mol HCl) = 15.0 g HCl
Limiting Reactant
 Limiting reactant – reactant that is
consumed first in a chemical reaction
 The amount of product formed is limited
when one reactant is completely
consumed before the others
 It is essential to determine which reactant
is the limiting reactant when using
stoichiometry to determine amount of
products formed in a reaction
How is a Limiting Reactant
Calculated?
Step 1: write and balance the equation for
the reaction
Step 2: determine moles of reactants from
the known masses
Step 3: use mole ratio to determine which
reactant is limiting
Limiting reactant is then used to calculate
moles of product formed.
Limiting Reactant Example
 10.00 g of O2 reacts with 20.00 g of Cu metal to form
copper (II) oxide. What is the limiting reactant?
Write balanced chemical equation:
O2(g) + 2Cu(s) → 2CuO (s)
Determine moles of reactant:
10.00 g O2 (1 mol O2/32.00 g O2) = 0.3125 mol O2
20.00 g Cu (1 mol Cu/63.55 g Cu) = 0.3147 mol Cu
Use mole ratio to determine which reactant is limiting:
0.3125 mol O2(2 mol Cu/1 mol O2) = 0.625 mol Cu
0.625 mol Cu>0.3147 mol Cu and Cu is limiting