Molecular Mass and Formula Mass
◆
for molecular compounds:
the molecular mass is the mass (in amu) of one
molecule of the compound
molecular mass = ∑atomic masses of elements present
Chapter 3:
ex. P2O5 molecular mass = 2(30.97 amu) + 5(16.00 amu)
= 141.94 amu
Calculations
with Chemical
Formulas and
Equations
◆
for ionic compounds:
the formula mass is the mass (in amu) of one formula
unit of the compound
formula mass = ∑atomic masses of elements present
ex. Ca(ClO3)2 formula mass = 40.08 amu + 2(35.45 amu)
+ 6(16.00 amu)
= 206.98 amu
The Mole
◆
◆
Using NA as a Conversion Factor
SI base unit for amount of substance
the mole is a counting unit
1 mole contains the same number of particles as
there are atoms in exactly 12 g of 12C.
1023
by definition: 1mol = 6.022 x
particles
Avogadro’s number, NA = 6.022 x 1023 mol!1
so: 1 mol Li = 6.022 x 1023 Li atoms
◆
NA relates amount of substance in mol to
number of particles*
* particles may be atoms, molecules, formula units or
ions depending on sample under consideration
examples:
Calculate the amount (in mol) of molybdenum in
a sample containing 6.295 x 1025 atoms Mo.
1 mol H2O = 6.022 x 1023 H2O molecules
1 mol NaNO3 = 6.022 x 1023 NaNO3 formula units
Calculate the number of SO3 molecules in
1.25 mol SO3.
Molar Mass
What Does 1 Mole Look Like?
◆
◆
Molar Mass of an Element - mass (in g) per mole
of an element.
units g/mol
Molar Mass of a Compound - mass (in g) per
mole of a compound.
units g/mol
molar mass compound = ∑molar mass of elements
present
Molar Mass of an Element
Molar Mass of an Element
◆
◆
calculate the molar mass of an element:
ex: Determine the mass (in g) of 0.775 mol Sn.
convert average atomic mass (in amu/atom) to
molar mass (in g/mol)
ex: Determine the molar mass of rhodium.
use molar mass as a conversion factor relating
sample mass (in g) to amount of substance (in mol)
◆
combine molar mass and NA as conversion factors ...
ex: Determine the number of atoms in 105.7 g Sn.
ex: Determine the mass (in g) of a single atom of Sn.
Molar Mass of a Compound
◆
◆
Molar Mass of a Compound - mass (in g) per
mole of a compound.
units g/mol
◆
molar mass compound = ∑molar mass of elements
present
What relationships exist between ...
mass in g
mass in amu
amount in mol
# of formula units
# of calcium ions
# of nitrate ions
mol of calcium ion
mol of nitrate ion
#’s of Ca, N, and O atom
mol of Ca, N, and O
ex: Calculate the number of molecules in
10.0 mg PCl5.
ex: Calculate the mass (in g) of a single
PCl5 molecule.
Compound Stoichiometry Calculations
a. molar mass
b. mol Ca(NO3)2
c. mol Ca2+
d. mol NO3!
e. mass (in g) of O
f. number of NO3! ions
g. number of formula units
h. number of O atoms
i. mass (in g) of 1 formula unit
use these ratios as conversion factors in
calculations
ex: Consider calcium nitrate. Write the
chemical formula.
ex: Calculate the molar mass of phosphorus
pentachloride.
Consider a 25.0 g sample of Ca(NO3)2.
Determine the following:
Compound Stoichiometry
the subscripts in a chemical formula tell us the
ratio of how elements combine
◆
Mass Percent Composition
composition of compound with respect to percent
mass of each element present
consider AxBy:
x(molar mass A)
mass% A = ———————— x 100
molar mass AxBy
y(molar mass B)
mass% B = ———————— x 100
molar mass AxBy
and: mass % A + mass % B = 100%
ex: Determine the mass % composition of
barium nitrate.
Chemical Formulas:
Empirical Formula vs. Molecular Formula
empirical formula:
relative numbers of atoms present
subscripts in chemical formula are smallest
possible whole number ratio
molecular formula:
exact numbers of atoms present
subscripts in chemical formula may or may not
be smallest possible whole number ratio
Determination of Empirical and Molecular Formulas:
Using Mass Percent Composition Data
ex: Vitamin C (molar mass = 176.1 g/mol) is
composed of 40.9% C, 4.57% H, and 54.5% O
by mass. Determine the empirical and
molecular formula of Vitamin C.
1. Determine the mol of each element present.
◆
assume a 100 g sample and use mass %
composition to interpret g of each element
present
◆
convert g element ! mol element
2. Determine the mol ratio of elements present
in the smallest whole numbers.
Determination of Empirical and Molecular Formulas:
Using Mass Percent Composition Data (Cont’d)
ex: Vitamin C (molar mass = 176.1 g/mol) is
composed of 40.9% C, 4.57% H, and 54.5% O
by mass. Determine the empirical and
molecular formula of Vitamin C.
3. Divide the actual molar mass of the compound
by the molar mass of the empirical formula.
This will equal a small whole number, n.
4. Multiply the subscripts of the empirical formula
by n ! molecular formula
Another Formula Determination Example:
Stoichiometry of Chemical
Reactions
A 500.0 mg sample of a compound composed of
only cadmium and phosphorus is analyzed and
found to contain 422.5 mg Cd.
Answer questions like:
Determine the empirical and molecular formulas of
this compound. The actual molar mass of the
compound is 399.1 g/mol.
Consider the reaction:
CaO (s) + 3 C (s) " CaC2 (s) + CO (g)
56.08 g/mol
12.01 g/mol
64.01 g/mol
28.01 g/mol
Calculate the mass (in g) of C required to react
completely with 155.2 g CaO.
How many g of reactant B are
required to react completely
with x g of reactant A?
◆
How many g of product C can
be formed from the complete
reaction of x g of reactant A?
Reaction Yields
theoretical yield: maximum amount of product that
can form in a reaction; typically reported in g or
mol
◆
◆
◆
Calculate the mass (in g) of CaC2 that can form
from the complete reaction of 155.2 g CaO.
◆
1
actual or experimental yield: actual amount of
product collected from a reaction
percent yield: percentage of the theoretical yield
that is actually collected
actual yield
% yield = ——————— " 100
theoretical yield
Identification of the Limiting Reactant:
(a not very scientific example)
Identification of the Limiting Reactant
The limiting reactant is the reactant that limits the
theoretical yield of product.
◆
◆
it is the reactant that is completely
consumed during the course of the chemical
reaction
as soon as one of the reactants is gone, the
reaction must stop!
TO CALCULATE A THEORETICAL YIELD,
YOU MUST ALWAYS START WITH THE
LIMITING REACTANT!!
to build a perfect s’more
you need:
2 graham crackers
2 marshmallows
1 square of chocolate
Consider 2 sets of ingredients:
A
B
50 graham crackers
60 graham crackers
50 marshmallows
75 marshmallows
25 squares chocolate
22 squares chocolate
Identification of the Limiting Reactant:
(a not very scientific example)
Identification of Limiting Reactant:
Calculations and Reaction Tables
set up s’more reaction tables:
2 graham
crackers
+
2 marshmallows
CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)
+
1 square
chocolate
"
1 s’more
before
50
50
25
0
change
-50
-50
-25
+25
after
0
0
0
25
2 graham
crackers
+
2 marshmallows
+
1 square
chocolate
"
1 s’more
before
60
75
22
0
change
-44
-44
-22
+22
after
16
31
0
22
100 g CaC2 and 100 g H2O are combined and
allowed to react.
◆
identify the limiting reactant
◆
calculate the theoretical yield of products
◆
◆
determine the mass of the excess reactant that
remains unconsumed when the reaction is
complete
complete the reaction table
CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)
CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)
Calculations to fill in the reaction table:
◆
determine mol of each reactant
Calculations to fill in the reaction table:
pick one reactant; calculate mol of 2nd reactant
required to react completely with the 1st reactant
◆
2 mol H2O
1.56 mol CaC2 " –––––––––––
= 3.12 mol H2O
1 mol CaC2
1 mol CaC2
100 g CaC2 " ––––––––––– = 1.56 mol CaC2
64.01 g
or conversely
1 mol H2O
18.02 g
100 g H2O " ––––––––––– = 5.55 mol H2O
1 mol CaC2
5.55 mol H2O " ––––––––––– = 2.78 mol CaC2
2 mol H2O
CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)
Calculations to fill in the reaction table:
◆
2nd
compare mol of
reactant required for complete
reaction to mol present in the system
CaC2 (s)
+
2 H2O (l)
"
Ca(OH)2 (s)
+
C2H2 (g)
before
1.56 mol
5.55 mol
0
0
change
!1.56 mol
!3.12 mol
+1.56 mol
+1.56 mol
after
0
2.43 mol
1.56 mol
1.56 mol
here - more H2O is present (5.55 mol) than is
required (3.12 mol) to completely consume CaC2
∴ H2O is the excess reactant
CaC2 (s)
OR
there is less CaC2 present (1.56 mol) than is
required (2.78 mol) to completely consume H2O
present
∴ CaC2 is the limiting reactant
+
2 H2O (l)
"
Ca(OH)2 (s)
+
C2H2 (g)
before
1.56 mol
5.55 mol
0
0
change
!1.56 mol
!(2∕1)1.56 mol
+(1∕1)1.56 mol
+(1∕1)1.56 mol
after
0
2.43 mol
1.56 mol
1.56 mol
CaC2 (s) + 2 H2O (l) " Ca(OH)2 (s) + C2H2 (g)
Calculations based on the reaction table:
◆
determine the theoretical yield of each product
Homework Problem: 3.95
3 TiO2 (s) + 4 C (s) + 6 Cl2 (g) ! 3 TiCl4 (g) + 2 CO2 (g) + 2 CO (g)
4.15 g
74.10 g
1 mol Ca(OH)2
1.56 mol Ca(OH)2 " –––––––––––– = 116 g Ca(OH)2
26.04 g
1.56 mol C2H2 " ––––––––––– = 40.6 g C2H2
1 mol C2H2
3 TiO2
before
change
after
◆
determine the mass of excess reactant unconsumed
when the reaction is complete
18.02 g
1 mol H2O
2.43 mol H2O " ––––––––––– = 43.8 g H2O
A Final Reaction Stoichiometry Example:
Gravimetric Analysis
A 27.2569 g sample containing some unknown amount
of potassium superoxide (KO2) and other substances is
treated with excess H2O; the following reaction occurs*:
4 KO2 (s) + 2 H2O (l) " 4 KOH (s) + 3 O2 (g)
*note: of the substances present in the original sample,
only the KO2 will react with water
Assuming that the you achieve complete conversion of
KO2 to KOH, determine the mass percent KO2 in the
original sample if the experiment yields13.6247 g KOH.
5.67 g
6.78 g
79.88 g/mol 12.01 g/mol 70.90 g/mol
0.0519 mol
+
4C
0.472 mol
+
6 Cl2
0.0956 mol
!
3 TiCl4
0
+
2 CO2
0
+
2 CO
0