Mathematical
Chemistry
Chemical Mathematics
Chapter 7 Homework (II)
Monday, November 3rd
 Pgs. 236-238
 Problems 30, 33, 40, 45, 46, 47,
50
 Due
Molar Mass
We will learn how to…
Calculate
the molar
mass
Calculate the percent
composition by mass
of a compound.
A really important formula!
mass(g)
number of moles 
MM
Practice Problems
 What
is the molar mass of Copper
(II) nitrate.
 187.566 g
 What is the mass in grams of 6.25
mole of Copper (II) nitrate.
 1170 g.
(cont.)
 How
many molecules are
contained in:
 25.0 g H2SO4?
 125 g. of sugar (C12H22O11)
1.53 x 1023 molecules of Sulfuric Acid
2.20 x 1023 molecules of Sugar
Percent
Composition
Percent Composition
The percent composition
of a compound is the
percent by mass of each
element in the
compound.
Percent Composition
mass of element
x 100 
molar mass of compound
% of element
Practice
Find the percent
composition to 4
significant figures...
PbCl2
74.50% Pb and 25.50% Cl
What’s the percent water in…
ZnSO4•7H2O
43.85%
Find the percent
composition of
(NH4)2CO3?
N=29.15%, H=8.39%,
C=12.50%,
O=49.95%
Practice Problem 7-9
Ibuprofen, C13H18O2, is the active
ingredient in many nonprescription
pain relievers. Its molar mass is
206.29 g/mol.
 If the tablets in a bottle contain a
total of 33 g. of ibuprofen, how
many moles of ibuprofen are in the
bottle?
0.16 mol
Practice Problem 7-9
Ibuprofen, C13H18O2, is the active
ingredient in many nonprescription
pain relievers. Its molar mass is
206.29 g/mol.
 How many molecules of ibuprofen
are in the bottle?
9.6 x 1022 Molecules
Practice Problem 7-9
Ibuprofen, C13H18O2, is the active
ingredient in many nonprescription
pain relievers. Its molar mass is
206.29 g/mol.
 What is the total mass in grams of
carbon in 33 g. of ibuprofen?
25 g
7.4 Determining Chemical
Formulas
The simplest formula or
empirical formula consists
of the symbols for the
elements combined with
subscripts showing the
smallest whole-number ratio
of the atoms.
Example
 Diborane
has a molecular
formula and in reality exists
as B2H6.
 The empirical formula is
BH3.
 Both are 78% Boron and
22% Hydrogen.
Simplest Formula (from
weight percents.)
 Step
1: Take percentages and
convert to a total of 100 g.
 Diborane is 78% by weight
Boron and 22% by weight
hydrogen.
 100 g would be 78 g. of Boron
and 22 g. of Hydrogen.
Simplest Formula (cont.)
 Step
2: Determine moles.
1 mol B
78 g B x
 7.1 mol B
11 g
1 mol H
22 g H x
 22 mol H
1.0 g H
Simplest Formula (cont.)
Step
3: Determine
simplest whole number
molar ratio.
Divide through by the
smallest number.
Simplest Formula (cont.)
 7.1
mol of Boron; 22 mol H
7.1 mol B 22 mol H
:
7.1
7.1
 1.0
mol B : 3.1 mol H
 Empirical formula is BH3
Practice Problem
 Analysis
shows a compound
to contain 26.56% potassium,
35.41% chromium, and 38.03
% Oxygen by weight. Find
the simplest formula for this
compound.
 K2Cr2O7.
Molecular Formula
 The
simplest or empirical formula
may or may not the correct
molecular formula.
 Is the correct (real) formula BH3,
B2H6, B3H9 ?
 Need to know the molecular mass
of the formula.
Molecular Formula (cont.)
If the formula mass of
the empirical formula
equals the known
molecular mass, then
it is the molecular
formula.
Molecular Formula (cont.)
 If
not, there is a whole number
multiplier, x, for all of the
subscripts.
(empirical formula mass ) ( X ) 
molecular mass
Empirical formula is BH3
The
BH3 has a formula
mass of 13.83 u.
The known formula
mass for Diborane is
27.67 u.
27.67 u
x
 2.000
13.83 u
(BH3)(2) = B2H6
Determine the molecular
formula of a compound
having the simplest
formula of CH and a
formula mass of 78.110
u.
Review
 4.04
g of N combine with
11.46 g O to produce a
compound with a formula
mass of 108.0 u. What is
the molecular formula?
 N2O5