Transcript Formula mass

Chapter 7
Mass Relationships
in Chemical
Reactions
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Introduction
In the margin of your notes record two
things a chemical formula can tell us
A
chemical formula indicates:
1. elements present in a compound
2. relative number of atoms/ions of each
element present in a compound
Introduction
 Chemical
formulas also allow chemists to
calculate a number of other characteristic
values for a compound:
 formula mass
 molar mass
 percentage composition
Formula mass
is the sum of the atomic masses of all
the atoms in a compound, in atomic
mass units (amu)
How do we find the atomic mass of an element?
• recall masses on P.T. are based on 12C
Formula mass for SO2
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
1 molecule SO2 = 64.07 amu
NO!!
Need to be able to relate amu to a
macroscale (grams)
Discovery of the MOLE

Amedeo Avogadro
 Italian Scientist
The mole: the amount of a
substance that contains as
many particles as there are
atoms in exactly 12.00 grams of
12C
1 mol = 6.02 x 1023 atoms, formula units, molecules
How do we measure an amount?
1 mol = 6.02 x 1023 particles
1 mole C atoms = 12.00 g C
1 mole C atoms = 6.022 x 1023 atoms
1 C atom = 12.00 amu
Relating Mass to Numbers of Particles
Molar Mass
the mass in grams of one mole (or approx.
6.02 × 1023 particles) of a substance
• calculated by adding the masses of the elements
present in a mole (same as recorded values on P.T.
compound’s molar mass is numerically equal to its formula mass
For any element
atomic mass (amu) = molar mass (grams/mol)
1S
SO2
2O
SO2
32.07 grams/mol
+ 2 x 16.00 grams/mol
64.07 grams/mol
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
Chemical Formulas Express Composition
It is often useful to know the percentage by mass of
a particular element in a chemical compound.
Mining for Diamonds!
Mining for Bauxite:
Al(OH)3
Percentage Composition of Iron Oxides
The mass percentage of an element in a compound
is the same regardless of the sample’s size.
There are 2 ways to determine
percent composition of a
compound:
• From a chemical formula
• From experimental data
Percent Composition:

The mass of each element in a compound
compared to the entire mass of the
compound and multiplied by 100 percent.
Percent composition
of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n = the number of moles of the element in 1
mole of the compound
Find the percent composition of the elements
in water (H2O).
H: 2 x 1.01 = 2.02
O: 1 x 16.00 = 16.00
18.02 g/mol
H:
2 x1.01 g/mol H
18.02 g/mol H2O X 100 = 11.21 % H
O:
16.00 g/mol O
18.02 g/mol H2O
X 100 = 88.79 % O
Find the percent composition of the elements
in Al2(SO4)3.
Calculate Total Mass:
Al: 2 x 26.98 = 53.96
S: 3 x 32.07 = 96.21
O: 12 x 16.00 = 192.00
342.17 g/mol
Find the percent composition of the elements
in Al2(SO4)3. - continued.
Al :
53.96 g/mol Al
X 100 = 15.77 % Al
342.17 g/mol Al2(SO4)3
S:
96.21 g/mol S
X
100
=
28.12
%
S
342.17 g/mol Al2(SO4)3
O:
192.00 g/mol O
342.17 g/mol Al2(SO4)3 X 100 = 56.11 % O
Find the percent composition of a
compound that contains 0.9480 g of C,
0.1264 g of O, and 0.0158 g H.
C = 0.9480 g
O = 0.1264 g
H = 0.0158 g
1.0902 g
C:
0.9480 g
1.0902 g
X 100 = 86.96 %
O:
0.1264 g
1.0902 g
X 100 = 11.59 %
H:
0.0158 g
1.0902 g
X 100 =
1.45 %
Mole Conversions
Mass to Mole Conversions
MOLAR MASS
11.2 g NaCl x 1 mol NaCl
58.44 g NaCl
Na: 1 X 22.99 = 22.99
Cl: 1 X 35.45 = 34.45
58.44 g/mol
0.192 mol NaCl
Mole to Mass Conversions
MOLAR MASS
3.2 mol Zn(NO3)2
Zn: 1 X 65.39 = 65.39
N: 2 X 14.07 = 28.14
O: 6 X 16.00 = 96.00
189.53g/mol
x
189.53 g Zn(NO3)2
1 mol Zn(NO3)2
606 g Zn(NO3)2
Particle to Mole Conversions
AVOGADRO’S NUMBER
8.74 x 1023 atoms CaCO3 x
1 mol CaCO3
6.02 x 1023 atoms CaCO3
1.45 mol CaCO3
Mole to Particle Conversions
AVOGADRO’S NUMBER
0.36 mol Al
x
6.02 x 1023 atoms Al
1 mol Al
2.2 x 1023 atoms Al
MASS
Mole Map
MOLE
6.02 x 1023
PARTICLES
What conversion
factor do I
use???
Multistep Conversions
250 g C12H22O11
X
1 mol C12H22O11
342.34 g C12H22O11
C: 12 X 12.01 = 144.12
H: 22 X 1.01 = 22.22
O: 11 X 16.00 = 176.00
342.34 g/mol
1 mol C12H22O11 X 6.02 x 1023 molec. C12H22O11
342.34 g C12H22O11 1 mol C12H22O11
4.4 x 1023 molecules C12 H22 O11
Remember Empirical Formula:
A chemical formula that gives the
simplest whole-number ratio of the
elements in the formula.
- Subscripts are used for these ratios.
Example Problem
Determine the empirical formula
of a compound found to have
13.5 g of Ca, 10.8 g of O, and
0.675 g of H.
1.
2.
3.
4.
Assume 100 g
Convert to mole
Divide by the smallest
Multiply ‘til whole
Ca :
13.5 g Ca x
10.8 g O x
1 mol
16.00 g
= 0.675 mol O
0.675 g H x
1 mol
1.01 g
= 0.668 mol H
O:
H:
1 mol = 0.337 mol Ca
40.08 g
Writing the complete formula:
a) Round to whole numbers
b) Put parentheses around polyatomic ions
c) Re-write the final formula.
Ca(OH)2
Remember Molecular Formula:
A chemical formula that gives the
actual number of the elements in
the molecular compound.
Example:
C2H4
C6H12O6
NOT CH2
NOT CH2O
Comparing Empirical and Molecular Formulas
• There is a direct relationship between
empirical and molecular formulas.
• There is a direct relationship between the
empirical formula mass and the molecular
formula mass.
FIND THE COMMON MULTIPLE!
The correct ratio can be found by:
dividing the experimental formula mass by
the empirical formula mass
Example Problem
The empirical formula of a compound of
phosphorus and oxygen was found to be P2O5.
Experimentation shows that the molar mass of
this actual compound is 283.89 g/mol. What is
the compound’s molecular formula?
1. Find Molar Mass of empirical formula.
P: 2 x 30.97 = 61.94
O: 5 x 16.00 = 80.00
141.94 g/mol
2. Divide the experimental formula mass by
the empirical formula mass.
283.89 g/mol
141.94 g/mol
= 2.00
3. Multiply the subscripts by
the common multiple.
2 × (P2O5) = P4O10