Transcript Chapter9
Chapter 9 – Chem 160
Chemical Calculations: The Mole
Concept and Chemical Formulas
The Law of Definite
Proportions
Compounds are pure substances and
They are a chemical combination
They can be broken down
They have a definite, constant
elemental composition
Problem 9.5 –
which two of three
experiments produced the same compound?
Compound
ExperimentX grams Q grams mass
1
3.37
8.90
12.27
2 0.561 1.711
2.272
3
26.9
71.0
97.9
Problem 9.5 cont’d
Compound
ExperimentX grams Q grams mass
1
3.37
8.90
12.27
2
0.561
1.711
2.272
3
26.9
71.0
97.9
X/Q
X/Q (s.f.)
0.378652
0.379
0.327878
0.328
0.378873
0.379
Calculation of Formula Masses
Definition
Calculation
Example 9.3 a) C7H6O2
a) benzoic acid formula mass
7*C
6*H
2*O
7 x 12.01 amu =
6 x 1.01 amu =
2 x 16.00 amu =
Sum
84.07
6.06
32.00
122.13
amu
amu
amu
amu
Significant Figures and
Atomic Mass
Uncertainty in data vs. uncertainty in
formula mass
We’ll use atomic masses rounded to
hundredths
Consider formula mass calculations as a
pure addition
Percent composition
Definition
Calculation - % composition of Au(NO3)3
Au
N
3*O
NO3 3*(NO3)
196.97 14.01 48.00 62.01 186.03
Formula Mass 383.00
% Au 51.43%
The Mole
Avogadro’s number
# of molecules in 1.20 moles of CO
1.20 moles CO x 6.022x1023 CO molecules
1 mole CO
Cancel “moles CO”
Answer 7.2264 x 1023 – what units?
Significant figures?
Mass of a Mole
Molar mass of an element is …
Molar mass of a compound is …
Problem 9.37 (d)
Mass of 1.357 moles of Na3PO4
3 x Na = 3 x 22.99 = 68.97 g
P = 30.97 g
4 x O = 4 x 16.00 = 64.00g
1 mole 68.97+30.97+64.00=163.94 g
1.357 mole x 163.94 g = 222.466 g
Note significant figures
Significant Figures and
Avogadro’s Number
The mole is the amount of substance …
Avogadro’s number should never be the
limiting factor in s.f. considerations
A.M.U. and Gram Units
6.022 x 1023 amu = 1.000 g
Proof
6.022x1023 atoms N x 1 mole N x 14.01 amu
1 mole N
14.01 g N 1 atom N
6.022 x 1023 amu/g
A.M.U. and Gram Units
(cont’d)
What is the mass, in grams, of a
molecule whose mass on the amu scale
is 104.00 amu? (Example 9.8)
104.00 amu x
1.000 g
6.022 x 1023 amu
1.7270009 x 10-22 g
1.727 x 10-22 g accounting for s.f.
Counting Particles by
Weighing
Atomic ratio to mass ratio
Table 9.2
Cl (35.45) / Na (22.99)
Extension to molecules
Counting Particles by
Weighing – Problem 9.58 b)
Grams of Cu that will contain twice as many
atoms as 20.00 g of Zn
20.00 g Zn x 6.022 x 1023 atoms x
65.41 g Zn
63.55 g Cu
6.02 x 1023 atoms
= 19.43128 g Cu contains …..
as many atoms as 20.00 g Zn
Answer is 19.43 x 2 = 38.86 g Cu
Mole and Chemical Formulas
Microscopic level interpretation
Macro level
Mole and Chemical Formulas –
Problem 9.60
6 mole to mole conversion factors from the
formula K2SO4
2 moles of K / 1 mole of K2SO4
1 mole of S / 1 mole of K2SO4
4 moles of O / 1 mole of K2SO4
2 moles of K / 1 mole of S
2 moles of K / 4 moles of O
1 mole of S / 4 moles of O
Mole and
Chemical Calculations
Moles of A
Formula subscript
Moles of B
Molar mass
Avogadro’s number
Molar mass
Particles of
A
Grams of A
Grams of B
Avogadro’s number
Particles of
B
Mole & Chemical Calculations
Problem 9.70 d)
Mass of 989 molecules of H2O
{(2x1.01)+16.00} g x 989 molecules
6.022 x 1023 molecules
= 2.95945 x 10-20 g
Purity of Samples
Definition
Problem 9.86 a) calculate the mass in
grams of Cu2S present in a 25.4 g
sample of 88.7% pure Cu2S
25.4 g of sample Cu2S x 88.7 g Cu2S
100 g of sample Cu2S
= 22.5298 g
Empirical and
Molecular Formulas
Empirical formula – smallest whole
number ratio of atoms
Molecular formula – actual number of
atoms in a formula unit
Empirical and
Molecular Formulas – cont’d
Problem 9.96 a) write empirical formula
for P4H10
P2H5
9.96 d) C5H12?
No change
Determination of
Empirical Formulas
Elemental composition data
Empirical formula + molecular mass
Empirical and
Molecular Formulas – cont’d
Problem 9.98 b) determine the empirical formula
if 40.27% K, 26.78% Cr and 32.96% O
K
Cr
O
Mass
Molar # of
(g)
mass
moles
40.27
39.10
1.030
26.78
52.00 0.5150
32.96
16.00
2.060
Empirical and
Molecular Formulas – cont’d
K
Cr
O
Mass
Molar # of
Divide
(g)
mass
moles by Cr
40.27
39.10
1.030
2.000
26.78
52.00 0.5150
1.000
32.96
16.00
2.060
4.000
Empirical formula for 9.98 b) is K2CrO4
Determination of
Molecular Formulas
Molecular mass information needed
molecular formula = (empirical formula)x;
where x is a whole number
x= molecular formula (experimental)
empirical formula(calculat’d from atomic masses)
For example (CH2)5 = C5H10
Determination of
Molecular Formulas – 9.116 b)
P2O3, empirical formula; molecular mass
is 220 amu
P: 2 x 30.97 = 61.94 amu
O: 3 x 16.00 = 48.00 amu
Empirical formula = 109.94 amu
Molecular/Empirical = 2.00
Molecular formula is P4O6
Determination of
Molecular Formulas – 9.120 b)
Citric acid, molecular mass 192 amu; 37.50% C,
4.21% H and 58.29% O
Citric acid
molar mass
Mole
units
Mole
Mass % of
comp.
Atomic mass Moles of
of comp.
comp.
g/mole
g/100 g
mole/g
moles
C
1
192
0.3750
12.01
5.995004
H
1
192
0.0421
1.01
8.003168
O
1
192
0.5829
16.00
6.9948
divide by