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