Download Mass Relationships of Atoms

Mass Relationships of Atoms
Atomic number and mass number
Atomic number (Z) = the number of protons in the nucleus.
Mass number (A) = the sum of the number of protons +
neutrons in the nucleus.
A
ZX
Some isotopes
1
H
1
12
C
6
16
O
8
235
U
92
2
H
1
13
C
6
17
O
8
238
U
92
3
H
1
14
C
6
18
O
8
Atomic masses
- synonymous with atomic weight
- is a relative scale
- the isotope of carbon with 6 p+ and 6
no(carbon-12) is the reference atom and assigned an
atomic mass of exactly 12
- one atomic mass unit (amu) is defined as a
mass exactly equal to 1/12th the mass of one
carbon-12 atom
- so 1 amu is about the mass of 1 ___
p+ or 1 ___
no
Average Atomic Mass and the
Periodic Table
Average Atomic Mass:
The Story of Boron
• Any chunk of Boron you dig up is a mixture
of 2 isotopes: on average ~20% will be 10B,
~80% will be 11B
• Say we get a small chunk of B.
• 100 atoms.
• 20 are B with a mass of 10 amu each
• 80 are B with a mass of 11 amu each
• What is the average mass of Boron?
10
11
•
The
Story
of
Boron
(cont.)
Average mass of a B atom
• = (total mass)/(total # of B atoms)
• Total mass for these 100 B atoms
• = 20 x 10 amu + 80 x 11 amu
• Average mass of a B atom
• = (20 x 10 amu + 80 x 11 amu)/100
20 x 10 amu + 80 x 11 amu
• = 100
100
• = 0.20 x 10 amu + 0.80 x 11 amu = 10.8 amu
20%
80%
So to get the average atomic mass, multiply %
abundance times each isotope mass and add up!
The Mole
- the fundamental SI measure of “amount of
substance”
- the amount of substance that contains as
many elementary entities as there are atoms in
exactly 12 g of carbon-12
- this number of atoms is 6.022045 x 1023
Avogadro’s number
The Mole vs. The Dozen
The Dozen - the amount of substance that
contains 12 entities.
The Mole - the amount of substance that
contains Avogadro’s number (6.022x1023) of
entities.
Dozen Apples = 10 Lbs.
Mole of Helium atoms = 4.0026g
Dozen Apples = 12 Apples
Mole of Helium atoms = 6.022x1023 atoms
Converting to Dozens
Example
How many dozens of apples are represented by
1.3 Lbs. of apples.
1.3 Lbs x
1 dozen
10 Lbs
= .13 dozen
Converting to Moles
Example
How many moles of He are in 6.46 g of He?
6.46 gHe x
1 mol
4.003 g
= 1.61 mol
Mass Relationships
of Atoms - Extended
Example
Calculate the number of grams of lead (Pb)
In 12.4 moles of lead.
12.4 mol x
207.2 g
1mol
= 2569 g
Example
What is the mass in grams of one silver atom?
Solution: Convert grams per mole to grams per atom.
107.9 g
1 mol
=
x
1 mol
6.022 x 1023 atoms
17.91 x 10–23 g
1 atom
Experimental Determination
Of Atomic & Molecular Masses
Atomic mass is measured
by mass spectrometry
e–
+
An atom is ionized in the instrument.
Ion is deflected by magnetic field
amount of
deflection depends
on mass to charge
ratio
highest m/z ratio
deflected ________
least
lowest m/z ratio
most
deflected ________
N
+
S
Ions are detected after passage
through magnetic field
target
N
+
+
+
+
+
+
S
Ions are detected after passage
through magnetic field
mixture of ions of
different mass gives
separate peak for each m/z
target
N
intensity of peak proportional
to percentage of each atom of
different mass in mixture
separation of peaks
depends on relative mass
S
++++
++
The mass spectrum of
the three isotopes of
neon.
90.92%
8.82%
0.26%
19
20
21
22
23
Percent Composition
of Compounds
Percent composition is the percent by mass
of each element the compound contains.
Obtained by dividing the mass of each
element in one mole of the compound by the
molar mass of the compound and
multiplying by 100%
Example
Calculate the percent composition by mass of H,P
and O for phosphoric acid (H3PO4)
Solution: Assume you have 1 mole of the compound.
Molar mass = 3(1.008g) + 30.97g + 4(16.00)
= 97.99
Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.99
%H =
%P =
%O =
3(1.008g)
97.99g
30.97g
97.99g
4(16.00)
97.99g
x
100%
=
3.086%
x 100%
=
31.61%
x 100%
=
65.31%
Determining Formula
Determining the Formula with a CHN Analyzer
Determining the Formula with a CHN Analyzer
O2, and other
gases
CO2 , H2O, O2 , and N2
O2
0.1156 g compound
containing only
C,H,&N
H2O absorber
CO2 absorber
Burning the sample completely
compound + O2
H2O + CO2 +
other
gasses
Determining the Formula
Reacting the C,H,N sample with O2 , we assume all of
the hydrogen and carbon present in the 0.1156g
sample is converted to H2O and CO2 respectively
grams collected
0.1638g CO2
0.1676g H2O
(27.3% C in CO2)
12.01g C
x
= 0.04470 g C
44.009 g CO2
x
2.016g H
= 0.01876 g H
18.015 g H2O
(11.1% H in H2O)
Determining the Formula
Remembering the original mass of the sample the
percentages of the components can be determined
0.04470 g C
x
100% =
x
100% =
38.67% C
0.1156 g sample
0.01876 g H
16.23% H
0.1156 g sample
38.67% C
+
16.23% H + % N = 100%
% N = 45.10%
Levels of Structure
Elemental Composition
Empirical Formula
Molecular Formula
Constitution
Configuration
Conformation
Elemental Composition
Examples:
Formaldehyde
Glucose
C: 40.00%
C: 40.00%
H: 6.73%
H: 6.73%
O: 53.27%
O: 53.27%
Levels of Structure
Elemental Composition
Empirical Formula
Molecular Formula
Constitution
Configuration
Conformation
√
Empirical Formula
The empirical formula tells us which elements
are present and the simplest
whole-number ratio of their atoms.
Empirical Formula
Examples: Formaldehyde and Glucose
Elemental Composition
C: 40.00% 40.00 g
H: 6.73% 6.73 g
O: 53.27% 53.27 g
assume a 100g sample
calculate atom ratios by dividing by atomic
weight
Calculating Empirical Formula
C:
H:
O:
40.00 g
6.73 g
53.27 g
100 g
x
x
x
1 mol
12.01 g
1 mol
1.00 g
1 mol
16.0 g
=
3.33 mol
=
6.73 mol
=
3.33 mol
Empirical Formula
Examples: Formaldehyde and Glucose
Elemental Composition
C: 40.00%
H: 6.73%
O: 53.27%
40.00 g
3.33 mol
6.73 g
6.73 mol
53.27 g
3.33 mol
assume a 100g sample
calculate atom ratios by dividing by atomic weight
determine the smallest whole number ratio by dividing
by the smallest molar value
Calculating Empirical Formula
C: 40.00 g x
H: 6.73 g
x
O: 53.27 g x
100 g
1 mol
12.01 g
1 mol
1.00 g
1 mol
16.0 g
=
3.33 mol
3.33 mol
=
6.73 mol
3.33 mol
=
3.33 mol
3.33 mol
= 1.00
= 2.02
= 1.00
Empirical Formula
Examples: Formaldehyde and Glucose
Elemental Composition
C: 40.00% 40.00 g
6.73 g
H: 6.73%
O: 53.27% 53.27 g
6.73 mol
1
2
3.33 mol
1
3.33 mol
Empirical Formula: CH2O
Example
A 1.723 g sample of aluminum oxide (which consists of
aluminum and oxygen only) contains 0.912g of Al.
Determine the empirical formula of the compound.
1.723 g sample - 0.912 g Al = 0.811 g O
0.912 g Al x
0.811 g O x
1 mol Al
26.98 g Al
1 mol O
16.0 g O
= 0.0338 mol
0.0338 mol
= 1.0 x 2 = 2
= 0.0507 mol =
0.0338 mol
Al2O3
1.5 x 2 = 3
Levels of Structure
Elemental Composition
Empirical Formula
Molecular Formula
Constitution
Configuration
Conformation
√
√
Molecular Formula (the REAL formula)
determined from empirical formula and
experimentally determined molecular mass
Compound
formaldehyde
glucose
True
Empirical
molar
Formula
mass
CH2O
CH2O
30
180
Calculation of empirical mass
1 mol C =
12.01 g
2 mol H x 2 =
2.016 g
1 mol O =
16.00g
30.026g
Molecular mass
glucose
formaldehyde
180g
=6
30g
30g
=1
30g
Empirical mass
Molecular Formula
determined from empirical formula and
experimentally determined molecular mass
Compound
formaldehyde
glucose
Empirical Molar
Formula mass
CH2O
CH2O
30
180
Molecular
formula
CH2O
C6H12O6
Mass Relationships
Stoichiometry
mass relationships
how much reactant is needed to yield a certain
amount of product
Amounts of Reactants
and Products
The mole method
1. Write and balance the equation.
2. Convert to moles.
3. Use the coefficients in the balanced equation
to relate the number of moles of known
substances to the unknown one.
4. Convert to desired units (g, L, etc.).
5. Check your answer.
Stoichiometry
Moles of X
Mass of X
(using
molar
ratio of
Y to X)
X
ny
nx
Moles of Y
Y
Mass of Y
Example
How many grams of nitrogen dioxide can be
formed by reaction of 1.44 g of nitrogen
monoxide with oxygen?
2 NO + O2
2NO2
Stoichiometry
Molar ratio of Y to X
Moles of X
1.44 g of NO
2 NO + O2
ny
nx
Moles of Y
Mass of NO2
2NO2
Stoichiometry
Molar ratio of Y to X
0.048 Moles NO
1.44 g of NO
2 NO + O2
ny
nx
Moles of Y
Mass of NO2
2NO2
Stoichiometry
Molar ratio NO2 to NO
0.048 Moles NO
1.44 g of NO
2 NO + O2
2
2
Moles of Y
Mass of NO2
2NO2
Stoichiometry
Molar ratio NO2 to NO
0.048 Moles NO
1.44 g of NO
2 NO + O2
2
2
0.048 Moles NO2
Mass of NO2
2NO2
Stoichiometry
Molar ratio NO2 to NO
0.048 Moles NO
1.44 g of NO
2 NO + O2
2
2
0.048 Moles NO2
2.21 g of NO2
2NO2
Example
2 NO + O2
1.44g NO x
x
46g NO2
1mol NO2
2NO2
1mol NO
30g NO
=
x
2mol NO2
2mol NO
2.21g NO2
Limiting Reagents
Limiting Reagent
Reactants are not always present (or available)
in “stoichiometric” quantities.
One reactant may be present in quantities such
that it is completely consumed while excess
amounts of other reactants remain.
- called “limiting reactant” or “limiting
reagent”
The limiting reagent will limit the amount of
product produced.
Example
How many moles of MgCl2 will be produced?
Mg + Cl2
MgCl2
Start 1 mol 1 mol
Finish
0
0
0
1 mol
Example
How many moles of MgCl2 will be produced?
Mg + Cl2
Start 1 mol 2 mol
Finish
0
1 mol
MgCl2
0
1 mol
magnesium is the limiting reagent
1 mol of chlorine will be left unchanged
Limiting Reagent
Molar ratio Y to X
Moles of X
Moles of W
Mass of X W + X
Mass of W
ny
nx
Moles of Y
Y
Mass of Y
Compare molar ratio W to X to their
coefficients in balancd equation; identify LR
Molar ratio Y to LR
Moles of X
Moles of W
Mass of X W + X
Mass of W
Moles of Y
Y
Mass of Y
Example
Determine the limiting reagent and the amount of PI3
produced when 6.00g P4 reacts with 25.0g of I2.
P4 + 6I2
6.00g P4 x
25.0g I2 x
1mol P4
4PI3
=
.0484mol P4
=
.0984mol I2
124g P4
1mol I2
254g I2
Example cont...
Determine how much I2 would be needed to react
completely with the available amount of P4.
P4 + 6I2
.0484mol P4 x
4PI3
6mol I2
=
1mol P4
but only .0984mol I2 is available
I2 is the limiting reagent
0.290mol I2
amount of I2
needed
Example cont...
…the amount of PI3 produced from the limiting
reagent...
.0984mol I2
x
4mol PI3
6mol I2
27.0g PI3
x
412g PI3
1mol PI3
=
another method
Determine the limiting reagent and the amount of PI3
produced when 6.00g P4 reacts with 25.0g of I2.
P4 + 6I2
4PI3
Compute the moles of product formed 2 ways (assuming
P4 is completely consumed and then assuming I2 is
completely consumed. See which reactant, if used up
completely, gives the least amount of product. It’s limiting.
Reaction Yield
Theoretical yield
the amount of product that would result if
all the limiting reagent reacted
Actual yield
the amount of product actually obtained
from the reaction
Almost always less than the
theoretical yield
Percent Yield
Actual yield
x 100%
%yield =
Theoretical yield
Determines how efficient a
reaction is
Example
In a certain industrial operation 3.54 x 107g of TiCl4 is
reacted with 1.13 x 107g of Mg. (a) Calculate the
theoretical yield of Ti in grams. (b) Calculate the percent
yield if 7.91 x 106g are actually obtained.
TiCl4 + 2Mg
Ti + 2MgCl2
Calculate theoretical yield
3.54 x 107g TiCl4
x
1.13 x 107g Mgx
1mol TiCl4
1.87 x 105mol
TiCl4
=
4.65 x 105 mol
Mg
187.7g TiCl4
1mol Mg
24.31g Mg
1.87 x 105mol TiCl4 x
=
2mol Mg
1mol TiCl4
= 3.74 x 105 mol Mg
there is more than enough
Mg
TiCl4 is limiting
3.54 x 107g TiCl4
x
47.88g Ti
1mol TiCl4
x
1mol Ti
x
1mol TiCl4
187.7g TiCl4
=
8.93 x 106g Ti
1mol Ti
%yield =
Actual yield
x 100%
Theoretical yield
7.91 x 106g Ti
100% x
8.93 x 106g Ti
=
= 88.6%