The mass of a single atom is far too small in grams to use
conveniently. Chemists use the unit called the unified atomic mass
unit (amu) or Dalton (Da). Definition of amu is exactly 1/12 the
mass of an atom of 12 C Amu (Da) = 1.660539 x 10 -24 g
Slide 3
Mass of one 12 C atom = 12 amu (exactly) 1 amu approximates the
mass of one proton or neutron. Mass of electron is neglible in
comparison. ParticleMass Charge gramsamucoulombse Electron9.109382
x 10 -28 5.485799 x 10 -4 -1.602176 x 10 -19 Proton1.672622 x 10
-24 1.007276+1.602176 x 10 -19 1 Neutron1.674927 x 10 -24 1.0086650
0
Slide 4
Elements differ in the number of protons in their atoms. The
atomic number Z All atoms of a given element have the same number
of protons. Number of electrons equals protons. Number of neutrons
= N Mass Number (A) = Z + N Mass number is the total number of
nucleons.
Slide 5
Why do all element not have atomic mass number listed in the
periodic table that is not a whole number or very close to it? Are
all atoms of an element the same?
Slide 6
Isotopes are atoms with the same atomic number, but different
mass number. The larger mass size is due to the difference in the
number of neutrons that an atom contains. Although mass numbers are
whole numbers, the actual masses of individual atoms are never
whole numbers (except for carbon-12). This explains how Lithium can
have an atomic mass of 6.941 Da.
Slide 7
The atomic masses on the periodic table take these isotopes
into account, weighing them based on their abundance in nature,
therefore, more weight is given to the isotopes that occur most
frequently in nature. Average mass of the element E is defined as:
m(E) = (m(I n ) * p(I n )) where represents a n-times summation
over all isotopes I n of element E, and p(I) represents the
relative abundance of the isotope I.
Slide 8
Find the average atomic mass of Boron Mass and abundance of
Boron isotopes n isotope I n mass m (Da) isotopic abundance p 1 10
B 10.013 0.199 2 11 B 11.009 0.801 Solution: The average mass of
Boron is: m(B) = (10.013 Da)(.199) + (11.009 Da)(.801) = 1.99 Da +
8.82 Da = 10.81 Da
Slide 9
Molecular mass: sum of atomic masses of all atoms in a molecule
Formula mass: sum of atomic masses of all atoms in a formula unit
of any compound, molecular or ionic.
Slide 10
What is a mole's favorite movie? The Green Mole What do you get
when you have a bunch of moles acting like idiots? Moleasses What
line from Shakespeare do high school moles have to memorize? To
mole or not to mole, that is the question. How much does Avogadro
exaggerate? He makes mountains out of molehills. What element do
moles love to study in chemistry? Moleybdenum
Slide 11
Copper (II) Nitrate Cu(NO 3 ) 2 63.5 + [(14 + {3 x 16}) x 2] =
187.5g Ca 3 (PO 4 ) 2 3 moles of Ca 2 moles of P 2 x 4 moles of O.
1 mole of Ca is 40.08g, so 3 moles are 120.24 g 1 mole of P is
30.9738g, so 2 moles are 61.9476g 1 mole of O is 15.9994g, so 8
moles are 127.9952g 1 mole of Ca 3 (PO 4 ) 2 is 310.18 g
Slide 12
From a balanced equation, the coefficients define the ratio of
reactants needed for the products that result from the reaction.
Counting atoms is impractical. Use a mass ratio: to predict mass of
products in ideal conditions. to calculate the percentage yield in
actual conditions. to obtain the mass of each reactant needed.
Slide 13
Balanced equation: C 2 H 4 + HCl C 2 H 5 Cl 1 : 1yields 1 for
ratio of molecules 28.0 : 35.5yields64.5 for mass ratio Ethylene:
Atomic mass of 2C = 2 x 12.0amu = 24.0amu Atomic mass of 4H = 4 x
1.0amu = 4.0amu Molecular mass of C 2 H 4 = 28.0amu Hydrogen
chloride: at. mass of H = 1.0amu at. Mass of Cl = 35.5amu Molecular
mass of HCl = 36.5amu Ethyl chloride: at. mass of 2C = 2 x 12.0amu
= 24.0amu at. mass of 5H = 5 x 1.0amu = 5.0amu at. mass of Cl =
35.5amu = 35.5amu Molecular mass of C 2 H 5 Cl = 64.5amu
Amadeo Avogadro was an Italian physics professor who proposed
in 1811 that equal volumes of different gases at the same
temperature contain equal numbers of molecules. Amadeo
Avogadro
Slide 16
If Avogadros hypothesis is true, then atomic weights for gases
can be derived by weighing equal volumes of different gases
(Cannizzaro). Johan Loschmidt (HS teacher) took the idea and
calculated the size of a molecule of air. He developed an estimate
for the number of molecules in a given volume of air. These three
ideas together lead to the number named for Avogadro. Loschmidt was
the first to calculate this number.
Slide 17
1 mole of a substance,mole N A = 6.02214179(30)10 23 is known
as the Avogadro constant.Avogadro constant For calculations please
use 6.02 x 10 23 http://www.youtube.com/watch?v=Hj83o
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Slide 18
Definition: A mole is the amount of substance that contains as
many elementary particles as there are atoms in exactly 12 grams of
carbon-12 ( 12 C). 1 Mole = 6.022045 x 10 23 particles (atoms,
molecules, ions, electrons, apples, wads of gum, elephants) = N A
particles ~100 million x 100 million x 100 million
Slide 19
6.022045 x 10 23 whatever kind of particles per mole One mole
of common substances: CaCO3 :100.09g Oxygen: 32.00g Copper: 63.55g
Water:18.02g
Slide 20
2 H 2 (g) + O 2 (g) 2 H 2 O(g) 2 dozen H 2 molecules react with
exactly 1 dozen O 2 molecules to give exactly 2 dozen H 2 O
molecules. 2 moles of H 2 molecules react with exactly 1 mole of O
2 molecules to give exactly 2 moles of H 2 O molecules. Why do we
do this? Because these last sizes are in the gram range and easy to
weigh. Conventions: 1 mole of 12 C atoms weighs 12 g exactly. 1
atom of 12 C weighs 12 amu exactly. (amu = atomic mass unit = ~mass
of a proton or neutron)
Slide 21
2 moles of H 2 molecules react with exactly 1 mole of O 2
molecules to give exactly 2 moles of H 2 O molecules. 2 moles of H
2 molecules 4 x 1.008g = 4.032g react with exactly 1 mole of O 2
molecules 2 x 15.994g = 31.988g to produce exactly 2 moles of H 2 O
molecules 2(2 x 1.008g + 15.994g) = 36.03g Sum of reactants = Sum
of products Law of Conservation of Mass
Slide 22
12 red marbles @ 7g each = 84g 12 yellow marbles @4e each=48g
55.85g Fe = 6.022 x 10 23 atoms Fe 32.07g S = 6.022 x 10 23 atoms
S
Slide 23
Element Atomic Mass Molar Mass Number of Atoms 1 atom of H =
1.008 amu 1.008 g = 6.022045 x 10 23 atoms 1 atom of S = 32.07 amu
32.07 g = 6.022045 x 10 23 atoms 1 atom of O = 15.994 amu 15.994 g
= 6.022045 x 10 23 atoms 1 molecule O 2 (15.994 x 2) 32.00amu 32.00
g = 6.022045 x 10 23 atoms 1 molecule S 8 (32.07 x 8) 256.56amu
256.56 g = 6.022045 x 10 23 atoms
Slide 24
Problem: Tungsten (W) is the element used as the filament in
light bulbs, and has the highest melting point of any element, 3680
o C. How many moles of tungsten, and atoms of the element, are
contained in a 35.0 mg sample of the metal? Plan: Convert Mass to
Moles Convert Moles to Atoms
Slide 25
Solution: Moles of W = 35.0x10 -3 g W x 1 mol W = 0.00019032
mol 183.9 g W Moles of W = 1.90 x 10 -4 mol No. of W atoms = 1.90 x
10 -4 mol W x 6.022 x 10 23 atoms 1 mole of W = 1.15 x 1020 atoms
of Tungsten
Slide 26
The molecular mass of a compound expressed in amu is
numerically the same as the mass of one mole of the compound
expressed in grams, called its molar mass.
Slide 27
Slide 28
Slide 29
Carbon6 x 12.01 amu = 72.06 amu Hydrogen12 x 1.008 amu = 12.096
amu Oxygen6 x 15.994 amu = 95.964 amu 180.12 amu
Slide 30
Carbon6 x 12.01 g/mol = 72.06 g/mol Hydrogen12 x 1.008 g/mol =
12.096 g/mol Oxygen6 x 15.994 g/mol = 95.964 g/mol 180.12
g/mol
Slide 31
Go from mass (g) to moles Go from moles to particles (N A ) Go
from moles to volume of gas (L) Go from moles to mass (g)
Slide 32
Mass (g) Number of particles (N A ) x g/mol : g/mol x 22.4L/mol
: 22.4L/mol : N A/ molx N A /mol
Slide 33
For a compound, the percent composition for a specific element
is the fraction of the compound mass that came from that element. A
n B m %A = n(A g/mol) x 100 A n B m g/mol
Slide 34
Slide 35
Mass of Red Balls = Mass Fraction Red = Mass % Red = Mass
Fraction Purple = Mass Fraction Yellow = Check: 56% + 25% + 19% =
100% 3.0g/ball x 3 balls = 9g 9.0g/16.0g total = 0.56 0.56 x 100% =
56% red 2.0g/ball x 2 balls = 0.25 16g total 0.25 x 100 = 25%
purple 1.0g/ball x 3 balls = 3.0g 16g total 0.19 x 100 = 19%
yellow
Slide 36
N__mol of N x _________ = _____g N H__mol of H x _________ =
_____g H C__mol of C x _________ = _____g C H__mol of H x _________
= _____g H O__mol of O x _________ = _____g O Molar mass = M =
_____ g
Slide 37
N 1 mol of N x 14.01g/mol = 14.01g N H 7 mol of H x 1.008 g/mol
= 7.056 g H C 2 mol of C x 12.011g/mol = 24.022g C C 2 H 3 O 2 O 2
mol of O x 15.994g/mol = 31.988g Molar mass = M = 77.076 g NH 4 C 2
H 3 O 2 %N = 14.01g N/77.076g = 18.18% %H = 7.056g H/77.076g =
9.15% %C = 24.022g C/77.076g = 31.17% %O = 31.988g O/77.076g =
41.50% 100.00% 100.00%
Slide 38
Empirical Formula - A formula that gives the simplest
whole-number ratio of atoms in a compound. Once the empirical
formula is found, the molecular formula for a compound can be
determined if the molar mass of the compound is known. Many
compounds can share the same empirical formula. Alkanes are C n H
2n +2
Slide 39
Start with the number of grams of each element, given in the
problem. If percentages are given, assume that the total mass is
100 grams so that the mass of each element = the percent given.
Convert the mass of each element to moles using the molar mass from
the periodic table. periodic table Divide each mole value by the
smallest number of moles calculated. Round to the nearest whole
number. This is the mole ratio of the elements and is represented
by subscripts in the empirical formula. If the number is too far to
round (x.1 ~ x.9), then multiply each solution by the same factor
to get the lowest whole number multiple. e.g. If one solution is
1.5, then multiply each solution in the problem by 2 to get 3. e.g.
If one solution is 1.25, then multiply each solution in the problem
by 4 to get 5.
Slide 40
A compound was analyzed and found to contain 13.5 g Ca, 10.8 g
O, and 0.675 g H. What is the empirical formula of the
compound?
Slide 41
Find the moles for each element
Slide 42
Divide each by the smallest number of moles present. Round to
nearest whole number.
Slide 43
This is the mole ratio of the elements and is represented by
subscripts in the empirical formula. Ca1 O2 H2 Therefore CaO 2 H 2
or with the correct formula, Ca(OH) 2.
Slide 44
NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O.
Calculate the empirical formula of NutraSweet and find the
molecular formula. (The molar mass of NutraSweet is 294.30
g/mol)
Slide 45
If percentages are given, assume that the total mass is 100
grams so that the mass of each element = the percent given. 57.14%
C, 6.16% H, 9.52% N, and 27.18% O.
Slide 46
Use the conversion factor: g/mol and divide or multiply by
1/g/mol.
Slide 47
Select the smallest number of moles to divide each element
(moles). Smallest one will equal 1.
Slide 48
This is the mole ratio of the elements and is represented by
subscripts in the empirical formula. If the number is too far to
round (x.1 ~ x.9), then multiply each solution by the same factor
to get the lowest whole number multiple.
Slide 49
Now, we can find the molecular formula by finding the mass of
the empirical formula and setting up a ratio:
Slide 50
A sample of a pure oxide of nickel was analyzed by heating to
drive off the oxygen. A team of students weighed an empty test
tube, recording a mass of 32.064 g. After adding a sample to the
tube, they measured a total mass of 33.076 g. The team then heated
the sample in an atmosphere of natural gas reducing it to pure
metal. The final mass after two heatings was 32.785 g for the tube
and the metal residue. Perform calculations necessary to find
results below, showing all of your work. Mass of nickel oxide
sample Mass of nickel present in sample Mass of oxygen present in
sample Mass percent of nickel Mass percent of oxygen Determine the
empirical formula of the oxide of nickel, showing your work
clearly. Name the compound according to IUPAC conventions.
Slide 51
Place the sample of oxide of nickel in the large test tube and
mount the tube at an angle. Attach gas supply to the tubing
entering the test tube. Connect the second tube to the burner and
the test tube. This produces a reducing atmosphere with very low
oxygen content.
Slide 52
Find the possible cations that nickel forms. Write balanced
equations. What are the molar ratios of reactant to elemental
nickel that you would expect to find? (Molar conversions) Predict
which ionic unit formula will result in describing the reaction.
Calculate the formula.