Transcript Chapter #3

Lecture #7 - (a) The Mole Concept,
(b) Formula of an Unknown
Chemistry 142 B
James B. Callis, Instructor
Autumn Quarter, 2004
Figure 3.1: Mass spectrometer
Figure
3.2 (a):
Peaks of
neon
injected
Figure 3.2 (b):
Bar graph of
neon injected
Problem 7-1: Isotopic Composition
The two isotopes of potassium with significant abundance in
nature are 39K (isotopic mass 38.9637 amu, 93.258%) and 41K
(isotopic mass 40.9618 amu, 6.730%). Fluorine has only one
naturally occurring isotope, 19F (isotopic mass 18.9984 amu).
Use this information to calculate the formula mass of potassium
fluoride.
Solution:
MOLE
• Definition: The amount of substance that
contains as many elementary particles (atoms,
molecules, ions, or other ?) as there are atoms in
exactly 12 grams of carbon 12.
• 1 Mole = 6.022145 x 1023 particles (atoms,
molecules, ions, electrons, or…) = NA particles
• NA is Avogado’s Number. (~100 million x 100
million x 100 million)
The Mole is a Chemical Concept
• It represents a fixed number of
chemical entities
• A mole of a chemical entity has a
fixed, unique mass. (Molar Mass)
• Thus, the mole allows the mass
balance to count chemical entities.
Counting objects of fixed relative mass
12 red marbles @ 7g each = 84g
12 yellow marbles @4e each=48g
55.85g Fe = 6.022 x 1023 atoms Fe
32.07g S = 6.022 x 1023 atoms S
Mole - Mass Relationships of Elements
Element
Atomic Mass
Molar Mass Number of Atoms
1 mole of H = 1.008 g = 6.022 x 1023 atoms
1 atom of H = 1.008 amu
1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023atoms
1 atom of S =
amu
1 atom of O =
amu
1 mole of S =
Molecular mass:
1 molecule of O2 =
1 mole of O2 =
1 molecule of S8 =
1 mole of S8 =
1 mole of O =
g =
atoms
g=
atoms
amu
g =
molecules
amu
g =
molecules
Mole - Mass Relationships of Elements
Element
Atomic Mass
1 atom of H = 1.008 amu
Molar Mass Number of Atoms
1 mole of H = 1.008 g = 6.022 x 1023 atoms
1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms
1 atom of S = 32.07 amu
1 mole of S = 32.07 g = 6.022 x 1023 atoms
1 atom of O = 16.00 amu
1 mole of O = 16.00 g = 6.022 x 1023 atoms
Molecular mass:
1 molecule of O2 = 16.00 x 2 = 32.00 amu
1 mole of O2 = 32.00 g = 6.022 x 1023 molecule
1 molecule of S8 = 32.07 x 8 = 256.56 amu
1 mole of S8 = 256.56 g = 6.022 x 1023 molecules
Molecular Mass - Molar Mass ( M )
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.
For water: H2O
Molecular mass = (2 x atomic mass of H ) + atomic mass of O
= 2(
amu) +
amu =
amu
Mass of one molecule of water =
amu
Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O)
=2(
g)+
g =
g
g H2O = 6.022 x 1023 molecules of water = 1 mole H2O
Molecular Mass - Molar Mass ( M )
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.
For water: H2O
Molecular mass = (2 x atomic mass of H ) + atomic mass of O
= 2 ( 1.008 amu) + 16.00 amu = 18.02 amu
Mass of one molecules of water = 18.02 amu
Molar mass = ( 2 x molar mass of H ) + (1 x molar mass of O)
= 2 ( 1.008 g ) + 16.00 g = 18.02 g
18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O
One mole
of
common
substances
CaCO3
100.09 g
Oxygen, O2
32.00 g
Copper
63.55 g
Water
18.02 g
Calculating the Number of Moles and Atoms
in a Given Mass of Element
Problem 7-2: Tungsten (W) is the element used as the filament in light
bulbs, and has the highest melting point of any element
3680oC. How many moles of tungsten, and atoms of the
element are contained in a 35.0 mg sample of the metal?
Plan: Convert mass into moles by dividing the mass by the atomic
mass of the metal, then calculate the number of atoms by
multiplying by Avogadro’s number.
Solution: Converting from mass of W to moles:
Moles of W =
No. of W atoms =
Calculating the Moles and Number of
Formula Units in a given Mass of Compound
Problem 7-3: Trisodium phosphate is a component of some detergents.
How many moles and formula units are in a 38.6 g sample?
Plan: We need to determine the formula, and the molecular mass from
the atomic masses of each element multiplied by the coefficients.
Solution: The formula is Na3PO4. Calculating the molar mass:
MM =
Converting mass to moles:
# Formula units =
Flow Chart of Mass Percentage Calculation
Moles of X in one
mole of Compound
Multiply by M (g / mol of X)
Mass (g) of X in one
mole of compound
Divide by mass (g) of one mole
of compound
Mass fraction of X
Multiply by 100 %
Mass % of X
Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem 7-4: Sucrose (C12H22O11) is common table sugar.
( a) What is the mass percent of each element in sucrose?
( b) How many grams of carbon are in 24.35 g of sucrose?
(a) Determining the mass percent of each element:
mass of C per mole sucrose =
mass of H / mol =
mass of O / mol =
total mass per mole =
Finding the mass fraction of C in Sucrose & % C :
Mass Fraction of C = mass of C per mole
=
mass of 1 mole sucrose
=
To find mass % of C =
Calculating Mass Percents and Masses of
Elements in a Sample of Compound - II
7-4 (a) continued
Mass % of H =
mol H x M of H x 100% =
mass of 1 mol sucrose
Mass % of O =
mol O x M of O x 100% =
mass of 1 mol sucrose
7-4 (b) Determining the mass of carbon:
Mass (g) of C = mass of sucrose x ( mass fraction of C in sucrose)
Mass (g) of C =
Empirical and Molecular Formulas
Empirical Formula - The simplest formula for a compound
that agrees with the elemental analysis. The
smallest set of whole numbers of atoms.
Molecular Formula - The formula of the compound as it
exists. It may be a multiple of the Empirical
formula.
Some Examples of Compounds with the same
Elemental Ratios
Empirical Formula
Molecular Formula
CH2(unsaturated Hydrocarbons)
C2H4 , C3H6 , C4H8
OH or HO
H 2O2
S
S8
P
P4
Cl
Cl2
CH2O (carbohydrates)
C6H12O6
Steps to Determine Empirical Formulas
Mass (g) of Element
÷ M (g/mol )
Moles of Element
Use no. of moles as subscripts.
Preliminary Formula
Change to integer subscripts:
÷ smallest, conv. to whole #.
Empirical Formula
Determining Empirical Formulas from
Masses of Elements - I
Problem 7-5: The elemental analysis of a sample compound gave the
following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the
empirical formula and name of the compound?
Plan: First we have to convert mass of the elements to moles of the
elements using the molar masses. Then we construct a preliminary
formula and name of the compound.
Solution: Finding the moles of the elements:
Moles of Na =
Moles of Cr =
Moles of O =
Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula:
Converting to integer subscripts (dividing all by smallest subscript):
Rounding off to whole numbers:
m
CnHm + (n+ )O2(g)
2
m
n CO2(g) + H2O(g)
2
Determining a Chemical Formula from
Combustion Analysis - I
Problem 7-6: Erythrose (M = 120 g/mol) is an important chemical
compound used often as a starting material in chemical
synthesis, and contains Carbon, Hydrogen, and Oxygen.
Combustion analysis of a 700.0 mg sample yielded:
1.027 g CO2 and 0.4194 g H2O.
From this data calculate the molecular formula.
Plan: We find the masses of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2. The mass of Carbon and
Hydrogen are subtracted from the sample mass to get the mass
of Oxygen. We then calculate moles, and construct the empirical
formula, and from the given molar mass we can calculate the
molecular formula.
Determining a Chemical Formula from
Combustion Analysis - II
Calculating the mass fractions of the elements:
Mass fraction of C in CO2 =
Mass fraction of H in H2O =
Calculating masses of C and H:
Mass of Element = mass of compound x mass fraction of element
Determining a Chemical Formula from
Combustion Analysis - III
Mass (g) of C =
Mass (g) of H =
Calculating the mass of O:
Calculating moles of each element:
C=
H=
O=
Answers to Problems in Lecture #7
1. 58.09 amu
2. 1.90 x 10 - 4 mol, 1.15 x 1020 atoms
3. 0.23545 mol, 1.46 x 1023 formula units
4. (a) 42.10% C, 6.479% H, 51.417% O; (b) 10.25 g C
5. Na2CrO4
6. C4H8O4