Transcript Lecture #7 - (a) The Mole Concept, Chemistry 142 B
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 fluoride. 39 K (isotopic mass 38.9637 amu, 93.258%) and naturally occurring isotope, 19 41 K (isotopic mass 40.9618 amu, 6.730%). Fluorine has only one F (isotopic mass 18.9984 amu). Use this information to calculate the formula mass of potassium 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 10 23 particles (atoms, molecules, ions, electrons, or…) = N A particles • N A 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 10 23 atoms Fe 32.07g S = 6.022 x 10 23 atoms S
Mole - Mass Relationships of Elements
Element Atomic Mass Molar Mass Number of Atoms
1 atom of H =
1.008 amu
1 mole of H =
1.008 g
= 6.022 x 10 23 atoms 1 atom of Fe =
55.85 amu
1 mole of Fe =
55.85 g
= 6.022 x 10 23 atoms 1 atom of S = 1 atom of O =
amu amu
1 mole of S = 1 mole of O =
g g
= atoms = atoms
Molecular mass:
1 molecule of O 2 = 1 mole of O 2 =
g amu
= molecules 1 molecule of S 8 =
amu
1 mole of S 8 =
g
= molecules
Mole - Mass Relationships of Elements
Element Atomic Mass Molar Mass Number of Atoms
1 atom of H =
1.008 amu
1 mole of H =
1.008 g
= 6.022 x 10 23 atoms 1 atom of Fe =
55.85 amu
1 mole of Fe =
55.85 g
= 6.022 x 10 23 atoms 1 atom of S =
32.07 amu
1 mole of S =
32.07 g
= 6.022 x 10 23 atoms 1 atom of O =
16.00 amu
1 mole of O =
16.00 g
= 6.022 x 10 23 atoms
Molecular mass:
1 molecule of O 2 1 molecule of S 8 = 16.00 x 2 =
32.00 amu
1 mole of O 2 = 32.07 x 8 = =
32.00 g 256.56 amu
= 6.022 x 10 23 1 mole of S 8 =
256.56 g
= 6.022 x 10 23 molecule 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: H 2 O 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 H 2 O = 6.022 x 10 23 molecules of water = 1 mole H 2 O
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: H 2 O 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 H 2 O = 6.022 x 10 23 molecules of water = 1 mole H 2 O
One mole of common substances CaCO 3 100.09 g Oxygen, O 2 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 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 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 Na 3 PO 4 . 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 (C 12 H 22 O 11 ) 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 :
= To find mass % of C = mass of 1 mole sucrose
Calculating Mass Percents and Masses of Elements in a Sample of Compound - II 7-4 (a) continued
mass of 1 mol sucrose Mass % 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 CH 2 (unsaturated Hydrocarbons) C 2 H 4 , C 3 H 6 , C 4 H 8 OH or HO H 2 O 2 S S 8 P P 4 Cl Cl 2 CH 2 O (carbohydrates) C 6 H 12 O 6
Steps to Determine Empirical Formulas
Mass (g) of Element ÷ M (g/mol ) Moles of Element Use no. of moles as subscripts.
Preliminary Formula Empirical Formula Change to integer subscripts: ÷ smallest, conv. to whole #.
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:
C n H m m + (n+ )O 2 2 (g) n CO 2 m 2 2 O(g)
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 CO 2 and 0.4194 g H 2 O.
From this data calculate the molecular formula.
Plan:
We find the masses of Hydrogen and Carbon using the mass fractions of H in H 2 O, and C in CO 2 . 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 CO 2 = Mass fraction of H in H 2 O = 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 10 20 atoms 3. 0.23545 mol, 1.46 x 10 23 formula units 4. (a) 42.10% C, 6.479% H, 51.417% O; (b) 10.25 g C 5. Na 2 CrO 4 6. C 4 H 8 O 4