Honors Chemistry
Chapter 12 notes—Chemical Quantities
I. Counting Particles of Matter
A. Stoichiometry
1. Defined-a. book—the study of quantitative relationships between reactants and products in a
chemical reaction
b. Hill—the measurement and calculation of the amounts of matter in chemical reactions
2. The mole
a. defined—the amount of substance contained in 6.02214179 x 1023 (6.022 x 1023)
(Avogadro’s number)
b. similar to a dozen is 12 units, or a pair is 2 units, or a ream is 500 units.
c. This is a large number
B. Hill method of solving for a variable
1. Get the variable on top
2. Get the variable by itself
a. whatever is attached get rid of it by doing the “opposite”
b. subtraction is the opposite of addition
c. division is the opposite of multiplication
C. Dimensional Analysis –a different form of doing math
1. Basis
a. One times anything is the original thing.
b. Anything divided by itself equals one.
c. One can look differently
1) 1 dozen eggs/ 12 eggs = 1
2) 12 inches/1 foot = 1
3) 1 pair of shoes/2 shoes = 1
2. The form
a. Horizontal line means divide
b. Vertical lines mean multiply
c. Math form
d. Dimensional Analysis Form
D. Molar Mass
1. Molar mass of an element
a. One dozen eggs do not have the same mass as one dozen oranges
b. One mole of an element does not have the same mass as one mole of another element
c. defined—the mass of one mole of an element
d. practicality—it is the average atomic mass expressed in grams instead of atomic mass units
e. examples
1) Sodium has an average atomic mass of 22.989770 amu
Sodium has a molar mass of 22.989770 grams (g)
2) Oxygen has an average atomic mass of 15.9994 amu
Oxygen has a molar mass of 31.9988 grams (g) (Remember it is diatomic)
2. Molar mass of a compound
a. molecular mass is the mass of the covalent compound in amu
b. formula mass is the mass of the ionic compound in amu
c. molar mass is the molecular mass or the formula mass in grams
d. Examples
1) water (H2O)
2 H atoms
2 x 1.00794 amu
=
2.01588 amu
1 O atom
1 x 15.9994 amu
=
15.9994 amu
_______________________________________________________________
Molecular mass of H2O
18.01528 amu
2) Table salt (NaCl)
1 Na atom
1 x 22.989770 amu
=
22.989770 amu
1 Cl atom
1 x 35.453 amu
=
35.453 amu
_______________________________________________________________
Formula mass of NaCl
58.44277 amu
e. See problems on pages 409-410.
E. Molar mass and stoichiometry
1. molar mass makes it easy to determine the number of particles in a sample of a substance
2. molar mass is useful in relating masses of reactants and products
II. Using Moles
A. Using Molar Masses in Stoichiometric Problems
1. We can use molar masses to convert from one substance to the next.
2. When converting from one substance to another you must convert using moles
B. Conversions
Number
of units
6.022 x 1023 units
1 mole
1.
2.
3.
4.
Mass
(grams)
Moles
Units to moles
Units to mass
Moles to moles
Moles to mass
gram formula mass
1 mole
III. Predicting yields and percents
A. Theoretical yield—the amount of product of a chemical reaction predicted by stoichiometry
1. actual yields are using less than predicted
2. reason for the reduction is caused by
a. collection techniques and apparatus used
b. time
c. skills of the chemist
B. Percent yield—the efficiency of the reaction
1. Formulated when the actual yield is less than the theoretical yield
2. Percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent
3. percent yield =(actual yield/theoretical yield) x 100
4. Percent composition
Part
Whole
X
100
C. Determining Mass Percents
1. Find the molecular formula of the compound
2. Find the molecular mass of the compound
3. Use the subscripts to represent the moles of the different elements
4. For example: Glucose C6H12O6
a. H is the most abundant element by number, but is it by mass?
b. 6 x 12.0107 + 12 x 1.00794 + 6 x 15.9994 = 180.15588
12 moles H
1.00794 g H
12.09528 g H
1 mole H
12.09528 g H
.06713785 x 100 = 6.71% Hydrogen
180.15588 g C6H12O6
6 moles C 12.0107 g C
72.0642 g C
1 mole C
6.71
40.00
72.0642 g C
.4000102578 x 100 = 40.00% Carbon
180. 15588 g C6H12O6
6 moles O 15.9994 g O
100.00
95.9964 g O
1 mole O
95.9964 g O
180.15588 g C6H12O6
53.29
.5328518836 x 100 = 53.29% Oxygen
IV. Formulas
A. Molecular formula—the normal chemical formula
1. Gives the elements in the compound
2. Gives the number of atoms per molecule (subscripts)
B. Emperical formula—the normal chemical formula in lowest terms
C. Structural formula—show the structure of the molecule
V. Determining Formulas
A. Emperical formula—formula of a compound in the smallest whole number ratio
1. Determine the composition by mass (ex. 3g H, 18g C, and 24g O)
2. Divide the grams by the grams/mole of the element (rounded numbers usually work)
a. 3g/1.00794g/mole = 2.97636741 moles of H
b. 18g/12.0107g/mole = 1.498663692 moles of C
c. 24g/15.9994g/mole = 1.500056252 moles of O
3. Divide the moles by the smallest number of moles
a. 2.97636741/1.498663692 = 1.986014224 =2
b. 1.498663692/1.498663692 = 1
=1
c. 1.500056252/1.498663692 = 1.000929201 =1
4. Use these values as subscripts (CH2O)
5. If the values are not whole numbers then (shortcut) (should convert to fractions and multiply
by common denominators)
a. if .2, .4, .6, or .8 multiply all by 5
b. if .25 multiply all by 4
c. if .3 or .7 multiply all by 3
d. if .5 multiply all by 2
B. Chemical formula (molecular formula)
1. If you know the gram molecular mass then (180.15588g)
2. you can divide the gram molecular mass by the empirical mass (30.02598g)
3. use this whole number to multiply all subscripts (6)
4. normally this isn’t done (most books don’t discuss)