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REVIEW
Dimensional Analysis (Factor labeling)
-
This is a problem solving method that focuses on units.
It uses conversion factors that are equivalents like 2 pints = 1 quart and 60 seconds = 1 minute.
MOST conversion factors have unlimited numbers of significant figures.
You must practice using dimensional analysis – we will use it all year and failure to use it usually results in a wrong answer.
THE UNITS MUST CANCEL – YOU MUST BE LEFT WITH THE UNITS YOU ARE SEEKING. WATCH
SGINIFICANT FIGURES.
SCIENTIFIC NOTATION
1. If the number is less than one, move the decimal to the right. The exponent is negative and equals the number of places you
moved the decimal point.
.00004567  4.567 x 10-5
2. If the number is greater than one, move the decimal to the left. The exponent is positive.
1234
 1.234 x 103
3. If the exponent is negative, move the decimal point to the left. The number is less than one.
4.567 x 10-5  0.00004567
4. If the exponent is positive, move the decimal point to the right. The number is greater than one.
1.234 x 103  1234
SIGNIFICANT FIGURES
1. Atlantic-Pacific Rule
This rule divides measurements into two kinds – those with a decimal point and those without. Imagine a map of the United
States and superimpose the measurement on top. If a decimal point is present in the number, count significant digits from the Pacific
side. (If decimal point is present, count from the Pacific side.) If a decimal point is absent, count from the Atlantic side.
You should start counting with the first nonzero digit you find. Thereafter, all digits, including zero, are significant.
2. Another Way
1.
2.
3.
4.
Digits other than zero are always significant.
One or more final zeros used after the decimal point are always significant.
Zeros between two other significant digits are always significant.
Zeros used solely for spacing the decimal point are not significant. The zeros are place holders only.
SC2. Students will relate how the Law of Conservation of Matter is used to determine chemical composition in compounds and
chemical reactions.
c. Apply concepts of the mole and Avogadro’s number to conceptualize and calculate
• Mass, moles and molecules relationships,
THE MOLE
CHAPTER 10
Introduction
Matter is made up of particles. “Particle” is a generic term. You must look at the formula to determine what type of particle it is
(atom, ion, molecule, etc.)
Atoms
Fe
He
(molecular)
Molecules
H2 O
N2
(ionic)
Formula units
NaCl
CaCl2
You should be able to tell how many atoms are in each chemical formula.
Ions
SO4-2
NH4+1
ATOMIC MASS, FORMULA MASS,
and MOLECULAR MASS
ATOMIC MASS
This is the mass of an atom in atomic mass units (amu). It is the same as the average mass of the atoms of an element found
on the periodic table.
Example: C - single atom – 12.01 amu
MOLECULAR MASS
This is the mass found by adding the atomic masses of the elements found in the molecule (molecular compound).
Example: H2O – molecular compound
H (2 x 1.01amu) + O (1x16.00amu) = 18.02 amu
FORMULA MASS
This is mass found by adding up the atomic masses of the elements found in a formula unit (ionic compound).
Example: NaCl – ionic compound
Na (1 x 22.99amu) + Cl (1 x 35.45amu) = 58.44 amu
Different atoms have different masses; THEREFORE different compounds have masses dependent upon the masses of the atoms that
make them up.
PRACTICE:
Calculate the formula or molecular mass of the following compounds. Show your work correctly.
A. H2SO4
C. Al(NO3) 3
B. C3H5N3O3
D. Fe(CH3COO)3
PERCENT COMPOSITION
Chocolate Chip Cookies:
450.0g sugar
20.0mg salt
450.0g Crisco
20.0mg baking soda
2 eggs (75.0g)
20.0mg vanilla
900.0g flour
Looking at the ingredients, what
percentage of the total
composition is eggs?
To Calculate:
1.
Get the total amount (add up all the ingredients).
Total = 450.0g + 450.0g + 75.0g + 900.0g + 0.002g + 0.002g + 0.002g = 1875.006g = 1875.0g
2.
Find the percentage of the total that is eggs.
part x 100
75.0g x 100 = 4.00 %
total
1875.0
Practice:
1. What is the mass percent of oxygen in H2SO4?
THE MOLE
The MOLE (mol) is the unit of measurement used to describe the amount of matter in a particular substance. (Just like dozen, gross.)
Also known as Avogadro’s number.
1 mole = 6.0213672 x 1023 particles = 6.02 x 1023 particles (This is a rounded number.)
(particles are atoms, ions, formula units, or molecules)
MOLES TO MOLES
These kinds of problems involve making a comparison of one item to another. In one formula unit of Ca(OH)2 there is one
atom of Ca, two atoms of O, and two atoms of H. If you have 2.0 moles of Ca(OH)2, then you would have 2.0 x 2 for the moles of O.
2.0 moles Ca(OH)2
2 mole O
1 mol Ca(OH)2
= 4.0 mol O
PRACTICE
Always start with the number you are given and then find the conversion factor that gets you to the desired outcome.
How many moles of oxygen are in 3.4 moles of H2SO4?
MOLES AND MASS
The mole is a conversion factor that relates the number of particles of matter to the matter’s mass. Molar mass = mass (g) of one mole
of atoms. We will use dimensional analysis (factor label) to make mole/mass conversions.
Example:
All have 6.02x1023 atoms
1 mole He = 4.00 g He
1 mole Na = 22.99 g Na
1 mole Mg = 24.31 g Mg
1 mole Al = 26.98 g Al
Use the conversion factor:
atomic
molecular MASS (g) = 1 MOLE = MOLAR MASS
formula
PRACTICE
A. Calculate the mass in grams of 2.0 moles of Na.
??g = 2.0 mol Na
22.99 g Na
1 mol Na
=
g Na
B. Calculate the mass in grams of 0.500 moles of CO2
You can also calculate the number of moles in a given mass.
C. How many moles are in 50.0 g of H2?
?? moles = 50.0 g H2
1 mole =
2.02g H2
D. How many moles are in 75.0 g of NaOH?
mol H2
MOLES AND PARTICLES
We use factor label to make mole/particles conversions.
Use the conversion factor:
1 mole = 6.02 x 1023 particles
(particles are atoms, formula units, or molecules)
PRACTICE:
E. Determine the number of molecules in 2.23 moles of nitrogen, N 2.
6.02x1023 molecules N2 =
1 mol N2
?? molecules = 2.23 mol N2
molecules N2
F.
Determine the number of atoms in 16.0 moles of sulfur.
G.
Determine the number of moles in 4.36 x 10 24 atoms of silver, Ag
MASS AND MOLE AND PARTICLE
You can also convert mass to particles by going through moles.
1 mole
PARTICLES
MOLE
(atoms, ions,
6.02
molecules, or funits)
H.
1 mole
x 10
23
MASS
molar mass (from
periodic table)
Determine the number of atoms in 16.0 g of sulfur.
?? atoms = 16.0 g S
1 mol S
32.06 g S
6.02x1023 atoms S
1 mol S
I.
Calculate the number of grams in 1.20 x 10 23 atoms of calcium.
J.
Calculate the number of molecules in 42.56g of water.
=
atoms S
SC2. Students will relate how the Law of Conservation of Matter is used to determine chemical composition in compounds and
chemical reactions.
a.
Apply concepts of the mole and Avogadro’s number to conceptualize and calculate
• Empirical/molecular formulas,
EMPIRICAL FORMULA
The simplest whole number ratio of elements in a compound is called the EMPIRICAL FORMULA.
In contrast, the molecular formula of a compound is the actual number of atoms of each element in the
compound.
The Empirical Formula can be calculated from the mole ratio or from percent composition data.
Mole Ratio Example:
What is the empirical formula for a compound if a 2.50 g sample contains 0.900 g of calcium and 1.60 g of
chlorine?
Step One – you must ALWAYS convert mass to moles:
Determine the number of moles of Ca and Cl. (You must first determine the atomic mass of each.)
Ca
0.900g Ca
Cl
1.60g Cl
1 mole Ca = 0.0225 mol
40.08g Ca
1 mole Cl = 0.0451 mol
35.45g Cl
Step Two:
To obtain the smallest ratio, divide both moles by the smaller number of moles (0.0225 mol).
Ca 0.0225 mol = 1.00
0.0225 mol
Cl 0.0451 mol = 2.00
0.0225 mol
Step Three:
Look at the ratio of the two numbers. Round off to whole numbers (there are exceptions to this).
Ca
Cl
1.00
2.00
becomes
becomes
Ca
Cl
1
2
The empirical formula is CaCl2
PRACTICE:
A. Find the empirical formula of the compound with 63.0g Rb and 5.90g O. Show your work correctly.
Percent Composition Example:
A compound has a percentage composition of 40.0% C, 6.71% H and 53.3% O. What is the empirical formula?
Assume that there is 100 g (from 100%). In a 100g sample, there would be 40.0g C, 6.71g H, and 53.3g O.
Then change the quantities to moles.
Step One:
C
H
O
40.0 g C 1 mole C = 3.33 mol C
12.01g C
6.71 g H 1 mole H = 6.64 mol H
1.01g H
53.3 g O 1 mole O = 3.33 mol O
16.00g O
Step Two:
To obtain the smallest ratio, divide both numbers of moles by the smaller number of moles (3.33 mol).
C
3.33 mol = 1.00
3.33 mol
H 6.64 mol = 1.99
3.33 mol
O
3.33 mol = 1.00
3.33 mol
Step Three:
Look at the ratio of the two numbers. Round off to whole numbers.
C
1.00 becomes
C
1
The empirical formula is CH2O
H
1.99 becomesH
2
O
1.00 becomes
O
1
PRACTICE:
A. What is the empirical formula for a compound that has 32.8% Cr and 67.2% Cl?
What happens when the mole ratios are not whole numbers?
B. What is the empirical formula of a compound that contains 53.73% Fe and 46.27% S?
MOLECULAR FORMULA
If you know the empirical formula of a substance and its mass, you can determine its molecular formula.
EXAMPLE
The empirical formula of a substance is CH2O. The mass of the molecule is 180.0 amu. What is the molecular
formula?
Step One:
Calculate the empirical mass for the empirical formula (CH2O) from the periodic table.
C
1 x 12.01 = 12.01
H
2 x 1.01 = 2.02
O
1 x 16.00 = 16.00
30.03 amu
Step Two:
Divide the molecular mass by the empirical mass. 180.0 amu = 6.000
30.03 amu
Step Three:
Multiply each element in the empirical formula by the number in step two to get the molecular formula.
6 x CH2O  C6H12O6
So C6H12O6 is the molecular formula.
PRACTICE:
What is the molecular formula of a compound whose empirical formula is P2O3? The molecular mass of this
compound is 220 g.
HYDRATES
-
Define:
-
Hydrates are solids with water molecules trapped in them.
Used in the desiccation of substances and to store energy in solar cells.
Usually have a specific ratio of water to the compound and are considered one single molecule.
The “” does not mean to multiply. It means that waters are attached.
The formula is written as:
(the number of waters)
XY  XH2O
(the formula)
CuSO4  5H2O, copper (II) sulfate pentahydrate
Use the same prefixes as with covalent compounds:
mono – 1
tri - 3
penta – 5
di - 2
tetra – 4
hexa – 6
hepta -7
octa - 8
nona -9
deca - 10
Practice Naming
Na3PO4  2H2O
Ba(OH)2  8H2O
SrCl2  6H2O
ANALYZING A HYDRATE
In order to analyze a hydrate, you must get rid of the water. This is usually done by heating the
compound so that the water evaporates off. The compound that is left is “anhydrous” – without water. The
compound usually changes color.
DETERMINING THE HYDRATE’S FORMULA
It is a lot like calculating the empirical formula only you are not using single elements. You are using water and
the compound. Follow these steps
You have a 5.00g sample of a barium chloride hydrate. Determine the correct formula is you are given 5.00g of
the hydrate and once you heat it and drive the water off, the new mass of the anhydrate is 4.26g
Step One: Find the individual masses if they are not given to you. In this case, you need to find water.
Hydrate mass – Anhydrate mass = water mass
Example:
5.00g
–
4.26g
=
0.74g
Step Two: Find the empirical formula
4.26g BaCl2
1 mol BaCl2 = 0.0205mol BaCl2
208.23g BaCl2
0.74g H2O
1 mol H2O
18.02g H2O
= 0.041mol H2O
Step Three: Divide the smallest mole amount into both.
BaCl2
0.0205mol = 1.00
0.0205mol
H2O
0.041mol = 2.0
0.0250mol
Practice:
Determine the formula of a hydrate with 48.8% MgSO4 and 51.2% H2O
BaCl2  2H2O