Unit 2 – Chemical Calculations
1. Using Chemical Formulas
I. Formula Masses
A. Formula Mass
1. The sum of the average atomic masses of all the atoms represented
in the formula of a molecule, formula unit, or ion
Formula Mass of glucose, C6H12O6 :
C = 12.01 amu
6 x 12.01 amu = 72.06 amu
H = 1.01 amu
12 x 1.01 amu = 12.12 amu
O = 16.00 amu
6 x 16.00 amu = 96.00 amu
Formula Mass = 180.18 amu
B.
Molar Masses
1. A compound's molar mass is numerically equal to it formula
mass, but expressed in units of grams/mole (g/mol)
Molar Mass of glucose, C6H12O6 = 180.18 g/mol
2.
Percentage Composition
Percentage Composition - The percentage by mass of each element in a compound
Mass of element in 1 mole of compound X 100 = % element in compound
Molar Mass of compound
Empirical and Molecular Formula
a. Empirical Formula
- When a new compound is prepared, chemists analyze the molecule to find its percentage
composition. This is then used to determine the empirical formula.
3.
Empirical formula: the simplest chemical formula. It gives the relative number of moles of each
type of atom in a compound.
Step 1: You are given the % composition of a compound.
92.3% carbon and 7.7% Hydrogen
Step 2: Assume that you have a 100.0 g sample of the compound, which would mean you had
92.3 g carbon and 7.7 g hydrogen
Step 3: Using the atomic mass of each element to determine the number of moles of each element.
92.3 g C
12.01 g/mol
= 7.69 mol C
7.7 g H
1.01 g/mol
= 7.70 mol H
1
Step 4: To obtain the simplest molar ratio (empirical formula) divide both by the smaller number of
moles present.
C:H
7.69 mol C: 7.70 mol H
7.69
1 mol C: 1 mol H
C1H1
Ex. What is the empirical formula of a compound which analysis shows 2.2% H, 26.7% C and
71.1% O?
b. Molecular Formula
- A formula that shows the actual number of atoms of each element in a molecule.
- Unlike an empirical formula, a molecular formula is a real or true chemical formula.
- There are two different ways to find a molecular formula depending on what type of information
you are given.
Type 1:
Step 1: You are given the % composition and the molar mass
21.9% Na, 45.7% C, 1.9% H, 30.5% O
molar mass = 210 g/mol
Step 2: Using the % composition, find the empirical formula (from method you did in a
previous class)
21.9g Na
22.99g/mol
= 0.95 mol Na
45.7g C
12.01g/mol
= 3.81 mol C
0.95 mol Na
3.81 mol C
0.95
0.95
= 1 mol Na
4 mol C
Empirical Formula NaC4H2O2
1.9g H
1.01g/mol
= 1.88 mol H
1.88 mol H
0.95
2 mol H
30.5g O
16.00g/mol
= 1.91 mol O
1.91 mol O
0.95
2 mol O
Step 3:
MM of NaC4H2O2
# of formula units =
Molecular formula
= 22.99 + 4(12.01) + 2(1.01) + 2(16.00)
= 105.04 g/mol
210 g/mol = 2 formula units
105.04 g/mol
= 2 x (NaC4H2O2) = Na2C8H4O4
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Type 2
Step 1: You are given the empirical formula and the molar mass
empirical formula – CH2O
molar mass – 180 g/mol
Step 2: Calculate the empirical formula mass
MM of CH2O = 12.01 + 2(1.01) + 16.00
= 30.03 g/mol
Step 3: Calculate the number of formula units
# of formula units = molar mass (step 1)
empirical formula mass (step 2)
# of formula units = 180 g/mol
= 6 formula units
30.03 g/mol
Step 4: Molecular formula = # of formula units x empirical formula
6 x (CH2O) = C6H12O6 (glucose)
4.
5.
Molar Mass as a Conversion Factor
A. Converting between moles of compound and grams of a compound
B.
Converting between moles of compound and particles of an element or compound
C.
Converting between moles of compound and volume of a gas
Hydrates
Crystalline compounds in which water molecules are bound in the crystal
structure Copper (II) sulfate pentahydrate CuSO4
5H20
a. The raised dot means "Water is loosely attached" It does NOT
mean multiply when determining formula weight
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