Lecture 3
Stoichiometry: Calculations with Chemical Formulas and Equations
Text: Sections E, F, and H
3.1 Atomic and Molecular Masses (sometimes called “Weights”)
The Atomic Mass Scale
1 amu = 1.66054x10
-24 g and 1 g = 6.02214x10
23 amu
1 H 1.6735x10
-24 g 1.0078 amu
Average Atomic Masses
Determine average atomic mass by using masses of various isotopes and their relative abundances. carbon is 98.892% 12 C and 1.108% 13 C
12 C is 12 amu (exactly) and 13 C is 13.00335 amu
(0.98892)(12 amu) + (0.01108)(13.00335 amu) = 12.011 amu
The average atomic mass of each element (expressed in amu) is also known as its atomic weight .
Formula and Molecular Weights
The formula weight of a substance is merely the sum of the atomic weights of each atom in its chemical formula. ex: H
2
SO
4
has a formula weight of 98.1 amu
FW = 2(AW of H) + (AW of S) + 4(AW of O)
= 2(1.0 amu) + 32.1 amu + 4(16.0 amu)
= 98.l amu
If the chemical formula is the molecular formula, then the formula weight is also called the molecular weight . ex: glucose, C
6
H
12
O
6
, has a molecular weight of 180.0 amu
MW = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu)
= 180.0 amu
Chapter 3 Stoichiometry - 1 -

With ionic substances such as NaCl, it is inappropriate to speak of molecules. We will use the formula weight
FW = 23.0 amu + 35.5 amu
= 58.5 amu
Percentage Composition from Formulas
Percent composition- the percentage by mass contributed by each element in the substance.
C
6
H
12
O
6
% C 
% H 
% O 
 amu

180.0
amu
 100  40.0%
 amu
180.0
amu
 amu

180.0
amu

 100  6.7%
 100  53.3%
3.2 The Mole
A mole is defined as the amount of matter that contains as many objects (atom, molecules, or whatever objects we are considering) as the number of atoms in exactly 12 g of 12 C.
That number is 6.0221421 x 10 23 .
This number is given a special name: Avagadro's number .
We will use 6.02 x 10 23 .
Molar Mass
The mass of single atom of an element (in amu) is numerically equal to the mass (in grams) of 1 mol of atoms of that element.
One 12 C atom weights 12 amu  1 mol of 12 C weighs 12 g.
One 24 Mg atom weights 24 amu  1 mol of 24 Mg weighs 24 g.
One 197 Au atom weights 197 amu  1 mol of 197 Au weighs 197 g.
Chapter 3 Stoichiometry - 2 -

The mass in grams of 1 mol of a substance is called its molar mass .
The molar mass (in grams) of any substance is always numerically equal to its formula weight
(in amu) :
One H
2
O molecule weighs 18.0 amu  1 mol of H
2
O weighs 18.0g.
One NO
3
– ion weighs 62.0 amu  1 mol of NO
3
– weighs 62.0 g
One NaCl unit weighs 58.5 amu  1 mol of NaCl weighs 58.5 g.
Interconverting Masses, Moles, and Numbers of Particles
Conversion of mass to moles and moles to mass are made easy by use of dimensional analysis.
Example: How many moles of trinitrotoluene, C
7
H
5
N
3
O
6
(TNT) are there in 214.0 g of TNT?
Solution: The molar mass can be used as a conversion factor for converting grams to moles. The molar mass of TNT is 227.0 g.
1 mol TNT = 227.0 g TNT.
Moles TNT  
214.0 g TNT

1 mol TNT
 227.0 g TNT


0.9427 mol TNT
Example: What is the mass of 3.125 mol of CO
2
?
Solution: The molar mass of CO
2
is 44.0 g. 1 mol of CO
2 grams CO
2
 
3.125 mol CO
2


44.0 g CO
2
1 mol CO
2


137.5 g CO
2
Conversion factors can also be used to find the number of atoms, molecules, or ions.
Example: Find the number of copper atoms in a copper penny, which weighs 3 g.
Cu atoms  
3 g Cu
 
1 mol Cu
 63.5 g Cu





6.02
 10
23
Cu atoms
1 mol Cu



 3  10
22
Cu atoms
Chapter 3 Stoichiometry - 3 -

3.3 Empirical Formulas from Analyses
Analysis gives the amount of each element as a percentage. If we assume the sample to be 100 g, we can divide these masses (the percentages in grams) by the appropriate atomic weight to obtain the number of moles of each element in 100 g.
We then divide the larger mole numbers by the smallest mole number to obtain the empirical formula. The ratios may not be exact due to experimental errors.
Chapter 3 Stoichiometry - 4 -

Molecular Formulas from Empirical Formulas
The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in the empirical formula.
Example: The empirical formula of ascorbic acid is C
3
H
4
O
3
, giving a empirical formula weight of 88.0 amu. The experimentally determined molecular weight is 176. amu. The molecule has twice the mass of the empirical formula. The molecule must contain twice as many atoms as the empirical formula.
Molecular Formula = n ( Empirical Formula)
? = n (C
3
H
4
O
3
)
Molecular Formula Weight = n ( Empirical Formula Weight)
176. amu = n (88.0 amu) n = 176. amu / 88.0 amu = 2
? = n (C
3
H
4
O
3
) = 2 (C
3
H
4
O
3
) = C
6
H
8
O
6
Chapter 3 Stoichiometry - 5 -

4
3.4 Solutions
3. Chemical Equations
2 H
2
 O
2 reactants
 2 H
2
O products
Symbols: + means "reacts with"  means "produces" coefficients - indicate amount of substance
Equations must be balanced. Equal amounts of each element on each side of the equation.
2 H
2
 O
2
 2 H
2
O
4 H 2O 4H, 2O
When balancing an equation, subscripts should never change, as this changes the chemical identity.
Changing coefficients only changes amount of a substance, not the identity.
O
H
2
O
H H
2 H
2
O
H
2
O
2
H
O
H H
O
H
H
O O
H
Chapter 3 Stoichiometry - 6 -






Balance those species that occur in the fewest formulas on each side.
First try:
C C H H O O
1 1 4 2 2 3
Double H in H
2
O 1 1 4 2(2) 2 4 (! 2 from both)
Double H in H
2
O 1 1
CH
4
  O
2
 very simple example!!! Do Self test H.1 !!!
4 2(2) 2(2) 4
CO
2


 H
2
O
Answers must have integer coefficients.
Chapter 3 Stoichiometry - 7 -