Chapter 3
CHEMICAL STOICHIOMETRY

^{I. CHEMICAL EQUATIONS}
A. Concise, precise method of showing how one set of substances, called REACTANTS , regroup into a new set of substances, called PRODUCTS, in a chemical reaction.

ACID + BASE ------> SALT + WATER

HCl_(aq) + NaOH_(aq) -----> NaCl_(aq) + H₂O_(l)

^{Equation shows change EXACTLY.}

HCl, NaOH are reactants. NaCl, H₂O are products.
--------> means: yields, forms, produces, reacts to form

(aq) = aqueous solution (l) = liquid

(g) or = gas (s) or = solid

D = heat
Numbers in front of formulas are coefficients. They are are necessary for eqtn. to obey Conservation of Mass.

2 Na(s) + 2 H₂O(l)-----> 2 KOH (aq) + H₂(g)

B. Balancing Chemical Equations

1. All atoms on both sides of the equation must balance.

2. You CANNOT change the subscripts to balance an equation! This changes the nature of the compound.

3. You must change the coefficients in front of the symbols to balance the equation.

Examples:


II. CLASSIFICATION OF CHEMICAL REACTIONS

A. COMBINATION REACTIONS (ALSO CALLED ADDITION REACTIONS)

S(s) + O₂(g) ----> SO₂(g)

6 Mg(s) + N₂(g) ----> 2 Mg₃N₂(s)

B. DECOMPOSITION REACTIONS (BREAKDOWN REACTIONS)

2 HgO(s) ----> 2 Hg(l) + O₂(s)

C. COMBUSTION REACTIONS

1. REACTIONS OF ELEMENTS OR COMPOUNDS WITH O₂ TO PRODUCE NEW COMPOUND “OXIDE”+ HEAT + LIGHT (OFTEN).

C(s) + O₂(g) ----> CO₂(g) + Heat


CH₄(g) + 2 O₂(g) ---->CO₂(g) + 2 H₂O(l) + Heat + Light

III. Chemical STOICHIOMETRY IS chemical BOOKKEEPING.

Chemical STOICHIOMETRY is the PIVOT POINT for the remainder of the study of chemistry.

A. ATOMIC MASS
Mass of the atom in amu

amu = a tomic m ass u nit = unit of mass for atoms

1 amu = EXACTLY equal to 1/12 of mass of

Found? in Periodic Table
Atomic weight? old term that is still used

B. MOLECULAR MASS -- MOLECULAR WEIGHT
FORMULA MASS -- FORMULA WEIGHT

MASS of MOLECULE in amu

MASS of IONIC FORMULA in amu

Molecular mass = sum of atomic masses of atoms in compound

Examples:



C. MOLES and Avogadro’s Number
1. MOLE? term designating a certain number of units.

trio?________ quartet?________
octet?_______ dozen?_________

MOLE? 6.022 x 10²³

2. One MOLE of atoms =# of atoms in 12.000g of

= 6.022 x 10²³ atoms

Therefore 1.00 MOLE of an element= 6.022 x 10²³atoms of that element

And 6.022 x 10²³ = AVOGADRO’S NUMBER

3. Masses vs Numbers

Mass of 1.00 atom = ^{Mass of 1.00 mole of atoms}
Avogadro’s number

Thus, mass of 1.00 mole =(mass of atom)(Avogadro’s #)

1.00 mole C-12 = 12.00 g = 6.022 x 10²³ atoms C-12

1.00 mole F-19 = 19.00 g = 6.022 x 10²³ atoms F-19
Thus, for any monatomic element :
1.00 mole = 6.022 x 10²³ atoms = atomic mass, g


D. COMPOUNDS
Molecular = covalent, composed of molecules

• MOLE of molecules = 6.022 x 10²³ molecules
  

Ionic = ionic, composed of formula units

• MOLE of ionic cmpd = 6.022 x 10²³ formula units
  

(Both can be represented by FORMULA, but both are not MOLECULAR; therefore FORMULA WEIGHT is the better term.)

EXAMPLE:
1. C₆H₁₂O₆ means 6 atoms carbon
12 atoms hydrogen
6 atoms oxygen

AND 6 MOLES carbon
12 MOLES hydrogen
6 MOLES oxygen

So MOLAR MASS =


2. NaCl means 1 sodium ion
1 chloride ion

AND 1 MOLE sodium ions
1 MOLE chloride ions

So FORMULA MASS =


E. Mass -----> Moles
To convert mass of sample in grams to moles:


Examples:
1. How many moles is 34.78 g of Mg?





2. How many moles is 47.28 g of NaNO₃?




F. Moles -----> Mass
To convert moles of sample to grams:


Examples:
1. How many grams is 5.65 moles of sulfur?





2. How many grams is 2.90 moles of H₂SO₄?

A. PERCENT COMPOSITION



1) From formula:
H₂O?
2 x (1.01 g) H + 1 x (16.0 g) O = 18.0 g H₂O

%H =

%O =

2) From empirical data:
Oxide of iron:
34.97 g Fe and 15.03 g O

% Fe =

% O =
This type of calculation gave rise to the Law of Definite Proportions .

B. DERIVATION OF EMPIRICAL FORMULA
Empirical Formula = the simplest possible whole number ratio of atoms which can be calculated for a compound .
1. From Percent Composition
a. % ----> grams (aasume 100 g of compound)
b. grams -----> moles
c. moles -----> molar ratio
d. molar ratio ----> Empirical Formula
2. From Empirical Data
Start with step b above:
b. grams ---> moles
c. moles ----> molar ratio
d. molar ratio ----> Empirical Formula

NOTES: Molar ratio? Divide all numbers of moles by SMALLEST number of moles

Not whole numbers? Multiply each term by a COMMON multiplier

Examples:
1. Calculate the empirical formula of the compound from the data:
Na 0.365 g
H 0.00800g
P 0.246 g
O 0.381 g



2. A 5.00 gram sample of an oxide of lead contains
4.53 g of lead. Determine the empirical formula for
the compound.





C. DERIVATION OF MOLECULAR FORMULA

Molecular Formula = actual composition of compound

To determine the molecular formula one must have
1. empirical formula
2. molecular mass
Example: What is the molecular formula of a cmpd.
whose empirical formula is CH and whose molecular mass is 78?
Solution
a. Calculate mass of empirical formula:
CH = 12.0 + 1.0 = 13.0

b. Divide molecular mass by mass of empirical formula:
78.0 / 13.0 = 6

c. Multiply all subscripts in empirical formula by answer above:
CH becomes C₆H₆

Example: What is the molecular formula of a cmpd.
whose empirical formula is CH₂ and whose molecular mass is 70?

Solution a. 12.0 + 2(1.0) = 14.0

b. 70 / 14 = 5

c. C₅H₁₀