Common Symbols
Used in a Chemical Reaction
(s) = solid
(l) = liquid
(g) = gas
(aq) = solution mixed with water
() = presence of heat
(catalyst) = compound which speeds up
reaction but is not produced or consumed
Types of Reactions
Synthesis - starts with two elements, not
in compounds, and forms only one
product.
-2 or more elements react to form 1
product.
A + B  AB
2Na(s) + Cl2(g)  2NaCl(s)
Types of Reactions
Decomposition - always starts with only
one reactant.
-a single compound breaks down into 2 or
more elements or compounds.
AB  A + B
2NaN3(s)  2Na(s) + 3N2(g)
Types of Reactions
Single-Replacement - always starts with
an element and a compound in the
reactants.
-atoms of one element replace atoms of
another element in a compound.
AB + C  AC + B
Cu(s) + 2AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq)
Types of Reactions
Double-Replacement – exchange of ions
between two compounds.
AB + CD  AD + CB
KCN(aq) + HBr(aq) KBr(aq) + HCN(g)
Types of Reactions
Combustion – oxygen combines with a
substance and release energy in the
form of heat and light.
2H2(g) + O2(g) 2H2O(g)
C(s) + O2(g) CO2(g)
Formula Mass (Molar Mass)
-amount of grams in one mole of that
substance.
-find molar mass from the Periodic Table
Percent Composition
-percent by mass of each element in a
compound
% Composition = mass of the element
total mass
of the compound
x 100
Example
MgCl2
Step #1 – calculate formula mass for compound
Mg – 1 x 24.305 = 24.305 g/mol
Cl - 2 x 35.453 = 70.906 g/mol
Formula mass = 95.211 g/mol
Step #2 – divide each component mass by the
formula mass and multiply by 100
Mg: 24.305/95.211 x 100 = 25.528%
Cl: 70.906/95.211 x 100 = 74.472%
**check that total is 100 when finished**
Empirical Formula
-symbols for the elements contained in a
compound with subscripts showing the
smallest whole number mole ratio
possible.
-may or may not be the same as the
actual molecular formula.
Empirical Formula
Example Problem:
36.5% Na, 25.4% S, 38.1% O;
Determine the empirical formula.
A. convert grams to moles
Na - 36.5 g Na x 1 mol Na = 1.58 mol Na
23.0 g Na
S - 25.4 g S x 1 mol S = 0.791 mol S
32.1 g S
O - 38.1 g O x 1 mol O = 2.38 mol O
16.0 g O
Empirical Formula
B. Divide all by smallest molar amount (in
example, 0.791 mol)
Na: 1.58 mol / 0.791 mol
= 2 Na
S: 0.791 mol / 0.791 mol
=1S
O: 2.38 mol / 0.791 mol
=3O
C. This whole number represents the smallest
mole ratio possible for that substance.
Na2SO3
Molecular Formula
-the actual number of atoms of each element in
one molecule or formula unit of the
substance.
Molecular Formula=(Empirical Formula)n
n= integer; factor by which the subscripts of the
empirical formula must be multiplied to
obtain molecular formula.
Molecular Formula
Example:
Molar Mass of Acetylene: 26.04 g/mol
Mass of empirical formula CH: 13.02 g/mol
experimentally determined molar mass of acetylene
mass of empirical formula CH
26.04 g/mol
13.02 g/mol
=
2.00
Molecular Formula of Acetylene= (CH)2
= C2H2