Types of Reactions
-Synthesis
A + B = AB
-Decomposition
AB = A + B
-Single Displacement
A + BC = AC + B
-Double Displacement
AB + CD = AD + CB
-Combustion (metal burns with oxygen)
C + O2 = CO2
Balancing Equations
Must have equal number of atoms on each side of the equation based on the Law of Conservation of Mass.
Steps to Balance an Equation
-write out unbalanced equation
-create a table with a column for reactants and products
-balance equations using coefficients
• balance elements other than O and H
• balance polyatomic ions that appear unchanged
• balance other elements
• balance H and then O
Isotopes
Isotopes- atoms that have the same number of protons but different numbers of neutrons
Radioisotopes- isotopes with unstable nuclei that decay into different often stable, isotopes
Isotopic abundance- the amount of a given isotope of an element that exists in nature, expressed as a percentage of the total amount of this element
Atomic Mass Unit (AMU)
AMU- one twelfth of the mass of a carbon-12 atom
Average Mass- sum of masses
/# of masses
Relative Atomic Mass (RAM)
RAM- the atomic mass of an element in relation to that of another element
To calculate RAM
-RAM = m1f1 + m2f2
-Answer expressed in AMU (u)
The Mole
Mole- the SI based unit that is used to measure the amount of a substance; symbol is n; unit is mol
Molar Mass- the mass of one mole of a substance; symbol is M; unit is g/mol
Avogadro’s Constant- the number of particles in one mole of a substance; equal to 6.02 x 1023 particles
Conversion Triangles








Law of Conservation of Mass
During a chemical reaction, the mass does not change, therefore the mass is conserved.
Law of Definite Proportions
Elements in a compound are always present in the same proportion by mass.
e.g. water = 11.2% Hydrogen by mass
88.8% Oxygen by mass
% mass of element=mass of element x100
mass of compound
Mass percent- the mass of an element in a compound, expressed as a percentage of the total mass of the compound
Percentage composition- the percent by mass of each element in a compound
Method 1: Chemical formula known
% composition of alanine (C3H7NO2)
% mass of element=total mass of element
molar mass
M= (12.0x3)+(1.0x7)+(14.0)+(16.0x2)
= 89.0 g/mol
% carbon=12.0x3 x100 = 40.4%
89.0g/mol
% hydrogen=1.0x7 x100 = 7.9%
89.0g/mol
% nitrogen=14.0 x100 = 15.7%
89.0g/mol
%oxygen=16.0x2 x100 = 36.0%
89.0g/mol
Method 2: Chemical formula unknown
A 24.5g sample of an unknown hydrocarbon is decomposed to yield 20.2g of pure carbon and 4.3g of hydrogen gas. Calculate % composition of this hydrocarbon.
m of compound = 24.5g
m of carbon = 20.2g
m of hydrogen = 4.3g
mass % C = mass of C x100
mass of compound
= 20.2g x100
24.5g
= 82.4%
mass % of H= 100% - 82.4%
= 17.6%
Method 3: Composition from ratio
When hydrogen combines with oxygen, it does so in a 1:8 ratio by mass. Calculate the % hydrogen.
G- 1.0g –H
8.0g –O
R- % composition
A- % H = mass of H x100
mass of H+O
M- % H = 1.0g x100
9.0g
= 11%
Empirical Formula
Empirical formula- a formula that shows the smallest whole number ratio of the elements in a compound
Determining Empirical Formula
-Convert percentage composition data into mass data by assuming that the total mass of the sample is 100g
-Determine the number of moles of each element in the sample by dividing the mass by the molar mass of each element
-Convert the number of moles of each element into whole numbers that become subscripts in the empirical formula by dividing each amount in moles by the smallest amount
-If the subscripts are not yet whole numbers, determine the least common multiple that will make the decimal values into whole numbers. Multiply all subscripts by this least common multiple. Use these numbers as subscripts to complete the empirical formula
Empirical Formula from % Composition
Calculate the empirical formula for a compound that is 85.6% carbon and 14.4% hydrogen
G- % carbon – 85.6%
% hydrogen – 14.4%
assume 100g of C?H?
R- empirical formula= ?
nC=? nH=?
A- nC= mC nH= mH
MC MH
M- nC= 85.6g
12.011g/mol
nC=7.1268004 mol
nH= 14.4g
1.00794 g/mol
nH=14.286565 mol
ratio of nC to nH = 7.1268004 mol
14.286565 mol
ratio of nC to nH = 1:2
S- therefore the empirical formula is CH4
Molecular Formula
Molecular formula- the formula for a compound that shows the number of atoms of each element that make up a molecule of that compound
Determining Molecular Formula
-Write the empirical formula
-Determine the integer that relates the empirical and molecular formulas
x= experimental molar mass
mass