9/13/2010
Unit 1
Chapter 3
Law of Conservation of Matter
• Matter cannot be created or destroyed
• Atoms are only rearranged during a chemical
reaction
• Reactions are already balanced in nature – you
just have to figure out how
NaCl + H2SO4 + MnO2 → Na2SO4 + MnSO4 + H2O + Cl2
calcium carbonate reacts with lithium chlorate to form
lithium carbonate and calcium chlorate
1
9/13/2010
Combination/Synthesis Reactions
metal + nonmetal  ionic compound
2 Na + Br2  2 NaBr
Decomposition Reactions
AB  A + B
Oxyacid (heated)  water + nonmetal oxide
• H2CO3 H2O + CO2
2
9/13/2010
metallic hydroxide (heated)
 metal oxide + water
• Ca(OH)2  CaO + H2O
metallic carbonates (heated)
 metallic oxide + carbon dioxide
• Li2CO3  Li2O + CO2
Metallic chlorates (heated)
 metal chloride + oxygen gas
• 2KClO3  2KCl + 3O2
3
9/13/2010
metallic oxide (heated)
 metal + oxygen gas
• 2HgO  2Hg +O2
Some decompose into their elements
• 2NaCl  2Na + Cl2
• 2H2O  2H2 + O2
Combustion Reactions – React with O2
• Hydrocarbons combine with oxygen to form
carbon dioxide and water.
• C3H8 + 5 O2  3 CO2 + 4 H2O
4
9/13/2010
Ammonia
• With LIMITED oxygen
▫ 4 NH3 + 5 O2  4 NO + 6 H2O
• With EXCESS oxygen
▫ 4 NH3 + 7 O2  4 NO2 + 6 H2O
Nonmetallic hydrides + O2  nonmetal
oxide + water
• 2 PH3 + 3 O2  P2O3 + 3 H2O
Nonmetallic sulfides + O2 
nonmetallic oxides + sulfur dioxide
• CS2 + 3 O2  CO2 + 2 SO2
5
9/13/2010
Molar Mass
• the mass of one mole of a substance (g/mol)
• atomic weight – the mass of one atom (amu)
• molecular weight – the mass of one molecule
(amu)
• formula weight – the mass of one formula unit
(ionic compounds; amu)
Find the molar mass
• C12H22O11
• Calcium nitrate
molar mass as a conversion factor
• mol  g
• What is the mass of 1.5 mol ammonium acetate?
• g  mol
• How many moles are in 3.5 g of potassium
chlorite?
6
9/13/2010
percent composition
• percent by mass of each element
• Find the percent composition of each element in
C12H22O11
Finding empirical formulas
• Assume 100 g of the substance OR find % of
each element
• %  g for each element
• g  moles for each element
• divide each mole value by the smallest value to
get subscripts for the formula
• watch out for common fractions – make sure
you multiply by denominators to get whole
number ratios
Finding empirical formulas
• Find the empirical formula for a substance that
is made up of
▫ 40.92 % carbon
▫ 4.58 % hydrogen
▫ 54.50 % oxygen
7
9/13/2010
Finding molecular formulas
• Find the number to multiply all subscripts by
• # = molecular wt. / empirical wt.
Finding molecular formulas
• What is the molecular formula of a substance
composed of 38.7% C, 9.7% H, and 51.6% O with
a molar mass of 62.1 g/mol?
Avogadro’s Number!!!!!!!!!
• 1 mole = 6.02 x 1023 “pieces”
pieces =
• atoms in a mole of an element
• molecules in a mole of a molecular compound
• formula units in a mole of an ionic compound
8
9/13/2010
Using Avogadro’s Number
• How many atoms are in 5.2 moles of sodium?
• How many molecules in 3.9 moles of water?
• How many atoms in 2.7 moles of barium sulfite?
• What is the mass of 3.82 x 1024 atoms calcium?
• What is the mass of 8.63 x 1022 molecules of
potassium permanganate?
Balance equations are conversion
factors
• Be sure to have a balance equation first!
• Only change substance by converting moles 
moles
• Any other conversion is for ONE substance only
Stoichiometry Examples
• 1.57 moles of hydrogen react with excess oxygen.
How many moles of water are formed?
• 3.52 g of methane are burned. What mass of oxygen
is used up?
• What mass of water is produced during the
oxidation of glucose (C6H12O6)?
• What is the mass of each product formed during the
decomposition of potassium chlorate?
9
9/13/2010
Combining some ideas…
To determine the empirical formula of a
substance, combustion analysis can be used.
0.255 g of the substance were combusted
producing 0.561 g of CO2 and 0.306 g of H2O.
What is the empirical formula?
Finding Limiting Reagents
• You will be given a mass for more than one reactant
(A and B)
• Choose the mass of one reactant (A) and convert it
to the mass of the other reactant (B)
• If the given mass of B is greater than the calculated
mass of B – B is the excess reagent (A is limiting)
• If the given mass of B is less than the calculated
mass of B – B is the limiting reagent (A is in excess)
• Use the given mass of the limiting reagent to
determine the mass of any product
Finding limiting reagents
• 3.50 g of sodium phosphate reacts with 6.40 g of
barium nitrate. What is the mass of barium
phosphate formed? (Other product is sodium
nitrate)
10
9/13/2010
% Yield
• theoretical yield - the amount of product
calculated from the given amount of limiting
reagent
• actual yield – the amount of product formed
during the actual experiment
• % yield = (actual/theoretical) x 100
% yield
• adipic acid (H2C6H8O4) is used to produce nylon
and is formed during the combustion of
cyclohexane (C6H12). In the lab 25.0 g of
cyclohexane oxidizes in an oxygen rich
environment. What is the percent yield if 33.5 g
of adipic acid are obtained?
11