QUANTITIES IN CHEMICAL REACTIONS
- The Mole
o The mole is the SI unit that is used to measure the amount of a substance
o 1 mole (mol) of a substance is the amount of substance that contains as many particles
[atoms, molecules, ions, or formula units] as the number of atoms in exactly 12 g of the
isotope C-12; or 6.02 x 1023 particles of the substance (which is called Avogadro's
constant)
 NA = 6.02 x 1023
o The representative particles for pure, monatomic elements are atoms
 Ex. 1 mol of Fe (s) has 6.02 x 1023 atoms
o The representative particles for diatomic molecules and molecular compounds are
molecules
 Ex. 1 mol of O2 has 6.02 x 1023 molecules
 Ex. 1 mol of H2O has 6.02 x 1023 molecules
o The representative particles for pure ionic compounds are formula units
 Ex. 1 mol of NaCl has 6.02 x 1023 formula units
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n = N/NA
 Where n is amount in moles
 N is number of individual particles
 NA is Avogadro constant
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Ex. Hydrazine, N2H4 (l), is used in airbags and we obtained a chemical sample of
3.65 mol.
 a. How many molecules are in the sample?
 b. How many atoms are in the sample?

Ex. If a teaspoon of [Mg(OH)2] stomach medicine contains 4.1 x 1021 formula
units, what amount in moles of magnesium hydroxide is in the teaspoon?
Mass and The Mole
o Recall: That 1 atom of C-12 has a mass of exactly 12 u, and 1 mol (6.02 x 1023 atoms) of
C-12 has a mass of exactly 12 g
 The atomic mass unit is defined using the C-12 isotope and the atomic masses of
all the other elements are defined using C-12 as the standard
 Therefore 1 mol of any element has a mass that is numerically equal to
the elements atomic mass expressed in grams
o
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o
Ex. An atom of iron has an atomic mass of 55.85 u, and 1 mole
of iron atoms has a mass of 55.85 g.
The term for the mass of 1 mol of substance is molar mass, given the symbol M,
and unit is g/mol
 i.e. 1 mol of mass of atoms is atomic molar mass, 1 mol of mass of
molecules is molecular mass, and 1 mol of mass of formula units is
formula unit molar mass
o Ex. What is the molar mass of water? of Al2(SO4)3?
The molar mass represents the mass of 6.02 x 1023 particles
 Therefore, m= n x M
--- If n = N/NA
then, m = (N x M)/NA
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 Ex. Calculate the mass of 1.28 x 10 mol of glucose, C6H12O6

Ex. What is the mass of 4.27 x 1023 formula units of CrI3 (s)?
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Chemical Proportions and Percent Composition
o The law of definite proportions states that a chemical compound always contains the
same proportions of elements by mass
 i.e. water is the same formula, H2O, if found in rain, snow, oceans, etc.
o This law deals with the proportion of each element in a compound by mass
 Expressed as a mass percent
 Mass of an element in a compound, expressed as a percent of the total
mass of the compound
o Ex. Calculate the mass percent of both elements in dinitrogen
tetroxide
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Empirical and Molecular Formulas
o A molecular formula for a compound shows the number of atoms of each element that
makes up a molecule of that compound
 Ex. Hydrogen peroxide, H2O2 --- 2 atoms of H and 2 atoms of O
o An empirical formula shows the smallest whole number ratio of the elements in a
compound
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o
o
o
Need to reduce ratios if possible
 Ex. H2O2 --- 2:2, becomes 1:1, so the empirical formula is HO
Note: for some compounds the empirical formula may be the same as the molecular
formula
 Ex. H2O - molecular formula
 Ratio is 2:1, therefore empirical is H2O
Recall ionic compound can only have 1 possible atomic configuration, therefore, only
represented by empirical formula's
 Ex. of 6 compounds with the empirical formula, CH2O
 CH2O - formaldehyde, C2H4O2 - acetic acid, C3H6O3 - lactic acid, C4H8O4 erythrose, C5H10O5 - ribose, C6H12O6 – glucose
o Molecular formulas
We use percent decomposition to determine the empirical formula
 Rules
 Convert percent into mass by assuming total mass of sample to be 100 g
 Determine number of moles of each element
 Convert moles of each element into whole numbers that become
subscripts by dividing each amount in moles by the smallest amount
 If subscripts are not whole numbers, determine the least common
multiple that will make the decimal values into whole numbers
o Multiply all subscripts by this common multiple
 Use these numbers as subscripts to complete empirical
formula
 Ex. Assume that a compound is 50.91% Zn, 16.04% P, and 33.15% O;
determine its empirical formula.

Ex. Determine the empirical formula for a compound that is found by
analysis to contain 27.37% Na, 1.2% H, 14.3% C, and 57.14% O

Ex. Chemical analysis indicates that a compound is 28.64% sulfur and
71.36% bromine. The molar mass of the compound is 223.94 g/mol.
Determine the molecular formula.
o
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A hydrate is a compound that has a specific number of water molecules bound to each
formula unit
 Often when a crystal forms from a water solution, water molecules are trapped
in the crystal in a specific arrangement
 Ex. CaSO4· 2H2O (s)
 Ionic compounds in solid state can be hydrates or anhydrous compounds
(without water)
 The water molecules usually do not interfere with the chemical activity of the
compound
 Ex. A 50.0 g sample of Ba(OH)2 · xH2O contains 27.2 g of Ba(OH)2.
Calculate the percent by mass of water in Ba(OH)2 · xH2O (s) and find the
value of x.
What is Stoichiometry?
o Stoichiometry is the study of the quantitative relationship among the amounts of
reactant used and the amounts of products formed in the chemical reaction
 A balanced chemical equation is required for any calculations; we use them in
ratios
o The coefficients in front of the chemical formulas in a balanced chemical equation
represent the relative numbers of particles involves in the reaction
 Ex. 2H2 (g) + O2 (g) → 2H2O (g)
 Ex. The combustion of octane, C8H18. If 450 molecules of water are produced,
how many molecules of carbon dioxide are produced

Ex. What amount in moles of CuO (s) forms when 0.0045 mol of malachite,
Cu2(CO3)(OH)2 (s) decomposes completely according to the following equation:
Cu2(CO3)(OH)2 (s) → CO2 (g) + H2O (g) + 2 CuO (s)

Ex. An astronaut produces an average of 1.0 x 103 g of CO2 each day. What mass
of oxygen would need to be produced by photosynthesis is each day? Mass of
water? Mass of glucose?
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Limiting and Excess Reactants
o A limiting reactant is completely consumed during a chemical reaction, limiting the
amount of product that is produced
o An excess reactant remains after a reaction is over
o The limiting reactant is not necessarily the reactant that is present in the smaller
amount, but the one that forms the smaller amount of product
 Ex. The production of ammonia is an important industrial process in the
manufacture of fertilizers. If 4.2 g of N2 (g) reacts with 0.75 (g) of H2 (g), which is
the limiting reactant

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Ex. 2 Al (s) + Fe2O3 (s) → Al2O3 (s) + 2 Fe (l) If 113 g of aluminum powder is
mixed with 279.50 g of iron (III) oxide, what mass of molten iron forms?
Reaction Yields
o Stoichiometric calculations allow you to calculate the amount of product that forms, in
theory, in a chemical reaction
 Called theoretical yield
o However the theoretical yield is not always the same as the amount of product that
actually forms
o The actual amount of product that forms in a chemical reaction during a lab is the actual
yield
o Factors that affect yield:
 Competing reaction
 A reaction that occurs along with the main reaction
 Involves the same reactants or products and therefore lowers the yield
of the main (principal) reaction
o Ex. Principal reaction
 C3H8 + 5 O2 → 3 CO2 + 4H2O
 Competing reaction
o 2 C3H8 + 7 O2 → 6 CO2 + 8H2O
[incomplete combustion]
o Ex. 2 P + 3 Cl2 → 2 PCl3
 PCl3 + Cl2 → PCl5
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o
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Reaction rates
 Speed of reaction can be affected by surface area, temperature,
pressure, and reaction vessel conditions
Purity of reactants
 Ex. Lab grade acetic acid compared to vinegar
o Vinegar is 5% acetic acid, 95% water
Laboratory techniques used to collect final products
 If product is soluble or clings to glassware
Calculate % yield
% yield :

Ex. If 20.0 g of glucose reacts but only 1.40 g of ethanol is produced, what is the
% yield of the reaction.
 C6H12O6 (aq) → 2 C2H5OH (aq) + 2 CO2 (g)

Ex. If 49.0 g of potassium phosphate is recovered after 49.0 g of phosphoric acid
reacts with 49.0 g of potassium hydroxide, what is the % yield of the reaction?