Transcript Quantitative Chemistry - Cathedral High School

QUANTITATIVE CHEMISTRY
Chapter 1
MOLE CONCEPT & AVOGADRO’S
CONSTANT
All units will be metric in nature
 Use SI system – System Internationale
 Volume – standard is m3 but dm3 and cm3 are
more commonly used in lab setting



1 L = 1 dm3
1 mL = 1 cm3
Mass – standard is kg, more common to use
grams in lab setting

1 kg = 2.2 lbs
1 lb = 454 g
MOLE CONCEPT & AVOGADRO’S
CONSTANT
If a substance is an atom we count atoms of it.
 If a substance is a compound we usually count
molecules of it.
 Mole (mol) – amount of a substance that contains
the same number of chemical species as there are
in exactly 12 grams of the isotope carbon-12.

MOLE CONCEPT & AVOGADRO’S
CONSTANT
Relative atomic mass (Ar) – average atomic mass
of the element, taking into account all its isotopes
and their relative abundance, compared to one
atom of carbon-12.
 Molar mass (M) – the mass of one mole of a
species. It is the mass expressed in grams and
has units of g mol-1. All molar masses will be
expressed and rounded, if necessary, to two
numbers after the decimal (16.00 would not be
rounded to 16).

MOLE CONCEPT & AVOGADRO’S
CONSTANT
Molecular formula – composition of a molecule
(elements that make up the compound).
 Relative molecular mass (Mr) – sum of the
relative atomic masses of the atoms in the
molecular formula.
 Example: C6H5OH
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MOLE CONCEPT & AVOGADRO’S
CONSTANT
If you are dealing with an ionic compound we
must call it the relative formula mass (even
though it’s really the same thing).
 Ionic compounds are made up of ions and
generally start with a metal.
 Relative formula mass – sum of the relative
atomic masses of the ions in a formula.
 Example: Ca(NO3)2

MOLE CONCEPT & AVOGADRO’S
CONSTANT
Acceptable shorthand: n = moles, m = mass and
M = molar mass
 # of moles (n) = mass (m) / molar mass (M)
 Avogadro’s constant (L) = 6.02 x 1023 mol-1; # of
atoms in 1 mol of an element and # of molecules
in 1 mol of a molecular compound.
 # of particles (N) = # of moles (n)
x Avogadro’s constant (L)

FORMULAS
Empirical formula – simplest ratio of the atoms
of different elements in a compound. It is the
molecular formula expressed as the simplest
ratio.
 Molecular formula – shows the actual number of
atoms of each element in the compound. You
must have the molar mass in order to determine
the molecular formula.

CHEMICAL EQUATIONS
The coefficients in a balanced chemical equation give
the molar ratios of the reactants and products.
 When balancing equations, first balance the elements
that appear only once on each side of the arrow.
 If a polyatomic ions appears on both sides of the
equation, do not separate it into individual elements.
 Balance O and H last as they are usually the most
difficult to balance and show up most in an equation.

MASS AND GASEOUS VOLUME
RELATIONSHIPS IN CHEMICAL REACTIONS
Mass (m) = # of moles (n) x molar mass (M)
 In stoichiometry problems , follow these steps:
1.Convert grams to moles by dividing by M.
2.Convert moles of what you start with to moles of
what you are trying to find using
the
coefficients of the balanced equation.
3.Convert moles of what you are trying to find
back to grams by multiplying by the molar
mass of what you are trying to find.

MASS AND GASEOUS VOLUME
RELATIONSHIPS IN CHEMICAL REACTIONS
Limiting reactant/reagent – reactant that is used
up first in a chemical reaction. It is the
substance that makes a smaller amount of
product.
 Amount of product produced (theoretical yield) –
comes from the limiting reactant and is the
smaller of the two amounts that could be
produced. It is the most product that could be
produced from the reactants you start with.
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MASS AND GASEOUS VOLUME
RELATIONSHIPS IN CHEMICAL REACTIONS
To find the amount of the excess reactant,
subtract the amounts of the two products and
convert the difference back to the excess reactant
(reactant the made the larger amount of product).
 % Yield = Experimental (Actual) Yield x 100
Theoretical Yield
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MASS AND GASEOUS VOLUME
RELATIONSHIPS IN CHEMICAL REACTIONS
Reasons % Yield may not be 100%
 Reaction is incomplete
 Side reaction occurring that is unwanted
 Complete separation of the product from the
reaction mixture is impossible (usually in
filtering)
 Product is lost during transfer process
 Mixture is not completely dried to evaporate off
all of the water (% yield would be over 100%).
 Why is % yield an important concept in industry?
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MASS AND GASEOUS VOLUME
RELATIONSHIPS IN CHEMICAL REACTIONS
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Kinetic Theory has 2 basic ideas
All matter consists of particles in motion
 As the temperature increases, the movement of
particles increases
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The state of matter at a given temperature and
pressure is determined by the strength of the
interparticle forces.
MASS AND GASEOUS VOLUME
RELATIONSHIPS IN CHEMICAL REACTIONS
Solid – interparticle forces restrict movement to
vibration about a fixed position.
 Liquid – interparticle forces are sufficiently weak
to allow particles to change places with each
other, but their movement is constrained to a
fixed volume.
 Gas – interparticle forces are negligible.
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MASS AND GASEOUS VOLUME
RELATIONSHIPS IN CHEMICAL REACTIONS
All temperatures in Kelvin (K = oC + 273)
 Absolute zero – lowest temperature attainable
and where all movement stops.
 Average kinetic energy is directly proportional to
Kelvin temperature.
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MASS AND GASEOUS VOLUME RELATIONSHIPS
IN CHEMICAL REACTIONS
Changes of state
 Below 0oC water is solid at normal pressure.
 When ice hits 0oC, it begins to melt, but
temperature does not increase until all the ice
melts.
 Water begins to boil at 100oC at normal pressure,
but temperature does not increase again until
after all water is turned to gas.
MASS AND GASEOUS VOLUME RELATIONSHIPS
IN CHEMICAL REACTIONS
Freezing – liquid to solid
 Melting – solid to liquid
 Boiling – liquid to gas
 Condensing – gas to liquid
 Deposition – gas directly to solid
 Sublimation – solid directly to gas

MASS AND GASEOUS VOLUME RELATIONSHIPS
IN CHEMICAL REACTIONS
 At
the melting point, the energy that is
added is used to break the intermolecular
bonds of the solid.
 At the boiling point, the energy that is
added is used to break the intermolecular
bonds of the liquid.
 Kinetic energy depends on mass (m) and
speed (v). All gases at the same
temperature have the same average
kinetic energy.
 KE = ½ mv2 , more massive particles at
MASS AND GASEOUS VOLUME RELATIONSHIPS
IN CHEMICAL REACTIONS
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Volatile substances have low boiling points, but
are not necessarily flammable.
Avogadro’s hypothesis – equal volume of different
gases contain equal numbers of particles at the
same temperature and pressure.
MASS AND GASEOUS VOLUME RELATIONSHIPS
IN CHEMICAL REACTIONS
Standard conditions or standard temperature
and pressure (STP) – 273 K and 100 kPa,
replaces previous standard pressure of 1 atm
(101.3 kPa).
 One mole of a gas at STP occupies 22.4 dm3 or
22,400 cm3 (22.4 L).
 Number of moles (n) =
volume (V)
molar volume (Vmol)

MASS AND GASEOUS VOLUME RELATIONSHIPS
IN CHEMICAL REACTIONS
Gas Laws
 An increase in the frequency or energy of gas
particle collisions will increase the pressure
 An increase in volume decreases frequency of
collisions so pressure decreases (assuming Kelvin
temp and amount of gas are constant.
 Pressure and volume are inversely proportional
(P = k1/V), k1 is a constant. This is Boyle’s law.
MASS AND GASEOUS VOLUME RELATIONSHIPS
IN CHEMICAL REACTIONS
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Pressure is directly proportional to Kelvin (absolute)
temperature assuming volume and amount of gas are
constant.
P = k2T where k2 is a constant, this is Charles’ law.
Volume is directly proportional Kelvin temperature
assuming pressure and amount of gas are constant.
V = k3T where k3 is constant, this is Gay-Lussac’s
law.
MASS AND GASEOUS VOLUME RELATIONSHIPS
IN CHEMICAL REACTIONS
Combined Gas Law – combines Boyle’s, Charles’
and Gay-Lussac’s laws holding only the amount
of a gas constant.
 P1V1/T1 = P2V2/T2

MASS AND GASEOUS VOLUME RELATIONSHIPS
IN CHEMICAL REACTIONS
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Ideal Gas Law is PV = nRT, where R is the ideal
gas constant (8.31 J K-1mol-1)
P should be in units of Pa (1 kPa = 1000 Pa) & V
should be in units of m3 (1m3 = 1000 dm3)
1 atm = 101.3 kPa = 760 mm Hg = 760 torr
29.92 inches of Hg
=
MASS AND GASEOUS VOLUME RELATIONSHIPS
IN CHEMICAL REACTIONS
PV = nRT can be used to find molar mass.
 n = m/M and can be rearranged in PV = nRT to
give M = mRT/PV
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Density: r = m/V
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M = rRT/P when density is in g m-3
SOLUTIONS
Various containers can be used to measure volumes,
including erlenmeyer flasks, beakers, graduated
cylinders, volumetric flasks and burettes.
 Solutions are made up of a solute and a solvent.
 Solute – less abundant part of solution
 Solvent – more abundant part of solution, usually
water if not specifically stated.

SOLUTIONS
Composition of a solution is expressed in concentration
(molarity).
 When solvent cannot hold any more solute it is said to
be saturated. If any more solute is added it will not
dissolve.
 Units for concentration are amount of solute in moles
dissolved in 1 dm3 of solvent, so units are mol dm-3 or g
dm.
 Square brackets are used to represent concentration.
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SOLUTIONS
Standard solution – solution of known
concentration.
 Dilution – standard solution that is diluted.
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# of moles = concentration x volume

Concentration1 x V1 = Concentration2 x V2
S
OLUTIONS
Steps for Titration
1. A known volume of one solution is
measured
into a flask using a pipette.
2. The other solution is then added from a
burette to find the equivalence point (volume
when reaction is complete).
 In an acid-base titration, an indicator is used to
determine when the equivalence point has been
reached. The most common acid-base indicator is
phenolphthalein.