Transcript Stoichiometry - COHS IB and CP Chemistry

Stoichiometry
Relationships between
reactants and products
in a chemical reaction
General definitions

Element- a pure substance which cannot be
broken down by any ordinary chemical means.
 Molecule- two or more atoms chemically bonded
together. Example O2 , H2O
 Compound- two or more elements chemically
bonded together. Physical and chemical
properties change.
 Solution: homogeneous,
containing 2 or more
substances. No reaction to make
 Mixtures:
Heterogeneous, The substances
are not evenly distributed. Ex. Flour in
water,
 Physical Change: There is no change in
the chemical formula of the substance. Ex.
H2O (s) H2O (l) have same formulas,
chemical and some physical properties.

Color, texture, density, taste, rate of
evaporation, boiling and melting points

Chemical Change: In a chemical change there is
a change in the arrangement of the atoms, that
is, the chemical formula changes.
 Chemical properties: flammability, temp. at
which something ignites, oxidation (rusting),
acidity, pH, reducing ability, etc.
 Some indications of chemical changes: heat
produced, change in color or smell, gas or
flames produced, precipitation.
Classification of Matter
Matter
Pure Substances
Elements
Compounds
Mixtures
Homogeneous
Mixtures
Solutions
Heterogeneous
Mixtures
The Mole
 What


is it?
A unit of measurement which equals 6.0 x
1023 of something
Similar to 1 dozen = 12 of something
 Why


do we use it?
Other conventional quantities such as “one
dozen” are too small a number to represent
atoms
Gives us a way to “count” atoms
ex. 1g H = 1mole of H = 6.0 x 1023 atoms
 How


the mole was developed.
The mass of each type of atom was determined using
a mass spectrometer.
ex. H = 1.66 x 10-24g/atom
The masses of the atoms were compared and ratios
were developed.
ex. He: 6.64 x 10-24 = 4.0 (amu) relative mass ratio
H 1.66 x 10-24
Mass Spectrometer
Mass spectrometer
The Mole cont’d.

The ratios were used to make standard
masses.
• Made “H” have a mass of 1g. Other elements
masses are determined by multiplying 1g by the
ratio.
ex. 1.0 g H
x 4.0 for He (since He is 4 times heavier)
4.0g for He

These masses represent one mole of the
element and are A.M.U. or atomic mass units.
x 1023 was determined by:
Atomic mass
= # of atoms in mass
Mass of one atom
ex. 1.00g H
= 6.02 x 1023atoms
1.66 x 10-24g/atom
 6.02
Amedeo Avogadro
Mole conversions
Chemical Formulas
Molecular Formula
-Elements are written
in their # of atoms
found in nature.
-This is often not the
lowest ratio.
-Hard to determine in
the lab.
C6H12O6
Empirical Formula
-Elements are
written in the
lowest ratio.
-Easy to
experimentally
determine.
CH2O
Empirical Formula

Ex. You are heating a 0.2636g sample of Ni and reacting
with oxygen. After reacting the nickel oxide weighed
0.3354g. Find the empirical formula.

-

Find the mass of each element
.3354g Ni ox.
.2636g Ni  masses
0.0718g O  mass
Convert to moles.
.2636g Ni x 1mol Ni = .004491mol Ni
58.69g
.0718g O x 1mol O = .004490mol O
16.00g
Empirical Formula cont’d.


Make ratio – use smallest mole value on the
bottom.
.004491mol Ni = 1mol Ni
.004490mol O
1mol O
Use the ratio to determine the formula.
NiO
Ex. An Aluminum oxide compound was found to
contain 4.151g Al and 3.692g O. What is
Aluminum oxides empirical formula?
• Determine masses.
4.151g Al, 3.692g O
• 4.151g Al x 1mol Al = 0.1539mol Al
26.98g
3.692g O x 1mol O = 0.2308mol O
1
16.00



Ratio
0.2308mol O = 1.5mol O
0.1539mol Al 1mol Al
Multiply the ratio by the smallest integer in
order to get whole #s.
2 x (1.5) = 3mol O
1
2mol Al
Formula.
Al2O3
Empirical Formula cont’d.
Ex. A Platinum compound was found to contain by mass:
65.02% Pt, 9.34% N, 2.02% H, and 23.63% Cl.
Determine its empirical formula.
 Find the masses.
65.02g Pt, 9.34g N, …
 65.02g Pt x 1mol Pt = 0.3333m Pt
195.1g
9.34g N x 1mol N = 0.667m N
14.0
2.02g H x 1mol H = 2.00m H
1.00
continued
 Ratios
N : 0.667m = 2 N
H = 2.00m H = 6H
Pt 0.333m 1 Pt
Pt 0.3333m Pt 1
Cl = 0.6667mol C = 2.0 Cl
Pt 0.3333mol Pt 1 Pt
 Formula?
PtN2H6Cl2
Percent Mass



Calculating the % by mass of an element in a molecular
formula/empirical formula.
Equation: Mass of element (in 1mole) x 100% =
Molecular mass
Find the % mass of Chromium in the molecular formula K2Cr2O7.




Find molecular mass:
K2Cr2O7 = 294.2g
Find mass of element in compound.
Cr2 = 104g
Substitute.
Mass of element x 100%
Molecular mass
104g Cr x 100 = 35.4% Cr
294.2g
% mass can be useful for determining the empirical formula of a
compound.
Chemical Equations

Definitions
ex. 2H2(g) + O2(g)  2H2O(g)
everything on the left of the arrow is a reactant,
everything on the right of the arrow is a product.
states of matter
(g) = gas
(l) = liquid
(s) = solid
(aq) = aqueous (dissolved in H2O)

symbol can be interpreted as makes,
produces, yields, decomposes
materials above and below the
arrow:
MnO2
ex. H2O2  H2O + O2
chemicals = catalysts
e- = electricity
Δ = heat
Types of Chemical Reactions




Single Displacement
Cu(s) + 2AgNO3(aq)  2Ag(s) + Cu(NO3)2(aq)
Double Displacement
CoCl2(aq) + 2NaOH(aq)  Co(OH)2(s) + 2NaCl(aq)
Decomposition
2H2O2  2H2O + O2
(NH4)2Cr2O7(s)  N2(g) + Cr2O3(s) + 4H2O(g)
Redox (Oxidation-reduction)
Fe2O3 + 3CO  2Fe + 3CO2

Synthesis
2H2(g) + O2(g)  2H2O(g)
Law of Conservation of Mass
 This
law states that matter is neither
created or destroyed in a chemical
reaction. The mass used in the beginning
of a reaction equals the mass made at the
end of a reaction.
 Balancing: This is a process by which the
written reaction is modified so the number
of atoms on the reactant side equals the
number of atoms on the product side.
Antoine Lavoisier
Percent Yield

Theoretical and actual yield



Percent yield


Theoretical yield – maximum amount of product that
can be produced. Determined by math.
Actual yield – the “actual” amount of product
produced.
Actual yield
x 100 = % yield
Theoretical yield
Problems with % yield. (Either the % yield or the
actual yield must be given in the problem.)