Transcript Chapter 3: Introduction to chemical formulas and reactivity

Stoichiometry
Chapter 2-6 – 2-12
Chapter 3-1 – 3-8
Key concepts
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Know what a mole is, and how to use it.
Understand the term molar mass and its relationship to
formula weight.
Interconvert between number of particles, moles, and mass.
Understand the term percent composition and know how to
calculate the percent composition of an element in a formula.
Use percentage composition to determine the empirical
formula of a substance.
Understand how to formulate a chemical equation.
Understand how to balance chemical equations by inspection.
Know how to use moles and a chemical formula to determine
quantitative information about chemical reactions.
Understand the terms theoretical yield and limiting reactant.
Introduction to solutions in chemical reactions
The mole
• THE MOLE IS A COUNTING UNIT!!!!
– You can have a mole of anything.
– 1 dozen = 12 “things”
– 1 gross = 144 “things” (a dozen dozen)
– 1 mol = 6.0221367  1023 “things”
Molar Mass
• 1 mole = number of atoms in 0.012 kg of carbon12.
• 1 amu = 1/12 mass of a carbon-12 atom.
• As a result, the atomic weight of an element (in
amu) is __________________ to the mass of
one mole of that element (in g/mol). This is
defined as the molar mass of the element.
• Review: Is molar mass an intensive or extensive
property?
Formula weight/molecular weight
• We may use the atomic
weight (in amu) of
elements in a molecular
formula (or formula unit)
to determine the
molecular weight (or
formula weight) of the
molecule (unit).
• Examples: Fe2(CO3)3,
C11H22O11
MW   ( AW )(# _ of _ atoms)
Converting from mass to moles
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Just as the atomic weight is numerically equal
to the molar mass of an atom, the molecular
weight (formula weight) is numerically equal to
__________________.
Now we have a powerful conversion between
the mass of a substance and the moles (or the
number of particles we have). Why is it so
important to have this kind of conversion? (or
is it important at all?)
Converting mass  moles
• We will need to determine the molar mass
of the substance(s) in question.
• Examples:
– How many moles of CO2 are in 0.56 g?
1.Determine molar mass
2.Convert from mass to moles
– What is the mass of 0.45 mol of lithium
carbonate?
% composition of a compound
• (we discussed this
before, but now we have
a better way to work it.)
• Use the chemical formula
and molar masses to
determine the relative
mass of each element in
the compound.
• What is the % C in
glucose?
%X 
( MM X )(# _ of _ X )
100%
MM compound
Empirical formula
• The relative number of each kind of atom
in a substance
• To determine empirical formulas, you must
count atoms. Therefore, you must use
moles, a counting unit, not grams.
Determining empirical formulas
1. assume 100 g of sample (when % is given). If
actual masses are known they may also be
used.
2. convert from grams of element to moles of
element
3. Ratio all other elements against the element
with the smallest number of moles (not mass).
• Example: a compound contains 52.9 % Al and
47.1 % O.
More examples
1. A certain compound is 73.9% Hg and 26.1% Cl
by mass. What is the empirical formula?
2. (BLB) A sample of a substance is analyzed
and found to contain 5.28 g Sn and 3.37 g F.
What is the empirical formula?
3. (BLB) A sample contains 11.66 Fe and 5.01 g
oxygen. What is the empirical formula?
• For ionic compounds, the empirical formula is
generally the formula unit.
• For molecular compounds, molecular formula is
not always the empirical formula, but will always
be some whole number multiple of the
subscripts of the atom.
• The relative ratio between the atoms must be
preserved (if you multiply one subscript by a
number, you must multiply all subscripts by the
same number)
– Law of constant composition.
• What would you need to know in order to find
the molecular formula using the empirical
formula?
Combustion analysis of compounds
• Done with hydrocarbons and other organic
compounds.
• We use deductive reasoning to determine
the % composition from the data.
• Example: A 1.000 g sample of an alcohol
is burned and produces 1.913 g CO2 and
1.174 g H2O. What is the empirical
formula?
Even more fun…
• (BLB) Nicotine contains C, H, and N. A
5.250 mg sample of nicotine was
combusted, and 14.242 mg CO2 and
4.083 mg H2O were produced. What is
the empirical formula for nicotine?
• The molar mass of nicotine is 160±5
g/mol. What is the molecular formula?
Formula of a hydrate
• A hydrate is a complex where an amount
of water is contained inside an ionic salt
crystal in a specific ratio.
• We may use the empirical formula
calculation process in determining the
formula of a hydrate as well.
Chemical equation:
a visual representation of a chemical reaction
• There are reactants and products in a
chemical reaction.
• Reactants:
• Products:
Conservation of matter
• Matter is not created or destroyed in a
chemical reaction.
• The number of each type of atom in a
chemical reaction must be the same for
reactants as it is for products.
• Sandwich example: write a “reaction” for a
sandwich made of two slices of bread and
one slice or cheese.
Balancing equations by inspection
1. Start with the element found in the fewest
chemical formulas on either side of the
reaction.
2. Fill out remaining elements, changing
coefficients as needed, until the number of
each type of atom is same on both sides of the
equation.
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When balancing equations, we change the
coefficients. we do not change the subscripts.
Changing the subscripts means we have changed
the molecule involved in the reaction (law of
constant composition).
More balancing examples
Using the chemical equation to
determine quantitative information
• A balanced chemical equation gives us the ratio
of reactants and products to each other, whether
in molecules or moles, but NOT the mass ratio.
• example: combustion of methane
• if we know how much we have of one reactant
(or product), we may determine how much of all
other reactants (and products) were used (or
produced) in the chemical reaction.
In general, comparing substance A
and substance B
1. Given mass of A, find moles of A using
molar mass of A
2. use ratio of A and B to find moles of B.
3. Use molar mass of B to find mass of B.
The ratio (called the stoichiometric ratio or
reaction ratio) is the key to solving these
types of problems.
• Let’s do some examples.
Limiting reactants.
• limiting reactant—the reactant that limits how
much product you can get. The limiting reactant
is entirely used up in the reaction.
• Back to the sandwich example: if I have 6 slices
of bread and 4 slices of cheese, which is the
limiting reactant?
• Note: it doesn’t matter if I start with the bread or
the cheese, I’ll still be able to determine the
answer if I use the ratio properly.
• Limiting reactants are always determined
by counting how many molecules are
needed; or, in other words, you must use
moles, not mass.
• Lets try some more examples.
How many grams of NH3 can be
prepared from 77.3 g N2 and 14.2 g H2?
N2 + H2  NH3
• (WGD)
12.6 g of AgNO3 and 8.4 g of BaCl2
are dissolved in water. A solid, AgCl,
forms. How much of this solid, in grams,
can form?
AgNO3 + BaCl2  AgCl + Ba(NO3)2
• (WGD)
Yield
• The yield is the amount produced in a
reaction.
• Theoretical yield:
• Actual yield:
• Percent yield:
Yields
Ethylene glycol (C2H6O2) is formed
according to the following rxn:
• (WGD)
C2H4Cl2 + Na2CO3 + H2O  C2H6O2 + 2 NaCl + CO2
(this is balanced)
• When 27.4 g C2H4Cl2 is used in the rxn,
10.3 g ethylene glycol is formed.
– What is the theoretical yield?
– What is the actual yield?
– What is the percent yield?
Sequential reactions
• The same concepts we have been
learning in the analysis of single reactions
can be applied to multiple reactions. Let’s
look at an example…
Sequential reactions
What mass KClO3 is needed to
provide enough O2 to react completely
with 66.3 g methane?
KClO3  KCl + O2
CH4 + O2  CO2 + H2O
• (WDP)
Break problem into steps; plot your course.
Solutions
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A solution is a ____________________.
A Solution contains a solute and a solvent.
Solute:
Solvent:
• We will discuss ways of expressing the
concentration of a solute in solution. We’ll talk
about the solvation process (process of
dissolving in solution) later in the course.
% by mass
msolute
% solute 
100%
msol'n
This is mass of the solution,
not mass of the solvent.
• In very dilute
solutions, % solute is
an awkward unit to
use. Often, parts-permillion (ppm) or partsper-billion (ppb) are
used instead.
msolute
6
ppm 
10
msol'n
msolute
9
ppb 
10
msol'n
Molarity
• Chemists will often
use molarity because
it is more directly
related to the moles
of solute (and
becomes more useful
in analyzing chemical
reactions).
M
____
nsolute

Vsol'n
Not volume of solvent.
Solution dilution…..
• Suppose we change the volume of a solution,
but we leave the amount of solute the same….
M1V1  M 2V2
Solutions in chemical reactions
• Just as we dealt with the mass of
reactants and products in a chemical
reaction by converting to moles, we may
also use the molarity of a solution in
analyzing chemical reactivity.
• The key is to always compare moles to
moles.