Transcript Ch. 2 notes

Chapter 2
Chemical formulas and
composition stoichiometry
AP Chemistry
Milam
2-1 chemical formulas
• Allotropes are elements with two different
forms like graphite and diamond are both
carbon
• Organic compounds all contain carbon and
usually contain hydrogen, oxygen, sulfur
and nitrogen
2-2 ions and ionic compounds
• Cations are ions with a positive charge
• Anions are ions with a negative charge
• Cations and ions form ionic compounds,
they form crystalline structures and the ratio
of cation to anion is how the chemical
formula is written
• The formula can be deduced using the
charges of the ions
2-2
• The smallest possible whole number ratio is
called the formula unit
• Ex. NaCl, CaBr2, Al2O3
• Polyatomic ions are groups of atoms
bonded together that have a net charge yet
are stable enough to act as a single unit
• It is to your advantage to memorize the
common polyatomic ions within a week
2-2
• Polyatomic ions in compounds have
parentheses when more than one is
necessary to balance charge
• Na2CO3, (NH4)2CO3, Al2(SO4)3
2-3 Names and formulas of some
ionic compounds
• You need to be able to write names and
formulas for ionic compounds without
thinking!
• The name of a compound always gives
enough information to figure out the
formula, and vice versa
• To write the formula, balance the + and charges
2-3
• If you have Na and Cl, Na has a +1 charge
and Cl has a –1 charge, therefore you need
equal numbers of Na and Cl – formula unit
NaCl
• If you have Ca and Cl, Ca has a +2 charge
so you need double the Cl ions – formula
unit CaCl2
• To name, you just give the names of the
ions, cation first and change the ending to
ide
2-3
• Polyatomic ions do not change their ending
to ide
• Some elements have multiple charges,
mostly transition and inner transition
metals, we indicate the charge in the name
with Roman numerals
• There is also a classic naming system that
uses an ic/ous ending to indicate the larger
charge
2-3
• Ex. Fe+2 and Fe+3
• Ferrous and Ferric ions – classic naming
• Iron (II) and Iron (III) – stock naming
system
• FeCl2 – because there are two chlorides
with a –1 charge, we can infer that iron has
a +2 charge, therefore this is Iron (II)
chloride or Ferrous chloride
2-4 Atomic Weights
• Atomic mass units are approximately the
mass of 1 proton or neutron
• The definition is 1/12th the mass of a C-12
atom
• The atomic weights are listed on the
periodic table
2-5 The Mole!
• We use moles all of the time in chemistry,
because we use them in complicated
situations, they seem complicated. Always
remember that moles are just like dozens!
They are a number, 6.02 x 1023 of anything
is a mol!
• A mol is a huge number,
602,000,000,000,000,000,000,000 or 602
sectillion
2-5
• Like amu’s, the mole is defined by the # of
particles in 12 grams of C-12, so the actual
number of 6.0221415 x 1023 but we will stick with
6.02 x 1023 in here.
• You must be able to convert between grams, moles
and # of particles and later we will use moles in
concentration, gas laws, heat transfer,
stoichiometry, electrochemistry, titrations and
other subjects
2-6 Formula weights, molecular
weights and moles
• Formula weight, molecular weight, atomic
weight are all terms that specify that you are
talking about molar mass and what type of
substance you have (ionic, molecular or
elemental)
• In here I will always say molar mass and
that’s what you will see for the most part
everywhere else as well
2-6
• Molar mass is the mass of one mole of a
substance
• Molar masses are obtained from the
periodic table and are equivalent
numerically to the amu’s of a molecule or
formula unit or atom
• To find the molar mass of something, take
the atomic mass of each element and
multiply by the subscripts and then add
them all
2-6 Molar mass of H2SO4
• Molar mass = 2*1.0 grams/mol + 1*32.1
g/mol + 4*16.0 g/mol = 98.1 g/mol
• This means that if I had 1 mol of sulfuric
acid I would have 98.1 grams of it as well
• Follow up, find the molar mass of Ca(OH)2.
• Ans. 74.1 g/mol
• In this class we will always round atomic
masses to the tenths place before calculating
2-7 Percent composition and
formulas of compounds
• To find the % composition, you take the
formula, find the molar mass, and the
masses of the individual elements taking
into account the subscripts then divide
elemental mass by total and convert to a
percentage.
• Ex. H2SO4 has a molar mass of 98.1 g/mol
2-7
• There are 2 hydrogens, so 2 g/mol divided
by 98.1 g/mol gives 0.0203 or 2.03%
• 1 Sulfur is 32.1 g/mol divided by 98.1 g/mol
gives .3272 or 32.72 %
• 4 oxygens are 64 g/mol divided by 98.1
g/mol gives .6524 or 65.24%
• The total is 2.03 + 32.72 + 65.24 = 99.99
which is close enough to 100%
2-8 Derivation of formulas from
elemental composition
• Empirical formulae are formulae with the
subscript ratios reduced to the lowest
possible whole number ratios, hydrogen
peroxide has a formula H2O2, and its
empirical formula is HO
• If you know the % composition of a
substance, you can calculate the empirical
formula and this can be slightly difficult
2-8
• Determine the empirical formula for a compound
with a % composition of 87.5 % nitrogen and
12.5% hydrogen
• Step 1 – assume you have 100 grams of total stuff,
so now you’ll have 87.5 grams of N and 12.5 g of
H
• Step 2 – convert each into moles
• 87.5 grams N = 6.25 moles N
• 12.5 g H = 12.5 mol H
2-8
• Step 3 – take the smallest # of moles and
divide all of the moles by that number
• 6.25 mol N/ 6.25 mol = 1 mol N
• 12.5 mol H/6.25 mol = 2 mol H
• These amounts become your subscripts
• Ans. NH2
• Most times the final answers will be
decimals such as 1.98 or 3.01, just round
them
2-8
• Also if you end up with numbers like 1 and
2.5 and 2, you need to double all of them to
get 2, 5 and 4 so that you only have whole
numbers
• Most people get stuck at step 3 and don’t know
what to do when they get decimal results,
remember that you are using an arbitrary starting
point (100 g) and you need to adjust by dividing
by the smallest number of moles to get whole
numbers
2-9 Determination of molecular
formulas
• Determining empirical formulas is difficult
and this is an extra step added on, but
empirical formulas are not what we always
want. Most of the time we want more
information so we want to find the
molecular formula. For example, if you tell
me something has an empirical formula of
CH, I don’t know whether it is benzene
(C6H6) or ethyne (C2H2) so we need more
information
2-9
• Once you find an empirical formula, the
extra information to find the molecular
formula is the molar mass
• For example, you determine CH to be the
empirical formula, which has a molar mass
of 13 g/mol
• If I tell you the molar mass if 78 g/mol, you
know there must be 6 times the mass, so
your molecular formula is C6H6
2-9
• Going from empirical formula to molecular
is a simple step, but when added into the
whole problem it makes steps 3 and 4 a bit
more confusing
2-10 Some other interpretations
of Chemical Formulas
• At this point there are now too many
problems to remember all of the steps to
follow for each type of problem, you need
to be able to solve what the steps are by
understanding the problem.
• In these situations it is useful to try and find
the first and last steps so you know what
you have and what you need to end up with
2-10
• Problem types include:
– Finding the mass of an element with the mass
of the compound and vice-versa
– Finding the formula of a hydrate*** this will be
done in lab and is likely to show up on an AP
exam
2-11 Purity of samples
• Not likely to end up on an AP exam, if it
does it should be solveable or qualitative in
nature