Each type of problem is worked out only once. Several other examples are
also given with answers in parentheses for you to practice. I will use
these in lecture as an instructional tool.
Naming Problems: See chart of ion names and activity for practice problems.
Generally, the main tool for ensuring that your formula or name is
correct is to be able to identify ions (monatomic and polyatomic) and
their charges. Charges must balance so that the ionic compound as a whole
has a charge of 0. Remember that metal ions often can have more than one
charge (but not at the same time!) so that you can have, for example,
Fe^{2+} [iron (II)] and Fe^{3+} [iron (III)]
Molar Mass: The molar mass of a compound
is the total of the molar masses of its constituent atoms. The molar mass
of an element is the mass given in the periodic table expressed in grams.
This is the average atomic mass and takes into account the mass of each
elements’ isotopes and their abundance.
The molar mass of NaCl is 58.44 g/mol because 22.99 g/mol (Na) + 35.45
g/mol (Ca) = 58.44 g/mol. CaCl_{2}: Ca, 40.08 g/mol; Cl, 35.45 g/mol
(1 × 40.08 g/mol) + (2 × 35.45 g/mol) = 110.98 g/mol
When a compound contains a polyatomic ion or a repeated unit, multiply
the subscripts inside the parentheses by the subscript outside the
parentheses: Ca_{3}(PO_{4})_{2}: Ca, 40.08 g/mol; P,
30.97 g/mol; O, 15.99 g/mol
(3 × 40.08 g/mol) + (2 × 1 × 30.97 g/mol) + (2 ×
4 × 15.99 g/mol) =
Percentage Composition: Each element in a
compound forms a percentage of the total molar mass of that compound. The
percentage composition of a compound gives the percent of the total mass
of the compound represented by a particular element. The percentage by
mass of an element can be calculated by dividing the molar mass of the
element, times its subscript in the formula, by the total molar mass of
the compound. To express the answer as a percent instead of a decimal
fraction, multiply the answer by 100. The procedure can be repeated for
each element in the compound and the resulting percentages should add up
to 100%.
H_{2}O
percentage hydrogen: mass of 2 mol H/molar mass of H_{2}O = 2
× (1.01 g/mol)/(18.02 g/mol) = 0.1121
multiply by 100 to get the answer as a percent: 0.1121 x 100% =
11.21% H
percentage oxygen: mass of 1 mol H/molar mass of H_{2}O = (16.00
g/mol)/(18.02 g/mol) = 0.8879
multiply by 100 to get the answer as a percent: 0.8879 x 100% =
88.79% O 88.79% O + 11.21% H = 100% H_{2}O
Note: once you figure out the percentage for all but one element in a
compound, just subtract the total percentages so far from 100% to find
the last.
Empirical Formulas: You can use
percentage composition data to find the formula of a compound. Rather, to
be more specific, you can find the empirical formula of the
compound. If you also have the total molar mass of the compound, you can
usually find the molecular formula, too. The simplest way to find the
formula is to assume you have 100 g of the compound and use the
percentages given as amounts in grams. Convert grams to moles for each
element and find the smallest whole number ratio of the moles of each
element.
Given: a compound consisting of 63% Mn and 37% O
Assuming 100 g of the compound, you have 63 g Mn and 37 g O
63 g Mn × (1 mol Mn)/(54.94 g Mn) = 1.1 mol Mn
37 g O × (1 mol O)/(16.00 g O) = 2.3 mol O
To find the smallest whole number ratio divide the number of moles of
each element by the number of moles found for the element present in the
smallest number of moles:
1.1 mol Mn/1.1 = 1 mol Mn 2.3 mol O/1.1 = 2.1 mol O
The best ratio is Mn:O::1:2 and the formula is MnO_{2}
Others to try: 39.32% Na, 60.66% Cl (NaCl); 30.45% N, 69.55% O
(NO_{2}); 34.97% Fe, 65.03% O (Fe_{2}O_{3});
40.00% C, 53.28% O, 6.72% H (CH_{2}O)
Molecular Formulas: Using an empirical
formula you can find the molecular formula of a compound if you know its
molar mass. All you need to do is find the formula weight of the
empirical formula and find out how many times this mass can be divided
into the given molar mass.
Given: empirical formula: CH_{2}O; molar mass: 60.06
g/mol
Find the molar mass of the empirical formula (this is called the formula
weight): it is 30.03 g/mol (see above for
method)
Divide the given molar mass by the formula weight: 60.06/30.03 = 2
The factor you are looking for is 2: multiply this factor by the
subscripts in the empirical formula to find the molecular formula:
C_{2}H_{4}O_{2}
Finding the Oxidation Number: Use the
following rules to assign oxidation numbers:
1) Free elements have an oxidation number of 0 (H_{2},
Br_{2}, Na, Be, K, P, Fe, etc. have an oxidation number of
0).
2) For ions made of only one atom, the oxidation number is the same as
the charge. All alkaline metals have an oxidation number of +1, alkaline
earth metals have an oxidation number of +2 and aluminum always has an
oxidation number of +3 in compounds.
3) Oxygen almost always has an oxidation number of –2 (exceptions
are in H_{2}O_{2} and O_{2}^{2–}
where it is –1)
4) Hydrogen has an oxidation number of +1 unless bonded to metals in
binary compounds when it is –1.
5) Fluorine always has an oxidation number of –1. Other halogens
generally have negative oxidation numbers except when they appear with O
in acids when they have positive oxidation numbers.
6) The sum of the oxidation numbers of the elements in a neutral
molecule must add up to zero. In a polyatomic ion the oxidation numbers
must add up to the charge on the ion.
7) Oxidation numbers are usually but do not have to be integers: oxygen
in O_{2}^{–} has an oxidation number of
-½
Count the atoms of each element in the equation in its unbalanced
form. Make a table for the reactant side (left) and one for the
product side (right) showing the quantity of each element or
polyatomic ion.
Balance the atom or polyatomic ion that appears only once on each
side by inserting a coefficient in front of the whole formula.
Note: you cannot change the subscripts in the formulas
since this would change the substance in the reaction! Sometimes an
element or polyatomic ion will appear more than once on one side.
Save these instances for later.
Balance the other elements and polyatomic ions one by one, leaving O
and H for last.
Update your table for each change you make to the equation.
When you are done, each element or polyatomic ion will have the same
number next to it in both the reactant table and the product table.
Some other things to keep in mind:
Treat polyatomic ions as units if they appear on both sides of an
equation, as they typically will. Do not count the atoms inside
the polyatomic ion since you might get them confused with the
same atoms that are not part of the ion.
Sometimes a change you make in the middle of the balancing
process will cause one of the other elements or polyatomic ions
to go out of balance. Don’t let this bother you: you will
just go back and update the coefficients as you go along.
Think in terms of the mathematical concept of least common
multiples (LCMs). You can figure out how many atoms or ions are
on each side of the equation; just find the LCM of the two
numbers to balance that unit.
The concept of the LCM is important because the ratios between
substances in the reaction should be expressed in lowest terms
Be observant: sometimes (for example, in redox reactions) the polyatomic
ions will not be on both sides of the equation. In those cases treat each element
as if it were not part of a polyatomic ion.
Finally, make sure you have the correct formulas for all of your
reactants: incorrect formulas lead to incorrect equations!
Stoichiometry: The key to
stoichiometry is to always convert everything to moles before you even
begin. When you have found the answer you can always convert to mass or
volume.
Here are some sequences with which to work in order to solve different
types of problems:
1. Starting with the number of moles of one of the substances in the balanced
chemical equation you can easily find the number of moles of one of the other
substances using the relevant molar ratio. Make
sure to put the substance you know on the bottom of the ratio and the substance
you need to find on the top of the ratio before you multiply.
2. Starting with a mass requires a further step: convert that mass into a number
of moles using the molar mass
of the substance. Then simply follow the same steps used above.
3. If the answer required is a mass, simply convert the number of moles
you find to the mass using the molar mass of the substance in
question.
4. If you are given a volume and a density, simply find the mass using D
= m/V and then calculate moles and complete the problem. Or simply use
this five-step method:
Write correct formulas for all reactants and products, and balance
the resulting equation.
Convert the quantities of some or all given or known substances
(usually reactants) into moles.
Use the coefficients in the balanced equation to calculate the number
of moles of the sought or unknown quantities (usually products) in
the problem.
Using the calculated numbers of moles and the molar masses, convert
the unknown quantities are required (usually grams).
Check that your answer is reasonable in physical terms.
C_{3}H_{8} + 5O_{2} → 3CO_{2} +
4H_{2}O Moles Example: You are told that a bottle contains 40 mol
C_{3}H_{8} (propane) and asked how many moles of
O_{2} you would need to burn it all.
The molar ratio is 5 mol O_{2} to 1 mol
C_{3}H_{8}.
40 mol C_{3}H_{8} × 5 mol O_{2}/1 mol
C_{3}H_{8} = 200 mol O_{2} Mass Example: Given 32 g of O_{2} and an excess of
C_{3}H_{8} how much water would the reaction produce, in
grams?
Convert mass to moles: 32 g O_{2} × 1 mol/16 g = 2 mol
O_{2}
Use molar ratio (4 mol H_{2}O to 5 mol O_{2}) to find
moles of water: 2 mol O_{2} × 4 mol H_{2}O/5 mol
O_{2} = 1.6 mol H_{2}O
Convert moles back to mass: 1.6 mol H_{2}O × 18.02 g/1 mol
= 28.83 g
Others to try: S + 6HNO_{3} → H_{2}SO_{4} +
6NO_{2} 2H_{2}O
starting w/14 g S; how many grams of HNO_{3} are needed? (165.07
g)
3 mol H_{2}O produced; how many grams of S reacted? (48.098
g)
10.0 L NO_{2} produced (D = 2.05 g/L); how many moles of S
reacted? (3.74 ×10^{-1} mol)
Limiting Reagents: In nearly any
reaction it will be the case that the physical quantities of the
reactants will not be present in a perfectly stoichiometric amount. One
of the reactants will limit the total output of the reaction because it
will be used up first. These examples will show how to figure out which
reactant is the limiting reactant and which is the excess reactant.
Generally, this is simply a special case of a stoichiometric calculation.
Given an amount for both/all reactants you need to find out which one
will run out first. The good thing about this is that it does not matter
which of two reactants you choose, you can find the limiting reactant
either way.
S + 3F_{2} → SF_{6} Four moles of S are added to 20
moles of F_{2}
The number of moles of F_{2} needed to completely react with four
moles of S is: 4 mol S × 3 mol F_{2}/1 mol S = 12 mol
F_{2}
Thus, F_{2} is the excess reactant since we have more than 12
moles of fluorine gas and S the limiting reactant
But say we started with F_{2}:
The number of moles of S needed to react with 20 moles of F_{2}
is: 20 mol F_{2} × 1 mol S/3 mol F_{2} = 6.7 mol
S
Thus, S is the limiting reactant since we do not have this much sulfur
and F_{2} is the excess reactant
Once you have found which is the limiting reactant you can use that molar amount to find the theoretical yield of the reaction. Remember the advice from the general stoichiometry problems: convert masses to moles before trying to solve the problem!
Others to try: 2Al + Fe_{2}O_{3} → Al_{2}O_{3} + 2Fe
100 g Al/150 g Fe_{2}O_{3} (1.85 mol Fe_{2}O_{3} needed: this is the limiting reactant since only 0.94 mol are present; 0.94 mol Al_{2}O_{3} produced; 95.84 g)
124 g Al/601 g Fe_{2}O_{3} (2.3 mol Fe_{2}O_{3} needed: this is the excess reactant; 2.3 mol Al_{2}O_{3} produced; 235 g)
Percent Yield: Limiting reagents are not the only thing that limits the yield of a given product. The actual yield of a reaction is always less than 100% of what the stoichiometry predicts. Sometimes how much less is very difficult to measure, so that it appears that 100% product formation has occurred. Generally, however, much less than 100% of the theoretical yield is achieved. Therefore it is useful to know how to find the percent yield.
5Ca + V_{2}O_{5} → 5 CaO + 2V Theoretical Yield
Calculate the theoretical yield when 1.54 × 10^{3} g of V_{2}O_{5} react with 1.96 × 10^{3} g of Ca.
First, we find the limiting reactant:
1.54 × 10^{3} g · 1 mol/181.881 g = 8.47 mol V_{2}O_{5}
1.96 × 10^{3} g · 1 mol/40.078 g = 48.9 mol
The number of moles of Ca to react completely with 8.47 mol V_{2}O_{5} is 8.47 mol V_{2}O_{5} · 5 mol Ca/1 mol V_{2}O_{5} = 42.4 mol
So Ca is in excess and we should use the molar amount of V_{2}O_{5} to determine the theoretical yield of V.
8.47 mol V_{2}O_{5} · 2 mol V/1 mol V_{2}O_{5} = 16.9 mol V Percent Yield
What is the percent yield if 803 g of V are obtained?
First, find out how many grams the theoretical yield of V represents: 16.9 mol · 50.941 g/mol = 863 g
Then just divide the actual yield (given) by the theoretical yield: 803g/863g = 0.930 × 100% = 93.0%
(notice how the units cancel out completely to give a unit-free ratio)