In This Lesson:
Chemical
Formulas and
Percent
Composition
(Lesson 2 of 4)
Today is Thursday,
November 5th, 2015
Stuff You Need:
Calculator
Periodic Table
Polyatomic Ion
List
Pre-Class:
Can you think of a simpler way to write this
formula? C6H12O6
(Imagine this were math class if it helps)
What is the compound in that formula,
anyway?
Today’s Agenda
•
•
•
•
Percent Composition
Hydrate Formula
Empirical Formula
Molecular Formula
• Where is this in my book?
– P. 305 and following…
By the end of this lesson…
• You should be able to determine the formula
of a compound based on its composition.
Percent Composition
• Rather than define it first, I’m just going to
have you guess at a problem.
– We’ll take a few different answers. No pressure.
• What percent of H2O’s mass is a result of
oxygen?
– Oxygen’s molar mass: 15.999 g/mol
– Water’s molar mass: 18.0148 g/mol
• 15.999/18.0148 = ~88%
Percent Composition
• Percent composition is simply how much of
each substance is in a given compound by
mass.
• In other words, I might ask what percentage of
oxygen is in one mole of CO2.
• Oxygen has a molar mass of 15.999 g/mol.
CO2 has a molar mass of 44.009 g/mol.
– Therefore, oxygen is 15.999 g (* 2) out of 44.009
g, or 72.7%.
Percent Composition
• I could also ask a general question like this:
• What is the percent composition of
magnesium carbonate, MgCO3?
• 24.3 + 12.011 + 3 (15.999) = 84.3 g/mol
24.31
Mg  (
)  100  28.83%
84.32
12.01
C(
)  100  14.24%
84.32
48.00
O(
)  100  56.93%
84.32
So…
• So there’s 28.83% Mg, 14.24% C, and 56.93% O.
• If I had 100 grams of MgCO3, then how much (in
grams) would I expect to have of Mg, C, and O?
• 28.83 g Mg
• 14.24 g C
• 56.93 g O
Percent Composition Practice
• Molar Mass and Percent Composition
worksheet.
– Try all of these.
Moving on…
• Remember when we talked about Hydrates?
– This thing? CoCl2  5H2O?
• Well, since today’s lesson focuses partially on
chemical formulas, we’re going to learn how
to calculate them.
• Think of it as figuring out exactly how much
water is present in the compound.
– First, a little review.
Fact Sheet
• Finding Empirical, Molecular, and Hydrate
Formulas
Naming Hydrates
• CoCl2 • 6H2O is “Cobalt II chloride
hexahydrate.”
• CoCl2 • 5H2O is “Cobalt II chloride
pentahydrate.”
– I think you get the idea.
• CoCl2 is “Cobalt II chloride anhydrous,” which
just means it is not hydrated.
• Oh, and the term “salt” means a general ionic
compound, not necessarily table salt (NaCl).
Hydrates
Formula of a Hydrate
• To find the formula of a hydrate, you need the
following information:
– How many moles of water evaporate when the
hydrated salt becomes anhydrous.
– How many moles of anhydrous salt you have.
• To solve, you need to find a ratio of moles of
water molecules to moles of salt.
– In other words, how much water do you have relative
to salt. The same amount? Five times as much?
• Then, divide moles of H2O by moles of salt.
Formula of a Hydrate Example
• Divide the next part of your notebook into
two columns.
• Solve the example with me on the left; solve
the next problem by yourself on the right.
Practice: Hydrate Formulas
Needed:
 Moles of H2O
☐
evaporated
☐ Moles of
anhydrous salt
• A 20 g sample of a hydrate of nickel (II) sulfate
weighed 10.37 g after heating. Determine the
hydrate’s formula.
• How much water did we burn off?
• 20 g (hydrated) – 10.37 g (anhydrous) = 9.63 g H2O
• How many moles of water is that?
• H2O molar mass = 18.0148 g/mol
• 9.63 / 18.0148 = 0.535 mol H2O
Practice: Hydrate Formulas
Needed:
 Moles of H2O
evaporated
 Moles of
☐
anhydrous salt
• A 20 g sample of a hydrate of nickel (II) sulfate
weighed 10.37 g after heating. Determine the
hydrate’s formula.
• How much anhydrous (dry) salt do we have?
• 10.37 g NiSO4
• How many moles of nickel (II) sulfate is that?
• NiSO4 molar mass = 154.756 g/mol
• 10.37 / 154.756 = 0.067 mol NiSO4
Practice: Hydrate Formulas
Needed:
 Moles of H2O
evaporated
 Moles of
anhydrous salt
• Now let’s make a ratio of water to salt.
• Make a fraction with moles of water in the
numerator and moles of anhydrous salt in the
denominator.
0.535 mol H 2O
(
)  7.98  8
0.067 mol NiSO 4
• This tells us there are approximately 8 times as
many moles of water are there are moles of salt.
• Therefore, our formula must be NiSO4  8H2O.
– Nickel II sulfate octahydrate
Now it’s your turn…
• A hydrated sample of barium chloride has a mass
of 4.13 grams. After heating, the mass of the
sample is 3.52 grams. What is the formula of the
hydrated barium chloride?
•
•
•
•
4.13 g – 3.52 g = 0.61 g H2O = 0.0339 mol H2O
3.52 g BaCl2 = 0.017 mol BaCl2
0.0339 mol / 0.017 mol = 1.994 ≈ 2
BaCl2  2H2O
– Barium chloride dihydrate
Other Hydrate Formula Problems
• Sometimes you’ll be given a combination
percent composition and hydrate formula
problem. Here’s what I mean:
• #8 from Empirical/Molecular/Hydrate
• What is the formula for a hydrate that is
76.9% La2(CO3)3 and 23.9% H2O?
– Solve this problem a similar way by determining
moles of each.
– The trick is to imagine you have 100 g of each.
Other Hydrate Formula Problems
• 76.9% La2(CO3)3 = 76.9 g La2(CO3)3
– That’s about 0.168 moles.
• 23.9% H2O = 23.9 g H2O
– That’s about 1.33 moles.
• Divide each by the smallest number (0.168).
– There’s 1 mole of La2(CO3)3 to every 8 moles of
H2O.
• The formula is La2(CO3)3 • 8H2O
– Lanthanum carbonate octahydrate
Hydrate Practice
• Hydrate Calculation Practice Sheet
• Lab – Formula of a Hydrate
Moving on…
• When describing a substance, there are two
kinds of chemical formulas (besides hydrates):
– Molecular Formula
• The “true” one.
– Empirical Formula
• The “simplified” one.
Molecular and Empirical Formula
• The molecular formula is the true number of
each atom in a compound.
– Example: Benzene’s molecular formula is C6H6
– There are 6 atoms of C and 6 atoms of H in every
benzene molecule.
• The empirical formula is the lowest whole
number ratio of atoms in the compound.
– Example: Benzene’s empirical formula is CH.
– Divide all the subscripts by 6, in this case.
Ionic Compounds
• Ionic compounds are always written as
empirical formulas.
• Examples:
– NaCl
– MgCl2
– Al2(SO4)3
– K2CO3
Molecular Compounds
• Molecular compounds may be written
empirically.
• Examples:
– H2O
– C6H12O6 
– C12H22O11
CH2O
Empirical/Molecular Practice
• Empirical, Molecular, Hydrate Formula
worksheet
– Try #1.
Determining Empirical Formula
• To figure out the empirical formula of a
compound based on its percent composition:
– Step 1: Pretend you have 100 grams of the
compound (if necessary).
– Step 2: Determine how many moles of each
element there are.
– Step 3: Divide each mole quantity by the smallest
mole quantity.
– Step 4: Round (if necessary) and multiply each
number by an integer to obtain all whole numbers
(if necessary). These are your subscripts.
Empirical Formula Memory Device
•
•
•
•
% to mass
Mass to mole
Divide by small
Multiply till whole
Determining Empirical Formula
Example
• Adipic acid contains 49.32% C, 6.85% H, and 43.84%
O by mass. What’s the empirical formula?
• Step 1: Pretend you have 100 grams.
• 49.32% C = 49.32 g C
• 6.85% H = 6.85 g H
• 43.84% O = 43.84 g O
• Step 2: Determine moles of each element.
• 49.32 g C = 4.107 mol C
• 6.85 g H = 6.78 mol H
• 43.84 g O = 2.74 mol O
Determining Empirical Formula
Example
• Step 3: Divide each mole value by the smallest.
• 4.107 mol C / 2.74 mol O = 1.50 C
• 6.78 mol H / 2.74 mol O = 2.47 H
• 2.74 mol O / 2.74 mol O = 1.00 O
• Step 4: Round and multiply to get integers.
• 1.50 * 2 = 3 C
• 2.47 ≈ 2.5 * 2 = 5 H
• 1.00 * 2 = 2 O
• The empirical formula of adipic acid is H5C3O2.
A Note About Rounding
• In this case, 2.47 (what we got for O) is not
close enough to either 2 or 3 to simply round.
– Thus, we multiplied all numbers by 2.
• If the value you get after the division step
(Step 3) is within .1 of an integer, you can
round (usually).
Empirical/Molecular Practice
• Empirical, Molecular, Hydrate Formula
worksheet
– Try #2-3, 6.
Determining Molecular Formula
• To find the molecular formula of a compound
from its empirical formula, follow these steps:
– Step 1: Determine the molar mass of the empirical
formula.
– Step 2: Find the molar mass of the molecular
formula, usually given in the problem.
– Step 3: Divide molecular molar mass by empirical
molar mass to get the ratio relating the two.
– Step 4: Multiply each subscript in the empirical
formula by the ratio to get the molecular formula.
Determining Molecular Formula
Example
• The empirical formula of adipic acid is H5C3O2.
What is the molecular formula if the
molecular mass is 146 g/mol?
• Step 1: Determine the empirical molar mass.
– Carbon: 3 * 12.011 g = 36.033 g
– Hydrogen: 5 * 1.0079 g = 5.0395 g
– Oxygen: 2 * 15.999 g = 31.998 g
– TOTAL: 73.0705 g/mol
Determining Molecular Formula
Example
• Step 2: Find the molecular molar mass (given).
• The problem says the molecular formula molar
mass is 146 g/mol.
• Step 3: Find the ratio relating the two.
• Molecular Molar Mass / Empirical Molar Mass
• 146 / 73.0705 = 1.998 ≈ 2
• Step 4: Multiply each subscript by the ratio.
• 2 (H5C3O2) = H10C6O4
Empirical/Molecular Practice
• Empirical, Molecular, Hydrate Formula
worksheet
– Try #4-5.
Closure
• What’s the empirical formula of C12H24O18 if
the molecular molar mass is 456.3 g/mol?
– C2H4O3