E-24
EMPIRICAL FORMULA
NGSS: HS: PS1-7 Use mathematical representations to support the claim that atoms, and therefore
mass, are conserved during a chemical reaction. CC Energy and Matter. SEP Using Mathematical
and Computational Thinking.
In E-14, you learned how to determine the molar mass of a compound if given its chemical formula. In E16, you then learned how to apply that information to find the number of moles that are in a given mass of a
compound. These are the main tools you will need to find the empirical and molecular formulas and
percent composition of compounds.
DEFINITIONS:
empirical (simplest) formula - the smallest whole number ratio between the atoms and/or moles of
elements in a compound.
molecular formula - the actual number of atoms in a compound. This may be a multiple of the
empirical formula. To determine the correct molecular formula, the molar mass (MM) and %
composition must be given.
hydrate - a special formula for a compound that has water molecules "trapped" in its crystalline
structure, e.g. CuCl2∙2 H2O, copper (II) chloride dihydrate (from the E-3 “Types of Observations”
experiment). The formula for the hydrate indicates that two moles of water are associated with
every one mole of cupric chloride. By heating a hydrate to a constant mass, the water is driven off
and the formula for the hydrate can be calculated.
anhydrous - the salt component of a hydrated compound, after the water has been removed.
The following problems illustrate five variations on the same theme: calculations involving
chemical formulas. Like molarity, it is important that each problem be analyzed first, to determine
which method of problem-solving can be used with the information given, and what is to be found.
In the problems where you are trying to find the empirical formula, you should see a common
theme. You will change the information you are given into moles first, then you will determine the
smallest whole number ratio between them.
The following chart will illustrate the relationships utilized here.
Molecular
formula
Percent
Composition
Mass of
elements
MOLES
Empirical
formula
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TYPE 1: PERCENT COMPOSITION OF ELEMENTS FROM A FORMULA
Example: Sodium bicarbonate, NaHCO3, is used in many over-the-counter products to relieve an
upset stomach. Determine the mass percent of all the elements in the compound.
Solution: Since "the whole (100%) is equal to the sum of its parts". This is the same method used
to find your % correct on a test.
Divide the mass contributed by each element by the total molar mass.
% Na in NaHCO3
=
22.99 g Na
84.01 g NaHCO3
X
100% =
27.36%
% H in NaHCO3
=
1.01 g H
84.01 g NaHCO3
X
100% =
1.20%
% C in NaHCO3
=
12.01 g C
84.01 g NaHCO3
X
100% =
14.30%
% O in NaHCO3
=
48.00 g O
84.01 g NaHCO3
X
100% =
57.14%
SUM (TOTAL) OF ALL ELEMENTS =
100.00%
PRACTICE PROBLEMS
Calculate the % composition of the elements in the following compounds
1. H2O
2. C6H12O6
3. NaClO4
4. CuSO45H2O
5. NiCl26H2O
Review of Common Fractions and Decimal Equivalents Needed to Find Empirical Formulas
Fraction
Equivalent Decimal
¼
0.25
0.33
0.50
0.67
0.75
1
3
½
2
3
¾
2
EMPIRICAL FORMULA OF A COMPOUND
TYPE 2A: EMPIRICAL FORMULA FROM % ANALYSIS
(This is basically the reverse of previous problem. Percent composition will be changed first to
mass, then to moles; the moles ratio in the compound will determine the formula.)
Example: Potassium dichromate, an orange solid, contains the elements potassium, chromium,
and oxygen. Analysis of a sample gives the following: K = 26.6%; Cr = 35.4%; and O = 38.0%.
Find the empirical formula.
Solution: When starting with percent mass always assume a 100.00 g sample.
Step 1: find the number of moles of each element
moles of K =
26.6 g K
X
1 mol K
=
39.10 g K
0.680 mol K
moles of Cr
=
35.4 g Cr
X
1 mol Cr
52.00 g Cr
=
0.681 mol Cr
moles of O
=
38.0 g O
X
1 mol O
16.00 g O
=
2.38 mol O
Step 2: To find the simplest whole number ratio, divide by the smallest number of moles, 0.680:
K
=
0.680 mol
Cr =
0.681 mol
O =
2.38 mol
0.680 mol
0.680 mol
0.680 mol
K
=
1.00
Cr =
1.00
O
=
3.50
Multiplying by 2 eliminates the decimal on the 3.50 (or written as a fraction, 3½) to obtain the
simplest whole number ratio, we get
K
=
2
Cr =
2
O =
7
** Note: you will NOT always multiply by 2!!! Multiply by the appropriate denominator to
eliminate the fraction!!
Step 3: Write the empirical formula
K2Cr2O7
Practice Problems:
1.
A compound is found to contain 40.00% carbon; 6.67% hydrogen; and 53.55% oxygen.
Find its empirical formula, CxHyOz.
2.
Analysis shows a compound is 92.3% carbon and 7.7% hydrogen. Calculate the simplest
formula, CxHy.
3.
One oxide of iron is called "magnetite." It is composed of 72.4% Fe and 27.6% oxygen.
What is the empirical formula of magnetite, FexOy?
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TYPE 2B: EMPIRICAL FORMULA FROM MASS ANALYSIS
Example: If 7.35 g of iron reacted directly with oxygen to form 10.17 g of a metal oxide, what is
the simplest formula of the oxide?
Solution:
Step 1 : determine the mass of each element
mass of Fe
mass of O
=
=
7.35 g (this was given)
10.17 g of metal oxide - 7.35 g Fe =
2.82 g O
Step 2: determine the number of moles of each element
moles of Fe
=
7.35 g Fe
X
1 mol Fe
55.85 g Fe
=
0.132 mol Fe
moles of O
=
2.82 g O
X
1 mol O
16.00 g O
=
0.176 mol O
Step 3: To find the simplest whole number ratio, divide by the smallest number of moles, 0.132
Fe
=
0.132 mol
O
=
0.176 mol
0.132 mol
0.132 mol
Fe
=
1.00
O
=
1.33
1
3
Multiplying by the denominator of the fractional form of 1.33 (1 ) to eliminate
fractions and obtain the whole number ratio:
Fe = 1.00 x 3 = 3
O
=
1.33 x 3 = 4
Step 4: Write the empirical formula
Fe3O4
Practice Problems:
1. Chromium forms several different compounds when it combines with oxygen. From the
following data, calculate the empirical formula for each oxide, CrxOy:
a.
0.765 g Cr and 0.235 g O
c.
1.24 g Cr and 0.76 g O
b.
5.60 g Cr and 2.62 g O
d.
0.52 g Cr and 0.48 g O
2. A 4.725 g sample of product was produced when 2.975 g of tin, Sn, were dissolved in an acidic
solution that contains chloride ions (Cl). Calculate the simplest formula for the compound,
SnxCly.
3. When 1.35 g of a silver oxide are decomposed, 1.26 g of silver, Ag, remain. Calculate the
simplest formula, AgxOy.
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TYPE 2C: EMPIRICAL FORMULA OF A HYDRATE
Example: Washing soda is a hydrate of sodium carbonate. Its formula is Na2CO3 X H2O. A
2.714 g sample of washing soda is heated until a constant mass of 1.006 g of Na2CO3 is reached.
What is x ?
Solution:
Step 1: Determine the mass of water and the anhydrous ("dry") salt separately.
mass of salt
=
1.006 g (this was given)
mass of H2O =
2.714 g sample - 1.006 g Na2CO3
=
1.708 g H2O
Step 2: Determine the number of moles of water and the anhydrous ("dry") salt separately.
moles of Na2CO3 =
1.006 g
X
1 mol Na2CO3
105.99 g Na2CO3
=
0.009491 mol Na2CO3
moles of H2O
1.708 g
X
1 mol H2O
18.02 g H2O
=
0.09478 mol H2O
=
Step 3: Determine the smallest whole number ratio by dividing by the smaller number of moles
Na2CO3
=
0.009491 mol
0.009491 mol
H2O
=
0.09478 mol
0.009491 mol
Na2CO3
=
1.000
H2O
=
9.986/1 or 10
Step 4: Write the empirical formula for the hydrate
Na2CO3 10 H2O
Practice Problems:
1. A 1.023 g sample of hydrated copper sulfate, CuSO4X H2O, is heated to remove all of the
water to produce 0.653 g of the anhydrous salt. What is the empirical formula of the hydrate?
2. Borax powder has the formula Na2B4O7X H2O. When 2.145 g of the hydrate is heated and the
water is removed, only 1.130 g of anhydrous salt are left. What is the empirical formula of
borax?
3. Calculate the % H2O in the compound NiCl26 H2O.
4. Calcium sulfate hydrate, CaSO4  X H2O, contains 20.93% water. Calculate X for the hydrate.
5
TYPE 3: MOLECULAR FORMULA USING MOLAR MASS
Example: The empirical formula of vitamin C is found by analysis to be C3H4O3. From another
experiment, its molar mass is found to be about 180 g/mol. What is its molecular formula?
Solution:
Step 1:
Calculate the molar mass for the empirical formula of the compound
C3H4O3
3(12.01 g) + 4(1.01 g) + 3(16.00 g) = 88.07 g
Step 2:
Divide the molar mass given in the problem by the answer from step 1
The ratio of the molar mass to formula mass is
180  88.07 = 2.0 (using significant figures)
This means that the molecular formula is 2 times the simplest formula.
Step 3:
Multiply the subscripts of the empirical formula by the answer from step 2.
The molecular formula is:
2 x C3H4O3 =
C6H8O6
Practice Problems:
1.
Hydrogen peroxide has the empirical formula, HO. If the molar mass is about 34 g/mol,
what is its molecular formula?
2.
Glucose has an empirical formula, CH2O. What is the molecular formula if the molar mass
is 180.18 g/mol?
3.
Maleic acid has an empirical formula of CHO. If the molar mass is 116 g/mol, what is the
molecular formula?
SAMPLE TEST QUESTIONS:
1. What percentage of BaCl22H2O is water?
2. What percentage of PbO2 is lead? How much lead can you get from 50.0 g of PbO2?
3. A compound contains 13.2% B and 86.8% Cl. What is the empirical formula of the compound,
BxCly? If the molar mass of the compound is 163 g/mol, what is the molecular formula?
4. A boron hydride, BxHy, consists of 78.14% B and 21.86% H. What is the empirical formula?
What is the molecular formula if the molar mass is 27 g/mol?
5. A sample contains 0.97 grams of phosphorus and 1.25 grams of oxygen. Further analysis
determines its molar mass to be 284 g/mol. Calculate its correct molecular formula.
6. A 1.256 g sample of sulfur is combined with fluorine to produce SFx. If 5.722 g of SFx was
produced, what is the empirical formula of the compound?
7. If 0.526 g of Xe is reacted with fluorine, 0.678 g of XexFy are produced. What is the empirical
formula of the compound?
8. If 1.687 g of epsom salt, MgSO4X H2O is heated to remove all water, 0.824 g of salt remain.
What is the empirical formula of epsom salt?
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