Louisiana Tech University, Chemistry 100. POGIL(Process Oriented
Guided Inquiry Learning) Exercise on Chapter 3. Chemical Formula:
Percent Composition, Molar Mass and Moles
Why?
In chemistry, the mole is the standard measurement of amount. However, in
experiments we use a balance to measure amounts in grams. Balances DO NOT give
readings in moles. So the problem is that, when we compare amounts of one substance to
another using mole, we must convert from grams which are actually measured to moles.
When substances react according to chemical equations, they do so in simple ratios of
moles. It is also important that we understand the mole as the basis for all calculations in
chemistry and have conceptual understanding of how mole is related to Avogadro’s
number.
Learning Objectives
The students should be able to understand the concepts and have working knowledge in
the following:
Mole of Compounds (Section 3.8)
1. Thoroughly explain the use of the mole concept for chemical compounds.
2. Calculate the molar mass of a compound (Section 3.8).
3. Calculate the number of moles of a compound given the mass, and vice versa.
4. Explain the formula of a hydrated ionic compound and calculate its molar mass.
Percent Composition (Section 3.9)
1. Express molecular composition in terms of percent composition
Determine Empirical and Mol. Formulas (Section 3.10)
1. Use percent composition and molar mass to determine the empirical and molecular
formulas of a compound
Success Criteria
Understand the mole concept for chemical compounds, calculate the molar mass of
a compound, calculate the number of moles of a compound given the mass, and vice
versa, explain the formula of a hydrated ionic compound and calculate its molar
mass, express molecular composition in terms of percent composition, and use
percent composition and molar mass to determine the empirical and molecular
formulas of a compound.
Resources
Chemistry: The Molecular Science 2 Edition, J.W. Moore, C. L. Stanitski and P.C. Jurs.
Prerequisites
High school chemistry: Definition of matter. Scientific method, Dalton’s Atomic theory,
Components of atoms: nucleus (proton and neutrons) and electrons. Atoms, molecules
and ions. Names and formula of chemical compounds.
New Concepts
Mole of a Compound
Chemists measure chemical in grams as the amount in the reaction. Therefore, we
need a conversion factor to convert grams to atoms or molecules. Mole is the connection
or the conversion factor between atoms and grams. Mole is just a large number 6.022 x
1023 for counting atoms like dozen -12 for counting to make counting of eggs easier.
Since atoms are so small, we need large number of them to make it physically observable
and able to weigh in gram quantities.
mole (6.022 x 1023 particles)-Chemists Dozen
Chemists measure chemical in grams as the amount in the reaction. Therefore, we
need a conversion factor to convert grams to atoms or molecules. Mole is the connection
or the conversion factor between atoms and grams. Mole is just a large number 6.022 x
1023 for counting atoms like dozen -12 for counting to make counting of eggs easier.
Since atoms are so small, we need large number of them to make it physically observable
and able to weigh in gram quantities.
Gram Mole and the Avogadro's Number
The gram mole is the grams of any chemical substance using the value atomic mass
obtained from the periodic table. E.g. For carbon gram mole is 12.01 grams of carbon
since its atomic mass is 12.01 amu in the periodic table. If you take atomic mass in grams
the number of atoms is simply 6.022 x 1023 atoms or particles. In other words gram
atomic weight or gram molecular weight contains 6.022 x 1023 atoms, molecules or
particles. This number is called Avogadro's number or mole of particles.
Molar mass of a compound
The first step in the determination of the molar mass of a molecular compound is to
determine the molecular or formula mass of the compound, which is the weighted
average mass of the compound’s naturally occurring molecules. This is found by adding
the atomic masses of the atoms in each molecule.
E.g. Calculate the molar mass of the following molecular compound: H3PO4 and
H2SO4.
Molecular weight of H3PO4: H = 1.01 amu, P = 30.97 amu, O = 16.00 amu
m.w. H3PO4 = 3 x 1.01 + 1 x 30.97 + 4 x 16.00 = 98.00 amu
amu =g/mol
Molar mass of H2SO4 = 98.00 g/mol
E.g. Calculate the molar mass of following ionic compound:Fe2(CO3)3.
F.W. of Fe2(CO3)3 = Fe2C3O9 = 2 x 55.85 + 3 x 12.01 + 9 x 16.00 = 294.73 amu
Molar mass of Fe2(CO3)3 = 294.73 g/mol
Molar mass of hydrated compounds
E.g. Calculate the molar mass of following ionic compound:CuSO45H2O
F.W. of CuSO45H2O = CuSO9H10 =1 x 63.55 +1 x 32.07+ 9 x 16.00 +10 x 1.01
=201.72 g/mol
Moles of a compound given the mass, and vice versa
How many moles of K2SO4 are present in 180.1g of potassium sulfate?
f.w. K2SO4 = 174.26 g/mol
180.1 g K2SO4
x 1 mol
= 1.033 mol K2SO4
174.26 g Fe
Percent Composition
The elemental composition of a compound, the relative amounts of elements in it, can be
expressed in two ways: atom ratio as in formula or percent mass of elements.
Converting molecular/ionic formula to percent composition
What is the atom/ion ratio and percent elemental composition in HgS?
Ion ratio in HgS = 1(Hg) :1(S)
The percentage composition of mercury (II) sulfide, HgS, is calculated as follows.
The molar mass of HgS is 200.59 + 32.06 = 232.65 g/mol.
% Hg = 100(1x 200.59 g/mol)/(232.65 g/mol) = 86.22%
%S = 100(1x 32.06 g/mol)/(232.65 g/mol) = 13.78%
Determine Empirical of and Molecular Formulas
Percent composition and molar mass is used to determine the empirical of ionic
compounds and molecular formulas of a covalent.
E.g. A compound with an empirical formula of CH has a molecular weight of 78
g/mol. What is the molecular formula?
The empirical formula weight of CH =is 13
Since molecular weight/ empirical formula weight = 78/13 = 6
the molecular formula is (CH)6 = C6H6, the molecular formula for benzene.
E.g. A compound containing 92.3 weight percent of carbon and 7.7 weight percent
of H. What is the empirical formula? If the molecular weight of the compound is 52
g/mole what is the molecular formula?
Assume that you have 100 g of the compound, then you have 92.3 g of carbon and 7.7 g
of hydrogen. Thus the mole ratio of C to H should be
Moles of C
Moles of H
Mole ratio C:H Empirical
Molecular
formula
formula
92.3
=7.7 7.7
=7.7
1:1
CH
C 4 H4
12
1.006
Combustion analysis of organic compounds
E.g. When 1.00 g of a compound containing only carbon and hydrogen is burned
completely, 3.14 g of CO2 and 1.29 g of H2O is produced. What is the empirical
formula? Hint:
Moles of carbon and hydrogen in 3.14 g of carbon dioxide and 1.29 g H2O
Moles of C
Moles of H
12 g C
1 mol C
3.14 g CO2 x -------x --------- = 0.0714 mol C
44 g CO2
12 g C
2gH
1 mol H
1.29 g H2O x ------------ x ------------ = 0.143 mol H
18 g H2O
1gH
0.143
-------- =2
0.0714
0.143
-------- = 1
0.0714
1C
2H
Therefore, the empirical formula is CH2
Key Questions (relatively simple to answer using the Focus Information)
1) What is the molecular mass (M.M.) of benzene (C6H6) and how many grams are there
in 1 more of benzene?
2) Find the formula weight (F.W.) of NaCl.
3) Calculate the molar mass of following hydrated ionic compound:FeCl3 9H2O.
4) Determine the number of moles of CO2 (M.W. 44.01 g/mole ) in 454 grams.
5) Calculate the grams present in 0.500 moles of C12 H22O11, sucrose (M.W.
342.3g/mole).
6) Calculate the grams present in 0.100 moles of Na2CO3 (F.W. 105.99).
7) Glucose has a molecular formula of C6H12O6 (M.W. 180.16 g/mol).
a) How many grams of C are available in 1 mole of glucose?
b) How many grams of H are available in 1 mole of glucose?
c) How many grams of O are available in 1 mole of glucose?
8) What are the mass % of carbon and hydrogen in carbon dioxide CO2?
(CO2; M.W. = 44.01 g/mole)
% C:
% O:
9) Bicarbonate of soda (sodium hydrogen carbonate) has ionic formula: NaHCO3. Find
the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate.
Molar mass:
%Na:
% C:
% O:
%H:
10) NH3 (M.W. 17.03 g/mole) and NH4NO3 (F.W. 80.05 g/mole) used as nitrogen
fertilizer. Which one will provide more nitrogen for the same weight?
11) A carbohydrate contains 38.71% weight of C, 9.71% weight of H and 51.58%
weight of O. What is the empirical formula?
Moles of C:
Moles of H:
Moles of O:
Mole ration of
C:=
H:=
O: =
Simple mole ratio of
C:=
H:=
O: =
Empirical formula =
12) Combustion Analysis gives the following mass %: 26.7% C; 2.2% H; 71.1% O
If a separate analysis determined the molecular mass of the compound to be 90
g/mole. Determine the Empirical Formula and the Molecular Formula of the
compound.
a) Moles of C:
b) Moles of H:
c) Moles of O:
d) Mole ration of
C:=
H:=
O: =
e) Simple mole ratio of
C:=
H:=
O: =
f) Empirical formula =
g) Empirical formula mass =
h) Molar mass/empirical formula mass =
i) Molecular formula =
13) When 5.00 g of a compound containing only carbon and hydrogen is burned
completely, 15.7 g of CO2 and 6.45 g of H2O is produced. What is the empirical
formula?
a) Moles of C:
b) Moles of H:
c) Mole ratio of
C:=
H:=
d) Simple mole ratio of C:=
H:=
e) Empirical formula of the compound =