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Chapter 7 The Mole & Chemical Composition
What is a Mole?
Molecular and formula masses are in atomic mass units. (a.m.u.). An atomic mass unit is only
1.66 x 10-24 grams. The mass of a single molecule is so small that it is impossible to measure in the
laboratory. So for everyday use in chemistry, a larger unit, such as a gram is needed.
We would like to choose a number of atoms that would have a mass in grams equivalent to the
mass of one atom in atomic mass units (a.m.u.). This same number would fit ALL elements, because
equal number of different atoms ALWAYS have the same mass ratio. Chemists have found that 6.02 x
1023 atoms of an element have a mass in grams equivalent to the mass of one atom in a.m.u. For
example, one atom of oxygen has a mass of 16 a.m.u.; 6.02 x 1023 atoms of oxygen have a mass of 16
grams.
The MOLE is a unit used to measure the number of particles of ANY KIND. One mole is equal
to 6.02 x 1023 particles. This number is called Avogadro's number (abbreviated N)(named after
Amadeo Avogadro). So one mole of particles (atoms, ions, molecules) has a mass in grams equivalent to
that of one particle in atomic mass units.
What is in a Mole?
It is important to note that;
one mole of atoms = 6.02 x 1023 atoms = molar mass in grams
ex) 1 mole of Cu = 6.02 x 1023 atoms of Cu = 64 g of Cu
one mole of molecules = 6.02 x 1023 molecules = molar mass in grams
ex) 1 mole of HF = 6.02 x 1023 molecules of HF = 20 g of HF
one mole of formula units =6.02 x 1023 formula units = molar mass in grams
ex) 1 mole of CaCl2 = 6.02 x 1023 formula units of CaCl2 = 111 g of CaCl2
one mole of ions = 6.02 x 1023 ions
ex) C6H12O6
= 180 amu
= 180 grams
= 1 mole of C6H12O6
= 6.02 x 1023 molecules of glucose
= contains 6 moles of C atoms
= 6(6.02 x 1023) atoms of C = 3.612 x 1024 atoms of C
= contains 12 moles of H atoms
= 12(6.02 x 1023) atoms of H = 7.224 x 1024 atoms of H
= contains 6 moles of O atoms
= 6(6.02 x 1023) atoms of O = 3.612 x 1024 atoms of O
ex) ZnBr2
= 225 amu
= 225 grams
= 1 mole of ZnBr2
= 6.02 x 1023 formula units of ZnBr2
= contains 1 mole of Zn2+ ions (6.02 x 1023 Zn2+ ions)
= contains 2 moles of Br2- ions [2(6.02 x 1023 Br2- ions)] = 1.204 x 1024 Br2- ions
= contains 3 moles of ions [3(6.02 x 1023 Br2- ions)] = 1.806 x 1024 ions
Molar Mass
 the mass in grams of one mole of a substance
 abbreviated M
 obtained by summing the masses of the component atoms
 numerically equal to the atomic mass or formula mass
Example: C6H12O6
C
H
O
number
of atoms
6
12
6
x
At. Wt.
x
x
x
12.01
1.01
16.00
Other examples: SO2, C2H3Cl, Pb3(PO4)2, CaSO4• 5H2O
Homework: Determine the molar mass.
1. Mn
2. Al
3. CuSO4
4. Na2CO3
5. Ba3(PO4)2
6. (NH4)3PO4
=
=
=
72.06
12.12
96.00
180.18 g/mol
Mole Conversions
Calculating Mass from Moles and Calculating Moles from Mass for Elements
In other words,
1) given the mass of a sample, calculate the number of moles
2) given the number of moles of a sample calculate the number of grams.
These problems can be solved in TWO ways:
1) Factor Label Method (conversion factors)(proportions)
We know that:
1 mole of a element = 6.022 x 1023 atoms of that element
= molar mass (atomic weight) in grams for that element
Given (units given)
1
x Conversion ( units of unknown) = desired answer
factor (units of given)
2) Using Formulas
Molar Mass = mass ÷ moles
Mass (grams of the substance) = (moles) (Molar Mass)
Moles = grams of the substance ÷ Molar Mass
Number of Particles = (Moles) (Avogadro’s #)
Ex) What is the mass in grams of 0.586 mol Zn atoms?
0.586 mol Zn X 65.38 g Zn
1
1 mol Zn
= 38.3 g Zn
OR
Mass = (moles)(molar mass)
= (0.586 mol Zn)(65.38 g Zn)
= 38.3 g Zn
Sample Problems
Ex) How many atoms are in 2.83 mol Cr?
Ex) How many moles are in 6.195 g of K?
Ex) How many grams are in 8.76 x 1024 atoms of Ba?
Homework
1. How many moles are in 600. g of calcium?
2. How many grams are in 14 moles of chromium?
3. How many atoms are in 7.5 moles of nitrogen?
4. How many moles are in 1.806 x 1023 atoms of nickel?
5. How many atoms are in 8.0 grams of magnesium?
6. How many grams are 3.01 x 1023 atoms of silver?
Calculating Mass from Moles and Calculating Moles from Mass for Compounds & Molecules
These problems can be solved in TWO ways:
1) Factor Label Method (conversion factors)(proportions)
1 mole of a compound = 6.022 x 1023 molecules for a compound
= molar mass (molecular weight) in grams for a compound
Given (units given)
1
x Conversion ( units of unknown)
factor (units of given)
= desired answer
2) Using Formulas
Molar Mass = mass ÷ moles
Mass (grams of the substance) = (moles) (Molar Mass)
Moles = grams of the substance ÷ Molar Mass
Number of Particles = (Moles) (Avogadro’s #)
Ex) Change 0.25 moles of aluminum sulfate to grams.
Al2(SO4)3
Molar Mass is
Al 2 x 26.98 = 53.96
S 3 x 32.06 = 96.18
O 12 x 16.00 = 192.00
_________
342.14 g/mol
0.25 mol Al2(SO4)3
1
X 342.14 g Al2(SO4)3 = 85.54 g Al2(SO4)3
1 mol Al2(SO4)3
OR
Mass = (moles)(M)
= [0.25 moles Al2(SO4)3][342.14 g Al2(SO4)3]
= 85.54 g Al2(SO4)3
Ex) Change 125.152 grams of stannic phosphate to moles.
Sn3(PO4)4
Molar Mass is
Sn 3 x 118.690 = 356.070
P 4 x 30.974 = 123.896
O 16 x 16.00 = 255.984
_________
735.950 g/mol
125.152 g Sn3(PO4)4
1
X
1 mol Sn3(PO4)4
= 0.170 moles Sn3(PO4)4
735.950 gSn3(PO4)4
OR
Moles = grams of the substance ÷ M
= [125.152 g Sn3(PO4)4] ÷ [735.950 g Sn3(PO4)4]
= 0.170 moles Sn3(PO4)4
Sample Problems
Ex) Given 4.86 moles of CaCO3 find the mass in grams of this sample.
Ex) Calculate the mass of 0.58 moles of lead chromate.
Ex) Determine the number of moles in 125 grams of Zn(ClO3)2.
Ex) Determine the number of formula units in 35 grams of barium sulfate.
Homework
1. Find the mass of 0.89 mol of CaCl2.
2. A bottle of PbSO4 contains 158.1 g of the compound. How many moles of PbSO4 are in the bottle?
3. Find the mass of 1.112 mol of HF.
4. Determine the number of moles of C5H12 that are in 362.8 g of the compound.
5. Determine the number of formula units that are in 1.48 mol of NaF.
6. How many formula units are in 3.5 g of NaOH?
7. What is the mass of 3.62 x 1024 molecules of CH3OH?
8. Determine the mass of 2.94 x 1024 molecules of C10H22.
9. How many moles of MgBr2 contain 5.38 x 1024 formula units?
10. How many moles of Ba(NO3)2 contain 6.80 x 1024 formula units?
Average Atomic Mass
Most Elements are a Mixture of Isotopes
•
•
•
Isotopes are atoms that have different numbers of neutrons than other atoms of the same element do.
Average atomic mass is a weighted average of the atomic mass of an element’s isotopes.
If you know the abundance of each isotope, you can calculate the average atomic mass of an element.
Atomic Mass = (Mass of Isotope1)
(% Isotope1 )
(% Isotope 2 )
+ (Mass of Isotope2)
+ ···
100
100
Calculating Average Atomic Mass
Sample Problem
The mass of a Cu-63 atom is 62.94 amu, and that of a Cu-65 atom is 64.93 amu. Using the data below, find the
average atomic mass of copper.
• abundance of Cu-63 = 69.17%
• abundance of Cu-65 = 30.83%
Homework
1. Calculate the average atomic mass of rubidium if 72.17% of its atoms have a mass of 84.91amu and
27.83% of its atoms have a mass of 86.91amu.
2. Calculate the average atomic mass of chlorine if 75.77% of its atoms have a mass of 34.97amu and
24.23% of its atoms have a mass of 36.96amu.
Formulas Express Composition
• A compound’s chemical formula tells you which elements, as well as how much of each, are present in a
compound.
• Formulas for covalent compounds show the elements and the number of atoms of each element in a
molecule.
• Formulas for ionic compounds show the simplest ratio of cations and anions in any pure sample.
• Any sample of compound has many atoms and ions, and the formula gives a ratio of those atoms or ions.
• Formulas for polyatomic ions show the simplest ratio of cations and anions.
• They also show the elements and the number of atoms of each element in each ion.
• For example, the formula KNO3 indicates a ratio of one K+ cation to one anion.
PERCENT COMPOSITION OF COMPOUNDS
(MASS PERCENT)
So far the compositions of compounds have been discussed in terms of the number of constituent atoms.
However, it is also useful to know a compounds composition in terms of the masses of the elements.
The percentage composition of a compound can be determined in one of two ways:
1. From the given chemical formula
2. Experimental analysis, given the mass of a sample
Method 1) Here we can compare the mass of each element present in 1 mole of the compound to the total mass
of 1 mole of the compound and then it is converted to mass percent by multiplying by 100.
% Composition = weight of the element
Molar Mass
Ex) KClO3
K 1 x 39.10 = 39.10
Cl 1 x 35.45 = 35.45
O 3 X 16.00 = 48.00
122.55 g/mol
x100
%K = 39.10g
X 100 = 31.91 %K
122.55 g/mol
%Cl = 35.45g
X 100 = 28.93 %Cl
122.55 g/mol
%O = 48.00 g
X 100 = 39.17 %O
122.55 g/mol
Other Examples: C10H14O (carvone), C14H20N2SO4 (penicillin)
Method 2) Here we can divide the mass of each element in the sample by the mass of the sample and
multiplying by 100.
Ex) A sample of an unknown compound with a mass of 0.2370 g is extracted from the roots of a plant.
Decomposition of the sample produces 0.0948 g of carbon, 0.1264 g of oxygen, and 0.0158 g of
hydrogen. What is the percentage composition of the compound?
%C = 0.0948 g
0.2370 g
X 100 = 40.00 %C
%O = 0.1264 g
0.2370 g
X 100 = 53.33 %O
%H = 0.0158
0.2370 g
X 100 = 6.67 %H
Ex) Find the percentage composition of a compound that contains 2.30 g of sodium, 1.60 g of oxygen,
and 0.100 g of hydrogen in a 4.00 g sample of the compound.
Ex) A sample of unknown compound with a mass of 0.562 g has the following percentage composition:
13.0% carbon, 2.20% hydrogen, and 84.8% fluorine. When this compound is decomposed into its
elements, what mass of each element would be recovered?
First solve the equation algebraically for the mass of the element
Mass of the element = (% Comp.)(mass of sample)
100
Mass of C = (13.0 %)(0.562g) = 0.0731 g C
100
Mass of H = (2.20 %)(0.562g) = 0.0124 g H
100
Mass of F = (84.8 %)(0.562g) = 0.477 g F
100
Homework
1. Determine the percentage composition, As2O5
2. Determine the percentage composition, Ni3(PO4)2
3. What is the percentage of carbon in calcium carbonate?
4. Find the percentage composition of a compound that contains 1.94 g of carbon, O.48 g of hydrogen, and
2.58 g of sulfur in 5.00-g sample of the compound.
5. A sample of an unknown compound with a mass of 0.847 g has the following composition: 50.51
percent fluorine and 49.49 percent iron. When this compound is decomposed into, its elements, what
mass of each element would be recovered?
6. Find the percentage composition of a compound that contains 2.63 g of carbon, 0.370 g of hydrogen,
and 0.580 g of oxygen in a 3.58 g sample of the compound.
7. A sample of an unknown compound with a mass of 2.876 g has the following composition: 66.07
percent carbon, 6.71 percent hydrogen, 4.06 percent nitrogen, and 23.16 percent oxygen. What is the
mass of each element in this compound?
8. A sample of a compound that has a mass of 0.432 g is analyzed. The sample is found to be made up of
oxygen and fluorine only. Given that the sample contains 0.128 g of oxygen, calculate the percentage
composition of the compound.
FORMULAS OF COMPOUNDS
Empirical Formula
The formula of a compound that expresses the smallest whole number ratio of atoms.
Molecular Formula
The chemical formula that indicates the numbers of each atom in a molecular compound.
The true or actual formula.
Calculating Empirical Formulas
The Empirical Formula of a compound can be determined in one of two ways:
1. From Percentage Composition data.
2. From Relative Mass data.
To determine an empirical formula, masses of elements are converted to moles and then a ratio of moles
is determined.
Method 1) From Percentage Composition data.
Assume a 100 g sample so that each percentage is numerically equal to the number of grams of the
element.
Ex) Find the empirical formula for sodium sulfite. Sodium sulfite contains 36.5% sodium, 25.4% sulfur,
and 38.1% oxygen.
Solution
The percentage composition data indicates that there are 36.5 grams Na, 25.4 grams S, and 38.1 grams 0
in 100 grams of compound.
(a) Number of moles.
Na
(36.5 g Na) 1 mol Na = 1.59 mol Na
23.0 g Na
S
(25.4 g S) 1 mol S = 0.791 mol S
32.06 g S
O
(38.1 g O) 1 mol O = 2.38 mol O
16.0 g O
(b) Ratio of Moles
Na
1.59 = 2.01
0.791
S
0.791 = 1.00
0.791
O
2.38 = 3.01
0.791
The empirical formula is Na2SO3.
Ex) What is the empirical formula of a compound which contains 53.73% Fe and 46.27% S?
Solution
There are 53.73 g Fe and 46.27g S in 100 g of the compound.
(a) Number of moles.
Fe
(53.73 g Fe) 1 mol Fe = 0.9620 mol Fe
55.85 g Fe
S
(46.27 g S) 1 mol S = 1.443 mol S
32.06 g S
(b) Ratio of Moles
Fe
0.9620 = 1.00
0.9620
S
1.443 = 1.50
0.9620
In the previous sample problem, the relative numbers of atoms were small whole numbers and we could
write the formula directly from them. The ratio 1 to 1.5 must be expressed in terms of whole numbers,
since a fractional part of an atom does not exist. By multiplying both numbers in the ratio by two, we
obtain two atoms Fe and three atoms S. The empirical formula is Fe2S3.
Ex) Nylon-6 has the following percentage composition data; 63.8% carbon, 12.38% nitrogen, 9.80%
hydrogen, and 14.14% oxygen, determine the empirical formula.
Method 2) From Relative Mass data.
Ex) An oxide of aluminum is formed by the reaction of 4.151 g of Al with 3.692 g oxygen. Calculate the
empirical formula.
(a) Number of moles
Al
(4.151 g Al) 1 mol Al = 0.1539 mol Al
26.98 g Al
O
(3.692 g O) 1 mol O = 0.2308 mol O
16.00 g O
(b) Ratio of moles
Al
0.1539 = 1.000
0.1539
O
0.2308 = 1.5
0.1539
The empirical formula is Al2O3.
Ex) A 0.35446 g of vanadium is heated in air, it reacts with oxygen to achieve a final mass of 0.6330 g.
Determine the empirical formula.
Homework
1) A sample of potassium sulfate has the following composition: 17.96 g of potassium, 7.35 g of sulfur,
and 14.70 g of oxygen. What is its empirical formula?
2) 11.0 g of a certain compound contains 2.82 g of magnesium and 8.18 g of chlorine. What is its empirical
formula?
3) A certain sample of a barium salt contains 8.57 g of barium and 4.43 g of chlorine. What is its empirical
formula?
4) 50.0 g of sulfur is mixed with 100.0 g of iron and the mixture is heated. When the reaction is completed,
12.7 g of iron remains. What is the formula of the compound formed?
5) Determine the empirical formula of a compound containing 20.23 percent aluminum and 79.77 percent
chlorine.
6) Determine the empirical formula of a compound containing 24.74 percent potassium, 34.76 percent
manganese, and 40.50 percent oxygen.
7) Find the empirical formula of a compound given that the compound is found to be 47.9 percent zinc and
52.1 percent chlorine by mass.
Calculating Molecular Formulas
It is impossible to determine the molecular formula unless the molar mass of the substance has been determined
first or given in the problem.
If the molar mass is known one can decide the molecular formula from the empirical formula by using the
equation;
(empirical formula weight) X = molar mass
OR
X = molar mass
formula weight
Ex) Hydrogen peroxide consists of 5.9% H and 94.1 % O. The molar mass is 34.0 grams. What is the molecular
formula?
Ex) A certain compound consists of 71.65% Cl, 24.27% C, and 4.07% H. The molar mass is 98.96 g. What is
the molecular formula?
Homework
1. A certain sugar has a chemical composition of 40 percent carbon, 6.6 percent hydrogen, and 53.3
percent oxygen. The molar mass is 180 g/mol. What is the molecular formula?
2. The neurotransmitter norepinephrine is 56.8 percent carbon, 6.5 percent hydrogen, 28.4 percent oxygen,
and 8.3 percent nitrogen. Its molar mass is 169 g / mol. Find the molecular formula of this substance.
3. Tooth enamel, or hydroxyapatite, has a molar mass of 422 g/mol. Its composition is 28.5 percent
calcium, 22 percent phosphorus, 49.3 percent oxygen, and 0.239 percent hydrogen. Calculate the
molecular formula.
Guide Sheet for Moles Problems
I. Calculating Molar Mass
1. multiply atomic mass of each element by number of atoms of that element in the formula
(shown by the subscript)
2. find the sum of all the atomic masses --this is formula mass (unit is a.m.u.)
3. express formula mass in grams (unit is g/mol). This is the Molar Mass.
II. Mole Conversions
1. Factor Label Method (conversion factors) (proportions)
Given (units given) x Conversion ( units of unknown) = desired answer
1
factor (units of given)
2. Using Formulas
Molar Mass = mass ÷ moles
Mass (grams of the substance) = (moles) (Molar Mass)
Moles = grams of the substance ÷ Molar Mass
Number of Particles = (Moles) (Avogadro’s #)
III. Calculating % Composition (from formula)
1. calculate formula mass
2. divide the total atomic mass of each element by the formula mass and multiply by 100
IV. Calculating % Composition (from masses of each element)
1. divide the mass of each element by the total mass of the compound and multiply by 100
V. Calculating Empirical Formula (from % Composition)
1.
2.
3.
4.
convert % of each element to grams based on 100 grams of the compound
multiply grams of each element by 1/molar mass that element
compare ratio of moles of each element and divide each by the smallest
if result in step 3 gives a ratio with decimal equivalent to 1/4, 1/3, 1/2, 2.3, 3/4 instead of
whole numbers, convert to the fraction and multiply all ratios by the denominator or the
fraction
VI. Calculating Empirical Formula (from experimentally determined masses)
1. multiply the mass of each element (in grams) by 1/molar mass of that element
2. continue with steps 3 & 4 from VI above.
VII. Finding Molecular Formulas (when molar mass is known)
1.
2.
3.
4.
calculate the empirical formula
use the equation : (empirical formula mass)x = molar mass
find value for x: x = molar mass/empirical formula mass
multiply each subscript in empirical formula by value for x