Topic 7: The Mole Concept - · PDF fileTopic 7: The Mole Concept ... A chemist does the same thing with atoms! Why? ... 2. What is the mass of 1.2 x 1024 atoms of magnesium? 3

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Topic 7: The Mole Concept Relating Mass to Numbers of Atoms (Chapter 3 in Modern Chemistry beginning on p.82)

In order to understand the quantitative parts of chemistry, there are three very important

concepts—the mole, Avogadro’s number, and molar mass. These provide the basis for relating

masses in grams to number of atoms.

The Mole

The mole is the SI unit for amount of substance. A mole (abbreviated mol) is the amount of a

substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. The

mole is a counting unit, just like a dozen is.

Avogadro’s number

The number of particles in a mole has been experimentally determined in a number of ways. The

best modern value is 6.022 1415 x 1023

. This is the number of particles (atoms, ions, molecules,

formula units) in a mole. It is called Avogadro’s number. Avogadro’s number—6.022 1415 x

1023—is the number of particles in exactly one mole of a pure substance. For most purposes,

Avogadro’s number is rounded to 6.022 x 1023.

Watch this video for an idea of how many particles Avogadro’s number really is.

http://www.youtube.com/watch?v=1R7NiIum2TI

Molar Mass

An alternative definition of mole is the amount of a substance that contains Avogadro’s number

of particles. The mass of one mole of a pure substance is called the molar mass of that

substance. Molar mass is usually written in units of g/mol. The molar mass of an element is

numerically equal to the atomic mass of the element in atomic mass units (which can be found in

the periodic table).

Using Chemical Formulas (Chapter 7 in Modern Chemistry beginning on p.237)

As you have seen, a chemical formula indicates the elements as well as the relative number of

atoms or ions of each element present in a compound. Chemical formulas also allow chemists to

calculate a number of characteristic values for a given compound. In this section, you will learn

how to use chemical formulas to calculate the molar mass, formula mass, and the percentage

composition by mass of a compound.

http://www.youtube.com/watch?v=1R7NiIum2TI

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Molar Mass/Formula Mass

Molecular and formula mass use the amu that we learned in Chapter 3. We use the atomic mass

units from the periodic table.

For example: Find the formula mass of potassium chlorate, KClO3.

1 K atom x 39.10 amu = 39.10 amu

1 Cl atom x 35.45 amu = 35.45 amu

3 O atoms x 16.00 amu = 48.00 amu

Formula mass of KClO3 = 122.55 amu

The term molecular mass is usually used for covalent compounds. The term formula mass is

usually used for ionic compounds.

Molar mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all

atoms represented in its formula. The units for these calculations are g/mole. This is calculated

the same way as molecular or formula mass.

Example: Find the molar mass of Ba(NO3)2.

1 mole Ba x 137.33 g/mole = 137.33 g Ba

2 moles N x 14.01 g/mole = 28.02 g N

6 moles O x 16.00 g/mole = 96.00 g O

Molar mass of 1 mole of Ba(NO3)2 = 261.35 g/ mole

Task 7a

1. Find the molar mass of the following.

a. NaCl

b. K2O

c. NaOH

d. Ca(OH)2

e. (NH4)3PO4

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Percent Composition

Sometimes you need to know the percentage by mass of a particular element in a chemical

compound. For example, suppose the compound potassium chlorate, KClO3, were to be used as

a source or oxygen. What is the percentage of oxygen in this compound? Use the following

formula:

mass of element in sample of compound x 100 =

% of element

in compound Mass of sample of compound

The percentage by mass of each element in a compound is known as the percentage

composition of the compound.

Find the percentage composition of copper(I) sulfide, Cu2S.

Find the molar mass of each element and of the compound.

2 Cu x 63.55 g/mol = 127.1 g/mole

1 S x 32.07 g/mol = 32.07 g/mole

Molar mass of Cu2S = 159.2 g/mole

Put these numbers into the formula for percent composition.

% Cu = 127.1 g/mol

x 100 = 79.85% 159.2 g/mole

% S = 32.07 g/mole

x 100 = 20.15% 159.2 g/mol

Task 7b 1. Find the percent composition of each element in the following compounds. (Note: You

already calculated the molar mass in Task 7a).

a. NaCl

b. K2O

c. NaOH

d. Ca(OH)2

e. (NH4)3PO4

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Hydrates

In some salts, water molecules from solution bind within their crystal structure. For example

sodium carbonate forms a hydrate, in which 10 water molecules are present for every formula

unit of sodium carbonate. The formula of the hydrate is Na2CO3 . 10 H2O. Note that the first

part of the formula is the formula of the compound, then a dot. Following the dot, the number of

water molecules that are attached to each formula unit is written. Hydrates are named by naming

the compound, prefix hydrate. Again, using the example above; the hydrate is named sodium

carbonate decahydrate.

The molar mass of a hydrate is calculated by the following formula:

MMhydrate = MMsalt + (# of water molecules x MMwater)

Na2CO3 . 10 H2O 2(23) + 12 + 3(16) + 10[2(1) + 16]

106 + 10(18)

106 + 180

286 g/mol

Task 7c

1. What is the name of MgSO4 . 5 H2O?

2. Write the formula of copper(II) chloride trihydrate.

3. Calculate the molar mass of CuSO4 . 5 H2O.

4. What is the percent of water in CuSO4 . 5 H2O?

The Mole

Why are M & Ms sold in a package rather than individually?

List other items we package and then refer to as a package rather than the individual item:

100 pennies =

2 socks =

12 eggs =

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A chemist does the same thing with atoms! Why?

A chemist’s package is a mole.

1 mole = 6.02 x 1023

particles.

6.02 x 1023

is called Avogadro’s number.

It has been experimentally determined that when one measures out the mass of an element equal

to its average atomic mass, the number of atoms contained in the sample is equal to 6.02 x 1023

atoms.

6.02 x 1023

atoms = 1 mole = Avogadro’s number = molar mass

Mole Conversions

Particles to moles o Convert representative particles to moles and moles to representative particles.

(Representative particles are atoms, molecules, formula units, and ions.)

Mass to moles

o Convert mass of atoms, molecules, and compounds to moles and moles of atoms,

molecules, and compounds to mass.

o Convert representative particles to mass and mass to representative particles.

Volume of a gas to moles

o Convert moles to volume and volume to moles at STP.

Mass Moles Particles

(grams) atoms, ions, molecules,

formula units

MM from periodic table Use 6.02 x 1023

(g/mol) Volume of Gas (at STP) (particles/mole)

(liters)

22.4 L/mole

The mole is the central unit in all mole conversions. The other units are used as needed for

conversion factors.

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The key to these problems is using dimensional analysis. Remember the key to dimensional

analysis is to cancel out the units you do not need and keep the units you do need.

Summary

If grams are in the problem: molar mass from the periodic table g/mole

If particles are in the problem: Avogadro’s number 6.02 x 1023

particles/mole

If liters of gas are in the problem: molar volume 22.4 L/mole

Examples:

Ibuprofen, C13H18O2, is the active ingredient in many nonprescription pain relievers.

1) If the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles of

ibuprofen are in the bottle?

2) How many molecules of ibuprofen are in the bottle?

3) What is the total mass in grams of carbon in 33 g of ibuprofen?

Answers

1) 33 g C13H18O2 1 mole C13H18O2 = 0.16 mole C13H18O2

206.31 g C13H18O2

2) 0.16 mole C13H18O2 6.02 x 1023

molecules = 9.6 x1022

molecules C13H18O2

mole

3) 0.16 mole C13H18O2 13 moles of C 12.01 g C = 25 g C

1 mole C13H18O2 mole

Task 7d

1. How many atoms in 400.0g of sulfur?

2. What is the mass of 1.2 x 1024

atoms of magnesium?

3. What is the mass of 2.5 moles of oxygen atoms?

4. Given 18 grams of water, how many molecules do you have?

5. What is the mass, in grams, of 1.2 x 1024

formula units of sodium chloride?

6. What is the volume of carbon dioxide gas in 589.4 g of carbon dioxide?

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Empirical & Molecular Formulas

Empirical Formula – the formula for a compound expressed as the smallest possible whole-

number ratio of subscripts of the elements in the formula.

Molecular Formula- the formula for a compound in which the subscripts give the actual

number of each element in the formulas it truly exists.

An empirical formula cannot be reduced any further. A molecular formula may or may not be

reduced.

Molecular Formula Empirical Formula

H2O

CH3COOH

CH2O

C6H12O6

Notice two things:

1. The molecular formula and the empirical formula can be identical.

2. You scale up from the empirical formula to the molecular formula by a whole number

factor.

Calculating Empirical Formulas

A Simple Rhyme for a Simple Formula by Joel S. Thompson

Percent to mass

Mass to mole

Divide by small

Multiply ‘til whole

In other words:

1. Since percent can be on any mass, assume a 100 g sample and change directly to

mass.

43 % of 100 g is 43 g.

2. Use molar mass to perform a mole conversion to change gram to moles.

3. Divide by the smallest number of moles to try to get a whole number ratio.

4. If you do not get a whole number, multiply by a whole number until you get a whole

number. Remember formulas cannot be in fractions.

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Example #1

Quantitative analysis shows that a compound contains 32.38 % sodium, 22.65 % sulfur,

and 44.99 % oxygen. Find the empirical formula of this compound.

Change % Use MM to change Divide by the Hopefully, get a

to grams g to moles smallest moles whole number

32.38 g Na mole Na = 1.408 mole Na = 2

22.99 g 0.7063 mole Write the formula

22.65 g S mole S = 0.7063 mole S = 1 Na2SO4

32.07 g 0.7063 mole

44.99 g O mole O = 2.812 mole O = 4 This is the

16.00 g 0.7063 mole empirical

formula

Example #2

Analysis of a 10.150 g sample of a compound known to contain only phosphorus and

oxygen indicates a phosphorus content of 4.433 g. What is the empirical formula of this

compound?

Do not have to Change g Divide by the Didn’t get a whole change to grams, to moles smallest number for both,

already in grams so multiply by a number

to get a whole number

4.433 g P mole P = 0.1431 mole P = 1 (2) = 2 Write the

30.97 g P 0.1431 mole formula

P2O5

5.717 g O mole O = 0.3573 mole O = 2.5 (2) = 5

16.00 g O 0.1431 mole The grams of O

came from subtracting

10.150 g – 4.433 g

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Task 7e

1. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the

empirical formula?

2. A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83%

oxygen. The molecular weight of the compound is known to be approximately 140 g/mol. What

is the empirical formula?

3. A compound is found to contain 63.52 % iron and 36.48 % sulfur. Find its empirical formula.

4. Find the empirical formula of a compound found to contain 26.56 % potassium, 35.41 %

chromium, and the remainder oxygen.

Calculating Molecular Formulas

To determine the molecular formula you must know the empirical formula and the molecular

mass. Remember the molecular formula is a multiple of the empirical formula.

To find the multiple: molar mass of the molecular formula = x

molar mass of the empirical formula

To write the molecular formula: multiply the subscripts in the empirical formula by x

Example #3

In example problem #2, the empirical formula of a compound of phosphorus and oxygen

was found to be P2O5. Experimentation shows that the molar mass of this compound is

283.89 g/mol. What is the compound’s molecular formula?

Molecular formula’s mass = 283.89 g/mol

Empirical formula’s mass = 141.94 g/mol

283.89 g/mol = 2 P2O5 x 2 = P4O10

141.94 g/mol

The empirical formula was

given in the problem.

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Example #4

A compound with a formula mass of 42.08 amu is found to be 85.64 % carbon and 14.36%

hydrogen by mass. Find the molecular formula.

Find the empirical formula first.

85.64 g C mole C = 7.131 mol C = 1 Empirical formula

12.01 g C 7.131 mol

CH2

14.36 g H mole H = 14.25 mol H = 2

1.008 g H 7.131 mol

42.08 amu = 3 CH2 (3) = C3H6

14.03 amu Molecular formula

Empirical formula mass

Task 7f 1. A 1.50 g sample of hydrocarbon undergoes complete combustion to produce CO2 and H2O.

The empirical formula of this compound is CH3. Its molecular weight has been determined to be

about 78. What is the molecular formula?

2. Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen

16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol. What is its molecular

formula?

Concentration of Solutions (Molarity) (Chapter 12 in Modern Chemistry beginning on p.418)

The concentration of a solution is a measure of the amount of solute in a given amount of

solvent or solution. In this section, we introduce two different ways of expressing the

concentrations of solutions: molarity and molality. Sometimes, solutions are referred to as

“dilute” or “concentrated,” but these are not very definite terms. “Dilute” just means that there

is a relatively small amount of solute in a solvent. “Concentrated,” on the other hand, means that

there is a relatively large amount of solute in a solvent. These terms are unrelated to the degree

to which a solution is saturated. A saturated solution of a substance that is not very soluble

might be very dilute.


Molarity

Molarity is the number of moles of solute in one liter of solution. Molarity is symbolized with a

“M” or brackets, [ ]. The formula for finding molarity is:

molarity (M) = moles of solute (mol)

volume of solution (L)

Note that the solute must be in moles and the volume of solution must be in liters.

It is also important to realize that making a 1 molar (1 M) solution is not made by adding 1 mol

of solute to 1 L of solvent. You need to dissolve the solute in a small amount of water then add

enough water to make the total volume of solution 1L.

Here are some examples of molarity problems.

Ex. 1 You have 3.50 L of solution that contains 90.0 g of sodium chloride. What is the molarity

of this solution?



First you have to change the 90.0 g of sodium chloride to moles of sodium chloride.

90.0 g NaCl moles of NaCl = 1.54 mol NaCl

58.5 g of NaCl

Then plug this information into the molarity formula.

M = 1.54 mol NaCl

= 0.440 M NaCl 3.50 L solution

OR

You can do the same problem using dimensional analysis.

90.0 g NaCl moles of NaCl = 0.440 M NaCl

58.5 g of NaCl 3.50 L solution


Ex. 2 You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution

contain?

Remember that the capital “M” means moles of solute per liters of solution so 0.5 M means 0.5

moles of HCl in 1 liter of solution. Writing the molarity this way allows you to see that you can

use dimensional analysis to cancel liters and solve for moles.

0.5 moles HCl 0.8 L solution = 0.4 mol HCl

L solution

You could also find the grams of HCl by using molar mass if needed.

Ex. 3 To produce 40.0 g of silver chromate, you will need at least 23.4 g of potassium chromate

in solution as a reactant. All you have on hand is 5 L of 6.0 M K2CrO4 solution. What volume

of the solution is needed to give you the 23.4 g K2CrO4 needed for the reaction?

First, it is important to realize that the 40.0 g of silver chromate is not important information.

We need to change 23.4 grams to moles.

23.4 g K2CrO4 1 mol K2CrO4 = 0.120 mol K2CrO4

194.2 g K2CrO4

Now plug the information into the molarity formula.



6.0 moles K2CrO4 = 0.120 mol K2CrO4

L solution x

x = 0.020 L K2CrO4 soln

OR


23.4 g K2CrO4 moles of K2CrO4 L of soln = 0.020 L K2CrO4 soln

194.2 g of K2CrO4 6.0 mol K2CrO4


Task 7g 1. What is the molarity of a solution composed of 5.85 g potassium iodide, KI, dissolved in

enough water to make 0.125 L of solution?

2. How many moles of H2SO4 are present in 0.500 L of a 0.150 M H2SO4 solution?

3. What volume of 3.00 M NaCl is needed for a reaction that requires 146.3 g of NaCl?

Molality

Molality is the number of moles of solute per kilogram of solvent. Molality is symbolized with a

“m”. The formula for finding molality is:

molality (m) = moles of solute (mol)

Kilograms of solvent (kg)

Note that the solute must be in moles and the solvent must be in kilograms.

It is also important to realize that making a 1 molal (1 m) solution is made by adding 1 mol of

solute to 1 kg of solvent.

Here are some examples of molality problems.

Ex. 1 A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g

of water. Find the molal concentration of this solution.

molality (m) = moles of solute (mol)

Kilograms of solvent (kg)

First you have to change the 17.1 g of sucrose to moles of sucrose.

17.1 g C12H22O11 moles of C12H22O11 = 0.0500 mol C12H22O11

342.34 g of C12H22O11

Then plug this information into the molality formula, remembering that you have to use kg of

solvent. 125 g of water is 0.125 kg of water.

m = 0.0500 mol C12H22O11 = 0.400 m C12H22O11 0.125 kg solvent

OR



17.1 g C12H22O11 moles of C12H22O11 = 0.400 m C12H22O11

342.34 g of C12H22O11 0.125 kg solvent

Ex. 2 A solution of iodine, I2, in carbon tetrachloride, CCl4, is used when iodine is needed for

certain chemical tests. How much iodine must be added to prepare a 0.480 m solution of iodine

in CCl4 if 100.0 g of CCl4 is used?

Remember that the “m” means moles of solute per kilograms of solvent so 0.480 m means 0.480

moles of I2 in 1 kg of CCl4. Writing the molality this way allows you to see that you can use

dimensional analysis to cancel kilograms and solve for moles.

0.480 moles I2 0.100 kg solvent = 0.480 mol I2

kg solvent

It is then possible to change moles of iodine to grams of iodine by using the molar mass of

iodine.

0.480 mole I2 253.8 g I2 = 12.2 g I2

mol I2

OR


0.480 moles I2 0.100 kg solvent 253.8 g I2 = 12.2 g I2

kg solvent

mol I2

Task 7h

1. What is the molality of acetone in a solution composed of 255 g of acetone, (CH3)2CO,

dissolved in 200. g of water?

2. What quantity, in grams, of methanol, CH3OH, is required to prepare a 0.244 m solution in

400. g of water?


Dilution

Sometimes you will need to dilute a more concentrated solution. The formula for dilution is:

M1V1 = M2V2

For example: What volume of 6.0 M KCl will be needed to make 300.0 mL of 2.4 M KCl?

(6.0 M)V1 = (2.4 M) (300.0 mL)

V1 = 120 mL

Task 7i

1. How would you make 300.0 ml of 2.4 M KCl from 6.0 M KCl?

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Topic 7: The Mole Concept - · PDF fileTopic 7: The Mole Concept ... A chemist does the same thing with atoms! Why? ... 2. What is the mass of 1.2 x 1024 atoms of magnesium? 3