| | |
SITE SEARCH
Learning Resource Centre
 
»Book a Tutor
»Become a Tutor
»Math Learning Centre
»Learning Strategies
»Learning Circles, Review Sessions & Open Tutoring9
»Tutoring Guidelines
»Practice and Review
Pathology Reviews
Pathology Worksheets/Quizzes
Anatomy Worksheets
»English Language Help
»Contact the Learning Centre
»eCentennial
»myCentennial
Font Size
Smaller
Restore Larger

Stoichiometry Review

A. What is a Mole?

The term mole, like a dozen, refers to a particular number of things.  While a dozen eggs contains 12 eggs, a mole of carbon contains 6.022 x 1023 molecules of carbon.

A mole is defined as the quantity of a given substance that contains 6.022 x 1023 molecules or formula units of that substance. The mass of individual molecules can be expressed using atomic mass units (amu). For example, 1 molecule of carbon weighs 12.01amu, and 1 molecule of water (H2O) weighs 14.03amu (2x1.01 + 12.01). Atomic mass units, however, are extremely small units that are not practical for use in a laboratory.  Instead, we use the much larger unit, a gram.  Because of this, the concept of a mole was developed: where avagadro's number 6.022 x 1023 is the number of atoms in 12.01g of carbon.

So, while one molecule of carbon has a mass of 12.01 amu, one mole (6.022 x 1023 molecules) of carbon has a mass of 12.01 g.  Likewise, one molecule of oxygen has a mass of 15.9994 amu, while one mole of oxygen weighs 15.9994g; one molecule of water has a mass of 14.03amu, while one mole has a mass of 14.03g.  This is true for any molecule; the mass written on the periodic table represents both the mass of one molecule in amu and the mass of one mole in grams.

The molecular mass of carbon can be written as:

12.01 amu/molecule    or   12.01 g/mole

 

 

 

Practice Problems

1. What is a mole?

2. What are the units for:

a) molecular mass

b) molar mass

3.  Give the molecular and molar masses for the following (be sure to include units):

a) Al     b) NaCl     c) SO2     d) Mg(OH)2     e) Ca3(PO4)2

4. How many molecules are in each of the following?

a) 3 moles of carbon (C)

b) 0.75 moles of oxygen (O)

c) 1.56 moles of boron (B)

 5.  How many molecules of hydrogen (H) are in the following?

a) 25 molecules of NH3

b) 10 molecules of H2O

c) 1 mole of NaOH

d) 3 moles of HCl

e) 1.5 moles of H2S

f)  2 moles of (NH4)2SO4

B. Introduction to Conversion Factors

In order to convert from one unit to another we simply multiply by a conversion factor, a fraction whose numerator and denominator represent two things that are equal. For example, we know that 1 mole = 6.022 x 1023 molecules, which can be used to make the following conversion factors:

 

              6.022 x 1023 molecules                 or                             1 mole                                     

                        1 mole                                                  6.022 x 1023 molecules  

 

But which conversion factor do we use? 

If we want to convert from moles to number of molecules, we use the first conversion factor. 

Example: How many molecules are in 5 moles of Carbon?

 

            5 moles of C   x      6.022 x 1023 molecules    =    3.011 x 1024 molecules of C

                                                      1 mole

Note: the units for moles cancel each other out, leaving us with molecules, and we simply multiply 5 by 6.022 x 1023.

Now, if we want to convert back to moles we use the second conversion factor.

Example: If we have      3.011 x 1024 molecules of C how many moles of carbon do we have?

            3.011 x 1024 molecules of C    x                   1 mole                    =    5 moles of C  

                                                                  6.022 x 1023 molecules   

Note: in this case the units for molecules cancel each other out leaving us with moles, and we simply divide 3.011 x 1024 by 6.022 x 1023.

 

So, to answer our question, we use the conversion factor that has the unit you are converting from in the denominator, and the unit you are converting to in the numerator.

 

Practice Problems

 

1. How many molecules of carbon (C) are in 6.0 moles of carbon?

2. If we have 1.8069 x 1024 molecules of aluminum (Al), how many moles of aluminum do we have?

3. How many molecules of methane (CH4) are there in 0.7 moles of methane?

4. If we have 9.033 x 1022 molecules of nitrogen dioxide (NO2), how many moles of nitrogen dioxide do we have?

5. If we have 2.4088 x 1021 molecules of neon (Ne), how many moles of neon do we have?

 

 

C. More Conversion Factors!

We can use the molecular mass as a conversion factor to convert between mass and moles.

Example:  The molecular mass of oxygen is 15.9994g/mol.  If we have 127.9952g of oxygen how many moles do we have?

We know from the molecular mass that 1mole of oxygen = 15.9994g of oxygen, and we can use this information to make a conversion factor.  In this case, we are converting from grams to moles.  So 15.9994g will become the denominator and 1 mole will become the numerator of our conversion factor.

   127.9952 g of oxygen   x       1mole of oxygen    =    8.0 moles of oxygen

                                               15.9994 g of oxygen

Note: the units for grams cancel each other out, leaving the answer in moles, and we simply divide 127.9952 by 15.9994 to get our answer.

Example:  The molecular mass of Nitrogen is 14.01g/mol.  If we have 3.5 moles of Nitrogen how many grams of Nitrogen do we have?

We know from the molecular mass that 1mole of nitrogen = 14.01g of nitrogen, and we can use this information to make a conversion factor.  In this case we are converting from moles to grams so 1.0 mole will become the denominator of our conversion factor, and 14.01g will be the numerator.

 

   3.5 moles of nitrogen   x       14.01 g of nitrogen    =    49.035 g of nitrogen

                                                  1 mole of nitrogen

 

Note: the units for moles cancel each other out, leaving the answer in grams, and we simply multiply 3.5 by 14.01 to get our answer.

 

We can use the molarity of a solution as a conversion factor to convert between moles and volume of solution.

 

Example: How many moles of NaCl are in 3.0L of a 0.75M solution of NaCl?

The symbol ‘M' simply means molar or moles per liter (moles/L), so writing 0.75M is the same as writing 0.75 moles/L.  The molarity of the solution tells us that 1.0L of NaCl solution = 0.75 moles of NaCl, so we can use this information to make our conversion factor.  In this case we are converting from liters to moles so 0.75 becomes the numerator of our conversion factor and 1.0L is the denominator.

 

   3.0 L of NaCl solution   x    0.75moles of NaCl     =   2.25 moles of NaCl

                                              1.0L of NaCl solution

 

Note: the units for liters cancel, each other out leaving the answer in moles, and we simply multiply 3.0 by 0.75 to get our answer.

 

We can use the concentration of a solution as a conversion factor to convert between moles and volume of solution.

Example:  If you wanted 2.45 grams of AgI, how many liters of a 0.35 g/L solution of AgI would you need?

The concentration of the solution tells us that 1.0L of AgI solution = 0.35 grams of AgI, so we can use this information to make our conversion factor.  We are converting from grams to liters so 0.35 grams is the denominator of our conversion factor and 1.0 L is the numerator.

 

   2.45 g of AgI   x       1.0L of AgI solution     =   7.0 L of AgI

                                         0.35g of Ag I

 

Note: the units for grams cancel each other out, leaving the answer in liters, and we simply divide 2.45 by 0.35 to get our answer.

 

Practice Problems

1.  Calculate the mass in grams of the following

a) 0.35 mole of Be

b) 3.6 mole of Li

c) 23 mole of Cu

2. Calculate the number of moles in the following

            a) 7.57 g of C

            b) 96.0 g of O2

            c) 75.8 g of Mg

3. Calculate the number of moles in the following

            a) 2.5 L of 6.0M NaCl

            b) 0.42 L of 0.8M KBr

            c)  0.22L of 4.0M CuSO4

4. Calculate the number of liters needed...

            a) to get 5.0 moles of HCl if you have a 2.0M solution of HCl

            b) to get 0.4 moles of Mg(SO4)2 if you have a 6M solution of Mg(SO4)2

            c) to get 0.860 moles of NaOH if you have a 0.215M solution of NaOH

5. Calculate the number of grams in the following

            a) 6.0L of a 0.95g/L solution of NaI

            b) 0.5L of a 0.23 g/L solution of K3PO4

            c) 0.2 L of a 7.5g/L solution of Fe(NO3)2

6. Calculate the number of liters needed...

            a) to get 2.0 moles of AgBr if you have a solution of AgBr with a conc. of 0.8g/L

            b) to get 0.48 moles of CaS if you have a solution of CaS with a conc. of 0.6g/L

            c) to get 6.2 moles of MgSO4 if you have a solution of MgSO4 with a conc. of 1.24g/L

 

D. Two-step Conversion Problems

 

It is not always possible to get our answer by doing one conversion; often we need to do two or more conversions to get our answer.

Example:  How many grams of NaCl are needed to make a 5.0 L of a 0.87M solution of NaCl?

We are starting with 5.0L and we know that we want to get an answer in grams, but we do not have a conversion factor that allows us to convert directly between liters and grams.  Instead we first have to convert from liters to moles, and then from moles to grams.

Step 1:

   5.0 L of NaCl solution   x    0.87moles of NaCl     =   4.35 moles of NaCl

                                               1.0L of NaCl solution

Step 2:

   4.35 mol of NaCl   x      58.44 g of NaCl    =    254.21 g of NaCl

                                          1 mole of NaCl

We can also write this out in one step as follows:

5.0 L of NaCl  x   0.87 moles of NaCl    x    58.44 g of NaCl    =   254.21 g of NaCl

                            1.0L of NaCl solution        1 mole of NaCl

Example:  How many molecules of fluorine are in 66.50g of fluorine?

Again, we don't have a conversion factor that allows us to convert directly from grams to number of molecules.  Instead, we must first convert from grams to moles, and then from moles to number of molecules.

   66.50 g of fluorine   x       1mole of fluorine    =    3.5 moles of oxygen

                                             19.00 g of oxygen

 

   3.5 moles of F   x      6.022 x 1023 molecules    =    7.38 x 1024 molecules of F

                                           1 mole

Practice Problems

1. How many grams of KOH are needed to make 2.0L of a 0.6M solution of KOH?

2. How many moles of HNO3 are in 4.2 L of a 3.0g/L solution of HNO3?

            (Hint: start with liters and convert to grams then to moles)

3. How many molecules are in 3.0g of Cobalt (Co)?

 

 

E. Stoichiometry:

Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction.

 

Stoichiometry is useful in answering important questions such as how much of each reactant is needed to make the desired amount of product. You want to be sure you make enough of the product, but reactants are often expensive, so you want to be sure that you don't make too much or you waste money.

A balanced chemical equation is like a recipe; it tells us how much of each reactant we need in order to get a certain amount of product.  For example, a cookie recipe tells you that you need 2 cups of flour, 1 cup of sugar, 2 eggs, 1 cup of chocolate chips etc. and you should get 3 dozen cookies.  In the same way a balanced chemical equation, like the one below, tells you that in order to make 2 moles of nitric acid (HNO3), you require 3 moles of nitrogen dioxide (NO2) and 1 mole of water (H2O).

            3NO2(g)  +  H2O(l)                                       2HNO3(aq)  +    NO(g)

The number of moles involved in a reaction is proportional to the coefficients in the balanced chemical equation.  As a result, the coefficients can be used to make a conversion factor, in order to convert between the different substances involved.

Example

According to the above equation, how many moles of HNO3 can be produced from 7.5 moles of NO2?

We have a choice between the following 2 conversion factors:

              3 moles NO2                or                     2 moles HNO3

            2 moles HNO3                                       3 moles NO2

Because we are converting from moles of NO2 to moles of HNO3, we will use the second conversion factor as follows:

            7.5 moles NO2  x  2 moles HNO3   =   5 moles HNO3

                                           3 moles NO2

Therefore 5 moles of HNO3 can be produced using 7.5 moles of NO2.

It is important to remember that a balanced chemical equation directly relates moles of a substance, not mass or volume of a substance.  Therefore, you cannot convert directly from the mass of substance A to the mass of substance B (or from volume of substance A to the volume of substance B).  Rather, you must first convert to moles.  There are three basic steps to any stoichiometry problem.

1.       convert from mass of volume of substance A to moles of substance A

2.       convert from moles of substance A to moles of substance B

3.       convert from moles of substance B to mass or volume of substance B

First, decide what piece of information you have to start with, and what piece of information you need to find, and follow the arrows to do the appropriate conversions.

 Example

According to the above equation, how many grams of nitrogen dioxide (NO2) are required to produce 10.0 g of Nitric Acid (HNO3)?

We are starting with 10.00 grams of HNO3 and need to find grams of NO2

Step 1: convert from grams of HNO3 to moles of HNO3.

            10.00g HNO3  x   1 mole HNO3  =  0.1587g HNO3

                                       63.02 g HNO3

Step 2: convert from moles of HNO3 to moles of NO2

            0.1587 mole HNO3  x  3 moles NO2_  =  0.2380 mole NO2

                                                2 moles HNO3

Step 3: convert from mole of NO2 to grams of NO2

            0.2380 mole NO2  x   46.01 g NO2  =   10.95g NO2

                                                 1 mole NO2

This can also be written all in one equation as follows:

10.00g HNO3  x   1 mole HNO3  x   3 moles NO2   x    46.01 g NO2  =   10.95g NO2

                           63.02 g HNO3       2 moles HNO3        1 mole NO2

Therefore 10.95g of NO2 is needed to prepare 10.00g of HNO3.

Practice Problems

1. Ethanol (C2H5O) burns in oxygen (O2) to give carbon dioxide (CO2) and water (H2O) according to the following equation

            C2H5O(l)  +  3O2(g)  -----------------  2CO2(g)  +  3H2O(l)

a.       How many moles of oxygen are needed to burn 5 moles of ethanol?

b.      If 6.0 g of ethanol is burned, how many grams of water will be produced?

c.       If 2.0 grams of ethanol is burned, how many liters of carbon dioxide gas at STP will be produced? ( Hint: 1 mole of carbon dioxide gas at STP occupies a  volume of 22.4L)

 

2. How many liters of 0.250 M HNO3 are needed to react with 1.6 liters of 0.150M Na2 CO3 according to the following equation

2HNO3(aq)  +  Na2CO3(aq) -------------  2 NaNO3(aq) + H2O(l)  + CO2(g)

 

3. When solutions of AgNO3 and NaCl are mixed, solid AgCl forms and precipitates out of solution

AgNO3 (aq)  +  NaCl(aq)  --------------  AgCl(s)  +   NaNO3(aq)

 

If 0.5 liters of 0.6M AgNO3 is combined with excess NaCl solution, how many grams of AgCl will be formed?

 

 

Answers

 

A. What is a Mole.

1. A mole is 6.022 x 1023 molecules of a substance.

2. a.  molecular mass  amu/molecule

    b.  molar mass  g/mole

3. a.  26.98 amu/molecule or 26.98 g/mole

    b.  58.44 amu/molecule or 58.44 g/mole

    c.  64.06 amu/molecule or 64.06 g/mole

    d.  58.33 amu/molecule or 58.33 g/mole

    e.  310.18 amu/molecule or 310.18 g/mole

 4. a.  3 x 6.022 x 1023 molecules = 1.8066 x 1024

    b.  0.75 x 6.022 x 1023 molecules = 4.5165 x 1023

    c.  1.56 x 6.022 x 1023 molecules = 9.3943x 1023

5. a.  25 molecules x 3 hydrogen atoms/molecule = 75 hydrogen atoms

    b.  10 molecules x 2 hydrogen atoms/molecule = 20 hydrogen atoms

    c.  6.022 x 1023 molecules x 1 H atom/molecule = 6.022 x 1023 H atoms

    d.  3 moles x 6.022 x 1023 molecules/mole x 1 H atom/molecule = 1.8066 x 1024 H atoms

    d.  1.5 moles x 6.022 x 1023 molecules/mole x 2 H atom/molecule = 1.8066 x 1024 H atoms

    d.  2 moles x 6.022 x 1023 molecules/mole x 8 H atom/molecule = 9.6352 x 1024 H atoms

 

B. Introduction to Conversion Factors

1.  6.0 moles of C   x      6.022 x 1023 molecules    =    3.6132 x 1024 molecules of C

                                           1 mole

2.  1.8069 x 1024 molecules of Al    x                1 mole                  =    3 moles of Al  

                                                 6.022 x 1023 molecules   

 

3.  0.7 moles of CH4   x      6.022 x 1023 molecules    =    4.2154 x 1023 molecules of CH4

                                               1 mole

 

4.  9.033 x 1022 molecules of NO2    x                1 mole                  =    0.15 moles of NO2  

                                                       6.022 x 1023 molecules   

 

5.  2.4088 x 1021 molecules of Ne    x                1 mole                  =    0.004 moles of Ne  

                                                       6.022 x 1023 molecules   

 

C. More Conversion Factors

 

1. a.   0.35 mole Be  x  9.01 g Be  =  3.15 g Be

                               1 mole Be

  

    b.  3.6 mole Li  x  6.94 g Li  =  25.0 g Li

                            1 mole Li

 

    c.  23 mole Cu   x   63.55 g Cu  =  1461.6 g Cu

                              1 mole Cu

2. a.  7.57 g C   x   1 mole C   = 0.63 mole Cu

                         12.01 g C 

 

    b.  96.0 g O2   x   1 mole O2 =  3.0 mole O2

                             32.00 g  O2 

 

    c.  75.8 g Mg   x   1 mole Mg =  3.12 mole Mg

                           24.30 g Mg 

 

 

3. a.   2.5 L NaCl  x  6.0 mole NaCl  =  15.0 g NaCl

                              1.0 L NaCl

  

    b.  0.45 L KBr  x  0.8 mole KBr  =  0.36 mole KBr

                                 1.0 L KBr

 

    c.  0.22 L CuSO4   x   0.4 mole CuSO4  =  0.88 mole CuSO4

                                     1.0L CuSO4

4. a.  5.0 moles HCl   x      1.0L HCl      = 2.5 L HCl

                                   2.0 mole HCl 

 

    b.  0.15 mole Mg(SO4)2   x       1.0 L Mg(SO4)2        =  0.025 L Mg(SO4)2  

                                             6.0 mol  Mg(SO4)2   

 

    c.  0. 860 mole NaOH   x        1.0 L NaOH        =   4.00 L NaOH

                                             0.215 mol  NaOH

 

5. a.   6.0 L NaI  x  0.95 g NaI  =  5.7 g NaI

                          1.0 L NaI

  

    b.  0.5 L K3PO4  x  0.23 g K3PO4  =  0.36 g K3PO4

                              1.0 L K3PO4

 

    c.  0.2 L Fe(NO3)2   x   7.5 g Fe(NO3)2     =  0.88 g Fe(NO3)2  

                                   1.0L Fe(NO3)2  

 

6. a.  2.0 g AgBr   x      1.0L AgBr      =  2.5 L AgBr

                                0.8 g AgBr

 

    b.  0.48 g CaS x   1.0 L CaS =  0.8 L CaS

                            0.6 g CaS

 

    c. 6.2 g MgSO4   x  1.0 L MgSO4     =  5.0 L MgSO4

                                1.24 g  MgSO4

D.  Two Step Conversion Problems

 

1.  2.0L KOH  x   0.6 mol KOH  x  56.01 g KOH  =  56.11 g KOH

                           1.0 L KOH        1 mol KOH

 

2.  4.2L HNO3  x  3.0 g HNO3   x  1 mol HNO3      =  0.20 mol HNO3

                              1 mol           63.02 g HNO3

 

3.  3.0 g Co  x  1 mole Co  x  6.022 x 1023 molecules Co  =  3.06 x 1022 molecules Co

                      58.93 g Co                  1.0 mole Co

E.  Stoichiometry

 

1. a. 5 moles C2H5O  x    3 moles O   =  15 moles O2

                                  1 mole C2H5O

 

    b. 6.0 g C2H5O  x   1 mol C2H5O    x   3 moles H2O    x   18.02 g H2O   =  9.8 g H2O

                            33.05 g C2H5O                  1 mole C2H5O        1 mole H2O

 

    c. 2.0 g C2H5O  x   1 mol C2H5O    x   2 moles CO2    x   22.4 L CO2   =  2.7 L CO2

                            33.05 g C2H5O                  1 mole C2H5O        1 mole CO2

2.  1.6L Na2 CO3  x  0.150 mol  Na2 CO3     x   1 moles Na2 CO3      x      1.0 L HNO3   =  0.18 L HNO3

                                1.0 L Na2 CO3                   2 mole HNO3             0.250 mole HNO3

3.  0.5 L AgNO3  x  0.6 mol  AgNO3     x   1 moles AgCl     x    143.32 g AgCl   =   42.996 g AgCl

                             1.0 L AgNO3             1 mole AgNO3        1.0 L mole AgCl

 

book a tutor