Chapter 5: Calculations and the Chemical Equation

 

The Mole Concept and Atoms

Atoms are exceedingly small, yet their masses have been experimentally determined for each of the elements. The periodic table provides atomic masses in atomic mass units (amu). A more practical unit for defining a "collection" of atoms is the mole, Avogadro's number of particles.

Why we need to use the concept of "mole" of atoms ( 6.022 x 1023 particles) in chemical stoichiometry?

            Chemical equations are written in terms of atoms and molecules. However, we cannot pick atoms individually and do reactions. Chemists always use mass in grams as the amount in the reaction. Therefore, we need a conversion factor to convert atoms and molecules to grams. Mole is the connection or the conversion factor between atoms and grams.

Avogadro's number

           The name "Avogadro's Number" is just an honorary name attached to the calculated value of the number of atoms, molecules, etc. in a gram mole of any chemical substance. Of course if we used some other mass unit for the mole such as "pound mole", the "number" would be different than 6.022 x 1023.

Calculations based on the chemical equation relate the number of atoms, moles and their corresponding mass. Conversion factors are used to relate the information provided in the problem to the information requested by the problem. It is often useful to map a pattern for the required conversion before beginning the problem.

1 mol = M.W. (molecular weight) taken in grams

1 mol = 6.022 x 1023 particles

1 mol = 6.022 x 1023 atoms

1 mol = 6.022 x 1023 molecules

1 mol = 6.022 x 1023 ions

 

Converting amu to g/mole

1 amu =  1 g/mole

An atom weighs 7.47 x 10-23 g. What is the name of the element this atom belongs to?

First convert g to amu and look up in the periodic table and find out the element. Conversion factor: 1g = 6.022 x 1023 amu

 

7.47 x 10-23 g   x

6.022 x 1023 amu      

  = 44.98 amu

 

1g

In the periodic table atomic masses increase generally with atomic number. Element with an atomic mass closer to the value calculated is Sc (Scandium).

The element is Sc.

 

 

 

 

Convert atomic mass of Hg in amu to g/mole

 

               

 

Relating Avogadro's number to molar mass:  calculation of the mass of Avogadro's number of sodium atoms

Calculate the number moles of Na in  9.03 x 1023 atoms Na.

               ?    mol Na =  9.03 x 1023 atoms Na x   = 1.50 mol Na

 

 

 

moles  =

 

Grams

Molecular weight

 

 

 

Molecular weight =

 

Grams

moles

 

Grams = moles x Molecular weight

 

 

Converting grams to atoms

O atoms = 40.0 g O x   = 1.51 x 1024 O atoms

The Mole and Avogadro's Number : Calculating Atoms, Moles, and Mass

 

5.24         15.0 mol C x  = 1.80 x 102 g C

 

Converting moles to atoms:    Moles x Avogadro’s number

Converting atoms to moles:    Atoms ¸  Avogadro’s number

Converting moles of a substance to mass in grams:  moles x molar mass

Converting grams to moles

How many moles of iron, Fe are in 4.00 g of Fe?

                              4.00 g Fe x  = 7.16 x 10–2 mol Fe

Converting kilograms to moles:  kg -> g  ->  moles

How many moles of N2   2.00 kg of N2?

                                        2.00 kg N2 x  = 71.4 mol N2

Converting pounds to moles:  lb -> g  ->  moles

How many moles of calcium, Ca is in 0.10 g of Ca?

 

                0.10 lb Ca x  = 1.1 mol Ca

 

Converting grams to number of atoms: grams ¸ atomic mass  ->  moles -> atoms

 

Chemical Compounds

In chemistry, a compound is a substance formed from two or more elements, with a fixed ratio determining the composition. For example, dihydrogen monoxide (water, H2O) is a compound composed of two hydrogen atoms for every oxygen atom.

 

The Chemical Formula

A chemical formula (also called molecular formula) is a concise way of expressing information about the atoms that constitute a particular chemical compound. It identifies each type of element by its chemical symbol and identifies the number of atoms of such element to be found in each discrete molecule of that compound. The number of atoms (if greater than one) is indicated as a subscript. For example methane, a simple molecule consisting of one carbon atom bonded to four hydrogen atoms has the chemical formula: CH4

and glucose with six carbon atoms, twelve hydrogen atoms and six oxygen atoms has the chemical formula: C6H12O6.

 

Calculating formula weight and molar mass (C2F2C2).

                  2 atoms of carbon   x 12.01 g/mole =               24.02

                  2 atoms of fluorine x 19.00 g/mole =                38.00

                  4 atoms of chlorine x 35.45 g/mole =             141.80

                                                                                       203.82 g/mole

The average mass of a single molecule of C2F2Cl4 is 203.82  g/mole.

Calculating formula weight and molar mass of following ionic compounds: NaCl and K2CO3

Formula weight of a ionic compound is the sum of all weights (i.e. atomic weight multiplied by their subscripts) in the chemical formula of the ionic compound.

 a) Calculating formula weight and molar mass (NaCl).

                  1 atoms of sodium   x 23.00 g/mole =              23.00

                  1 atoms of chlorine x 35.45 g/mole =               35.45

                                                                                         58.45 g/mole

                                                                                            

The average mass of a single formula unit of NaCl is 58.45 g/mole.

b) Calculating formula weight and molar mass (K2CO3).

                  2 atoms of potassium   x 39.10 g/mole   =     78.20

                  1 atoms of carbon x 12.01 g/mole            =     12.01                                 

                  3 atoms of oxygen x 48.00 g/mole             =     48.00

                                                                                  =   138.21 g/mole

The Mole Concept Applied to Compounds

Just as a mole of atoms is based on the atomic mass or atomic weight, a mole of a compound is based upon the molar mass/formula mass or formula weight. To calculate the moles from formula weight, the formula unit must be known.

 

How many moles of MgCl2 are present in 35.0 g of  MgCl2?  

                               35.0 g MgCl2 x  = 0.368 mol MgCl2  

How many moles of K2SO4 are present in 180.1g of potassium sulfate?

180.1 g K2SO4 x  = 1.033 mol K2SO4  

The Chemical Equation and the Information It Conveys

In a chemical equation, the identity of reactants and products must be specified. Reactants are written to the left of the reaction arrow (®) and products to the right. The physical states of reactants and products are shown in parentheses. The symbol D over the reaction arrow means that heat energy is necessary for the reaction to occur. The equation must be balanced to reflect the law of conservation of mass, which states that matter can neither be gained nor lost in the process of a chemical reaction.

Chemical Equations: An equation with coefficients and formulas showing the starting and final substances maintaining atomic balance.

 

Features of a Chemical Equation

Chemical Equations show:

1.      The reactants which enter into a reaction.

2.      The products which are formed by the reaction.

3.      The amounts (moles) of each substance used and each substance produced.

The Numbers in a Chemical Equation:

1.      Subscripts: The small numbers to the lower right of chemical symbols. Subscripts represent the number of atoms of each element in the molecule

2.      Stoichiometric Coefficients: The large numbers in front of chemical formulas. Coefficients represent the number of molecules of the substance in the reaction.

Physical State of reactants and products

1.      If a reactant or product is a solid, (s)   and liquid, (l) is placed after the formula.

2.      If a reactant or product is a gas, (g) is placed after it.

3.      If a reactant or product is in water solution, (aq) is placed after it.

 

A Recipe for Chemical Change

A chemical equation is similar to a cookbook recipe in that it shows how many units of each substance is required to give the desired result. It shows the combination of various elements and/or molecules and then the resulting elements and/or molecules.

 

Balancing Chemical Equations

Chemical equations do not come already balanced. This must be done before the equation can be used in a chemically meaningful way. All chemical calculations to come must be done with a balanced equation. A balanced equation has equal numbers of each type of atom on each side of the equation.  The Law of Conservation of Mass is the rationale for balancing a chemical equation.

For a chemical equation to be balanced, the same number of each kind of atom must be present on both sides of the chemical equation. The French chemist Antoine Lavoisier described the law of conservation of matter -- in a chemical reaction matter can neither be created nor destroyed. From Dalton's atomic theory we know that all substances are composed of atoms. During a chemical reaction atoms may be combined, separated, or rearranged, but not created or destroyed.

Equations Must Be Balanced Because: Atoms Can Be Neither Created Nor Destroyed By Ordinary Chemical Means, so there must be the same number of atoms on both sides of the equation.

Seven Steps to Balance Equations By Inspection

1. Check for Diatomic Molecules - H2 - N2- O2 - F2 - Cl2 - Br2 - I2

    If these elements appear By Themselves in an equation, they Must be written with a subscript of 2

2. Balance Metals

3. Balance Nonmetals

4. Balance Oxygen

5. Balance Hydrogen

6.Recount All Atoms

7. If EVERY coefficient will reduce, rewrite in the simplest whole-number ratio.

 

Balance following chemical equations and give the sum of stiochiometric coefficients:

            It is of most importance for a chemist to be able to write correctly balanced equations and to interpret equations written by others. It is also very helpful if he/she knows how to predict the products of certain specific types of reactions.

1)         Heating sodium hydroxide

Sodium Hydroxide --> Sodium Oxide + Water

             2NaOH --> Na2O + H2O

2)         Rusting of iron

Iron + Oxygen --> Iron (III) Oxide

            4Fe + 3O2 --> 2 Fe2O3

3)         Carbonation of water

Carbon Dioxide + Water --> Glucose + Oxygen

            6CO2 + 6 H2O --> C6H12O6 + 6 O2

4)         Double dispalcement

Iron (II) Sulfide + Hydrochloric Acid --> Iron (II) Chloride + Hydrogen Sulfide

            FeS + 2 HCl --> FeCl2 + H2S

5)         Burning of hydrogen in oxygen

Oxygen + Hydrogen --> Water

            O2 + 2 H2 --> 2 H2O

6)         Single displacement

Chlorine + Sodium Iodide -->  Sodium Chloride + Iodine

            Cl2 + 2 NaI --> 2 NaCl + I2

7)         Double dispalcement

Aluminum Nitrate + Sulfuric Acid -->

            2 Al(NO3)3 + 3 H2SO4 --> Al2(SO4)3 + 6 HNO3

8)         Silver smelting

Silver Oxide --> Silver + Oxygen

            2 Ag2O --> 4 Ag + O2

9)         Double dispalcement

Ammonium Phosphate + Barium Hydroxide -->

            2 (NH4)3PO4 + 3 Ba(OH)2 --> Ba3(PO4)2 + 6 NH4OH

10)       Acid Base reacxtions

Calcium Hydroxide + Nitric Acid --> Salt + Water

            Ca(OH)2 + 2 HNO3 --> Ca(NO3)2 + 2 H2O

11)       Single replacement

MgO(s) + Si(s) = Mg(s) + SiO2(s)

            2 MgO (s) + Si (s) = 2 Mg (s) + SiO2(s)

 12)      Acid formation: Non-metal oxides with water

P4O10(s) + H2O(l) = H3PO4(l)

            P4O10 (s) + 6H2O (l) = 4 H3PO4 (l)

 13)      Double dispalcement

K2CO3(aq) + BaCl2(aq) = KCl(aq) + BaCO3(s)

            K2CO3(aq) + BaCl2(aq) = 2 KCl(aq) + BaCO3(s)

 14)      Double displacement

NaOH(aq) + Al(NO3)3(aq) = NaNO3(aq) + Al(OH)3(aq)

            3 NaOH (aq) + Al(NO3)3 (aq) = 3 NaNO3 (aq) +Al(OH)3 (aq)

15)       Gas formation

Fe2(CO3)3(s) + H2SO4(aq) = Fe2(SO4)3(aq) + H2O(l) + CO2(g)

            Fe2(CO3)3 (s) + 3 H2SO4 (aq) = Fe2(SO4)3 (aq) + 3 H2O(l) + CO2 (g)

 16)      The combustion of propane

C3H8 (g) + O2 (g) = CO2 (g) + H2O (l)

            C3H8 (g) + 5 O2 (g) = 3 CO2 (g) + 4 H2O (l)

 

 

Calculations Using the Chemical Equation

General Principles

Limiting reactant

  The reactant used up first in a  chemical reaction.

In real life it is rare for a chemical reaction to use up all the reactants in the formation of products. When    reactants are not used up in equal amounts, we say that the reactants are not present in stoichiometric    quantities. Problems of this type are referred to as limiting reactant problems. Reactions in one reactant in limited Supply.

 

Analogy in Recipe : Making cheese sandwiches

You were given 20 slices bread, 5 slices of cheese, 4 slices of ham

 

If you want to make sandwiches containing  two slices bread and one slice of cheese and one slice of ham, how many sandwiches you could make? What is the limiting ingredient?

Stoichiometry

The quantitative relationship among reactants and products is called stoichiometry. The term stoichiometry is derived from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). On this subject, you often are required to calculate quantities of reactants or products.

We are already familiar with converting grams to moles.

 

Calculating reacting quantities: 

Use of Conversion Factors

Mole conversion factors for the following equation:

         4 NH3(g) + 5 O2(g) ------> 4 NO(g) + 6 H2O(g)

Mole ratios of reactants and products

 

4 mol NH3 = 5 mol O2

5 mol O2 = 6 mol H2O

4 mol NH3 = 4 mol NO; 1mole NH3 = 1 mole NO

1 mol NH3 = 1 mol NO

 Converting moles of Reactants to Products

Calculate the following using the chemical equation given below:

                      4 NH3(g) + 5 O2(g) ------> 4 NO(g) + 6 H2O(g)

a)      moles of NO(g) from 2 moles of NH3(g) and excess O2(g).

b)      moles of H2O(g) from 3 moles of O2(g) and excess NH3(g).

c)      Moles of NO(g) from 2 mole of NH3(g) and excess O2(g)

How many Moles of NO(g) are produced from 2 mole of NH3(g) and excess O2(g)

4 NH3(g) + 5 O2(g) ------> 4 NO(g) + 6 H2O(g)

There are several conversion factors that you can obtain from S.C. of the balanced chemical equation

4 mol NH3        = 5 mol O2

5 mol O2          = 6 mol H2O

4 mol NH3        = 4 mol NO; 1mole NH3 = 1 mole NO

1 mol NH3        = 1 mol NO

Now the problem is to convert 0.80 moles of O2 to H2O

2.00 mol NH3 --> ? mol H2O

We need the conversion factor: 1 mol NH3 = 1mol NO

2.00 mol NH3       x    1 mol NO                               = 2 mol NO

                                   1 mol NH3 

                                  

b) Moles of H2O(g) from 3 moles of O2(g) and excess NH3(g).

This problem is to convert 3 mole O2 to mole of H 2O

We need the conversion factor: 5 mol O2 = 6 mol H2O

x = 3.6 mol H2O

How many moles of H2O will be produced by 0.80 mole of O2, according to the equation?

                    2H2(g) + O2(g) = 2 H2O(l)

2H2(g) + O2(g) = 2 H2O(l)

2H2(g) + O2(g) = 2 H2O(l)

0.80 mol O2     =   ? mol H2O

There are several conversion factors that are coming from the chemical equation:

2 mol H2 = 1 mol O2

2 mol H2 = 2 mol H2O ; 1 mol H2 = 1 mol H2O

1 mol O2 = 2 mol H2O

We need the conversion factor is:

1 mol O2 = 2 mol H2O

x = 1.6  mol H2O

Calculating grams of product from moles of reactant

 

Calculate the mass of CO2 produced burning C2H5OH in excess oxygen.

    C2H5OH  + 5 O2   --> 2 CO2   +   6 H2O

1 mol C2H5OH x  = 88.0 g CO2 

 

Relating masses of reactants and products

Calculating grams of product from grams of reactant

 

Barium carbonate is BaCO3.

a. Write a chemical equation for the decomposition of BaCO3.

b.  How many  grams of CO2 will be produced  from 50.0 g BaCO3?

 

a.        

 

b.         50.0 g BaCO3 x  =  11.2 g CO2 

 

Theoretical and Percent Yield

 

Evaluating Success of Synthesis

One of the most important contribution of chemistry to other areas of sciences providing chemicals or raw materials for building various components of complex structures. Chemists come up with chemical reactions and optimize the conditions at which highest yield of the products could be synthesized. The amount of product measured in grams is called the yield. There are several descriptions of yield: theoretical and actual. Theoretical and actual  yields will be discussed below. Therefore, yield of a chemical reaction plays an important role evaluating the success of a synthesis. If reactant are in stoichiometric amounts and the reaction take place as written in the chemical equation the percent yield should be 100%. However, there are side reactions and reaction yield are always less than 100%.  Side reaction are competing chemical reactions that take place other than the one you think is taking place. They produce by products.  By products are the products of the unwanted competing reactions. Situations occur where yield could be more than 100% pointing to products containing moisture and other contaminants.

 

 

Theoretical Yield

The amount of product predicted by the balanced equation when all of the limiting reagent has   reacted.

 

Actual Yield

The amount of product actual obtained in a real chemical reaction that is carried out in the laboratory.

 

Percent Yield or % Yield

Percent yield compares the actual yield to theoretical yield and shows the efficiency of a chemical synthesis

The % yield is given by the equation:    

Example: Calculate the percent yield of chloroform, CHCl3, produced in a reaction excess methane, CH4 is reacted with  105 g of chlorine, Cl2. In this reaction 10.0 g of  chloroform, CHCl3 is actually produced.

a.         Step 1. Write down information about the reaction:

                                 CH4(g) + 3 Cl2(g) ® 3 HCl(g) + CHCl3(g)

                                 (excess)     105 g

 

                  Step 2.    Convert the mass of Cl2 to moles of Cl2:

 

                                 105 g Cl2  x  = 1.48 mol Cl2 

 

                  Step 3.   The reaction states that 3 moles of Cl2 will react to form one mole of CHCl3, so the mole ratio is 3:1. Use this conversion factor to calculate the mass of product:

                                 1.48 mol Cl2 x  = 58.9 CHCl3

 

            b.   % yield =  x 100 %

 

                  % yield =  x 100 %  = 17.0% yield