Atomic Weight -- the weighted average of the naturally occurring isotopes of an element.

            (amu = atomic mass unit)

·        Found on the periodic table

 

Molecular Weight  -- the sum of the masses of all of the atoms in a molecular formula.

            (Formula weight for ionic compounds)

 

Example 1: Calculate the molecular weight of an H2O molecule:

            1.01 + 1.01 + 16.00 = 18.02 amu

 

Example 2: Calculate the formula weight of one unit of NaCl:

            22.99 + 35.45 = 58.44 amu

 

One mole = 6.022 ´ 1023 particles  (Avogadro's number)

 

Defined as the number of atoms in exactly 12 grams of Carbon-12.

 

Molar mass: the mass in grams of one mole of a substance.  Expressed in the unit of grams/mole.

•     1 mole of carbon atoms weighs 12.01 g.

      Molar mass = 12.01 grams/mole

•     Molar mass of oxygen atom = ___ g/mole

•     Molar mass of carbon monoxide = ___ g/mol

 

Calculations using molar mass:

 

1.  How many moles of carbon monoxide molecules are present in a 14.0 g sample?

 

14.0 grams CO          x          1mole  =                                 moles

                                                28.01 g

 

2.  How many moles of carbon tetrachloride are present in a 25.0 gram sample?

 

 

 

3.  How many molecules of carbon tetrachloride are present in this sample?

 

 

 

4.  How many chlorine atoms are present?

 

 

 

 

5.  How many grams of magnesium chloride are present in 2.50 moles?

 

 

 

 

 

6.  How many chloride ions are present?

 

 

 

 

7.  What is the mass in grams of one oxygen molecule?

 

 

 

 

 

A hydrate is a compound which contains a set number of molecules of water associated with it in its crystalline structure.

 

Copper (II) sulfate pentahydrate    CuSO4·5H2O

 

The water molecules must be taken into consideration when calculating molar mass and performing mole conversions.

 

Hydrates can become anhydrous when heated.

CuSO4·5H2O(s) à CuSO4(s) + 5H2O(g)

 

Molar mass of hydrate =                                         

Molar mass of anhydrous substance =                              

 

What mass of the hydrate would be equivalent to 2.75 moles?

 

 

Composition of Matter

 

Composition -- the identity and amounts of all components of matter in a substance.

·        Calculations of composition can be performed to determine the percent composition of each component.     

·        Percent Composition =  part    x 100

                                                whole

 

Calculate the percent composition by mass of a water molecule.

% Hydrogen = 2.02  amu     x          100 =  11.2%

                           18.02 amu

 

% Oxygen =   16.00 amu      x          100 = 88.79%

                        18.02 amu

 

Calculate the percent composition by mass of C2H6O:

 

            Carbon:                      24.02               x 100   =

                                                46.08  

 

            Hydrogen:      6.06                x 100   =

                                    46.08  

 

            Oxygen:                      16.00               x 100   =

                                                46.08  

 

 

Empirical Formula and Molecular Formula

 

Empirical Formula: the simplest ratio of atoms in a compound; determined by percent composition in moles.

 

Steps to calculation:

1.      Find grams of each component

2.      Convert grams to moles

3.      Divide all by smallest number to find ratio

4.      If needed, multiply to get whole numbers

5.      Use whole numbers as subscripts for empirical formula

 

Ex. 1) Analysis of an unknown liquid showed that it consisted of 11.0% hydrogen and 89.0% oxygen by mass.  Determine the empirical formula of the substance.

1.      Find grams of each component

Assuming 100 gram sample…

100 grams ´ 11.0 % = 11.0 g H

100 grams ´ 89.0 % = 89.0 g O

 

2.      Convert grams to moles

            11.0 grams H ´         1mole  =         10.9 moles H

                                                1.01 g

 

            89.0 grams O ´         1mole  = 5.56 moles O

                                                16.00 g

 

3.      Divide all by smallest number to find ratio

10.9 moles H    =       2 H                  5.56 moles O      =     1 O

5.56 moles O                                      5.56 moles O

 

4.      If needed, multiply to get whole numbers

 

5.      Use whole numbers as subscripts for empirical formula

            H2O

 

Ex. 2) A forensics lab attempting to determine the cause of death for a police file identified the presence of a foreign substance in the blood stream of the deceased.  Chemical analysis of a sample of the substance revealed that it consisted of 53.81% carbon, 6.453% hydrogen, 8.969% nitrogen, 10.27% sulfur, and 20.49% oxygen by mass.  Determine the empirical formula of the substance.

1.      Find grams of each component

Assuming 100 gram sample…

 

 

 

 

2.      Convert grams to moles

 

 

 

 

 

 

 

3.      Divide all by smallest number to find ratio

 

 

 

 

 

4.      If needed, multiply to get whole numbers

5.      Use whole numbers as subscripts for empirical formula

 

 

Ex. 3) A forensics lab attempting to determine the cause of death for a police file identified the presence of a foreign substance in the blood stream of the deceased.  Chemical analysis of a sample of the substance revealed that it consisted of 18.0% Carbon, 79.7% Chlorine and 2.29% Hydrogen.  Determine the empirical formula of the substance.

 

 

 

 

 

 

 

 

 

 

 

 

 

Molecular formula represents both the type and the quantity of each atom in one molecule.  It is always a whole number multiple of the empirical formula. (Sometimes same as empirical formula.)

 

Ethane has the empirical formula CH3.  If one mole of ethane weighs 30.08 grams, determine the molecular formula of ethane.

 

a.  Calculate molar mass for empirical formula:

                        CH3      =         15.04 grams/mole

 

b.  Divide molecular weight by empirical weight

 

            30.08/15.04   =         2

 

c.  Multiply subscripts of empirical formula by number obtained in part b:

            CH3      x          2          =         C2H6

 

A forensics lab attempting to determine the cause of death for a police file identified the presence of a foreign substance in the blood stream of the deceased.  Chemical analysis of a sample of the substance revealed that it consisted of 49.5% Carbon, 28.8% Nitrogen and 5.20% Hydrogen, and 16.5% Oxygen.  Further analysis indicated that the compound had a molecular weight of 194.2 grams/mole.  Determine the empirical formula and the molecular formula of the substance. 

 

 

 

 

 

 

 

 

 

 

 

 

 

Balancing Equations

 

Chemical equation: Shorthand way of describing chemical change, using symbols and formulas to represent elements and compounds involved

·        General format:

                                    C         +         O2                    à                    CO2

                                                Reactants                                              (yield)                    Products

·        A balanced equation shows the mole relationships between the substances involved in a particular reaction.  Coefficients represent the number of moles of each substance involved in the reaction.

·        Why do we need to balance equations? 

            To show that all matter is conserved

 

Perform an inventory of all atoms present on each side of equation.

 

Example 1:

                                    H2        +         O2        à        H2O

                                    H = 2                                      H = 2

                                    O = 2                                      O = 1

 

Balance equations by adding in coefficients.

                                    2 H2     +         O2        à        2 H2O

                                    H = 4                                      H = 4

                                    O = 2                                      O = 2

 

Example 2:

C3H8   +         O2        à        CO2     +         H2O

C = 3                                                  C = 1 

H = 8                                                  H = 2

O = 2                                                  O = 2 + 1 = 3

 

C3H8   +         O2        à                    3 CO2  +         4 H2O

C = 3                                                  C = 3 

H = 8                                                  H = 8

O = 2                                                  O = 6 + 4 = 10

 

C3H8   +         5 O2     à        3 CO2  +         4 H2O

 

C = 3                                                  C = 3 

H = 8                                                  H = 8

O = 10                                                O =10

Balance the following equations:

MgSO4  +       Na3PO4  à    Mg3(PO4)2 +   Na2SO4

 

 

 

Al        +         HCl     à        AlCl3   +         H2

 

 

 

Na       +         H2O    à        H2                    +         NaOH

 

 

 

(NH4)2Cr2O7  à  Cr2O3        +         N2 +   H2O

 

 

 

NH3  +            O2  à             NO      +         H2O

 

 

 

H2SO4  +        Al(OH)3 à    Al2(SO4)3 +   H2O

 

 

 

Stoichiometry

 

A balanced equation shows the mole relationships between the substances involved in a particular reaction through the use of coefficients.

Ex: BaCl2  +  2 NaOH  à  2 NaCl  +  Ba(OH)2

 

1 mole of barium chloride reacts with 2 moles of sodium hydroxide to give 2 moles of sodium chloride and 1 mole of barium hydroxide.

 

What is the molar ratio for sodium hydroxide to barium hydroxide?

 

What is the molar ratio for sodium hydroxide to sodium chloride?

 

What is the molar ratio for barium chloride to barium hydroxide?

 

What is the molar ratio for barium chloride to sodium chloride?

 

Using stoichiometry: A balanced equation can be used to predict the mass of products gained from a reaction or the mass of reactants needed for a reaction. (Mass-mass calculations)

 

Example 1:     KClO3 à        KCl     +         O2

If 500.0 g of potassium chlorate decompose, what mass of potassium chloride will be produced?

 

1.  Balance the equation.

2.  Start with the given and convert to moles:

500.0 g KClO3 x        1 mol KClO3  = 4.080 mol KClO3                                                  

            122.6 g KClO3                      

 

3.  Use molar ratio to convert to moles of new substance:

4.080 moles KClO3  x           2 mol KCl     = 4.080 mol KCl

                                                2 mol KClO3                        

 

4.  Convert to units required for new substance:

4.080 mol KCl           x          74.6 g  KCl     =         304.2 g KCl

                                                1 mol KCl

 

The thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid fuel rocket motors.  The reaction is:                  Fe2O3(s)  +  2 Al(s)       à        2 Fe(l)  +  Al2O3(s)

What mass of iron (III) oxide and aluminum must be used to produce 15.0 g iron?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Alka-Seltzer uses the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz:

            3 NaHCO3 (aq)            +  C6H8O7 (aq)   à      3 CO2 (g)  +  3 H2O (l)  +  Na3C6H5O 7 (aq)

(a)  What mass of citric acid should be used for every 1.0 x 102 mg NaHCO3?

 

 

 

 

 

 

 

 

(b) What mass of CO2 could be produced from such a mixture?

 

 

 

 

 

 

 

 

 

 

Limiting Reactants and Percent Yield

 

Consider the following reaction:

4 Al     +         3 O2     à        2 Al2O3

 

If 4.00 grams of aluminum are mixed with 3.00 grams of oxygen gas, how many grams of aluminum oxide will form?

 

Remember:  A balanced reaction shows mole relationships between the substances involved.

 

·        According to the reaction, the following ratio is needed:                             

4 mol Al               =         1.33 mol Al

      3 mol O2                                    1 mol O2

·        How many moles of each reactant is available?

Aluminum:

 

 

 

Oxygen:

 

 

Compare the available ratio to the needed ratio.

Avail.:                                                                        Needed:

 

 

 

Which reactant will be present in excess?

 

Which reactant will run out first?

 

The reactant which runs out first is called the limiting reactant.  Once the limiting reactant is gone, the reaction must stop. 

 

·        Determine the number of grams of product which would form from the limiting reactant.  This is referred to as the theoretical yield.

 

Theoretical yield =

 

 

 

 

 

How many grams of silver chloride can be produced from 16.99 g of silver nitrate and 9.92 g of calcium chloride?

 

2 AgNO3        +         CaCl2 à        Ca(NO3)2        +         2 AgCl

 

Determine limiting reactant:

 

 

 

 

 

 

 

 

 

 

 

Calculate theoretical yield in grams:

 

 

 

 

Often side reactions or less than ideal reaction/ collection conditions lead to a recovered mass of product which is less than the theoretical yield.  The amount obtained is referred to as the actual yield and can be used to calculate the percent yield.

 

Percent yield  =         actual yield                 x 100

                                    theoretical yield

                                   

 

If 12.14 g of AgCl was actually obtained under the conditions shown above, calculate the percent yield.

A solution containing 3.50 g of sodium phosphate is mixed with a solution containing 6.40 g of barium nitrate.  After vacuum filtration, 4.07 g of solid barium phosphate is isolated.  Calculate the percent yield.

 

2 Na3PO4        + 3 Ba(NO3)2 à Ba3(PO4)2 + 6 NaNO3

 

 

 

 

 

 

 

1.00 g of magnesium is allowed to react with 20.00 g of iodine.  Calculate the theoretical mass of magnesium iodide formed.

Mg  +  I2  à  MgI2

 

 

 

 

 

 

 

 

 

 

2.0 g of solid silver is allowed to react with 2.0 g of solid sulfur.  Calculate the theoretical mass of silver sulfide formed.

16 Ag  + S8  à  8 Ag2S