Chapter Three:

Stoichiometry: Calculations with Chemical Formulas and Equations

Overview

• Chemical Equations
• Patterns/Reactions
• Atomic/Molecular Weights
• Moles/Molar Mass
• Empirical/Molecular Formulas
• Quantitative Relationships
• Limiting Reactants/Theoretical Yields

Chemical Equations

• chemical 'sentences'

reactants and products described by formulas or symbols combined with "punctuation"

2 H_2(g) + O_2(g) è 2 H₂O_(l)

• "atoms can be neither created nor destroyed"

all equations must be 'balanced' with the same number of atoms on both sides of the reaction arrow

H2O + O2     goes to          H2O2

unbalanced

2H2O + O2      goes to          2H2O2

balanced

Examples

• CH3OH(l) + O2(g)      goes to         CO2(g) + H2O(l)
  
• Na(s) + H2O(l)            goes to              NaOH(aq) + H2(g)
  
• HBr(aq)+ Ba(OH)2(aq)      goes to       H2O(l) + BaBr2(aq)

Patterns of Chemical Reactivity

• Because elements are grouped by chemical
  properties, their reactions can also be grouped:

• alkali metals and water

specific

2K(s) + 2H2O(l)      goes to         2KOH(aq) + H2(g)

general

2M(s) + 2H2O(l)      goes to        2MOH(aq) + H2(g)

• Combustion in air
  

specific

C3H8(g) + O2(g)       goes to         CO2(g) + H2O(l)

general

CxHy + O2(g)         goes to           CO2(g) + H2O(l)

• Combination Reactions

specific

2Mg(s) + O2(g)                    goes to                     2MgO(s)

general

X + Y            goes to               XY

• Decomposition Reactions

specific

CaCO3(s)        goes to         CaO(s) + CO2(g)

general

XY        goes to              X + Y

Name the Reaction

PbCO3(s)        goes to           PbO(s) + CO2(g)             decomposition

C(s) + O2(g)       goes to             CO2(g)                                    combination

2NaN3(s)         goes to            2Na(s) + 3N2(g)                decomposition

2C2H6(g) + 7O2(g)      goes to    4CO2(g) + 6H2O(l)         combustion

Atomic and Molecular Masses

• Amu scale
• defined by assigning the mass of ¹²C as 12 amu exactly
• 1 amu = 1.66054 x 10^-24 g
• 1 g = 6.02214 x 10²³ amu
• Average Atomic Masses

¹²C 98.892% abundant ¹³C 1.1108% abundant

(0.98892)(12 amu) + (0.01108)(13.00335 amu) = 12.011 amu

• Formula and Molecular Masses

sum of all atomic masses in the formula of an ionic or molecular compound

vitamin C C6H8O6

6 x 12.0 = 72.0 amu

8 x 1.0 =    8.0 amu

6 x 16.0 = 96.0 amu

               176.0 amu

[formula mass of vitamin C (often called molecular mass)]

Percentage Composition

• Calculate the percent mass that each type of atom contributes to a molecule
• % X = (no. X atoms)(X amu) x 100
              formula mass cmpd

C6H8O6

% C = (6)(12.01amu) x 100 = 40.94% C
             176.0 amu

% H = (8)(1.01amu) x 100 = 4.59% H
            176.0 amu

% O = (6)(16.00 amu) x 100 = 54.55% O
            176.0 amu

The Mole

We can measure masses in amu but how do we relate that to mass in grams? We define a quantity of atoms - a mole - which has the same mass in grams as the mass of the element in amu.

So how many atoms does it take to make, say, 1.00 g of H?

1.0 g H x        1 atom H            @ 6.0 x 10²³ atoms of H
                 1.7 x 10^-24 g H
12.0 g C x        1 atom C           @ 6.0 x 10²³ atoms of C
                   2.0 x 10^-23 g C

Avogadro's Number

• 6.02214 x 10²³ units/mole
  
• No. of atoms per mole of an element
• No. of molecules per mole of molecular cmpd.
• No. of formula units per mole of ionic cmpd.
• No. of cows per mole of cows

Memorize this number & what it means!

• 1 C atom = 12 amu 1 mole C atoms = 12 g
• 1 Mg atom = 24 amu 1 mole Mg atoms = 24 g
• 1 CO molecule = 28 amu 1 mole CO molecules = 28
• 1 NaCl fm. unit = 58 amu 1 mole NaCl fm.units = 58g
  
  Molar Mass
• From this information we can define something called the molar mass (MM) of an atom (or molecule or formula):
• from the equality: 1 mole C = 12.0 g C
• we define the molar mass of a substance
  
  12.0 g C = MM            or             Molar Mass
  1 mole C                                     (Atomic Mass)
                                                    (Molecular Mass)
                                                    (Formula Mass)

• Problems

Practice Ex. 3.9:

Given: MM = 84.02 g/mol NaHCO3 508 g NaHCO₃?
How many mole in 508 g of NaHCO3?

508 g NaHCO3 x         1 mole              =        6.05 mole NaHCO3
                              84.02 g NaHCO3

How many formula units of NaHCO3?

Given: 6.02 x 10²³   form. units/mole NaHCO₃

6.05   mole NaHCO3    x     6.02 x 1023 fm. units          =      3.64 x 10²⁴ fm. Units NaHCO₃
                                                1  mole

• Molar Mass converts between moles and grams of a substance
• Avogadro's number converts between moles of a substance and atoms (or molecules or formula units) of that substance
  
• These are very important conversion factors, know & understand them!

Problems

• How many moles of vitamin C are contained in 5.00 g of vitamin C? C6H8O6 176.0 g/mol
  
• 17.5 mg of cocaine (C17H21NO4) per kg of body weight is a lethal dose. How many moles is that? How many molecules?
  
• In 25 g of C12H30O2 THC (tetrahydrocannibinol) how many moles are there? How many molecules are there? How many C atoms are there?
• How many moles of O are contained in 1.50 moles of C6H5NO3?
  
• How many grams of nitrogen are contained in 70.0 g of C6H5NO3? How many atoms?
  

Determination Empirical Formulas

• simplest ratio of atoms
• change g of each element to moles or
• assume 100 grams of substance & change the % of each element to moles
• change the mole ratio of atoms to the simplest ratio by dividing by the smallest number of moles

• Practice Ex. 3.12:

5.325 g methyl benzoate contains 3.758 g C, 0.316 g H, 1.251 g O. Determine empirical formula.

3.758 g C x 1 mole = 0.313 mol C
                   12.01 g

0.316 g H x 1 mole = 0.313 mol H
                      1.01 g

1.251 g O x 1 mole = 0.0782 mol O
                   16.00 g

C0.313H0.313O0.0782                        C4H4O

Determination of Molecular Formulas

• actual ratio of atoms
• determine the empirical formula
• divide the actual molar mass by the empirical formula mass to get 'n'
• multiply the mole ratio in the empirical formula by 'n'

• Practice Ex. 3.13:

Ethylene glycol is composed of 38.7% C, 9.7% H & 51.6% O by mass. Its true molar mass is 62.1 g/mol. What are the empirical and molecular formulas?

38.7 g C x 1 mole = 3.23 mole C
                  12.0 g

9.7 g H x 1 mole = 9.60 mole H
                 1.01 g

51.6 g O x 1 mole = 3.22 mole O
                    16.0 g

C3.23H9.60O3.22                                   CH3O                      C2H6O2
                                                                             empirical fm. (n=2)                                                     molecular fm.

Formulas from Combustion Data

• Formulas determined from products of combustion products

Menthol is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g H2O. What is the empirical formula?

CxHyOz + O2        goes to           CO2 +    H2O
0.1005 g                                       0.2829 g      0.1159 g

Calculate moles CO2 & C; moles H2O & H

0.2829g CO2 x 1 mol     x     1 mol C     =      0.00643 mol C
                        44.0 g          1 mol CO2

0.1159g H2O x     1 mol     x      2 mol H      =    0.0129 mol H
                             18.0 g            1 mol H2O

total mass of C + H = 0.0902 g

mass of O = 0.1005 g - 0.0902 g = 0.0103 g O x 1 mol =        6.44 x 10^-4 mol O
                                                                       16.0 g

C0.00643H0.0129O0.000644                                       C10H20O
                                                                                         (empirical formula mass 156 g/mol)

If the MM is 156 g/mol, what is the molecular formula?

n=1 therefore molecular formula is C10H20O

Quantitative Stoichiometry

• Determination of quantities from balanced chemical reaction equations
• mole ratios from balanced chemical equation convert between species
• if quantities are given for more than one reactant, the limiting reactant must be determined

• Given the following balanced equation:
  
  1Mg(OH)2 + 2HCl       goes to            1MgCl2 + 2H2O
  
• Calculate the number of moles of HCl required to react completely with 0.42 mol of Mg(OH)2
         0.42 mol Mg(OH)2 x      2 mol HCl         =       0.84 mol HCl
                                         1 mol Mg(OH)2
  
  The mole ratio comes from the balanced chemical equation
• How many grams of MgCl2 can be produced?
  
  0.42 mol Mg(OH)2    x    1 mol MgCl2      x        95.3 g MgCl2         =          40.0 g MgCl2
                                   1 mol Mg(OH)2                  1 mol                             Theoretical Yield
  
  

Conversion sequence:

g reactant molar mass moles reactant mole ratio moles product molar mass g prod.

Practice Ex. 3.14:

How many grams of O2 can be prepared from 4.50 g of  KClO3?

2KClO3      goes to            2KCl + 3O2       

                4.50 g KClO3      x         1 mol x         3 mol O2        x          32.0 g O2         =         1.76 g O2
                                                   122.6 g          2 mol KClO3           1 mol

• Limiting Reactantgiven a non-stoichiometric amount of both reactants, you will have to determine which is the limiting reagent or reactant
  

example: you have 10 bicycle frames and 16 bicycle wheels and you need to put them together to produce as many bicycles as possible, how many bicycles can be produced, what is the limiting "reagent", and how much excess "reagent" do you have left over?

Balanced 'Equation'

1 (mole) frame + 2 (moles) wheels         goes to            1 (mole) bicycles
   [10 (moles) frames]             [16 (moles) wheels]                                   [8(moles) bicycles]

Limiting Reactant -- will produce the least amount of product

10 mol frames       x         1 mol bicycles         =         10 bicycles
                                      1 mol frames

16 mol wheels        x       1 mol bicycles          =          8 bicycles          Limiting
                                       2 mol wheels

Practice Ex. 3.16:

A mixture of 1.5 mol of Al and 3.0 mol of Cl2 react. What is limiting & how many moles of AlCl3 are formed?
2Al(s) + 3Cl2(g)       goes to          2AlCl3(s)
1.5 mol        3.0 mol

1.5 mol Al      x       2 mol AlCl3       =          1.5 mol AlCl3            Limiting
                              2 mol Al

3.0  mol Cl2      x      2 mol AlCl3        =         2.0 mol AlCl3
                              3 mol Cl2