Chapter 8 - Quantitative Calculations and the Mole
I. Relative Mass of elements
-Don’t want to have to count molecules - too small
- For example - buy oranges by the pound
- Same for compounds - weigh out instead of count
1 atom He = 4.00 amu
1 atom C = 12.0 amu
12.0 g of Carbon = same number of atoms 4.00 g He
45.0 g of Carbon = ? g He
II. THE MOLE
mole (SI symbol mol) - the number of atoms represented by the atomic mass of an element expressed in grams.
1.00 mol = 6.02 x 1023 objects
Avogadro’s Number
MOLAR MASS - the atomic mass expressed in grams. Contains 6.02 x 1023 atoms or molecules.
III. Compounds
Molecular weight (formula weight) - molecular weight of a compound is determined from the number of atoms and the atomic mass of each element indicated by a chemical formula.
Example:
1. CO2
Atom |
# of Atoms |
Atomic Mass |
Total Mass |
C |
1 |
X 12.0 amu |
= 12.0 amu |
O |
2 |
X 16.0 amu |
= 32.0 amu |
| 44.0 amu |
Each molecule of CO2 weighs 44.0 amu
2. Fe(NO3)2
Molar Mass of a compound - the mass of one mole (6.02 x 1023 molecules) - formula weight expressed in grams.
IV. Diatomic elements:
Some atoms existed as diatomic molecules in nature
| element | Molar Mass |
| O2 | 32.0 g/mole |
| N2 | 28.0 g/mole |
| H2 | 2.02 g/mole |
| Cl2 (and all halogens) | 71 g/mol |
V. Composition of Compounds
Sulfuric Acid H2SO4
1 mole of sulfuric acid contains
2 mol of H atoms
1 mol of S atoms and
4 mol of O atomsTo calculate components must use conversion factor
1 mole H2SO4 = 2 mole H atoms
1 mole H2SO4 = 1 mole S atoms
1 mole H2SO4 = 4 mole H atoms
VI. Percent composition
expresses the mass of each element per 100 mass units of compound.
VII. Empirical and Molecular Formulas
Many molecules have same %mass of components, but are very different.
ex: The two compounds below have a mass % of C = 92.3%
| acetylene | benzene |
| C2H2 | C6H6 |
| gas used in welding | solvent used in industry |
Also have the same Empirical formula: CH
Empirical formula - the simplest whole number ratio of atoms in a molecule.
Determining empirical formula from % mass.
1. Convert % composition to an actual mass.
2. Convert mass to moles of each element
3. Find the whole -number ratio of the moles diff. elements.
Examples:
What is the empirical formula of laughing gas, which is 63.6% Nitrogen
and 36.4% oxygen?
Molecular formula - the actual whole number ratio of atoms in a molecule.
Determine molecular formula using the molar mass - multiply subscripts by a,
where
VIII. Stoichiometry
Stoichiometry - the quantitative relationships between reactants and/or products.
Production of ammonia
| N2(g) | + | 3H2(g) | ® | 2NH3(g) |
| 1 molecule | + | 3 molecule | ® | 2 molecule |
| 12 molecules | + | 36 molecules | ® | 24 molecules |
| 1 dozen molecules | + | 3 dozen molecules | ® | 2 dozen molecules |
| 6.02 x 1023 molec. | + | 18.1 x 1023 molec. | ® | 12.0 x 1023 molec. |
| 1 mol | + | 3 mol | ® | 2 mol |
| 28 g | + | 6 g | ® | 34 g |
Mole relationships or mole mole ratios.
1 mol of N2 produces 2 mol of NH3
1 mol of N2 reacts with 3 mol of H2
3 mol of H2 produces 2 mol of NH3
How to solve stoichiometry problems
1. Write down what is given and what is requested
2. If mass is given covert to moles ( Molar Mass)
3. If # molecules is given convert to moles (Avogadro’s number)
4. Using the mole ratio convert what is given to what is requested (in moles)
5. If mass is requested convert moles to mass (Molar Mass)
6. If molecules is requested convert moles to molecules (Avogadro’s Number)
IX. Limiting Reactant
Limiting reactant - if specific amounts of each reactant are mixed, the reactant that produces the least amount of product.
ex: 2H2 + O2 ® 2H2O
1. Determine the number of moles of reactants
2. Determine the number of moles of product produced by each reactant (using mole mole ratios)
3. The reactant producing smallest # moles product = limiting reactant
X. Percent Yield
Actual yield - the measured amount of product obtained in any reaction
Theoretical yield - the calculated amount of product that would be obtained if all of the reactant were converted to a given product.
Percent Yield - the actual yield in grams divided by the theoretical yield in grams times 100%.
XI. Heat energy in chemical Reactions
Chemical Thermodynamics -the study of heat and its relationship to chemical changes.
All elements and compounds have Potential energy (PE)
| Heat is evolved if PE of reactants > PE products | EXOTHERMIC |
| Heat is absorbed if PE of reactants < PE products (heat must be applied for reaction to occur) | ENDOTHERMIC |
Thermochemical equation - a balanced equation that includes heat energy.
1. Heat energy listed at end of equation
2H2(g) + O2(g) ® 2H2O(l) DH = -572 kJ
DH is the change in enthalpy of reaction - Negative Þ Exothermic
Positive Þ Endothermic
Or may be Enthalpy per mole: DH = -286 kJ/mole H2O
2. Heat energy listed as product or reactant
2H2(g) + O2(g) ® 2H2O(l) + 572 kJ Exothermic
N2(g) + O2(g) + 181 kJ ® 2NO(g) Endothermic
