| Chapter 3 |
| Definition of a Mole- The # equal to the # of carbon atoms in exactly 12 grams of pure Carbon 12. |
| Avogadro's number- 6.022x10^23 |
| Molar Mass- the mass in grams of one mole of the substance. |
| Writing and Balancing the Equation for a Chemical Reaction: |
| 1) Determine what reaction is occuring. What are the reactants, the products, and the physical states involved? 2) Write the unbalanced equation that summarizes the reaction described in step one. 3) Balence the equation by inspection starting with the most complicated molecule. |
| Steps for Solving Stoichiometry Problem Involving Masses of Reactants and Products: |
| 1) Write and balence the equation for the reaction 2) Convert the known masses of substances to moles. 3) Determine which reactant is limiting 4) Using the amount of limiting reactant and the appropriate mole ratios, compute the # of moles of the desired product. 5) Convert the moles to grams using molar mass. |
| Examples: |
| Example 1- Compute both the # of moles of atoms and the # of atoms in a 10 gram sample of Aluminum. 10.0g Al x ____1 mol Al____ = 0.371 mol Al 26.98 g Al 0.371 mol Al x __6.022x10^23atoms___ = 2.23 x10^23 atoms 1 mol Al .371 mol Al 2.23 x 10^23 atoms |
| Example 2: Determine the empirical and molecular formulas for a compound that gives the following mass percents: 71.65% Cl 24.27% C 4.07% H The molar mass is known to be 98.96g/mol. 71.65g Cl x __ 1mol Cl__ = 2.021mol Cl 35.45g Cl 24.27g C x __1mol C__ = 2.021 mol C 12.01g C 4.07g Hx __1mol H__ = 4.04 mol H 1.008g H Empirical Formula Equals ClCH2 molar mass 98.96g/mol ------------------------- = ------------------ = 2 empirical formula mass 49.48g/mol Molcular Formula Equals (ClCH2)2= Cl2C2H4 |