 The first step to doing conversions with proportions is to understand what an equality is.  In most cases, an equality compares two things where one of the numbers is 1.  Examples: ◦ 1 foot = 12 inches ◦ 1 meter = 100 cm ◦ 1 year = 365 days  Note that these equalities are reversible ◦ 1 foot = 12 inches = 1 foot

 Given value (unit) Unknown value (unit) Standard value (unit) Standard value (unit)  How many inches are in 4.9 feet? ◦ The given value is 4.9 and the unit is feet ◦ The unknown value is unknown and the units are inches. ◦ The equality we use for the standard value is  1 ft = 12 inches =

 Given value (unit) Unknown value (unit) Standard value (unit) Standard value (unit)  4.9 feet ? Inches 1 foot 12 inches  4.9 feet ? Inches 1 foot 12 inches To solve: Cross multiply and divide. = = =

 4.9 feet ? Inches 1 foot 12 inches  4.9 = ? 1 12  12 X 4.9 = XX 12 =

 1 MOLE = MOLAR MASS ◦ Molar mass = the atomic mass of an element from the Periodic Table or the mass of all the elements in a compound. ◦ 1 mole Ne = 20.18g ◦ 1 mole NaCl = 58.44g  1 MOLE = 6.02 X 10 23

8 moles CO 2 ______g CO 2 ______p CO 2 ______mole H g H 2 ______ p H 2 ___mole C 3 H 8 _____ g C 3 H x10 23 pC 3 H 8 3 mole NaCl_____ g NaCl_____ p NaCl _____mole H 2 SO g H 2 SO 4 _____ p H 2 SO 4

8 moles CO g CO x p CO 2 1 mole H g H x p H mole C 3 H g C 3 H x10 23 pC 3 H 8 3 mole NaCl g NaCl1.8 x p NaCl 4 mole H 2 SO g H 2 SO x p H 2 SO 4

 How many grams of CH 4 are in 8.16 moles? 8.16 mole CH 4 ? grams CH 4 1 mole CH g CH 4  8.16mole x 16.05g = 131g CH 4  How many particles are in 8.16 moles CH4? 8.16 mole CH 4 ? particles CH 4 1 mole CH x p CH 4  8.16mole x 6.02 x p = 4.91 x p = =

 2 1/4 cups all-purpose flour  1 teaspoon baking soda  1 teaspoon salt  1 cup (2 sticks) butter, softened  3/4 cup granulated sugar  3/4 cup packed brown sugar  1 teaspoon vanilla extract  2 large eggs  2 cups (12-oz. pkg.) NESTLÉ® TOLL HOUSE® Semi-Sweet Chocolate MorselsNESTLÉ® TOLL HOUSE® Semi-Sweet Chocolate Morsels  1 cup chopped nuts  Yields 60  What amounts would I use if I wanted to make 120?

 In a recipe, the amounts called for can be doubled or halved. As long as all the amounts are treated the same way, the recipe should turn out.  Balanced equations can be treated the same way.

 When an equation is balanced, the coefficient numbers represent numbers of atoms, molecules or formula units.  As long as the coefficient numbers remain proportional, the equation remains balanced.

 2H 2 + O 2  2H 2 O  Two molecules H 2 + one molecule O 2 forms two molecules H 2 O.  What if I changed the 2 in front of H 2 to a 4? What would happen to the other coefficient numbers?  What if I changed the 2 in front of the H 2 O to 12?

 What if I changed the coefficient in front of O 2 to 6.02x ?  As long as the proportions within the equation remain the same, the equation remains balanced.

 Therefore, balanced equations can be used to determine how much of one substance would be needed or produced from another substance in the equation.  This is called stoichiometry.

# of atoms 2 atoms N6 atoms H2 atoms N 6 atoms H # of moles 1 moles N 2 3 mole H 2 2 moles NH 3 # of particles 6.02x x x10 24 Mass g34.08g Total mass reactants 34.08g Total mass products 34.08g

 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O  If 4 moles of C 2 H 6 are reacted, how many moles of O 2 are needed? ◦ Since C 2 H 6 doubled, O 2 will also double.  If 8.94 mole H 2 O are produced, how many mole, O 2 reacted? ◦ 8.94 mole H 2 O = ? Mole O 2 6 mole H 2 O 7 mole O 2 = 10.4 mole O 2

 Using Proportions Given amount (g A) = Unknown amount (g B) Total molar mass A Total molar mass B (molar mass A x # moles) (molar mass B x # moles)

 EXAMPLE: How many grams of O 2 are needed to produce 135 g H 2 O 135g H 2 O = ? g O 2 = 599 g O g H 2 O 160g O 2 Total molar mass (2 x 18.02g) (5 x 32.00g)