1. C. Johannesson Ch. 4 - Molar Relationships III. Percent Composition (p. 143-144)

2. C. Johannesson A. Definition n the percent by mass of each element in a compound n to distinguish between similar compounds n to determine elemental mass in a sample

3. C. Johannesson  100 = B. Calculation of % Comp n Find the % composition of FeO. %Fe = 55.85 g 71.85 g  100 = %O = 16.00 g 71.85 g 77.73% Fe 22.27% O

4. C. Johannesson  100 = B. Calculation of % Comp n Find the % composition of Fe 2 O 3. %Fe = 111.70 g 159.70 g  100= %O = 48.00 g 159.70 g 69.944% Fe 30.06% O

5. C. Johannesson C. Compound Identification n A compound contains 28 g Fe and 8.0 g O. Is it FeO or Fe 2 O 3 ? %Fe = 28 g 36 g  100 =  The compound is FeO. 78% Fe

6. C. Johannesson D. Elemental Mass in a Sample n How many grams of iron are in a 38.0-gram sample of iron(III) oxide? 38.0 g Fe 2 O 3 69.944 g Fe 100 g Fe 2 O 3 = 26.6 g Fe Fe 2 O 3 is 69.944% Fe - or - (38.0 g Fe 2 O 3 )(0.69944) = 26.6 g Fe

7. C. Johannesson CHALLENGE PROBLEM n A mining company needs to choose a mine site. Which would provide a better source of iron? Ore containing 32% FeO or ore containing 48% Fe 2 O 3 ? FeO is 77.73% Fe Fe 2 O 3 is 69.944% Fe SOLUTION

8. C. Johannesson CHALLENGE PROBLEM (con’t) Fe 2 O 3 (100 g ore)(0.48) = 48 g Fe 2 O 3 (48 g Fe 2 O 3 )(0.69944) = 34 g Fe FeO (100 g ore)(0.32) = 32 g FeO (32 g FeO)(0.7773) = 25 g Fe Fe 2 O 3 is 69.944% Fe FeO is 77.73% Fe BETTER SOURCE!

9. C. Johannesson  100 = E. % Water in a Hydrate n CaCl 2 ·2H 2 O  calcium chloride dihydrate %H 2 O = 36.04 g 147.02 g 24.51% H 2 O