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Ch. 4 - Molar Relationships

Ch. 4 - Molar Relationships. III. Percent Composition (p. 143-144). A. Definition. the percent by mass of each element in a compound. to distinguish between similar compounds to determine elemental mass in a sample. WHY?. 55.85 g 71.85 g. 16.00 g 71.85 g. B. Calculation of % Comp.

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Ch. 4 - Molar Relationships

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

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

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

  4. 111.70 g 159.70 g 48.00 g 159.70 g B. Calculation of % Comp • Find the % composition of Fe2O3. 100= 69.944% Fe %Fe =  100 = 30.06% O %O = C. Johannesson

  5. 28 g 36 g C. Compound Identification • A compound contains 28 g Fe and 8.0 g O. Is it FeO or Fe2O3?  100 = 78% Fe %Fe = The compound is FeO. C. Johannesson

  6. D. Elemental Mass in a Sample • How many grams of iron are in a 38.0-gram sample of iron(III) oxide? Fe2O3 is 69.944% Fe 38.0 g Fe2O3 69.944 g Fe 100 g Fe2O3 = 26.6 g Fe - or - (38.0 g Fe2O3)(0.69944) = 26.6 g Fe C. Johannesson

  7. CHALLENGE PROBLEM • 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% Fe2O3? FeO is 77.73% Fe Fe2O3 is 69.944% Fe SOLUTION C. Johannesson

  8. 36.04 g 147.02 g E. % Water in a Hydrate • CaCl2·2H2O • calcium chloride dihydrate  100 = 24.51% H2O %H2O = C. Johannesson

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