1. convert the following: 1. Find the % composition of Al (NO 3 ) 3 2. If 8.52 grams of aluminum nitrate are dissolved in exactly 250 cm 3, what is the molarity? 3. ACT-prep question: Which chart would you use to answer the following question? Which chart would you use to answer the following question? Do the data support the hypothesis that frogs in warmer climates will be noisier than frogs in colder climates?: Quiz: March 15, 2005 1. Al = 27 amu, N = 14 amu 0 = 16 amu (27.0) + 3(14.0) + 9(16) =213 amu or 213 g/mole 27/213 x 100% = 12.7% Al 42/213 x 100% = 19.7 % N 144/213 x100% = 67.6 % O 2. 8.52 gAl(NO 3 ) 3 mole_Al(NO 3 ) 3 _1000 cm 3 250 cm 3 213 g Al(NO 3 ) 3 1 dm 3 0.160 M Al(NO 3 ) 3 Time of Day (P.M.) Total number of predators A. Population size Average call rate Average Call volume C. Time of Day (P.M.) Average call rate Average Call volume B. Water Temperature Average call rate Average Call volume D. Test This Thursday

2. HW 18: #56  64 p210; Honor’s #103 a,b and 104 a, b p 218 Empirical Formula, first introduced in Chapter 7, refers to the experimental data you obtain when trying to find out how much of each element is in a sample of a compound. Basically, you do the opposite of the procedure to find % composition. Use grams to find a ratio of part:whole, convert to moles, find a mole ratio. Write an empirical formula

3. .57 g of Magnesium burns in air. 0.96 g of magnesium oxide is measured after combustion is complete. What is the empirical formula of magnesium oxide?.57 g of Magnesium burns in air. 0.96 g of magnesium oxide is measured after combustion is complete. What is the empirical formula of magnesium oxide? Underline important information List what you have, and what you want. 0.57 g Mg 0.96 g Mg ? O ? ____g O find mole ratio Mg: O 0.57 g Mg 0.96 g Mg ? O ? 0.39 g O Subtract 0.96 – 0.57 to find g O Find the # moles of each element 0.57 g Mg 24.3 g/mole Mg 0.39 g O 16.0 g/moleO 0.0235 mole Mg 0.0244moles O 0.0235 = 1 mole Mg = 1 mole O MgO

4. A compound has a percentage composition of 40% C, 6.71 % H, and 53.3% O 40 g C 12 g/mole C 6.71 g H 1.01 g/mole C 53.3 g O 16.0 g/mole O  3.33 mole C  6.64 mole H  3.33 mole O ____________ 3.33 ____________ 3.33 ____________ 3.33  1 mole C  2 mole H  1 mole O CH 2 O Step 1: assume % = grams Step 2: change grams  moles Step 3: Find whole number ratio Step 4: Use ratio to write formula

5. Analysis of a 10.150 g sample of a compound known to contain only phosphorous and oxygen yields 5.717 g of oxygen. What is the empirical formula of this compound? 10.150 – 5.717 = 4.433 g P 4.433 g P 30.97 g/mole P 5.717 g O 16.00 g/mole O  0.1431 mole P  0.3573 mole O ____________ 0.1431 ____________ 0.1431  1 mole P  2.497 mole O = 2 ½ mole O = 2 ½ mole O P 2 O 2(2.5) Step 1:get grams of each thing. Step 2: change grams  moles Step 3: Find whole number ratio Step 4: Use ratio to write formula P2O5P2O5

6. A compound has a percentage composition of 40% C, 6.71 % H, and 53.3% O, for the following molecular masses, what would the actual formula be? 40 g C 12 g/mole C 6.71 g H 1.01 g/mole C 53.3 g O 16.0 g/mole O  3.33 mole C  6.64 mole H  3.33 mole O ____________ 3.33 ____________ 3.33 ____________ 3.33  1 mole C  2 mole H  1 mole O CH 2 O a.30 g/mole b.60 g/mole c.120 g/mole d.240 g/mole  C = 12  2H = 2  O = 16_________  CH 2 O = 30 g/mole a. 30/ 30 = 1 CH 2 O b. 60/30 = 2 C 2 H 4 O 2 c. 120/30 =4 C 4 H 8 O 4 d. 240/30 = 8 C 8 H 16 O 8

7. % water in a hydrate We have a 10.407 g sample of hydrated barium iodide. The sample is heated to drive off the water. The dry sample has a mass of 9.520 g. What is the formula of the hydrate? We have a 10.407 g sample of hydrated barium iodide. The sample is heated to drive off the water. The dry sample has a mass of 9.520 g. What is the formula of the hydrate? 10.407 g hydrated BaI 2 9.520 g BaI 2 g water driven off 0.887 Ba = 137 2 I = 2(127) =254 391g/mole 2H = 2(1.01) O = 16.0 18.0 g/mole water 9.520 g BaI 2 = 0.0243 mole BaI 2 391g/mole 0.887 g water = 0.493 mole H 2 O 18.0 g/mole water 0.0243 1 BaI 2 : 2 H 2 O so… BaI 2. 2 H 2 O