1. convert the following: 1. (4.0 x 10 6 ) X 0.0040 = _____ 2. Find the gfw of Al (NO 3 ) 3 3. How many moles are in 0.0426 grams of aluminum nitrate? 4. If 0.852 grams of aluminum nitrate are dissolved in exactly 500 cm 3, what is the molarity? 5. 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 small populations call more frequently than frogs in large populations: Quiz: March 14, 2005 1. 1.6 x 10 4 2. Al = 27 amu, N = 14 amu 0 = 16 amu (27) + 3(14) + 9(16) =213 amu or 213 g/mole 3. 0.0426 g mole_Al(NO 3 ) 3 213 g 2.00 x 10 -4 mole aluminum nitrate 4. 0.852 gAl(NO 3 ) 3 mole_Al(NO 3 ) 3 _1000 cm 3 500 cm 3 213 g Al(NO 3 ) 3 1 dm 3 0.00800 M Al(NO 3 ) 3 note,.994 ~ 1 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.

2. Moles mass in grams [concentration] ? M x GFW  dm 3 X dm 3  GFW

3. Find the molarity of the following solutions: ConcentrationMolescubic centimetersgrams name of compound Formula 1.042 M.21 mole 5,0000dm 3 12g sodium chloride NaCl 2 0.1000 1.0000 =.1000M 0.1000mol e 1,000. dm 3 (.10mole)(56g) (mole) = 5.6 g calcium oxide 40 + 16 = 56 g/mole CaO 30.1M 0.1x 0.05 =.005mole 0.05000dm 3.005 mole ( 133 g) mole = 0.7 g copper (II) chloride 63 + 2(35) = 133g/mole CuCl 2 40.02M 0.2x 0.05 =.01mole0.05000dm 3.01mole ( 169 g) mole = 2 g silver nitrate 107 + 14 + 3(16) 169 g/mole AgNO 3 50.666M3.00mole 0.666 mole/dm 3 = 4.50 dm 3 = 4500 cm 3 3.00 mole ( 169 g) mole = 507 g silver nitrate 107 + 14 + 3(16) 169 g/mole AgNO 3 6 0.0033mole 0.333 dm 3 0.0099 M.0033mole 0.333.00dm 3.0033mole ( 56 g) mole = 0.18 g Sodium chloride NaCl 70.50 M.0010mole 0.50 mole 0.0010 mole/dm 3 = 500 dm 3 = 5 x 10 5 cm 3.0010mole ( 124 g) mole = 0.18 g copper (II) carbonate 64 + 12 + 3(16) = 124 g/mole CuCO 3

4. To Find % composition find gfw Divide mass of each part by the whole gfw. Ca 3 (PO 4 ) 2 3 Ca 3(40.1) 2 P 2(31.0) 8 O 8(16.0) = 120. = 62.0 = 128 3 Ca 120. 2 P 62.0 8 O 128 310 g/mole 310 g/mole 3 Ca 120.g/310.g (100%) = 38.7% 2 P 62.0g/ 310g (100%) = 20.0% 8 O 128g/310 g (100%) = 41.3 % The sum should be within 1% of 100%

5. 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

6. .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

7. HW 17: #50-55 page 208; #101p218 % = the proportion out of 100 pieces % = the proportion out of 100 pieces If there were 100, how much would be … If there were 100, how much would be … In a package of M&Ms 15 were brown, 6 were yellow, and 4 were blue. What % were brown? In a package of M&Ms 15 were brown, 6 were yellow, and 4 were blue. What % were brown? 15 + 6 + 4 = total number of M&Ms = 25 15 + 6 + 4 = total number of M&Ms = 25 15/25 = the fraction which are brown. 15/25 = the fraction which are brown. 60% are brown 60% are brown 15 x 100% 25

8. Find the % nitrogen in ammonia. NH 3 = ammonia NH 3 = ammonia http://www.ausetute.com.au/percentc.htmlhttp://www.ausetute.com.au/percentc.html Tutorial for extra help N = 14.0 3H = 3 x 1.0 17.0 amu 14.0 amu N 17.0 amu NH 3 100% = 82.4% N in NH 3

9. To find the % composition of H 2 O H = 2(1.01) H = 2(1.01) O = 16.0 O = 16.0 H 2 O = 18.0 g/mole H 2 O = 18.0 g/mole % H = 0.112 x 100% 11.2% H % O = 0.889 x 100% 88.9% O %H = 2.02 g H 2 18.0 g/mole H 2 O 18.0 g/mole H 2 O %O = 16.0 g O 18.0 g/mole H 2 O 18.0 g/mole H 2 O