1. MOLES REVIEW, MAY 28 Al 2 O 3  O 2 + Al O 2 = O 2 = Al = Al = 2 22 23 4 66 4 4 2 Al 2 O 3  3 O 2 + 4 Al a) What is the mole ratio of O 2 to Al? O 2 = Al 3 4 THE MOLE RATIOS OF SUBSTACES IN A REACTION ARE THE BALANCED REACTION COEFFICIENTS.

2. 2 Al 2 O 3  3 O 2 + 4 Al b) If 6 mol of O 2 is produced, how many moles of Al are produced? O 2 = = Al 3 4 6 X X = 8 MOL of Al 2 Al 2 O 3  3 O 2 + 4 Al c) If 6 mol of O 2 is produced, how many grams of Al are produced? Moles = mass GFM 8 = mass = 215.2g Al 26.9

3. 2 Al 2 O 3  3 O 2 + 4 Al d) What is the % by mass composition of Al 2 O 3 ? ELEMENT SUBSCRIPTS MULTIPLY ATOMIC MASS EQUALS MASS SUBTOTAL Al 2 X26.98g/ mol =53.96 g O3 X15.99g/ mol =47.97 g + 101.93 g/mol % COMP (MASS) = PART X 100 WHOLE First you need to find the molar mass (gfm) of the substance NEVER USE COEFFICENTS FOR MOLAR MASS CALCULATION S

4. % COMP (MASS) = PART X 100 WHOLE % COMP Al = 53.96 g X 100 = 52.93 % 101.93 g ELEMENT SUBSCRIPTS MULTIPLY ATOMIC MASS EQUALS MASS SUBTOTAL Al 2 X26.98g/ mol =53.96 g O3 X15.99g/ mol =47.97 g 101.93 g/mol

5. REVIEW CLASS MULTIPLE CHOICE QUESTIONS, FRIDAY JUNE 1 2007 #7) CALCULATE THE MOLECULAR FORMULA OF A COMPOUND THAT IS CH, GIVEN MOLECULAR MASS IS 78 G/MOL. ELEMENT SUBSCRIPTS MULTIPLY ATOMIC MASS EQUALS MASS SUBTOTAL C 1 X12.01g/ mol =12.01 g H1 X1.007g/ mol =1.007 g + 13.017 g/mol THIS IS THE MASS (GFM)OF THE EMPIRICAL FORMULA

6. REVIEW CLASS MULTIPLE CHOICE QUESTIONS, FRIDAY JUNE 1 2007 #7)CONTINUED 13.017 g/mol 78. g/mol MOLAR MASS EMPIRICAL MASS = 5.99 = 6 THIS MEANS THAT THE MOLECULAR(TRUE) FROMULA IS 6 TIMES AS LARGE AS THE EMPIRICAL(SIMPLEST) FORMULA. MULTIPLY THE SUBSCRIPTS OF THE EMPIRICAL FORMULA BY 6. CH X 6 = C 6 H 6 GIVEN IN PROBLEM =

7. REVIEW CLASS MULTIPLE CHOICE QUESTIONS, FRIDAY JUNE 1 2007 #10) CALCULATE THE FORMULA OF A COMPOUND THAT IS 85% SILVER and 15% FLUORINE BY MASS. ELEMENT MASS/ ATOMIC MASSRAW RATIO DIVIDE BY SMALLEST SUBSCRIPT RATIO Ag 85 g/ 107.86g/mol = 0.7880 / 0.7880 = 1.0 F 15 g/ 18.99 g/mol=0.7898 / 0.7880 = 1.0 Assume 100g of the sample, this will allow you to assume 85% is 85 grams. Total mass does NOT affect % composition. Ag 1 F 1 AgF

8. #23) CALCULATE THE LITERS OF AMMONIA (NH 3 ) GAS FORMED FROM 20 LITERS OF N 2 GAS REACTED in the reaction given below. 3 H 2 (g) + N 2 (g)  2 NH 3 (g) X = 40 Liters of ammonia gas 1 2NH 3 (g) N 2 (g) == X liters 20 GAS VOLUMES CAN EXIST IN A RATIO AS DEFINED BY THE COEFFICIENTS OF THE BALANCED EQUATION, RATIO AS YOU WOULD RATIO MOLES! THIS ASSUMES P AND T ARE CONSTANT.

9. #28) WHAT IS THE MOLARITY OF A SOLUTION OF KNO 3 (MOLECULAR MASS=101) THAT CONTAINS 404 GRAMS OF KNO 3 IN 2.00 LITERS OF SOLUTION? MOLARITY = MOLES VOLUME MOLES = MASS GFM MOLES = 404. g = 4mol KNO 3 101. g/mol MOLARITY = 4 mol = 2.00M 2.00L

10. #15) The density of a gas is 3.00 g/L at STP, WHAT IS THE GFM OF THE GAS? 22.4 LITERS X 3.00 g 1 LITER = = 67.2 g/mol DENSITY RATIO, GIVEN 1 MOLE = 22.4L THE MASS OF 22.4 L IS THE MOLAR MASS AT STP