PRINCIPLES OF CHEMISTRY
A CAR TRAVELING 350 MILES (MI) IN 7 HOURS (H) HAS A SPEED OF SPEED 5 350 MI 7 H 5 50 MI 1 H (OFTEN WRITTEN 50 MI?H21) IN 3 HOURS, THE CAR TRAVELS A DISTANCE OF DISTANCE 5 3 H 3 50 MI 1 H 5 150 MI CONSTRUCTING A CONVERSION FACTOR CONVERSION FACTORS ARE RATIOS USED TO EXPRESS A QUANTITY IN DIFFERENT UNITS SUPPOSE WE WANT TO KNOW THE DISTANCE OF THAT 150-MILE CAR TRIP IN FEET TO CONVERT MILES TO FEET, WE USE EQUIVALENT QUANTITIES, 1 MI 5 5280 FT 1 3 CHEMICAL PROBLEM SOLVING FROM WHICH WE CAN CONSTRUCT TWO CONVERSION FACTORS DIVIDING BOTH SIDES BY 5280 FT GIVES ONE CONVERSION FACTOR (SHOWN IN BLUE): 1 MI 5280 FT 5 5280 FT 5280 FT 5 AND, DIVIDING BOTH SIDES BY 1 MI GIVES THE OTHER CONVERSION FACTOR (THE INVERSE): 1 MI 1 MI 5 5280 FT 1 MI 5 SINCE THE NUMERATOR AND DENOMINATOR OF A CONVERSION FACTOR ARE EQUAL, MULTIPLYING A QUANTITY BY A CONVERSION FACTOR IS THE SAME AS MULTIPLYING BY 1 THUS, EVEN THOUGH THE NUMBER AND UNIT CHANGE, THE SIZE OF THE QUANTITY REMAINS THE SAME TO CONVERT THE DISTANCE FROM MILES TO FEET, WE CHOOSE THE CONVERSION FACTOR WITH MILES IN THE DENOMINATOR, BECAUSE IT CANCELS MILES AND GIVES THE ANSWER IN FEET: DISTANCE (FT) 5 150 MI 3 5280 FT 1 MI 5 792,000 FT MI 1 FT CHOOSING THE CORRECT CONVERSION FACTOR IT IS EASIER TO CONVERT IF YOU FIRST DECIDE WHETHER THE ANSWER EXPRESSED IN THE NEW UNITS SHOULD HAVE A LARGER OR SMALLER NUM- BER IN THE PREVIOUS CASE, WE KNOW THAT A FOOT IS SMALLER THAN A MILE, SO THE DISTANCE IN FEET SHOULD HAVE A LARGER NUMBER (792,000) THAN THE DISTANCE IN MILES (150) THE CONVERSION FACTOR HAS THE LARGER NUMBER (5280) IN THE NUMERATOR, SO IT GAVE A LARGER NUMBER IN THE ANSWER MOST IMPORTANTLY, THE CONVERSION FACTOR YOU CHOOSE MUST CANCEL ALL UNITS EXCEPT THOSE YOU WANT IN THE ANSWER THEREFORE, SET THE UNIT YOU ARE CONVERTING FROM (BEGIN- NING UNIT) IN THE OPPOSITE POSITION IN THE CONVERSION FACTOR (NUMERATOR OR DENOMINATOR) SO THAT IT CANCELS AND YOU ARE LEFT WITH THE UNIT YOU ARE CONVERTING TO (FINAL UNIT): BEGINNING UNIT FINAL UNIT BEGINNING UNIT 5 FINAL UNIT AS IN MI 3 FT MI 5 FT OR, IN CASES THAT INVOLVE UNITS RAISED TO A POWER: (BEGINNING UNIT 3 BEGINNING UNIT) AS IN MI H 3 FT MI 5 FT H CONVERTING BETWEEN UNIT SYSTEMS WE USE THE SAME PROCEDURE TO CONVERT BETWEEN SYSTEMS OF UNITS, FOR EXAMPLE, BETWEEN THE ENGLISH (OR AMERICAN) UNIT SYS- TEM AND THE INTERNATIONAL SYSTEM (A REVISED METRIC SYSTEM; SECTION 1 4) SUPPOSE WE KNOW THAT THE HEIGHT OF ANGEL FALLS IN VENEZUELA (THE WORLD’S HIGHEST) IS 3212 FT, AND WE FIND ITS HEIGHT IN MILES AS HEIGHT (MI) 5 3212 FT 1 MI 5280 FT 5 0 6083 MI FT 1 MI NOW, WE WANT ITS HEIGHT IN KILOMETERS (KM) THE EQUIVALENT QUANTITIES ARE 1 609 KM 5 1 MI BECAUSE WE ARE CONVERTING FROM MILES TO KILOMETERS, WE USE THE CONVERSION FACTOR WITH MILES IN THE DENOMINATOR IN ORDER TO CANCEL MILES: HEIGHT (KM) 5 0 6083 MI 3 1 609 KM 1 MI 5 0 9788 KM MI 1 KM CHAPTER 1 KEYS TO THE STUDY OF CHEMISTRY NOTICE THAT KILOMETERS ARE SMALLER THAN MILES, SO THIS CONVERSION FACTOR GAVE US AN ANSWER WITH A LARGER NUMBER (0 9788 IS LARGER THAN 0 6083) IF WE WANT THE HEIGHT OF ANGEL FALLS IN METERS (M), WE USE THE EQUIVALENT QUANTI- TIES 1 KM 5 1000 M TO CONSTRUCT THE CONVERSION FACTOR: HEIGHT (M) 5 0 9788 KM 3 1000 M 1 KM 5 978 8 M KM 1 M IN LONGER CALCULATIONS, WE OFTEN STRING TOGETHER SEVERAL CONVERSION STEPS: HEIGHT (M) 5 3212 FT 1 MI 5280 FT 3 1 609 KM 1 MI 3 1000 M 1 KM 5 978 8 M FT M A SYSTEMATIC APPROACH TO SOLVING CHEMISTRY PROBLEMS THE APPROACH USED IN THIS BOOK TO SOLVE PROBLEMS EMPHASIZES REASONING, NOT MEMO- RIZING, AND IS BASED ON A SIMPLE IDEA: PLAN HOW TO SOLVE THE PROBLEM BEFORE YOU TRY TO SOLVE IT, THEN CHECK YOUR ANSWER, AND PRACTICE WITH A SIMILAR FOLLOW-UP PROBLEM IN GENERAL, THE SAMPLE PROBLEMS CONSIST OF SEVERAL PARTS: 1 PROBLEM ?THIS PART STATES ALL THE INFORMATION YOU NEED TO SOLVE THE PROBLEM, USUALLY FRAMED IN SOME INTERESTING CONTEXT 2 PLAN ?THIS PART HELPS YOU THINK ABOUT THE SOLUTION BEFORE JUGGLING NUMBERS AND PRESSING CALCULATOR BUTTONS THERE IS OFTEN MORE THAN ONE WAY TO SOLVE A PROBLEM, AND THE GIVEN PLAN IS ONE POSSIBILITY THE PLAN WILL CLARIFY THE KNOWN AND UNKNOWN: WHAT INFORMATION DO YOU HAVE, AND WHAT ARE YOU TRYING TO FIND? SUGGEST THE STEPS FROM KNOWN TO UNKNOWN: WHAT IDEAS, CONVERSIONS, OR EQUATIONS ARE NEEDED? PRESENT A ROAD MAP (ESPECIALLY IN EARLY CHAPTERS), A FLOW DIAGRAM OF THE PLAN THE ROAD MAP HAS A BOX FOR EACH INTERMEDIATE RESULT AND AN ARROW SHOWING THE STEP (CONVERSION FACTOR OR OPERATION) USED TO GET TO THE NEXT BOX 3 SOLUTION THIS PART SHOWS THE CALCULATION STEPS IN THE SAME ORDER AS IN THE PLAN (AND THE ROAD MAP) 4 CHECK THIS PART HELPS YOU CHECK THAT YOUR FINAL ANSWER MAKES SENSE: ARE THE UNITS CORRECT? DID THE CHANGE OCCUR IN THE EXPECTED DIRECTION? IS IT REASONABLE CHEMICALLY? TO AVOID A LARGE MATH ERROR, WE ALSO OFTEN DO A ROUGH CALCULATION AND SEE IF WE GET AN ANSWER “IN THE SAME BALLPARK” AS THE ACTUAL RESULT HERE’S A TYPICAL “BALLPARK” CALCULATION FROM EVERYDAY LIFE YOU ARE AT A CLOTHING STORE AND BUY THREE SHIRTS AT $14 97 EACH WITH A 5% SALES TAX, THE BILL COMES TO $47 16 IN YOUR MIND, YOU KNOW THAT $14 97 IS ABOUT $15, AND 3 TIMES $15 IS $45; WITH THE SALES TAX, THE COST SHOULD BE A BIT MORE SO, YOUR QUICK MENTAL CALCULATION IS IN THE SAME BALLPARK AS THE ACTUAL COST 5 COMMENT THIS PART APPEARS OCCASIONALLY TO PROVIDE AN APPLICATION, AN ALTERNATIVE APPROACH, A COMMON MISTAKE TO AVOID, OR AN OVERVIEW 6 FOLLOW-UP PROBLEM THIS PART PRESENTS A SIMILAR PROBLEM THAT REQUIRES YOU TO APPLY CONCEPTS AND/OR METHODS USED IN SOLVING THE SAMPLE PROBLEM OF COURSE, YOU CAN’T LEARN TO SOLVE CHEMISTRY PROBLEMS, ANY MORE THAN YOU CAN LEARN TO SWIM, BY READING ABOUT IT, SO HERE ARE A FEW SUGGESTIONS: FOLLOW ALONG IN THE SAMPLE PROBLEM WITH PENCIL, PAPER, AND CALCULATOR TRY THE FOLLOW-UP PROBLEM AS SOON AS YOU FINISH THE SAMPLE PROBLEM A FEATURE CALLED BRIEF SOLUTIONS TO FOLLOW-UP PROBLEMS APPEARS AT THE END OF EACH CHAPTER, ALLOWING YOU TO COMPARE YOUR SOLUTION STEPS AND ANSWER READ THE SAMPLE PROBLEM AND TEXT AGAIN IF YOU HAVE TROUBLE A MEASURED QUANTITY CONSISTS OF A NUMBER AND A UNIT A CONVERSION FACTOR IS A RATIO OF EQUIVALENT QUANTITIES (AND, THUS, EQUAL TO 1) THAT IS USED TO EXPRESS A QUANTITY IN DIFFERENT UNITS SAMPLE PROBLEM 1 3 CONVERTING UNITS OF LENGTH PROBLEM ?TO HANG SOME PAINTINGS IN YOUR DORM, YOU NEED 325 CENTIMETERS (CM) OF PICTURE WIRE THAT SELLS FOR $0 15/FT HOW MUCH DOES THE WIRE COST? PLAN ?WE KNOW THE LENGTH OF WIRE IN CENTIMETERS (325 CM) AND THE PRICE IN DOLLARS PER FOOT ($0 15/FT) WE CAN FIND THE UNKNOWN COST OF THE WIRE BY CONVERTING THE LENGTH FROM CENTIMETERS TO INCHES (IN) AND FROM INCHES TO FEET THE PRICE GIVES US THE EQUIVALENT QUANTITIES (1 FT 5 $0 15) TO CONVERT FEET OF WIRE TO COST IN DOLLARS THE ROAD MAP STARTS WITH THE KNOWN AND MOVES THROUGH THE CALCULATION STEPS TO THE UNKNOWN SOLUTION ?CONVERTING THE KNOWN LENGTH FROM CENTIMETERS TO INCHES: THE EQUIVALENT QUANTITIES ALONGSIDE THE ROAD MAP ARROW ARE NEEDED TO CONSTRUCT THE CONVERSION FACTOR WE CHOOSE 1 IN/2 54 CM, RATHER THAN THE INVERSE, BECAUSE IT GIVES AN ANSWER IN INCHES: LENGTH (IN) 5 LENGTH (CM) 3 CONVERSION FACTOR 5 325 CM CHECK ?THE UNITS ARE CORRECT FOR EACH STEP THE CONVERSION FACTORS MAKE SENSE IN TERMS OF THE RELATIVE UNIT SIZES: THE NUMBER OF INCHES IS SMALLER THAN THE NUMBER OF CENTIMETERS (AN INCH IS LARGER THAN A CENTIMETER), AND THE NUMBER OF FEET IS SMALLER THAN THE NUMBER OF INCHES THE TOTAL COST SEEMS REASONABLE: A LITTLE MORE THAN 10 FT OF WIRE AT $0 15/FT SHOULD COST A LITTLE MORE THAN $1 50 COMMENT ?1 WE COULD ALSO HAVE STRUNG THE THREE STEPS TOGETHER: COST ($) 5 325 CM COST ($) OF WIRE ROAD MAP