Transcript Les Brown
Shoot for the moon. Even if you miss, you'll land among the stars (space debris). - Les Brown - LET’S REVIEW WHAT FORMULAS TELL US. IF WE WRITE THE FOLLOWING: 2NaCO3 THE 2 TELLS US THAT WE HAVE 2 FORMULA UNITS OF SODIUM CARBONATE. EACH FORMULA UNIT CONTAINS 1 ATOM OF SODIUM, 1 ATOM OF CARBON, AND 3 ATOMS OF OXYGEN. IN A LABORATORY, WE WANT TO BE ABLE TO DETERMINE THE FORMULAS OF COMPOUNDS AND DETERMNE HOW MANY FORMULA UNITS OR MOLECULES WE ARE DEALING WITH. ATOMS ARE TOO TINY TO BE ABLE TO DETERMINE THEIR MASS EASILY. EVEN MOLECULES (COMBNATIONS OF SEVERAL ATOMS ARE TOO SMALL) 1 ATOMIC MASS UNIT (AMU) = 1.6 X 10-24 g THE BEST LAB BALANCES WILL ONLY MEASURE MASS TO 0.000001 g (1 microgram). WE NEED SOME OTHER WAY. YOU CAN COUNT OBJECTS BY WEIGHING. AN EXAMPLE THAT IS SOMETIMES USED IS SELLING JELLY BEANS. IF YOU NEEDED 1000 JELLY BEANS, AND EACH JELLY BEAN WEIGHED 5 GRAMS, IT WOULD BE MUCH EASIER TO WEIGHT 5000 GRAMS RATHER THAN COUNTING 1000. THIS IS BASICALLY WHAT WE DO IN CHEMISTRY. WE WANT TO WORK WITH EQUAL NUMBERS OF ATOMS OR WHOLE NUMBER RATIOS OF ATOMS, SO WE USE THE MOLE CONCEPT. IN THIS CONCEPT, 12 GRAMS OF CARBON 12 CONTAIN 6.02 X 1023 ATOMS OF CARBON (ONE MOLE). IF WE TAKE THE ATOMIC WEIGHT OF ANY ELEMENT IN GRAMS (THE GRAM ATOMIC WEIGHT), IT CONTAINS 6.02 X 1023 ATOMS. OR, IF WE TAKE THE MOLAR MASS IN GRAMS OF A COMPOUND, WE HAVE 6.02 X 1023 MOLECULES. 6.02 X 1023 IS AVOGADRO’S NUMBER, NAMED IN HONOR OF AVOGADRO, AND WE USE THE SYMBOL N. N = 6.02 X 1023 = ONE MOLE CONVERSIONS WE CAN DO: 1)THERE ARE 6.02 X 1023 PARTICLES IN ONE MOLE. IF WE KNOW THE NUMBER OF MOLES, WE CAN CALCULATE THE NUMBER OF PARTICLES – MOLECULES, ATOMS, OR FORMULA UNITS – OR VICE VERSA. 2)THE NUMBER OF MOLES OF A SUBSTANCE IN A GIVEN MASS CAN BE DETERMINED BY DIVIDING THE MASS IN GRAMS BY THE FORMULA MASS. EXAMPLES: 1)CONVERT 60.0 g of NaOH TO MOLES. 2)CONVERT 3.00 MOLES OF ZnCl2 TO GRAMS. 3)HOW MANY MOLECULES ARE IN 0.0045 MOLES OF Al2(CO3)3? 4)HOW MANY MOLES ARE PRESENT IN 1.00 X 1025 MOLECULES OF H2? 5)HOW MANY MOLECULES (ATOMS) ARE PRESENT IN 8.00 g He? 6)HOW MANY GRAMS OF AlPO4 ARE IN 1.00 X 1023 MOLECULES? PERCENT COMPOSITION IS IMPORTANT IN CHEMISTRY. WE HAD AN EXAMPLE IN NUCLEAR CHEMISTRY. IF YOU ARE MINING FOR URANIUM, AN ORE THAT CONTAINED 1% URANIUM WOULD BE FAR MORE VALUABLE THAN A ORE CONTAINING ONLY 0.01%. THEN, ONCE YOU PURIFY THE URANIUM, IF IT CONTAINED 1% U-235, IT WOULD BE FAR MORE VALUABLE THAN IF IT CONTAINED 0.5%. THE NATURAL AVERAGE IS 0.7%. PERCENT COMPOSITION IS ALSO IMPORTANT IN DEALING WITH COMPOUNDS. COMPOUNDS ARE COMPOSED OF TWO OR MORE ELEMENTS CHEMICALLY COMBINED. A GIVEN COMPOUND ALWAYS HAS THE SAME PERCENT COMPOSITION BECAUSE THE ELEMENTS ALWAYS COMBINE IN THE SAME WAY – THE FORMULA DOES NOT CHANGE. NOTE: YOU ARE COMBINING GIVEN NUMBERS OF WHOLE ATOMS. THIS IS ANOTHER WAY OF STATING THE LAW OF DEFINITE PROPORTIONS, SOMETIMES REFERRED TO AS THE LAW OF DEFINITE COMPOSITION. THE LAW OF DEFINITE COMPOSITION STATES THAT CHEMICAL COMPOUNDS ARE COMPOSED OF A FIXED RATIO OF ELEMENTS AS DETERMINED BY MASS. PERCENT COMPOSITION CAN BE DETERMINED BY EXPERIMENT, AND, IN TURN, CAN BE USED TO DETERMINE THE MOLE RATIO (EMPIRICAL FORMULA) OF THE COMPOUND. FOR EXAMPLE, IN THE LAB, THE MASS OF EACH ELEMENT IN A COMPOUND COULD BE DETERMINED BY DECOMPOSING A COMPOUND. EXAMPLE: LAB PROCEDURES SHOW THAT 50.0 g OF AMMONIA YIELD 41.0 g NITROGEN AND 9.00 g HYDROGEN. WHAT IS THE PERCENT COMPOSITION OF AMMONIA? NOTE: EMPIRICAL FORMULAS CAN BE DETERMINED FROM EXPERIMENTAL DATA OR FROM PERCENT COMPOSITION. EXAMPLE: A 2.50 G SAMPLE OF A COMPOUND CONTAINS 0.900 g CALCIUM AND 1.60 g CHLORINE. DETERMINE THE EMPIRICAL FORMULA. STEP 1: DETERMINE THE # OF MOLES Ca = 0.900 g/40.1 g/mole = 0.0224 mole Cl = 1.60 g/35.5 g/mole = 0.0451 mole THE MOLE RATIO IS 0.0224 mole Ca TO 00451 mole Cl. STEP 2: CALCULATE THE SIMPLEST WHOLE NUMBER RATIO. Ca = 0.0224/0.0224 = 1 Cl = 0.0451/0.0224 = 2.01 or 2 THE EMPIRICAL FORMULA IS CaCl2 YOU COULD DO THE SAME THING FROM PERCENT COMPOSITION. REMEMBER: PERCENT IS PARTS PER HUNDRED. EXAMPLE: YOU HAVE A COMPOUND WHOSE PERCENT COMPOSITION IS 40% C, 6.71% H, AND 53.3% O. SO, IF YOU HAD A 100 g OF THE COMPOUND, YOU WOULD HAVE 40.0 g C, 6.71 g H, and 53.3 g O. C = 40 g/12.0 g/mole = 3.33 mole H = 6.71 g/1.01 g/mole = 6.64 mole O = 53.3 g/16.0 g/mole = 3.33 mole OR C = 3.33/3.33 = 1 H = 6.64/3.33 = 2 THE EMPIRICAL FORMULA IS CH2O O = 3.33/3.33 = 1 IF YOU KNOW THE MOLECULAR MASS, YOU CAN CONVERT THE EMPIRICAL FORMULA TO THE MOLECULAR FORMULA. IN THE PREVIOUS EXAMPLE, IF YOU KNEW THE MOLECULAR MASS WAS 60: THE EMPIRICAL FORMULA MASS IS: C = 12.0 2H = 1.01 O = 16.0 total = 30.0 SO, 60/30 = 2 THE MOLECULAR FORMULA IS 2 TIMES THE EMPIRICAL FORMULA OR: C2H4O2 = MOLECULAR FORMULA THE MOLECULAR FORMULA IS ALWAYS A WHOLE NUMBER MULTIPLE OF THE EMPIRICAL FORMULA.