THE MOLE

 

1.         SINCE ATOMS ARE TOO SMALL TO BE MEASURED ONE AT A TIME, CHEMISTS USE A UNIT WITH THE SAME NUMBER OF ATOMS TO MEASURE WEIGHT.  THINK OF THIS AS A BAKERS DOZEN.  THIS UNIT IS CALLED A MOLE.  SINCE THE NUMBER OF ATOMS FOR A MOLE OF ANY SUBSTANCE IS ALWAYS THE SAME, PROPORTIONS ALWAYS STAY THE SAME AND IT WORKS AS IF YOU WERE WORKING WITH ONE ATOM.

 

2.         THE ATOMIC MASSES OF ALL THE ELEMENTS HAS BEEN ADJUSTED SO THEY EQUAL THE WEIGHT OF A MOLE OF THE SUBSTANCE.

EXAMPLE:     THE ATOMIC MASS OF CARBON IS 12.011 THE WEIGHT OF ONE MOLE OF CARBON 12 IS ALSO 12 GRAMS.

 

3.         THE ATOMIC MASS UNITS OF CARBON IS 12.011 AMU.  NOTICE THESE ARE JUST RELATIVE UNITS WITH NO WEIGHT.  SO IF CARBON IS 12.011 AMU AND MAGNESIUM IS 24.305 AMU, IT MEANS Mg IS APPROXMATELY TWICE AS HEAVY AS C.

 

4.         BY USING AMU NUMBERS WHICH ARE RELATIVE TO EACH OTHER AS THE WEIGHT OF MOLE OF A SUBSTANCE, THE NUMBER OF ATOMS WEIGHED IN GRAMS ALWAYS REMAINS THE SAME.

            EXAMPLE:     APPLES = 12 AMU

                                    ORANGES – 24 AMU

THIS MEANS ORANGES ARE TWICE AS HEAVY AS APPLES.  IF 12 LBS. OF APPLES EQUALS A DOZEN, THEN 24 LBS. OF ORANGES WOULD ALSO BE A DOZEN.

 

 

 

 

 

 

 

 

 

 

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5.         IN PLACE OF A DOZEN CHEMISTS USE THE MOLE WHICH EQUALS 6.02 x 1023 ATOMS OR MOLECULES DEPENDING OF IF YOU ARE WORKING WITH ELEMENTS OR COMPOUNDS.

 

6.         6.02 x 10 ATOMS PER MOLE IS CALLED AVOGADRO’S NUMBER SINCE HE HAS MOST ACCURATELY MEASURED THIS AMOUNT.  THIS IS BASED ON CARBON 12 (THE ISOTOPE) HAVING 6.02 x 1023 ATOMS IN 12 GRAMS.

 

7.         A MOLE OF A SUBSTANCE CAN BE MEASURED IN THREE WAYS.

           

            A.        THE NUMBER OF PARTICLES = 6.02 x 1023

B.         THE MOLAR MASS (MASS OF A MOLE = ITS ATOMIC MASS IN GRAMS

C.        THE VOLUME OF A MOLE OF GAS AT STANDARD CONDITIONS (0 C AND 1 ATM) = 22.4 LETERS

 

8.         WHEN FINDING THE WEIGHT OF MOLE OF MOLECULES THE ATOMS IN THE MOLECULE MUST BE ADDED TOGETHER.

            EXAMPLE:   WHAT IS THE MOLAR WEIGHT OF CO2

                       

                        C x 12.011 gms/mole x 1 ATOM =      12.011

                        O x 15.999 gms/mole x 2 ATOMS =   31.998

                            TOTAL MOLAR WEIGHT      =    44.009 GRAM

 

 

 

 

 

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MOLE CONVERSIONS

 

1.         YOU CAN CONVERT BETWEEN ANY OF THE THREE MEASUREMENTS OF A MOLE (WEIGHT, PARTICLES, OR VOLUME) BY USING CONVERSION FACTORS

 

                        1 MOLE OF A SUBSTANCE     =     ITS MOLECULAR WEIGHT

                        1 MOLE OF A SUBSTANCE     =    6.02  x  1023

                        1 MOLE OF A SUBSTANCE     =    22.4 LITERS

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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3.         BY USING THE SAME MECHANICAL PROCESS FOR CONVERSION THAT WE LEARNED EARLIER THIS YEAR, YOU CAN CONVERT FROM ANY ONE OF THE MEASUREMENTS OF THE MOLE TO THE OTHER BY GOING THROUGH THE MOLE.

 

            EXAMPLE:

 

            IF YOU HAVE 11.2 GRAMS OF NaCl, HOW MANY MOLES DO YOU HAVE?

            A.        1 MOLE OF NaCl EQUALS ITS MOLECULAR WEIGHT

                                    Na       =          23.

                                    Cl         =          35.5

                                                            58.5 GRAM/MOLE

 

            B.         1 MOLE OF NaCl       =          58.5 GRAMS

 

SINCE YOU ARE FINDING MOLES AND ELIMINATING GRAMS, THE GRAMS GO ON THE BOTTOM AND MOLES GO ON TOP

 

            C.        11.2 GRAMS  x          1 MOLE/58.5 GRAMS           =          /191MOLES

 

YOU CAN CONVERT IN THE OPPOSITE DIRECTION

 

HOW MANY GRAMS ARE IN 2.50 MOLES OF NaCl?

 

USE THE SAME CONVERSION FACTOR BUT THIS TIME GRAMS IS WHAT IS BEING FOUND AND NEEDS TO BE ON THE TOP

 

                        2.50 MOLES   x          58.5 GRAMS/1MOLE            =          146 GRAMS

 

 

 

 

 

 

 

 

JUST AS YOU CAN CONVERT FROM MOLES TO MASS AND BACK, YOU CAN CONVERT FROM MOLES TO PARTICLES OR VOLUME AND BACK

 

            EXAMPLE:     A MARBLE CONTAINS 8.74   x   10 MOLECULES OF CaCO3.  HOW MANY MOLES ARE PRESENT?

 

            A.        USE THE CONVERSION FACTOR THAT RELATES MOLES TO PARTICLES. MOLES GOES ON TOP, SINCE THAT IS WHAT IS BEING FOUND.

 

                        1 MOLE = 6.023 x 1023 MOLECULES

           

            8.74 x 1022 MOLECULES x 1 MOLE/6.023 x 1023 MOLECULES

                                    = 1.45 MOLES OF CaCO3

 

WHEN ANY OF THE THREE MESURES OF MOLES IS USED IT CAN BE CHANGED TO ONE OF THE OTHER MEASURES BY FIRST GOING TO MOLES

 

                        MASS                   PARTICLES

 

 


                                       

   MOLES

 


                     

 

                          VOLUME 22.4 L AT STP

 

 

 

 

 

 

 

 

 

 

 

 

 

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EXAMPLE:     A STUDENT FILLS A 1.0 L FLASK WITH CO2 AT STANDARD TEMPERATURE AND PRESSURE.  HOW MANY MOLECULES OF GAS ARE IN THE FLASK?

 

THE MOLE IS THE CENTRAL UNIT.  THEREFORE, THE MOLUME IS FIRST CHANGED TO MOLES THEN MOLES ARE CHANGED TO PARTICLES

 

CONVERT TO MOLES USING       

                        1 MOLE – 22.4 L AT STP

THEN CONVERT TO PARTICLES USING

                        1 MOLE = 6.0 x 1023 MOLECULES

 

1.0 L x 1 MOLE/22.4L x 6.02 x 1023 MOLECULES/1MOLE

                        = 2.7 x 1022 MOLECULES

 

 

 

 

 

 

 

 

 

 

 

 

 

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PERCENT COMPOSITION PROBLES

 

1.         THESE PROBLEMS ARE WORKED JUST LIKE ANY PERCENTAGE PROBLEM.

 

EXAMPLE:                 PERCENT = THE PART/WHOLE      x     100

 

2.         EXAMPLE:  WHAT IS THE PERCENT OF OXYGEN AND HYDROGEN IN H20?

 

FROM THE PERIODIC CHART WE FIND THAT THE MOLAR WEIGHT OF 1 MOLE OF H = 1 GRAM.  IF THERE ARE 2 MOLES OF H THEN THE MOLAR WEIGHT OF H WOULD BE 2 GRAMS.

 

THE MOLAR WEIGHT OF H2O       IS         H = 2 x 1 = 2 GRAMS

                                                                        O = 16     = 16 GRAMS

TOTAL MOLAR WEIGHT FOR H2O                          = 18 GRAMS

 

PART    x      100              =          H     =     2 GRAMS      x     100  = 11%

WHOLE                                        H2O        18 GRAMS

 

 

 

 

 

 

 

 

3.         THE SAME THING IS TRUE IF YOU WERE GIVEN WEIGHT AMOUNTS THAT DIDN’T CORRESPOND WITH THE MOLECULAR WEIGHT.

 

EXAMPLE:     WHEN HEATING A SAMPLE OF MG, THE MG WEIGHED 48 GRAMS.  AFTER HEATING THE SAMPLE, THE MGO THAT WAS FORM WEIGHED 90 GRAMS.  WHAT IS THE PERCENTAGE COMPOSITION OF T MB AND O IN MGO?

 

 

 

 

 

 

 

 

 

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EMPIRICAL FORMULAS

 

1.         AN EMPIRICAL FORMULA IS THE SIMPLEST NUMBER RATION BETWEEN THE ELEMENTS IS A COMPOUND.

 

EXAMPLE:  THE EMPIRICAL FORMULA FOR A COMPOUND IS ClH201.  ITS ACTUAL MOLECULAR FORMULA IS C6H1206.  THE EMPIRICAL FORMULA IS IN THE SIMPLEST SAME RATIO OF THE ACTUAL MOLECULE.

 

2.         IN FINDING THE EMPIRICAL FORMULA YOU FIRST MUST GET TO MOLES SINCE THE WEIGHT OF ELEMENTS IS LIKE COMPARING APPLES TO ORANGES.

 

3.         EXAMPLE:     IN A CARBON HYDROGEN COMPOUND CARBON WEIGHS 80 GRAMS AND THE HYDROGEN WEIGHS 20 GRAMS.  WHAT IS THE EMPIRICAL FORMULA?

 

FIRST GET TO MOLES SO WE ARE COMPARING THE SAME AMOUNTS (SINCE C & H WEIGH DIFFERENT FOR THE SAME AMOUNT)

 

            20 GRAMS OF H   x   1 MOLE/GRAM     =     20 MOLES OF H

 

            80 GRAMS OF C    x   1 MOLE-12GRAMS = 6.7 MOLES OF CARBON

 

NOW THAT WE HAVE COMMON AMOUNTS FOR EACH SUBSTANCE THE SIMPLE RATIO CAN BE CALCULATED BY DIVIDING EACH ELEMENTAL MOLAR AMOUNT BY THE SMALLEST NUMBER.

 

            C    =    6.7/6.7  =  1             H   =    20/6.7  = 2.98    =   CH3

 

 

 

 

 

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DETERMINING THE MOLECULAR FORMULA

FROM THE EMPIRICAL FORMULA

 

 

THIS IS AN EASY STEP.  FOR INSTANCE IF YOU KNEW THAT THE ACTUAL FORMULA WAS TWICE AS HEAVY AS THE EMPIRICAL FORMULA FOR THE COMPOUND JUST COMPLETED, YOU JUST NEED TO MULTIPLY BY TWO TO GET THE ACTUAL MOLECULAR FORMULA

 

                        2        x         CH3           =       C2H6

 

THE PROBLEM IS A LITTLE HARDER IF YOU WERE TOLD THE ACTUAL MOLECULAR WEIGHTS IS 30 GRAMS.

 

YOU THEN FIND THE MOLECULAR WEIGHT OF THE EMPIRICAL FORMULA AND DIVIDE IT INTO THE ACTUAL MOLECULAR WEIGHT OF THE COMPOUND

 

            CH3    =    15 GRAMS

 

30 GRAMS ACTUAL/15 GRAMS EMPIRICAL  = 2 TIMES AS HEAVY

 

IN WORKING PROBLEMS LIKE THIS REMEMBER GETTING TO MOLES SO EVERYTHING IS IN COMMON IS ALWAYS ONE OF THE FIRST STEPS BEFORE GOING ON.