Stoichiometry

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A quantitative approach to chemical reactions.
**More good stuff available at:
www.wsautter.com
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Stoichiometry

  1. 1. Copyright Sautter 2015
  2. 2. STIOCHIOMETRY “Measuring elements” Determing the Results of A Chemical Reaction
  3. 3. PREDICTING HOW MUCH OF A SUBSTANCE CAN BE MADE BY A CHEMICAL REACTION BEFORE IT IS CARRIED OUT!! • STEP I • ALWAYS WRITE THE EQUATION USING CHEMICAL FORMULAE AND BALANCE IT. • (REMEMBER TO BE SURE YOU USE THE CORRECT SUBSCRIPTS FOR THE FORMULAE AND CHANGE ONLY THE COEFFICIENTS WHEN BALANCING THE EQUATION) • EXAMPLE: • CALCIUM CARBONATE (LIMESTONE) WHEN HEATED GIVES CALCIUM OXIDE (LIME) AND CARBON DIOXIDE • CALCIUM = Ca (+2) • CARBONATE = CO3 (-2) • CALCIUM CARBONATE = CaCO3 • OXIDE = O (-2) • CALCIUM OXIDE = CaO • CARBON DIOXIDE = CO2 • CaCO3(S)  CaO(s) + CO2(g) • THERE IS ONE Ca ON EACH SIDE OF THE EQUATION, ONE C ON EACH SIDE AND THREE O ON EACH SIDE. • THE EQUATION IS BALANCED! 3
  4. 4. IN ORDER TO PREDICT THE RESULTS OF A CHEMICAL REACTION WE MUST BE GIVEN THE QUANTITIES OF MATERIALS THAT ARE TO BE USED IN THE REACTION SUPPOSE WE ARE GIVEN 200 GRAMS OF CALCIUM CARBONATE AND ASKED TO DECIDE HOW MUCH CALCIUM OXIDE AND CARBON DIOXIDE CAN BE MADE?? WE ALREADY HAVE THE BALANCED EQUATION BUT WHAT DO WE DO NEXT?? WELL, SINCE BALANCED EQUATIONS SHOW THE NUMBER OF ATOMS AND MOLECULES INVOLVED, WE MUST WORK WITH NUMBERS OF ATOMS AND MOLECULES. REMEMBER WE COUNT ATOMS AND MOLECULES WITH MOLES. (6.02 X 1023 = 1 MOLE) 4
  5. 5. **REMEMBER ** TO CONVERT GRAMS (MASS) TO MOLES WE DIVIDE THE WEIGHT OF ONE MOLE FROM THE PERIODIC TABLE INTO THE GIVEN NUMBER OF GRAMS • FOR EXAMPLE: • SINCE CaCO3 CONTAINS • 1Ca FROM THE PERIODIC TABLE WE USE 40grams x 1 • 1 C FROM THE PERIODIC TABLE WE USE 12 grams x1 • 3 O FROM THE PERIODIC TABLE WE USE 16 grams x 3 • (1x 40) + (1x 12) + (3 x 16) = 100 • THE MASS OF ONE MOLE OF CALCIUM CARBONATE IS 100 grams • IN OUR PROBLEM WE ARE USING 200 grams of CaCO3 • DIVIDING 200grams by 100grams per mole of CaCO3 WE FIND THAT WE HAVE EXACTLY 2 MOLES OF THE CALCIUM CARBONATE REACTANT. 5
  6. 6. STEP II IN SOLVING STIOCHIOMETRY PROBLEMS THEN IS TO CONVERT THE GIVEN NUMBER OF GRAMS TO MOLES. .
  7. 7. CaCO3(S)  CaO(s) + CO2(g) • THE EQUATION SAYS THAT ONE CALCIUM CARBONATE MAKES ONE CALCIUM OXIDE. HOW MANY CALCIUM OXIDES WOULD TEN CALCIUM CARBONATES MAKES? • HOW ABOUT TEN? • WHAT ABOUT ONE HUNDRED CALCIUM CARBONATES? • WOULDN’T THEY MAKE ONE HUNDRED CALCIUM OXIDES? • OF COURSE !! • THEN WHAT ABOUT A MOLE OF CALCIUM CARBONATE? WOULDN’T THEY BE EXPECTED TO MAKE A MOLE OF CALCIUM OXIDE? • AND OF COURSE THEN TWO MOLES WOULD MAKE TWO MOLES! 7
  8. 8. • STEP III IS PROBABLY THE MOST DIFFICULT ONE! • IT REQUIRES US TO PUT TOGETHER THE BALANCED EQUATION FROM STEP I AND THE MOLES THAT WE CALCULATED IN STEP II !! • ** REMEMBER ** • HERE’S OUR BALANCED EQUATION • CaCO3(S)  CaO(s) + CO2(g) • AND HERE’S THE MOLES WE CALCULATED • DIVIDING 200grams by 100grams per mole of CaCO3 WE FIND THAT WE HAVE EXACTLY 2 MOLES OF THE CALCIUM CARBONATE REACTANT 8
  9. 9. CaCO3(S)  CaO(s) + CO2(g) WHAT ABOUT THE CARBON DIOXIDE? THE BALANCED EQUATION ALSO SAYS THAT ONE CALCIUM CARBONATE MAKES ONE CARBON DIOXIDE TOO. • SO USING THE SAME LOGICAL THAT WE APPLIED TO THE CALCIUM OXIDE, IT IS OBVIOUS THAT TWO MOLES OF CALCIUM CARBONATE WILL PRODUCE EXACTLY TWO MOLES OF CARBON DIOXIDE ALSO. • STEP III – USING THE BALANCED EQUATION RATIOS FOUND IN STEP I AND THE MOLES DETERMINED IN STEP II, FIND THE MOLES OF EACH PRODUCT MATERIAL FORMED. 9
  10. 10. STEP IV – CONVERT THE MOLES FOUND IN STEP III TO GRAMS (MASS) • ** REMEMBER** TO CONVERT MOLES TO GRAMS, MULTIPLY THE MASS OF ONE MOLE FROM THE PERIODIC TABLE BY THE NUMBER OF MOLES. • EXAMPLE: • CaO CONTAINS 1 Ca (1 X 40grams from the Periodic Table) and 1 O (1 x 16 from the Periodic Table) • THE MASS OF ONE MOLE OF CaO IS (1 x 40) + (1 x 16) = 56 grams per mole • TWO MOLES OF CaO ARE FORMED THEREFORE, 2 MOLES x 56 grams per mole = 112 gram of CaO ARE FORMED IN THE REACTION • CO2 CONTAINS 1 C (1 x 12grams from the Periodic Table) and 2 O (2 x 16 grams from the Periodic Table) • THE MASS OF ONE MOLE OF CO2 IS (1 x 12) + (2 x 16) = 44 grams per moles • TWO MOLES OF CO2 ARE FORMED THEREFORE, 2 MOLES x 44 grams per mole = 88 grams of CO2 ARE FORMED IN THE REACTION 10
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