Stoichiometry

Slide 1: 

Stiochiometry Copyright Sautter 2003

Slide 2: 

The next slide is a quick promo for my books after which the presentation will begin Thanks for your patience! Walt S. Wsautter@optonline.net More stuff at: www.wsautter.com

STIOCHIOMETRY : 

STIOCHIOMETRY “Measuring elements” Determing the Results of A Chemical Reaction

PREDICTING HOW MUCH OF A SUBSTANCE CAN BE MADE BY A CHEMICAL REACTION BEFORE IT IS CARRIED OUT!! : 

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!

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 : 

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)

**REMEMBER ** TO CONVERT GRAMS (MASS) TO MOLES WE DIVIDE THE WEIGHT OF ONE MOLE FROM THE PERIODIC TABLE INTO THE GIVEN NUMBER OF GRAMS : 

**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.

STEP II IN SOLVING STIOCHIOMETRY PROBLEMS THEN IS TO CONVERT THE GIVEN NUMBER OF GRAMS TO MOLES. : 

STEP II IN SOLVING STIOCHIOMETRY PROBLEMS THEN IS TO CONVERT THE GIVEN NUMBER OF GRAMS TO MOLES. .

CaCO3(S)  CaO(s) + CO2(g) : 

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!

Slide 10: 

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

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. : 

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.

STEP IV – CONVERT THE MOLES FOUND IN STEP III TO GRAMS (MASS) : 

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

CaCO3(S)  CaO(s) + CO2(g) 200 grams 0 grams 0 grams 2 moles 0 moles 0 moles before reaction occurs0 grams 112 grams 88 grams 0 moles 2moles 2molesafter reaction occursSTARTING TOTAL MASS = 200 +0 +0 = 200 GRAMSFINAL TOTAL MASS = 0 + 112 + 88 = 200 GRAMS(CONSERVATION OF MASS) : 

CaCO3(S)  CaO(s) + CO2(g) 200 grams 0 grams 0 grams 2 moles 0 moles 0 moles before reaction occurs0 grams 112 grams 88 grams 0 moles 2moles 2molesafter reaction occursSTARTING TOTAL MASS = 200 +0 +0 = 200 GRAMSFINAL TOTAL MASS = 0 + 112 + 88 = 200 GRAMS(CONSERVATION OF MASS) QUESTIONS ?????

NOW FOR A HARDER ONE! : 

NOW FOR A HARDER ONE! HYDROGEN GAS REACTS WITH OXYGEN GAS TO FORM WATER. HOW MUCH HYDROGEN AND OXYGEN MUST BE COMBINED TO MAKE 45 GRAMS OF WATER? STEP I – WRITE AND BALANCE THE EQUATION HYDROGEN = H2 (DIATOMIC ELEMENT) OXYGEN = O2 (DIATOMIC ELEMENT) WATER = H2O 2 H2(g) + 1 O2(g)  2 H2O(g) THERE ARE 4 ATOMS OF HYDROGEN ON EACH SIDE AND TWO ATOMS OF OXYGEN ON EACH SIDE. THE EQUATION IS BALANCED !!

STEP II – CONVERT THE GIVEN GRAMS TO MOLES : 

STEP II – CONVERT THE GIVEN GRAMS TO MOLES WATER HAS A MOLAR MASS (MASS OF ONE MOLE) FROM THE PERIODIC TABLE IS 18 GRAMS. { (2 x 1) FOR HYDROGEN AND (1 x 16) FOR OXYGEN } 45 grams of water DIVIDED BY 18 grams per mole = 2.5 moles We want to make 2.5 moles of water in our reaction.

STEP III – USE THE BALANCED EQUATION TO FIND THE MOLES OF REACTANT OR PRODUCT REQUIRED : 

STEP III – USE THE BALANCED EQUATION TO FIND THE MOLES OF REACTANT OR PRODUCT REQUIRED Here’s our balanced equation from STEP I 2 H2(g) + 1 O2(g)  2 H2O(g) TWO MOLES OF WATER ARE MADE FROM TWO MOLES OF HYDROGEN. HOW MANY MOLES OF HYDROGEN WOULD BE NEEDED TO MAKE 2.5 MOLES OF WATER? (A ONE FOR ONE RATIO) HOW ABOUT 2.5 ?? NOW FOR THE OXYGEN ! IT TAKES ONLY 1 MOLE OF OXYGEN TO MAKE 2 MOLES OF WATER, HALF AS MUCH ! (A TWO FOR ONE RATIO) THEREFORE TO MAKE 2.5 MOLES OF WATER NEEDS ONLY ½ OF 2.5 MOLES OF OXYGEN! THEREFORE 1.25 MOLES OF OXYGEN IS REQUIRED.

STEP IV – CONVERT MOLES OF UNKNOWN TO GRAMS : 

STEP IV – CONVERT MOLES OF UNKNOWN TO GRAMS SINCE WE NOW KNOW THAT 2.5 MOLES OF HYDROGEN IS REQUIRED, WE CAN MULTIPLY 2.5 TIMES 2.0 GRAMS, THE MOLAR MASS OF HYDROGEN TO GET 5.0 GRAMS OF H2(g) NEEDED! THE MOLAR MASS OF OXYGEN IS 32 GRAMS. MULTIPLYING 32 TIMES 1.25 MOLES GIVES 40 GRAMS OF OXYGEN ARE NEEDED.

THEN, IN ORDER TO MAKE 45.0 GRAMS OF WATER, USING HYDROGEN GAS AND OXYGEN GAS WE HAVE CALCULATED THAT: : 

THEN, IN ORDER TO MAKE 45.0 GRAMS OF WATER, USING HYDROGEN GAS AND OXYGEN GAS WE HAVE CALCULATED THAT: THE MOLES OF HYDROGEN NEEDED ARE 2.5 MOLES OR 5.0 GRAMS AND THE MOLES OF OXYGEN NEEDED ARE 1.25 MOLES OR 40.0 GRAMS AND ALL OF THIS HAS BEEN DETERMINED WITHOUT EVER PICKING UP A TEST TUBE BY USING STOICHIOMETRY!!!

Slide 19: 

LIMITING REACTANT THE REACTANT THAT RUNS OUT FIRST!

Limiting Reactant Calculations : 

Limiting Reactant Calculations In the reactions previously discussed, an amount of only one of the reactants was given. We assumed that we could use as much of the other reactant as we needed. Unfortunately, this is not always the case. Situations where specific amounts of both of the reactants are given are called “limiting reactant” problems. The limiting reactant is the one that runs out first! In limiting reactant problems three possibilities exist. Possibility one is that the first reactant is used up first. Possibility two is that the second reactant runs out first and possibility three is that both reactants run out at the same time.

Slide 21: 

Understanding Limit Reactant Concepts Containers of nuts and bolts are to be threaded together. One nut threaded on one bolt. How many combinations Can be made ? bolts nuts Only four nut – bolt combinations can be made. The bolts have run out. The bolts are the limiting factor +

Slide 22: 

Understanding Limit Reactant Concepts Containers of nuts and bolts are to be threaded together. Two nuts threaded on one bolt. How many combinations Can be made ? bolts nuts Only three nut – bolt combinations can be made. The nuts have run out. The nuts are the limiting factor The one with the smallest number is not always the limiting factor. It depends on the ratio of combinations! +

Limiting Reactant : 

Limiting Reactant Now, let’s try a limiting factor (reactant) problem using a chemical reaction! Remember, numbers of atoms and molecules are measured in moles. The balanced equation tells us the ratio of combination of the atoms and molecules that are used to make the products. In the reaction: H2 + I2  2 HI, one molecule of hydrogen is combined with one molecule of iodine to give two molecules of hydrogen iodide. It is also true to say that one mole of hydrogen is combined with one mole of iodine to give two moles of hydrogen iodide. All we have really done is multiply the entire equation through by 6.02 x 1023 (1 mole).

Limiting Reactant : 

Limiting Reactant H2 + I2  2 HI Suppose that exactly one mole of H2 and exactly one mole of I2 are available. In this case we can make exactly two moles of HI and no H2 or I2 will be left. Now, suppose that we have one mole of H2 and two moles of I2. Once the H2 is used up, no more I2 can be reacted. One mole of H2will use exactly one mole of I2 leaving an extra mole of I2 unused. The limiting reactant is H2. It ran out first. The excess reactant is I2. We have extra I2. Only two moles of HI can be made. Accordingly to our balanced equation, for each H2 used, two HI are formed. Only one mole of H2 was used so only two moles of HI are produced. The quantity of products formed is based on the limiting reactant!

Limiting Reactant : 

Limiting Reactant Problem: Given the reaction: 1Ca + 1Cl2  1CaCl2. If we mix 120 grams of calcium and 71 grams of chlorine, which reactant is the limiting factor? How many grams of CaCl2 can be made? Solution: Remember that balanced equations are based on moles. We must first convert the given grams to moles. Moles for Ca = 120 / 40 = 3.0, Moles for Cl2 = 71 / (2 x 35.5) = 1.0 From the balanced equation, 1 Ca requires exactly 1 Cl2. Since only 1.0 mole of Cl2 is available only 1.0 mole of Ca can be consumed and 2.0 moles of Ca remain unused. Chlorine (Cl2) is the limiting reactant. The amount of product that can be formed is based on the limiting reactant. The equation tells that for each Cl2 used, one CaCl2 is made. Since 1.0 moles Cl2 are used, 1.0 moles of CaCl2 are produced. Grams CaCl2 = 1.0 moles x (40 +2(35.5)) grams per mole = 111 g

Limiting Reactant : 

Limiting Reactant Problem: Given the equation: 2Na + Cl2  2NaCl. How many grams of NaCl can be made by reacting 69 grams of Na with 5.0 moles of Cl2 ? Solution: Again we must work in moles. Cl2 is already moles but Na must be converted (69 / 23 = 3.0 moles of Na) From the equation, 2Na needs 1Cl2 (half the moles of Na) so 3 Na needs 1.5 moles of Cl2. We have 5.0 moles of Cl2, more than enough. Therefore all of the Na is used and Na is the limiting reactant! The amount of product is based on the limiting reactant. Since 2Na make 2NaCl, 3.0 Na will make 3.0 NaCl Grams of NaCl = 3.0 moles x (23 + 35.5) gram per mole of NaCl = 175.5 grams of NaCl are formed.

STOICHIOMETRY : 

STOICHIOMETRY THE END