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"When I play with my cat, who knows whether she is not amusing herself with me more than I with her." Michel Eyquem De Montaigne Michel Eyquem De Montaigne.

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Presentation on theme: ""When I play with my cat, who knows whether she is not amusing herself with me more than I with her." Michel Eyquem De Montaigne Michel Eyquem De Montaigne."— Presentation transcript:

1 "When I play with my cat, who knows whether she is not amusing herself with me more than I with her." Michel Eyquem De Montaigne Michel Eyquem De Montaigne

2 A + 2B  3C + D SOMETIMES, BUT NOT ALWAYS, WHEN DIFFERENT SUBSTANCES ARE BROUGHT TOGETHER, A CHEMICAL CHANGE CAN OCCUR. CHEMICAL BONDS THAT BIND THE ATOMS TOGETHER CAN BREAK, THE ATOMS CAN REARRANGE THEMSELVES, AND NEW CHEMICAL BONDS CAN FORM. WE CALL THIS A CHEMICAL REACTION. WE DESCRIBE THIS IN WORDS OR WE CAN USE THE CHEMISTS’ SHORTHAND METHOD AND WRITE A CHEMICAL EQUATION.

3 LET’S TAKE A LOOK AT A TYPICAL CHEMICAL EQUATION AND SEE WHAT SORT OF INFORMATION IT GIVES US. Zn (s) + 2HCl (aq)  ZnCl 2 (aq) + H 2 (g) IF WE STATED THIS IN WORDS, WE MIGHT SAY “SOLID ZINC METAL WHEN ADDED TO AN AQUEOUS SOLUTION OF HYDROCHLORIC ACID PRODUCES A SOLUTION OF ZINC CHLORIDE AND HYDROGEN GAS. ONE MOLE OF ZINC WILL REACT WITH TWO MOLES OF HYDROCHLORIC ACID TO PRODUCE ONE MOLE OF ZINC CHLORIDE AND ONE MOLE OF HYDROGEN GAS. THERE IS A GREAT DEAL OF INFORMATION HERE. 1)THE FORMULAS TELL US WHAT THE REACTANTS AND PRODUCTS ARE. 2)THE PREFIXES TELL US HOW MANY MOLECULES OR MOLES OF EACH ARE REACTED OR FORM. 3)THE LETTERS IN PARANTHESIS TELL US THE STATES OF EACH. 4)KNOWING THE NUMBER OF MOLES AND THE MOLAR MASSES, WE COULD ALSO INFER THE MASSES INVOLVED.

4 THE ENERGY LOST OR GAINED COULD ALSO BE INCLUDED IN THE EQUATION. THE POTENTIAL ENERGIES OF THE CHEMICAL BONDS INVOLVED IS RARELY THE SAME. SO, YOU OFTEN HAVE ENERGY LOST OR GAINED. MASS IS NOT LOST OR GAINED IN A CHEMICAL REACTION, AND ATOMS OF ONE ELEMENT ARE NOT CONVERTED INTO ATOMS OF OTHER ELEMENTS. SO, THE LAW OF CONSERVATION OF MASS ALWAYS HOLDS. FOR A CHEMICAL EQUATION TO BE TRUE, IT MUST BE BALANCED – YOU HAVE THE SAME NUMBER OF ATOMS OF EACH KIND ON BOTH SIDES OF THE EQUATION. YOU ALSO HAVE THE SAME MASS ON EACH SIDE OF THE EQUATION.

5 PROBLEMS YOU SHOULD BE ABLE TO SOLVE WITH EQUATIONS 1)MOLE RELATIONSHIPS IN A REACTION EXAMPLE: GIVEN THE FOLLOWING REACTION 2Fe(NO 3 ) 3 + 3Na 2 S  Fe 2 S 3 + 6NaNO 3 HOW MANY MOLES OF Na 2 S WOULD BE REQUIRED TO MAKE 15 moles OF Fe 2 S 3 ?

6 WE WANT 15 moles Fe 2 S 3, AND WE WANT TO KNOW HOW MUCH Na 2 S IS NEEDED. X moles 15 moles 2Fe(NO 3 ) 3 + 3Na 2 S  Fe 2 S 3 + 6NaNO THE EQUATION TELLS US THAT 3 moles OF Na 2 S WILL GIVE US 1 mole OF Fe 2 S 3. SO X / 3 = 15 / 1 SOLVING FOR X X = 3 x 15 = 45 moles

7 2) GIVEN THE AMOUNTS OF STARTING MATERIAL YOU HAVE, HOW MUCH PRODUCT CAN YOU MAKE? EXAMPLE: UNDER HIGH TEMPERATURE AND PRESSURE, NITROGEN GAS CAN BE MADE TO REACT WITH HYDROGEN GAS TO FORM AMMONIA. YOU HAVE 2,000 KG OF HYDROGEN, HOW MUCH AMMONIA CAN YOU FORM? FIRST, YOU NEED THE EQUATION, SO YOU’D WRITE N 2 + H 2  NH 3 IS IT BALANCED?

8 TO GET IT TO BALANCE N 2 (g) + 3H 2 (g)  2NH 3 (g) SO, THIS SAYS THAT ONE MOLE OF NITROGEN GAS REACTS WITH 3 MOLES OF HYDROGEN GAS TO FORM 2 MOLES OF AMMONIA GAS. THE MOLAR MASS OF HYDROGEN IS 2 X 1.01 = 2.02 AND 3 MOLES ARE INVOLVED OR 3 X 2.02 = 6.06 THE MOLAR MASS OF AMMONIA IS X 1.01 = 17.0 SO, PLUGGING IN WHAT YOU KNOW AND YOUR UNKNOWN: 2,000 kg X kg N 2 (g) + 3H 2 (g)  2NH 3 (g) OR 2000/6.06 = X/ g 34.0 g X = (2000/6.06) X 34.0 = 11,200 kg AMMONIA

9 3) HOW MUCH STARTING MATERIAL DO YOU NEED TO PRODUCE A GIVEN AMOUNT OF PRODUCT? EXAMPLE: YOU WANT TO USE THE THERMITE REACTION TO PRODUCE 500 g OF IRON FOR A WELD BETWEEN TWO RAILS ON A RAILROAD. HOW MUCH Fe 2 O 3 DO YOU NEED TO START WITH? THE FIRST THING YOU NEED IS A BALANCED EQUATION: Fe 2 O 3 + 2Al  Al 2 O 3 + 2Fe THE EQUATION TELLS YOU THAT FOR EVERY MOLE OF IRON (III) OXIDE YOU START WITH, YOU END UP WITH 2 MOLES OF IRON. SO, LET’S CHANGE OUR MOLE RELATIONSHIPS TO MASS AND PUT OUR VALUES IN.

10 FIRST, WE NEED OUR FORMULA MASSES: 2 Fe = 55.8 x 2 = O = 16.0 x 3 = 48.0 formula mass = 160 g/fm X g 500 g Fe 2 O 3 + 2Al  Al 2 O 3 + 2Feso, our equation is 160 g 112 g X/160 = 500/112 and X = (500/112) x 160 = 714 g

11 SOME OTHER PROBLEMS YOU SHOULD BE ABLE TO SOLVE 4) PERCENT COMPOSITION – CALCULATE PERCENT COMPOSITION FROM FORMULA MASS EXAMPLE: CALCULATE THE % OF EACH ELEMENT PRESENT IN Na 3 PO 4 FIRST CALCULATE FORMULA MASS 3 Na = 3 x 23 = 69 1 P = 1 x 31 = 31 4 O = 4 x 16 = 64 formula mass = 164 g/mole

12 % Na = (69/164) x 100 = 42% % P = (31/164) x 100 = 19% % O = (64/164) x 100 = 39%

13 5) GIVEN % COMPOSITION, CALCULATE THE EMPIRICAL FORMULA EXAMPLE: A molecule with molecular weight of g/mol is analyzed and found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen. THIS MEANS THAT IF YOU HAD 100 grams, YOU WOULD HAVE g CARBON 6.72 g HYDROGEN 53.8 g OXYGEN

14 NEXT, TAKE THESE AMOUNTS AND CALCULATE MOLES C = 40.0 g/12 g/mole = 3.33 moles H = 6.72 g/1.0 g/mole = 6.72 moles O = 53.8 g/16 g/mole = 3.36 moles NOW WE DIVIDE BY THE SMALLES NUMBER TO TRY TO GET A WHOLE NUMBER RATIO. C = 3.33/3.33 = 1 H = 6.72/3.33 = 2 0 = 3.36/3.33 = 1 EMPIRICAL FORMULA = CH 2 O

15 5) CALCULATE THE MOLECULAR FORMULA IF WE KNOW THE EMPIRICAL FORMULA EXAMPLE: If the empirical formula is CH 2 O and the molecular mass is 90 g/mole, what is the molecular formula. WE KNOW THE MOLECULAR FORMULA IS SOME WHOLE NUMBER MULTIPLE OF THE EMPIRICAL FORMULA, SO CALCULATE THE EMPIRICAL MASS 1C + 2H + 1O = = 30 DIVIDE 90 BY 30 = 90/30 = 3, SO MOLECULAR FORMULA = C 3 H 6 O 3

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