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Stoichiometry Applying equations to calculating quantities.

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Presentation on theme: "Stoichiometry Applying equations to calculating quantities."— Presentation transcript:

1 Stoichiometry Applying equations to calculating quantities

2 Learning objectives  Use chemical equations to predict quantity of one substance from given quantity of another  Determine percentage yield  Identify limiting reactant

3 The chemical equation aA + bB = cC + dD The Law of Conservation of Matter states that matter is neither created nor destroyed All atoms on left must be same as those on right Reactant side Product side coefficient ELEMENT or COMPOUND

4 Working with equations: STOICHIOMETRY  Predict how much product is obtained from given amount of reactant  Predict how much reactant is needed to give required amount of product  Predict how much of one reactant is required to give optimum result with given amount of another reactant

5 Stoichiometry with equations: The roadmap  Equations are in moles, but we measure in grams  Three conversions required:  A is given substance; B is target substance 1.Must convert grams A to moles A using molar mass 2.Use coefficients in equation to get moles B from moles A 3.Convert moles B to grams B using molar mass

6 Mole:mole ratio: the central step  Tells us molar ratio of substances in balanced chemical equation aA + bB = cC + dD  Mole:mole ratio B:A = b/a

7 Types of problems: Moles A → moles B (A is given; B is target) aA + bB = cC + dD  Single step  Require mole:mole ratio:  Always target/given (B/A)  In balanced equation a mol A ≡ b mol B

8 Moles A → mass B aA + bB = cC + dD Two steps on road map 1.Convert moles A → moles B: Mole:mole ratio (target/given): 2.Convert moles B → mass B using molar mass B

9 Mass A → mass B aA + bB = cC + dD All three steps 1.Mass A → moles A using molar mass A 2.Moles A → moles B using mole:mole ratio 3.Moles B → mass B using molar mass B

10 Summary of stoichiometry problems  Maximum of three conversions required 1.Must convert grams A to moles A using molar mass 2.Use coefficients in equation to get moles B from moles A 3.Convert moles B to grams B using molar mass  Maximum of three pieces of information required 1.Molar mass of given substance (maybe) 2.Molar mass of target substance (maybe) 3.Balanced chemical equation (always)

11 Work this example  CH 4 + 2O 2 = CO 2 + 2H 2 O  What mass of CO 2 is produced by the complete combustion of 16 g of CH 4  Atomic weight H = 1, C = 12, O = g  Do stoichiometry exercises Do stoichiometry exercises Do stoichiometry exercises

12 Reaction Yield  The actual yield from a chemical reaction is normally less than predicted from the stoichiometry.  Incomplete reaction  Product lost in recovery  Competing side reactions  Percent yield is:

13 Worked example  Actual yield of product is 32.8 g after reaction of 26.3 g of C 4 H 8 with excess CH 3 OH to give C 5 H 12 O.  What is theoretical yield? Use stoichiometry to get mass of product: convert mass  moles  moles  mass  Theoretical yield = 41.4 g  Percent yield = 32.8/41.4 x 100 %

14 Limiting reactant  Balanced equations involve precisely the right amounts of each reactant. None is left over  Real-life situations usually involve an excess of one reactant  Burning CH 4 in excess O 2  Reacting Zn with excess HCl

15 Identifying the limiting reactant  In balanced equations each reactant will give the same amount of products  When one reactant is limiting (deficient) the amount of product will be determined by its amount

16 Identifying the limiting reactant  Which reactant gives the least amount of product CO(g) + 2 H 2 (g) = CH 3 OH(g)  Given 3 mol CO and 5 mol H 2  3 mol CO → 3 mol CH 3 OH  (mol:mol ratio CH 3 OH:CO = 1)  5 mol H 2 → 2.5 mol CH 3 OH  (mol:mol ratio CH 3 OH:H 2 = 0.5)  H 2 is limiting  Maximum product = 2.5 mol CH 3 OH

17 Solution stoichiometry  In solids, moles are obtained by dividing mass by the molar mass  In liquids, it is necessary to convert volume into moles using the concentration

18 Solution stoichiometry  How much volume of one solution to react with another solution  Given volume of A with molarity M A  Determine moles A  Determine moles B  Find volume of B with molarity M B

19 Titration  Use a solution of known concentration to determine concentration of an unknown  Must be able to identify endpoint of titration to know stoichiometry  Most common applications with acids and bases

20 Example  How much M NaHCO 3 is required to neutralize 18.0 mL of M HCl?


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