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SToichiometry. Law of Conservation of Mass Antoine Lavoisier Studied combustion Showed it involved oxygen & life is supported by oxygen in a process similar.

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Presentation on theme: "SToichiometry. Law of Conservation of Mass Antoine Lavoisier Studied combustion Showed it involved oxygen & life is supported by oxygen in a process similar."— Presentation transcript:

1 SToichiometry

2 Law of Conservation of Mass Antoine Lavoisier Studied combustion Showed it involved oxygen & life is supported by oxygen in a process similar to combustion Carefully weighted reactants & products In an (ordinary) chemical reaction, mass is neither created nor destroyed. Wrote first modern chemistry book Worked as a tax collector during French Revolution Considered enemy of the peopleSent to guillotine

3 Law of Conservation of Mass All atoms you start with you end up with Need to show conservation of mass with balanced equation Reactant + Reactant --> Product Combustion: Burning in the presence of oxygen Heat is evolved Combustion of hydrocarbonProduces CO2 & Water Combustion:It is usually easiest to balance carbon first hydrogen second count how many oxygen needed in products C5H12 + O2 --> H2O + CO2 5 6 Need 16 oxygen for products 8

4 Mass-Mass & Mole-Mole Relations in Reactions Suggested method: 1. write balanced equation CO + 2 H 2 --> CH 3 OH 2. Make Chart Grams Molar Mass Moles have Given 100.0 g of each reactant, a. which is limiting reactant b. how much of reactant in excess is left c. How much product, in grams 100.0 g 28.0 g/mol2.00 g/mol ______ / / / / 3.57 mol50.0 mol 3. Pick one take moles have & divide by balanced coefficient This gives you a ratio to multiply other coefficients by 3.57/1 = 3.57 Moles use3.57 mol 4. Take answer from step 3 multiply other coefficient by it 3.57 x 2 = 7.14 7.14 mol Moles left -- 042.9 mol Grams left085.8 g 3.57 x 1 = 3.57 3.57 mol 32.0 g/mol 114. g CO 85.8 g 114. g Theoretical yield Percent yield = (actual yield/ theoretical yield) 100 Let’s say you got 100.0 g for this reaction Percent yield = (100.0 g / 114. g) 100 = 87.7 %

5 You might have your own method for stoichiometry-- that’s fine too.


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