stoichiometry. law of conservation of mass antoine lavoisier studied combustion showed it involved...

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SToichiometry SToichiometry

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Page 1: SToichiometry. Law of Conservation of Mass Antoine Lavoisier Studied combustion Showed it involved oxygen & life is supported by oxygen in a process similar

SToichiometrySToichiometry

Page 2: SToichiometry. Law of Conservation of Mass Antoine Lavoisier Studied combustion Showed it involved oxygen & life is supported by oxygen in a process similar

Law of Conservation of

Mass

Antoine LavoisierStudied 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 RevolutionConsidered enemy of the peopleSent to guillotine

Page 3: SToichiometry. Law of Conservation of Mass Antoine Lavoisier Studied combustion Showed it involved oxygen & life is supported by oxygen in a process similar

Law of Conservation of MassAll 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 evolvedCombustion of hydrocarbonProduces CO2 & Water

Combustion:It is usually easiest to balancecarbon first

hydrogen secondcount how many oxygen needed in products

C5H12 + O2 --> H2O + CO256Need 16 oxygen for products

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Page 4: SToichiometry. Law of Conservation of Mass Antoine Lavoisier Studied combustion Showed it involved oxygen & life is supported by oxygen in a process similar

Mass-Mass & Mole-Mole Relations in ReactionsSuggested method:

1. write balancedequation

CO + 2 H2 --> CH3OH2. 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 leftc. How much product, in grams

100.0 g 100.0 g

28.0 g/mol 2.00 g/mol

______ ______/

//

/

3.57 mol 50.0 mol

3. Pick onetake moles have

& divide by balanced coefficient

This gives you a ratioto multiply other

coefficients by

3.57/1 = 3.57

Moles use 3.57 mol4. Take answer from step 3 multiply other

coefficient by it

3.57 x 2 = 7.14

7.14 mol

Moles left

- -

0 42.9 mol

Grams left 0 85.8 g

3.57 x 1 = 3.57

3.57 mol

32.0 g/mol

114. g

CO85.8 g

114. g

Theoretical yieldPercent yield = (actual yield/ theoretical yield) 100Let’s say you got 100.0 g for this reaction

Percent yield = (100.0 g / 114. g) 100 = 87.7 %

Page 5: SToichiometry. Law of Conservation of Mass Antoine Lavoisier Studied combustion Showed it involved oxygen & life is supported by oxygen in a process similar

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