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 Stoichiometry Chemistry S.Fleck 2015-2016. Objectives  9.1 – The Arithmetic of Equations  Interpret balanced chemical equations in terms of interacting.

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Presentation on theme: " Stoichiometry Chemistry S.Fleck 2015-2016. Objectives  9.1 – The Arithmetic of Equations  Interpret balanced chemical equations in terms of interacting."— Presentation transcript:

1  Stoichiometry Chemistry S.Fleck 2015-2016

2 Objectives  9.1 – The Arithmetic of Equations  Interpret balanced chemical equations in terms of interacting moles, representative particles, masses, and gas volume at STP.  9.2 – Chemical Calculations  Construct mole ratios from balanced chemical equations and apply these ratios in mole-mole stoichiometric calculations.  Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP.

3 Stoichiometry  Calculation of quantities in chemical equations.  Bookkeeping  Cookie analogy

4 How many tricycle seats, wheels, and pedals are needed to make 288 tricycles?  You know how many seats, wheels, and pedals are needed to make 1 tricycle, use those conversion factors to come up with the answers!

5 Interpreting Chemical Formulas

6 Mole to Mole Calculations  Relationships between reactants and products can be achieved by using properly balanced formulas.  N 2 (g) + 3H 2 (g)  2NH 3 (g)  Coefficients are used to write conversion factors!  Called Mole Ratios 2 mol NH 3 3 mol H 2 1 mol N 2 1 mol N 2 1 mol N 2 2 mol NH 3

7 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s)  Formation of Aluminum Oxide  Write 6 mole ratios that can be derived from this equation.  How many moles of aluminum are needed to form 3.7 mol Al 2 O 3 ?

8 Gram – Gram Calculations  From the mass of a reactant or product, the mass of any other reactant or product can be calculated.  You will use the same mole ratios from the balanced equation but you will need to go through more steps to get your answer.

9 Gram-Gram Example  Calculate the number of grams of NH 3 produced by the reaction of 5.40g of hydrogen with an excess of nitrogen.  N 2 (g) + 3H 2 (g)  2NH 3 (g)  Knowns  Mass of H 2 = 5.40 g H 2  3 mol H 2 = 2 mol NH 3 (balanced equation, conversion factor)  1 mol H 2 = 2.0 g H 2 (molar mass)  1 mol NH 3 = 17.0 g NH 3 (molar mass)

10 Mole Island!!!

11 Gram – Gram Example Continued  Mass of H 2 = 5.40 g H 2  3 mol H 2 = 2 mol NH 3 (balanced equation, conversion factor)  1 mol H 2 = 2.0 g H 2 (molar mass)  1 mol NH 3 = 17.0 g NH 3 (molar mass)


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