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Stoichiometry: The study of quantitative measurements in chemical formulas and reactions Chemistry – Mrs. Cameron

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**Review of Chemical Equations**

The Law of Conservation of Mass applies to all chemical equations. All equations must be balanced to have the same number and type of atoms on both the product and reactant sides of an equation. Equations are balanced by adding coefficients in front of compound formulas in chemical equations.

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**Consider the reaction of ethylene with oxygen:**

C2H O2 → 2CO2 + 2H2O Can be read as: 1 molecule C2H4 + 3 molecules O2 → 2 molecules CO2 + 2 molecules H2O OR: 1 mole C2H4 + 3 moles O2 → 2 moles CO2 + 2 moles H2O

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C2H O2 → 2CO2 + 2H2O The coefficients in a chemical equation provide the ratio in which moles or molecules of one substance react with moles or molecules of another. The coefficients in a chemical equation provide the ratio in which moles or molecules of reactants relate to the moles or molecules of the product(s).

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C2H O2 → 2CO2 + 2H2O If the coefficients represent moles rather than molecules, everything in the equation is enlarged by a factor of Avogadro’s #. For example: Every 3 moles of O2 require 1 mole of C2H4 Every 3 moles of O2 produce 2 moles of CO2

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C2H O2 → 2CO2 + 2H2O The Coefficients provide the relational quantities of reactant(s) and product(s). Within an equation, they can be used to form a “molar ratio” between any two substances in the reaction. Every 3 moles of O2 produce 2 moles of CO2 OR

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C2H O2 → 2CO2 + 2H2O Stoichiometry uses larger or smaller quantities – like cutting recipe in half or doubling or tripling a recipe Stoichiometry allows for calculations of multiples of the standard equations.

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Molar Ratio (63.4 g O2) = 58.1 g CO2

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**Calculating volume of a gas from a mass of another reactant or product.**

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**Simplification of Volume-Volume Problems**

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**Applications of Stoichiometry Problems**

Limiting Reactants The limiting reactant “limits” the amount of product that can be formed It is related to the molar ratio like the measurements in a recipe The limiting reactant is totally consumed or “used up” in a chemical reaction Excess Reactants - the reactant left over.

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N2(g) + 3H2(g) 2NH3(g) In this reaction, nitrogen reacts with hydrogen to form ammonia. For every one molecule (or mole) of nitrogen, you need 3 molecules (or moles) of hydrogen. Since there are only enough hydrogen molecules (3 sets of 3) to make 3 “recipes” of the ammonia, and not enough to make another, the hydrogen is the limiting reactant.

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N2(g) + 3H2(g) 2NH3(g) In this reaction, nitrogen reacts with hydrogen to form ammonia. For every one molecule (or mole) of nitrogen, you need 3 molecules (or moles) of hydrogen. Since there are two molecules of nitrogen left over, it is the excess reactant.

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**Calculating Limiting Reactant Problems**

Balance the equation Complete a mass-mass problem for each reactant and the same product. Whichever reactant makes the least product is the limiting reactant. Whichever reactant makes the most product will have some leftover, or is the excess reactant.

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Example: C2H O2 → 2CO2 + 2H2O Ethylene reacts with oxygen to produce carbon dioxide and water. If 8.0 g of ethylene and 16.3 g of oxygen react, how many grams of water can be produced? 8.00 g C2H4 = 10.3 g H2O 16.3 g O2 = g H2O Since the oxygen produces less product, it is the limiting reactant. The ethylene is the excess reactant. From this basic calculation, a chemist could also calculate how much ethylene was left over.

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**Percent Yield Problems**

Theoretical Yield The maximum amount of a given product that can be formed when the limiting reactant is completely consumed. The actual yield (amount produced) of a reaction is usually less than the maximum expected (theoretical yield). Percent Yield The actual amount of a given product as the percentage of the theoretical yield.

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**Percent Yield Problems**

An example: In the reaction above, it was found that only 5.20 grams of water were produced. What is the percent yield of this reaction? X 100 % = 85.1% yield

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Calculate empirical formula from mass percent : Find the molecular formula of a compound has 20 % H, 80 % C, if its Mw = 30 g/mol.

Calculate empirical formula from mass percent : Find the molecular formula of a compound has 20 % H, 80 % C, if its Mw = 30 g/mol.

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