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Chapter 9: Stoichiometry Coach Kelsoe Chemistry Pages 298–325 Section 9-1: Introduction to Stoichiometry Coach Kelsoe Chemistry Pages 299–301 Section 9-1 Objectives n Define stoichiometry n Describe the importance of the mole ratio in stoichiometric calculations. n Write a mole ratio relating two substances in a chemical equation Stoichi-what? n Stoichiometry (pronounced “sto-key-ometry”) comes from two Greek words: n Stoicheion means “element” n Metron means “measure” Introduction to Stoichiometry n A lot of our knowledge of chemistry is based on the careful quantitative analysis of substances involved in chemical reactions. n Composition stoichiometry deals with the mass relationships of elements in compounds. n Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction. Introduction to Stoichiometry n The good news: this is not entirely new material. We’ve already learned about composition stoichiometry in Section 3-2. We just never called it that! n Reaction stoichiometry is based on chemical equations and the law of conservation of matter. n ALL reaction stoichiometry calculations start with a balanced equation. Reaction-Stoichiometry Problems n The problems in this chapter can be classified according to the information given in the problem and the information you are expected to find, called the unknown. n The given and unknown may both be reactants, they may both be products, or one may be a reactant and the other a product. n Masses should be expressed in grams. n Stoichiometric problems are solved by using ratios from the balanced equation to convert the given quantity using the following methods. Reaction-Stoichiometry Problems n Problem Type 1: Given and unknown quantities are in moles. n When you are given the amount of substance in moles and asked to calculate the amount in moles of another substance in the chemical reaction, the general plan is n Amount of given Amount of unknown substance (in mol) substance (in mol) Reaction-Stoichiometry Problems n Problem Type 2: Given is an amount in moles and the unknown is a mass that is often expressed in grams. n When you are given the amount in moles of one substance and asked to calculate the mass of another substance in the chemical reaction, the general plan is n Amount of Amount of Mass of unknown given substance unknown substance substance (in mol) (in mol) (in grams) Reaction-Stoichiometry Problems n Problem Type 3: Given is a mass in grams and the unknown is an amount in moles. n When you are given the mass of one substance and asked to calculate the amount in moles of another substance in the chemical reaction, the general plan is n Mass of Amount of Amount of unknown given substance given substance substance (in grams) (in mol) (in mol) Reaction-Stoichiometry Problems n Problem Type 4: Given is a mass in grams and the unknown is a mass in grams. n When you are given the mass of one substance and asked to calculate the mass of another substance in a chemical reaction, the general plan is n Mass of Amount of Amount of Mass of given given unknown unknown substance subtance substance substance (in grams) (in mol) (in mol) (in grams) Mole Ratio n Solving any reaction-stoichiometry problem requires the use of a mole ratio to convert from moles or grams of one substance in a reaction to moles or grams of another substance. n A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. It is obtained directly from the balanced chemical equation. Mole Ratio n Consider the following equation: 2Al2O3(l) à 4Al(s) + 3O2(g) n Remember that the coefficients in a chemical equations satisfy the law of conservation of matter and represent the relative amounts of moles of reactants and products. n Therefore, 2 mol of aluminum oxide decompose to produce 4 mol of aluminum and 3 mol of oxygen gas. See board for the six ratios. Mole Ratio n For the decomposition of aluminum oxide, the appropriate mole ratio would be used as a conversion factor to convert a given amount in moles of one substance to the corresponding amount in moles of another substance. n For example, if given 13 mol of aluminum oxide and need to find out how much oxygen it decomposes into: 13.0 mol Al2O3 x 3 mol O2 = 19.5 mol O2 2 mol Al2O3 Mole Ratio n Think about this in terms of making a PB&J sandwich: 4 slices of bread + 1 cup PB + 1 cup J à 2 PB&J sandwiches n Let’s assume this equation is balanced. n If we have 16 slices of bread, how much PB will we need? n 16 bread x 1 cup PB = 4 cups PB 4 Bread n We can do this for any ingredient! Mole Ratio n Mole ratios are exact, so they do not limit the number of significant figures in a calculation. n The number of significant figures in the answer is therefore determined only by the number of significant figures of any measured quantities in a particular problem. n In layman’s terms, you should have the same number of sig figs when you multiply by your mole ratio as you had in your given amount. Molar Mass n Remember that the molar mass is the mass, in grams, of one mole of a substance. n The molar mass is the conversion factor that relates the mass of a substance to the amount in moles of that substance. n To solve reaction-stoichiometry problems, you will need to determine the molar masses using the periodic table. Molar Mass n Returning to the aluminum oxide problem: 2Al2O3(l) à 4Al(s) + 3O2(g) n The molar masses can be expressed as so: Aluminum 101.96 g 1 mol Al2O3 n 1 mol Al2O3 OR 101.96 g oxide n Aluminum 26.98 g OR 1 mol Al 1 mol Al 26.98 g 32.00 g OR 1 mol O2 n Oxygen 1 mol O2 32.00 g Molar Mass n To find the number of grams of aluminum equivalent to 26.0 mol of aluminum, the calculation would be as follows: n 26.0 mol Al x 26.98 g Al = 701 g Al 1 mol Al Vocabulary n Composition stoichiometry n Mole ratio n Reaction stoichiometry

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posted: | 7/18/2013 |

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