Unit 2: Introduction to Stoichiometry

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Mind Map on Unit 2: Introduction to Stoichiometry, created by Alexia Charlize on 24/02/2017.
Alexia Charlize
Mind Map by Alexia Charlize, updated more than 1 year ago
Alexia Charlize
Created by Alexia Charlize over 8 years ago
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Unit 2: Introduction to Stoichiometry
  1. Composition stoichimotry
    1. deals with the mass relationships of elements in compounds.
    2. Ratios of substances in chemical reactions can be used as conversion factors
      1. the reaction stoichiometry problems in this chapter can be classified according to the information given in the problem and the information you are expected to find, the unknown.
        1. The given and the unknown may both be reactants or products, as well as one can be a reactant and the other can be a product. The masses are generally expressed in grams, but they can also be expressed as kg or mg.
          1. Problem type 1: Given and Unknown quantities are amounts in moles.
            1. When you are given the amount of a substance in moles and asked to calculate the amount in moles of another substance in the chemical reaction, the general plan is = amount of given substance (mol) -> amount of unknown substance (mol).
            2. Problem type 2: Given is an amount in moles and unknown is a mass that is often expressed in grams.
              1. 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 = amount of given substance (mol)-> amount of unknown substance (mol) -> mass of unknown substance (g)
              2. Problem type 3: Given is a mass in grams and unknown is an amount in moles.
                1. 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 = mass of given substance (g) -> amount of given substance (mol) -> amount of unknown substance (mol) -> mass of unknown substance (g).
                2. Problem type 4: Given is a mass in grams and unknown is a mass in grams.
                  1. When you are given the mass of one substance and asked to calculate the mass of another substance in the chemical reaction, the general plan is = mass of given substance (g) -> amount of given substance (mol) -> amount of unknown substance (mol) -> mass of unknown substance (g).
              3. Stoichiometric problems are solved by using ratios from the balanced equation to convert the given quantity.
              4. Reaction stochiometry
                1. involves the mass relationships between reactants and products in a chemical reaction
                  1. based on chemical equations and the law of conservation of mass. all reaction stoichiometry calculations start with a balanced chemical equation. this equation gives the relative numbers of moles of reactants and products
                  2. Mole Ratio
                    1. 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.
                      1. A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. This information is obtained directly from the balanced chemical equation.
                    2. Molar Mass
                      1. molar mass is the mass, in grams, of one mole of substance. The molar mass is the conversion factor that relates the mass of a substance to the amount in moles of that substance. To solve reaction stoichiometry problems, you will need to determine molar masses using the periodic table.
                        1. returning to the previous example, the decomposition of aluminium oxide, the rounded masses from the periodic table are the following:
                          1. 1 mol Al2O3 = 101.96 g
                            1. 1 mol Al = 26.98 g
                              1. these molar masses can be expressed by the following conversion factors.
                                1. 101.96 g Al2O3/ 1 mol Al2O3 OR 1 mol Al2O3/ 101.96 g Al203
                                  1. 26.98 g Al/ 1 mol Al OR 1 mol Al / 26.98 g
                                    1. 32.00g O2 / 1 mol O2 / 1 mol O2 / 32.00 g O2
                                  2. 1 mol O2 = 32.00 g
                                  3. To find the number of grams of aluminum equivalent to 26.0 mol of aluminium, the calculation would be as follows: 26.0 mol Al X 26.98 G Al/ 1 mol Al = 701 g Al
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