{"id":620,"date":"2018-03-20T15:00:27","date_gmt":"2018-03-20T15:00:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/?post_type=chapter&p=620"},"modified":"2018-08-10T18:24:59","modified_gmt":"2018-08-10T18:24:59","slug":"5-3-quantitative-relationships-based-on-chemical-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-monroecc-orgbiochemistry\/chapter\/5-3-quantitative-relationships-based-on-chemical-equations\/","title":{"raw":"5.3 Quantitative Relationships Based on Chemical Equations","rendered":"5.3 Quantitative Relationships Based on Chemical Equations"},"content":{"raw":"
\r\n
\r\n
\r\n

Learning Objectives<\/h3>\r\n
\r\n
\r\n
\r\n
    \r\n \t
  1. Calculate the amount of one substance that will react with or be produced from a given amount of another substance.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n

    A balanced chemical equation not only describes some of the chemical properties of substances\u2014by showing us what substances react with what other substances to make what products\u2014but also shows numerical relationships between the reactants and the products. The study of these numerical relationships is called stoichiometry<\/span><\/span>. The stoichiometry of chemical equations revolves around the coefficients in the balanced chemical equation because these coefficients determine the molecular ratio in which reactants react and products are made.<\/p>\r\n\r\n

    \r\n
    \r\n

    Note<\/h3>\r\n

    The word stoichiometry<\/em> is pronounced \u201cstow-eh-key-OM-et-tree.\u201d It is of mixed Greek and English origins, meaning roughly \u201cmeasure of an element.\u201d<\/p>\r\n\r\n<\/div>\r\n

    \r\n
    \r\n\r\nLooking Closer: Stoichiometry in Cooking<\/span>\r\n\r\n<\/div>\r\n
    \r\n

    Let us consider a stoichiometry analogy from the kitchen. A recipe that makes 1 dozen biscuits needs 2 cups of flour, 1 egg, 4 tablespoons of shortening, 1 teaspoon of salt, 1 teaspoon of baking soda, and 1 cup of milk. If we were to write this as a chemical equation, we would write<\/p>\r\n2 c flour + 1 egg + 4 tbsp shortening + 1 tsp salt + 1 tsp baking soda + 1 c milk \u2192 12 biscuits<\/span><\/span>\r\n

    (Unlike true chemical reactions, this one has all 1 coefficients written explicitly\u2014partly because of the many different units here.) This equation gives us ratios of how much of what reactants are needed to make how much of what product. Two cups of flour, when combined with the proper amounts of the other ingredients, will yield 12 biscuits. One teaspoon of baking soda (when also combined with the right amounts of the other ingredients) will make 12 biscuits. One egg must be combined with 1 cup of milk to yield the product food. Other relationships can also be expressed.<\/p>\r\n

    We can use the ratios we derive from the equation for predictive purposes. For instance, if we have 4 cups of flour, how many biscuits can we make if we have enough of the other ingredients? It should be apparent that we can make a double recipe of 24 biscuits.<\/p>\r\n

    But how would we find this answer formally, that is, mathematically? We would set up a conversion factor, much like we did in Chapter 1 \"Chemistry, Matter, and Measurement\"<\/a>. Because 2 cups of flour make 12 biscuits, we can set up an equivalency ratio:<\/p>\r\n[latex]\\frac{\\text{12 biscuits}}{\\text{2 c flour}}[\/latex]<\/span>\r\n

    We then can use this ratio in a formal conversion of flour to biscuits:<\/p>\r\n [latex]\\text{4c flour}\\times\\frac{\\text{12 biscuits}}{\\text{2c flour}}=\\text{24 biscuits}[\/latex] <\/span>\r\n

    Similarly, by constructing similar ratios, we can determine how many biscuits we can make from any amount of ingredient.<\/p>\r\n

    When you are doubling or halving a recipe, you are doing a type of stoichiometry. Applying these ideas to chemical reactions should not be difficult if you use recipes when you cook.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    Consider the following balanced chemical equation:<\/p>\r\n

    2C2<\/sub>H2<\/sub> + 5O2<\/sub> \u2192 4CO2<\/sub> + 2H2<\/sub>O<\/span><\/span><\/p>\r\n

    The coefficients on the chemical formulas give the ratios in which the reactants combine and the products form. Thus, we can make the following statements and construct the following ratios:<\/p>\r\n\r\n

    \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Statement from the Balanced Chemical Reaction<\/th>\r\nRatio<\/th>\r\nInverse Ratio<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    two C2<\/sub>H2<\/sub> molecules react with five O2<\/sub> molecules<\/td>\r\n[latex]\\frac{2C_{2}H_{2}}{5O_{2}}[\/latex]<\/span><\/td>\r\n [latex]\\frac{5O_{2}}{2C_{2}H_{2}}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n
    two C2<\/sub>H2<\/sub> molecules react to make four CO2<\/sub> molecules<\/td>\r\n [latex]\\frac{2C_{2}H_{2}}{4CO_{2}}[\/latex] <\/span><\/td>\r\n [latex]\\frac{4CO_{2}}{2C_{2}H_{2}}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n
    five O2<\/sub> molecules react to make two H2<\/sub>O molecules<\/td>\r\n [latex]\\frac{5O_{2}}{2H_{2}O}[\/latex]<\/span><\/td>\r\n [latex]\\frac{2H_{2}O}{5O_{2}}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n
    four CO2<\/sub> molecules are made at the same time as two H2<\/sub>O molecules<\/td>\r\n [latex]\\frac{2H_{2}O}{4CO_{2}}[\/latex]<\/span><\/td>\r\n [latex]\\frac{4CO_{2}}{2H_{2}O}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n

    Other relationships are possible; in fact, 12 different conversion factors can be constructed from this balanced chemical equation.\u00a0 Notice that the numbers in the rations have not been reduced to lowest terms in order to make it clear that these numbers\u00a0are<\/em> the coefficients from the balanced equation. In each ratio, the unit is assumed to be molecules because that is how we are interpreting the chemical equation.\u00a0 In future sections, the unit will be moles.<\/p>\r\n

    Any of these fractions can be used as a conversion factor to relate an amount of one substance to an amount of another substance. For example, suppose we want to know how many CO2<\/sub> molecules are formed when 26 molecules of C2<\/sub>H2<\/sub> are reacted. As usual with a conversion problem, we start with the amount we are given\u201426C2<\/sub>H2<\/sub>\u2014and multiply it by a conversion factor that cancels out our original unit and introduces the unit we are converting to\u2014in this case, CO2<\/sub>. That conversion factor is [latex]\\frac{4CO_{2}}{2C_{2}H_{2}}[\/latex], <\/span> which is composed of terms that come directly from the balanced chemical equation. Thus, we have<\/p>\r\n

    \u00a0 [latex]26C_2H_2(\\frac{4CO_{2}}{2C_{2}H_{2}})[\/latex]<\/span><\/span><\/p>\r\n

    The C2<\/sub>H2<\/sub> labels cancel leaving<\/p>\r\n

    [latex]26\\left ( \\frac{4CO_{2}}{2} \\right )[\/latex]<\/span><\/span><\/p>\r\n

    Thus, 52 molecules of CO2<\/sub> are formed.\u00a0 Notice that the word molecules was assumed but not written in the original labels.\u00a0 Stoichiometry is most valuable when it is scaled up from molecules, which we cannot see or weigh, to moles.\u00a0 Moles <\/strong>will be discussed in Chapter 6, at which time the equation coefficients will be interpreted as moles of the different substances, rather than molecules.<\/p>\r\n

    This application of stoichiometry is extremely powerful in its predictive ability, as long as we begin with a balanced chemical equation. Without a balanced chemical equation, the predictions made by simple stoichiometric calculations will be incorrect.<\/p>\r\n\r\n

    \r\n

    Example 2<\/h3>\r\n

    Start with this balanced chemical equation.<\/p>\r\nKMnO4<\/sub> + 8HCl + 5FeCl2<\/sub> \u2192 5 FeCl3<\/sub> + MnCl2<\/sub> + 4H2<\/sub>O + KCl<\/span><\/span>\r\n

      \r\n \t
    1. Verify that the equation is indeed balanced.<\/li>\r\n \t
    2. Give 2 ratios that give the relationship between HCl and FeCl3<\/sub>.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"906280\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"906280\"]\r\n
        \r\n \t
      1. Each side has 1 K atom and 1 Mn atom. The 8 molecules of HCl yield 8 H atoms, and the 4 molecules of H2<\/sub>O also yield 8 H atoms, so the H atoms are balanced. The Fe atoms are balanced, as we count 5 Fe atoms from 5 FeCl2<\/sub> reactants and 5 FeCl3<\/sub> products. As for Cl, on the reactant side, there are 8 Cl atoms from HCl and 10 Cl atoms from the 5 FeCl2<\/sub> formula units, for a total of 18 Cl atoms. On the product side, there are 15 Cl atoms from the 5 FeCl3<\/sub> formula units, 2 from the MnCl2<\/sub> formula unit, and 1 from the KCl formula unit. This is a total of 18 Cl atoms in the products, so the Cl atoms are balanced. All the elements are balanced, so the entire chemical equation is balanced.<\/li>\r\n \t
      2. Because the balanced chemical equation tells us that 8 HCl molecules react to make 5 FeCl3 formula units, we have the following 2 ratios: [latex]\\frac{8HCl}{5FeCl_{3}}[\/latex] <\/span> and [latex]\\frac{5FeCl_{3}}{8HCl}[\/latex]<\/span>. There are a total of 42 possible ratios. Can you find the other 40 relationships?[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n
        \r\n
        \r\n

        Skill-Building Exercise<\/h3>\r\n

        Start with this balanced chemical equation.<\/p>\r\n3 CH2<\/sub>=CH2<\/sub>(g) + 2 KMnO4<\/sub> (aq) + 4 H2<\/sub>O (l) \u2192 3 CH2<\/sub>OHCH2<\/sub>OH (aq) + 2 MnO2<\/sub> (s) + 2 KOH (aq)\r\n

          \r\n \t
        1. \r\n
          \r\n

          Verify that the equation is balanced.<\/p>\r\n\r\n<\/div><\/li>\r\n \t

        2. \r\n
          \r\n

          Give 2 ratios that give the relationship between KMnO4<\/sub> and CH2<\/sub>=CH2<\/sub>. (A total of 30 relationships can be constructed from this chemical equation. Can you find the other 28?)<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

          \r\n
          \r\n
          \r\n

          Concept Review Exercises<\/h3>\r\n
            \r\n \t
          1. \r\n
            \r\n

            Explain how stoichiometric ratios are constructed from a chemical equation.<\/p>\r\n\r\n<\/div><\/li>\r\n \t

          2. \r\n
            \r\n

            Why is it necessary for a chemical equation to be balanced before it can be used to construct conversion factors?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n

            \r\n
            \r\n

            [reveal-answer q=\"379011\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"379011\"]<\/p>\r\n\r\n

              \r\n \t
            1. Stoichiometric ratios are made using the coefficients of the substances in the balanced chemical equation.<\/li>\r\n \t
            2. A balanced chemical equation is necessary so the ratios from the coefficients are true..[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
              \r\n
              \r\n
              \r\n

              Key Takeaway<\/h3>\r\n