{"id":694,"date":"2018-03-20T15:23:44","date_gmt":"2018-03-20T15:23:44","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/?post_type=chapter&p=694"},"modified":"2018-08-14T01:31:21","modified_gmt":"2018-08-14T01:31:21","slug":"6-5-mole-mass-and-mass-mass-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-monroecc-orgbiochemistry\/chapter\/6-5-mole-mass-and-mass-mass-problems\/","title":{"raw":"6.5 Mole-Mass and Mass-Mass Problems","rendered":"6.5 Mole-Mass and Mass-Mass Problems"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  1. Convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWe have established that a balanced chemical equation is balanced in terms of moles as well as atoms or molecules. We have used balanced equations to set up ratios, now in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that ability to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:\r\n\r\n<\/div>\r\n<\/div>\r\n
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    Collectively, these conversions are called mole-mass calculations<\/span><\/strong><\/span>.<\/p>\r\n

    As an example, consider the balanced chemical equation<\/p>\r\n

    Fe2<\/sub>O3<\/sub> + 3SO3<\/sub> \u2192 Fe2<\/sub>(SO4<\/sub>)3<\/sub><\/span><\/span><\/p>\r\n

    If we have 3.59 mol of Fe2<\/sub>O3<\/sub>, how many grams of SO3<\/sub> can react with it? Using the mole-mass calculation sequence, we can determine the required mass of SO3<\/sub> in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO3<\/sub> needed. Then using the molar mass of SO3<\/sub> as a conversion factor, we determine the mass that this number of moles of SO3<\/sub> has.<\/p>\r\n

    The first step resembles the exercises we did in Section 6.4 \"Mole-Mole Relationships in Chemical Reactions\"<\/a>. As usual, we start with the quantity we were given:<\/p>\r\n\r\n<\/div>\r\n

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    [latex]3.59\\cancel{\\text{ mol Fe}_2\\text{O}_3}\\times\\frac{3\\text{ mol SO}_3}{1\\cancel{\\text{ mol Fe}_2\\text{O}_3}}[\/latex] = 10.8 mol SO3<\/sub><\/p>\r\n

    The mol Fe2<\/sub>O3<\/sub> units cancel, leaving mol SO3<\/sub> unit, and based on 3 sig figs in the 3.59 mol SO3<\/sub>, the answer was rounded to 3 sig figs.\u00a0 As you will see soon, such intermediate rounding introduces rounding error in the final answer.<\/p>\r\n

    Next convert this answer to grams of SO3<\/sub>, using the molar mass of SO3<\/sub> as the conversion factor.<\/p>\r\n

    [latex]10.8\\cancel{\\text{ mol SO}_3}\\times\\frac{80.06\\text{ g SO}_3}{1\\cancel{\\text{ mol SO}_3}}[\/latex] = 865 g SO3<\/sub><\/p>\r\n

    The answer to this step is expressed to three significant figures. Thus, in a two-step process, we find that 865 g of SO3<\/sub> will react with 3.59 mol of Fe2<\/sub>O3<\/sub>. Many problems of this type can be answered in this manner.<\/p>\r\n

    To avoid rounding error, it is better to do multi-step problems all at once, keeping intermediate answers in the calculator and rounding only once at the end.:<\/p>\r\n\r\n

    \"image\"<\/div>\r\n

    Notice that, when rounding after each step, the answer was 865 g SO3<\/sub>.\u00a0 But when the calculator was used to hold onto all of the decimal places of the intermediate number, the answer came out to 862 g SO3<\/sub>.\u00a0 This should call to mind the fact that the last digit of a measured number or of a calculated number based on measurements is an uncertain estimate.\u00a0 However, the most accurate answer is the one obtained by doing all of the steps at once and rounding for the correct number of sig figs at the very end.\u00a0 In this case the most accurate answer is 862 g SO3<\/sub>.<\/p>\r\n\r\n

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    Example 8<\/h3>\r\n

    How many grams of CO2<\/sub> are produced if 2.09 mol of HCl are reacted according to this balanced chemical equation?<\/p>\r\nCaCO3<\/sub> + 2HCl \u2192 CaCl2<\/sub> + CO2<\/sub> + H2<\/sub>O<\/span><\/span>\r\n

    Solution<\/p>\r\n

    Our strategy will be to convert from moles of HCl to moles of CO2<\/sub> and then from moles of CO2<\/sub> to grams of CO2<\/sub>. We will need the molar mass of CO2<\/sub>, which is 44.01 g\/mol. Performing these two conversions in a single-line gives 46.0 g of CO2<\/sub>:<\/p>\r\n\r\n

    \"image\"<\/div>\r\n

    The molar ratio between CO2<\/sub> and HCl comes from the balanced chemical equation.<\/p>\r\n\r\n<\/div>\r\n

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    Skill-Building Exercise<\/h3>\r\n
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      How many grams of glucose (C6<\/sub>H12<\/sub>O6<\/sub>) are produced if 17.3 mol of H2<\/sub>O are reacted according to this balanced chemical equation?<\/p>\r\n6CO2<\/sub> + 6H2<\/sub>O \u2192 C6<\/sub>H12<\/sub>O6<\/sub> + 6O2<\/sub><\/span><\/span>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n

      It is a small step from mole-mass calculations to mass-mass calculations<\/span><\/strong><\/span>. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance\u2019s molar mass as the conversion factor. Then\u2014and only then\u2014we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:<\/p>\r\n\r\n

      \"image\"<\/div>\r\n

      This three-part process should be combined into a single calculation that contains three conversion factors as shown in the following example.<\/p>\r\n\r\n

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      Example 9<\/h3>\r\n

      Methane can react with elemental chlorine to make carbon tetrachloride (CCl4<\/sub>). The balanced chemical equation is as follows:<\/p>\r\n

      CH4<\/sub> + 4Cl2<\/sub> \u2192 CCl4<\/sub> + 4HCl<\/span><\/span><\/p>\r\n

      How many grams of HCl are produced by the reaction of 100.0 g of CH4<\/sub>?<\/p>\r\n

      Solution<\/p>\r\n

      The first conversion factor converts the mass of CH4<\/sub> to moles of CH4<\/sub>, using the molar mass of CH4<\/sub> (16.05 g\/mol).\u00a0 Since the unit \"g CH4<\/sub>\"\u00a0 must cancel, the molar mass is inverted to create the first conversion factor:<\/p>\r\n

      [latex]\\frac{1\\text{ mol CH}_4}{16.05\\text{ g CH}_4}[\/latex]<\/p>\r\n

      The second conversion must bridge from mol CH4<\/sub> to mol HCl. \u00a0 The coefficients in the balanced chemical equation represent the ratio of moles HCl and moles CH4, <\/sub>with mol CH4<\/sub> in the denominator so it will cancel:<\/p>\r\n

      [latex]\\frac{4\\text{ mol HCl}}{1\\text{ mol CH}_4}[\/latex]<\/p>\r\n

      The third and final conversion factor uses the molar mass of HCl (36.46 g\/mol) , this time without inversion because mol HCl must be cancelled:<\/p>\r\n

      [latex]\\frac{36.46\\text{ g HCl}}{1\\text{ mol HCl}}[\/latex]<\/p>\r\n

      Combining the three conversion factors into a single, multi-step problem:<\/p>\r\n

      [latex]100.0\\cancel{\\text{ g CH}_4}\\times{\\frac{1\\text{ mol CH}_4}{16.05\\cancel{\\text{ g CH}_4}}}\\times{\\frac{4\\text{ mol HCl}}{1\\cancel{\\text{ mol CH}_4}}}\\times{\\frac{36.46\\text{ g HCl}}{1\\cancel{\\text{ mol HCl}}}}[\/latex] = 908.7 g HCl<\/p>\r\n

      There were 4 sig figs in the measured number 100.0 g CH4<\/sub> and the molar masses each had 4 sig figs as well.\u00a0 The whole numbers from the equation coefficients are exact numbers and do not limit the sig figs.\u00a0 Therefore, the final answer is rounded to have 4 sig figs.<\/p>\r\n\r\n<\/div>\r\n

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      Skill-Building Exercise<\/h3>\r\n
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        The oxidation of propanal (CH3<\/sub>CH2<\/sub>CHO) to propionic acid (CH3<\/sub>CH2<\/sub>COOH) has the following balanced chemical equation:<\/p>\r\n3CH3<\/sub>CH2<\/sub>CHO + K2<\/sub>Cr2<\/sub>O7<\/sub> +4H2<\/sub>SO4<\/sub> \u2192 3CH3<\/sub>CH2<\/sub>COOH + Cr2<\/sub>(SO4<\/sub>)3<\/sub> + 4H2<\/sub>O + K2<\/sub>SO4<\/sub>\r\n<\/span><\/span>\r\n

        How many grams of propionic acid are produced by the reaction of 135.8 g of K2<\/sub>Cr2<\/sub>O7<\/sub>?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

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        Concept Review Exercises<\/h3>\r\n
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          What is the general sequence of conversions for a mole-mass calculation?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

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          What is the general sequence of conversions for a mass-mass calculation?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n

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          Answers<\/h3>\r\n[reveal-answer q=\"275842\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"275842\"]\r\n
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          1. mol first substance \u2192 mol second substance \u2192 mass second substance<\/li>\r\n \t
          2. mass first substance \u2192 mol first substance \u2192 mol second substance \u2192 mass second substance[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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            To Your Health: The Synthesis of Taxol<\/h3>\r\n

            Taxol is a powerful anticancer drug that was originally extracted from the Pacific yew tree (Taxus brevifolia<\/em>). As you can see from the accompanying figure, taxol is a very complicated molecule, with a molecular formula of C47<\/sub>H51<\/sub>NO14<\/sub>. Isolating taxol from its natural source presents certain challenges, mainly that the Pacific yew is a slow-growing tree, and the equivalent of six trees must be harvested to provide enough taxol to treat a single patient. Although related species of yew trees also produce taxol in small amounts, there is significant interest in synthesizing this complex molecule in the laboratory.<\/p>\r\n

            After a 20-year effort, two research groups announced the complete laboratory synthesis of taxol in 1994. However, each synthesis required over 30 separate chemical reactions, with an overall efficiency of less than 0.05%. To put this in perspective, to obtain a single 300 mg dose of taxol, you would have to begin with 600 g of starting material. To treat the 26,000 women who are diagnosed with ovarian cancer each year with one dose, almost 16,000 kg (over 17 tons) of starting material must be converted to taxol. Taxol is also used to treat breast cancer, with which 200,000 women in the United States are diagnosed every year. This only increases the amount of starting material needed.<\/p>\r\n

            Clearly, there is intense interest in increasing the overall efficiency of the taxol synthesis. An improved synthesis not only will be easier but also will produce less waste materials, which will allow more people to take advantage of this potentially life-saving drug.<\/p>\r\n\r\n

            \r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"1525\"]\"image\" Figure 6.4 The Structure of the Cancer Drug Taxol. <\/em>Because of the complexity of the molecule, hydrogen atoms are not shown, but they are present on every atom to give the atom the correct number of covalent bonds (four bonds for each carbon atom).[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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            Key Takeaway<\/h3>\r\n<\/div>\r\n
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