{"id":7586,"date":"2021-06-08T21:56:56","date_gmt":"2021-06-08T21:56:56","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/concentrations-as-conversion-factors\/"},"modified":"2021-10-04T22:06:44","modified_gmt":"2021-10-04T22:06:44","slug":"concentrations-as-conversion-factors","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/concentrations-as-conversion-factors\/","title":{"raw":"Concentrations as Conversion Factors","rendered":"Concentrations as Conversion Factors"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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  1. Apply concentration units as conversion factors.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nConcentration can be a conversion factor between the amount of solute and the amount of solution or solvent (depending on the definition of the concentration unit). As such, concentrations can be useful in a variety of stoichiometry problems. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor.\r\n\r\nA simple example of using a concentration unit as a conversion factor is one in which we use the definition of the concentration unit and rearrange; we can do the calculation again as a unit conversion, rather than as a definition. For example, suppose we ask how many moles of solute are present in 0.108 L of a 0.887 M NaCl solution. Because 0.887 M means 0.887 mol\/L, we can use this second expression for the concentration as a conversion factor:\r\n

    [latex]0.108\\text{ \\cancel{L NaCl}}\\times\\dfrac{0.887\\text{ mol NaCl}}{1\\text{ \\cancel{L NaCl}}}=0.0958\\text{ mol NaCl}[\/latex]<\/p>\r\n(There is an understood 1 in the denominator of the conversion factor.) If we used the definition approach, we get the same answer, but now we are using conversion factor skills. Like any other conversion factor that relates two different types of units, the reciprocal of the concentration can be also used as a conversion factor.\r\n

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    Example 11.6<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nUsing concentration as a conversion factor, how many litres of 2.35 M CuSO4<\/sub> are needed to obtain 4.88 mol of CuSO4<\/sub>?\r\n\r\nSolution<\/em>\r\n\r\nThis is a one-step conversion, but the concentration must be written as the reciprocal for the units to work out:\r\n

    [latex]4.88\\cancel{\\text{ mol }\\ce{CuSO4}}\\times\\dfrac{1\\text{ L}}{2.35\\cancel{\\text{ mol }\\ce{CuSO4}}}=2.08\\text{ L of solution}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\n\r\nUsing concentration as a conversion factor, how many litres of 0.0444 M CH2<\/sub>O are needed to obtain 0.0773 mol of CH2<\/sub>O?\r\n\r\nAnswer<\/em>\r\n\r\n1.74 L\r\n\r\n<\/div>\r\n<\/div>\r\nOf course, once quantities in moles are available, another conversion can give the mass of the substance, using molar mass as a conversion factor.\r\n

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    Example 11.7<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat mass of solute is present in 0.765 L of 1.93 M NaOH?\r\n\r\nSolution<\/em>\r\n\r\nThis is a two-step conversion, first using concentration as a conversion factor to determine the number of moles and then the molar mass of NaOH (40.0 g\/mol) to convert to mass:\r\n

    [latex]0.765\\text{ \\cancel{L}}\\times\\dfrac{1.93\\text{ \\cancel{mol NaOH}}}{\\text{\\cancel{L} solution}}\\times\\dfrac{40.0\\text{ g NaOH}}{1\\text{ \\cancel{mol NaOH}}}=59.1\\text{ g NaOH}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\n\r\nWhat mass of solute is present in 1.08 L of 0.0578 M H2<\/sub>SO4<\/sub>?\r\n\r\nAnswer<\/em>\r\n\r\n6.12 g\r\n\r\n<\/div>\r\n<\/div>\r\nMore complex stoichiometry problems using balanced chemical reactions can also use concentrations as conversion factors. For example, suppose the following equation represents a chemical reaction:\r\n

    2AgNO3<\/sub>(aq) +\u00a0CaCl2<\/sub>(aq) \u2192 2AgCl(s) + Ca(NO3<\/sub>)2<\/sub>(aq)<\/p>\r\nIf we wanted to know what volume of 0.555 M CaCl2<\/sub> would react with 1.25 mol of AgNO3<\/sub>, we first use the balanced chemical equation to determine the number of moles of CaCl2<\/sub> that would react and then use concentration to convert to litres of solution:\r\n

    [latex]1.25\\cancel{\\text{ mol }\\ce{AgNO3}}\\times\\dfrac{1\\cancel{\\text{ mol }\\ce{CaCl2}}}{2\\cancel{\\text{ mol }\\ce{AgNO3}}}\\times\\dfrac{1\\text{ L solution}}{0.555\\cancel{\\text{ mol }\\ce{CaCl2}}}=1.13\\text{ L }\\ce{CaCl2}[\/latex]<\/p>\r\nThis can be extended by starting with the mass of one reactant, instead of moles of a reactant.\r\n

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    Example 11.8<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat volume of 0.0995 M Al(NO3<\/sub>)3<\/sub> will react with 3.66 g of Ag according to the following chemical equation?\r\n

    3Ag(s) + Al(NO3<\/sub>)3<\/sub>(aq) \u2192 3AgNO3<\/sub> + Al(s)<\/p>\r\nSolution<\/em>\r\nHere, we first must convert the mass of Ag to moles before using the balanced chemical equation and then the definition of molarity as a conversion factor:\r\n\r\n[latex]\\begin{multline*}\r\n3.66\\text{ \\cancel{g Ag}}\\times\\dfrac{1\\text{ \\cancel{mol Ag}}}{107.97\\text{ \\cancel{g Ag}}}\\times \\dfrac{1\\cancel{\\text{ mol }\\ce{Al(NO3)3}}}{3\\text{ \\cancel{mol Ag}}}\\times \\dfrac{1\\text{ L solution}}{0.0995\\cancel{\\text{ mol }\\ce{Al(NO3)3}}} \\\\ \\\\\r\n=0.114\\text{ L}\\hspace{5mm}\r\n\\end{multline*}[\/latex]\r\n\r\nTest Yourself<\/em>\r\nWhat volume of 0.512 M NaOH will react with 17.9 g of H2<\/sub>C2<\/sub>O4<\/sub>(s) according to the following chemical equation?\r\n

    H2<\/sub>C2<\/sub>O4<\/sub>(s) + 2NaOH(aq) \u2192 Na2<\/sub>C2<\/sub>O4<\/sub>(aq) + 2H2<\/sub>O(\u2113)<\/p>\r\nAnswer<\/em>\r\n0.777 L\r\n\r\n<\/div>\r\n<\/div>\r\nWe can extend our skills even further by recognizing that we can relate quantities of one solution to quantities of another solution. Knowing the volume and concentration of a solution containing one reactant, we can determine how much of another solution of another reactant will be needed using the balanced chemical equation.\r\n

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    Example 11.9<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nA student takes a precisely measured sample, called an aliquot<\/em>, of 10.00 mL of a solution of FeCl3<\/sub>. The student carefully adds 0.1074 M Na2<\/sub>C2<\/sub>O4<\/sub> until all the Fe3+<\/sup>(aq) has precipitated as Fe2<\/sub>(C2<\/sub>O4<\/sub>)3<\/sub>(s). Using a precisely measured tube called a burette, the student finds that 9.04 mL of the Na2<\/sub>C2<\/sub>O4<\/sub> solution was added to completely precipitate the Fe3+<\/sup>(aq). What was the concentration of the FeCl3<\/sub> in the original solution? (A precisely measured experiment like this, which is meant to determine the amount of a substance in a sample, is called a titration<\/em>.) The balanced chemical equation is as follows:\r\n

    2FeCl3<\/sub>(aq) + 3Na2<\/sub>C2<\/sub>O4<\/sub>(aq) \u2192 Fe2<\/sub>(C2<\/sub>O4<\/sub>)3<\/sub>(s) + 6NaCl(aq)<\/p>\r\nSolution<\/em>\r\nFirst we need to determine the number of moles of Na2<\/sub>C2<\/sub>O4<\/sub> that reacted. We will convert the volume to litres and then use the concentration of the solution as a conversion factor:\r\n

    [latex]9.04\\text{ \\cancel{mL}}\\times \\dfrac{1\\text{ \\cancel{L}}}{1000\\text{ \\cancel{mL}}}\\times \\dfrac{0.1074\\text{ mol }\\ce{Na2C2O4}}{\\text{ \\cancel{L}}}=0.000971\\text{ mol }\\ce{Na2C2O4}[\/latex]<\/p>\r\nNow we will use the balanced chemical equation to determine the number of moles of Fe3+<\/sup>(aq) that were present in the initial aliquot:\r\n

    [latex]0.000971\\cancel{\\text{ mol }\\ce{Na2C2O4}}\\times \\dfrac{2\\text{ mol }\\ce{FeCl3}}{3\\cancel{\\text{ mol }\\ce{Na2C2O4}}}=0.000647\\text{ mol }\\ce{FeCl3}[\/latex]<\/p>\r\nThen we determine the concentration of FeCl3<\/sub> in the original solution. Converting 10.00 mL into litres (0.01000 L), we use the definition of molarity directly:\r\n

    [latex]\\text{M}=\\dfrac{\\text{mol}}{\\text{L}}=\\dfrac{0.000647\\text{ mol }\\ce{FeCl3}}{0.01000\\text{ L}}=0.0647\\text{ M }\\ce{FeCl3}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nA student titrates 25.00 mL of H3<\/sub>PO4<\/sub> with 0.0987 M KOH. She uses 54.06 mL to complete the chemical reaction. What is the concentration of H3<\/sub>PO4<\/sub>?\r\n

    H3<\/sub>PO4<\/sub>(aq) + 3KOH(aq) \u2192 K3<\/sub>PO4<\/sub>(aq) + 3H2<\/sub>O<\/p>\r\nAnswer<\/em>\r\n0.0711 M\r\n\r\n[caption id=\"attachment_3251\" align=\"aligncenter\" width=\"400\"]\"When When a student performs a titration, a measured amount of one solution is added to another reactant.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\nWe have used molarity exclusively as the concentration of interest, but that will not always be the case. The next example shows a different concentration unit being used.\r\n

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    Example 11.10<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nH2<\/sub>O2<\/sub> is used to determine the amount of Mn according to this balanced chemical equation:\r\n

    2MnO4<\/sub>\u2212<\/sup>(aq) + 5H2<\/sub>O2<\/sub>(aq) + 6H+<\/sup>(aq) \u2192 2Mn2+<\/sup>(aq) + 5O2<\/sub>(g) + 8H2<\/sub>O(\u2113)<\/p>\r\nWhat mass of 3.00% m\/m H2<\/sub>O2<\/sub> solution is needed to react with 0.355 mol of MnO4<\/sub>\u2212<\/sup>(aq)?\r\n\r\nSolution<\/em>\r\nBecause we are given an initial amount in moles, all we need to do is use the balanced chemical equation to determine the number of moles of H2<\/sub>O2<\/sub> and then convert to find the mass of H2<\/sub>O2<\/sub>. Knowing that the H2<\/sub>O2<\/sub> solution is 3.00% by mass, we can determine the mass of solution needed:\r\n

    [latex]\\begin{multline*}\r\n0.355\\cancel{\\text{ mol }\\ce{MnO4^-}}\\times\\dfrac{5\\cancel{\\text{ mol }\\ce{H2O2}}}{2\\cancel{\\text{ mol }\\ce{MnO4^-}}}\\times\\dfrac{34.02\\cancel{\\text{ g }\\ce{H2O2}}}{1\\cancel{\\text{ mol }\\ce{H2O2}}}\\times\\dfrac{100\\text{ g solution}}{3\\cancel{\\text{ g }\\ce{H2O2}}} \\\\ \\\\\r\n=1006\\text{ g solution}\\hspace{5mm}\r\n\\end{multline*}[\/latex]<\/p>\r\nThe first conversion factor comes from the balanced chemical equation, the second conversion factor is the molar mass of H2<\/sub>O2<\/sub>, and the third conversion factor comes from the definition of percentage concentration by mass.\r\n\r\nTest Yourself<\/em>\r\nUse the balanced chemical reaction for MnO4<\/sub>\u2212<\/sup> and H2<\/sub>O2<\/sub> to determine what mass of O2<\/sub> is produced if 258 g of 3.00% m\/m H2<\/sub>O2<\/sub> is reacted with MnO4<\/sub>\u2212<\/sup>.\r\n\r\nAnswer<\/em>\r\n7.28 g\r\n\r\n<\/div>\r\n<\/div>\r\n

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    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

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