{"id":6368,"date":"2020-05-14T20:24:47","date_gmt":"2020-05-14T20:24:47","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/?post_type=chapter&p=6368"},"modified":"2020-05-14T20:34:19","modified_gmt":"2020-05-14T20:34:19","slug":"gravimetric-analysis-combustion-reactions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/chapter\/gravimetric-analysis-combustion-reactions\/","title":{"raw":"Gravimetric Analysis - Combustion Reactions","rendered":"Gravimetric Analysis – Combustion Reactions"},"content":{"raw":"

Gravimetric Analysis<\/h2>\r\nA gravimetric analysis<\/strong> is one in which a sample is subjected to some treatment that causes a change in the physical state of the analyte that permits its separation from the other components of the sample. Mass measurements of the sample, the isolated analyte, or some other component of the analysis system, used along with the known stoichiometry of the compounds involved, permit calculation of the analyte concentration. Gravimetric methods were the first techniques used for quantitative chemical analysis, and they remain important tools in the modern chemistry laboratory.\r\n\r\nThe required change of state in a gravimetric analysis may be achieved by various physical and chemical processes. For example, the moisture (water) content of a sample is routinely determined by measuring the mass of a sample before and after it is subjected to a controlled heating process that evaporates the water.\r\n
\r\n

Example 2:\u00a0Gravimetric Analysis<\/strong><\/h3>\r\nA 0.4550-g solid mixture containing CaSO4<\/sub> is dissolved in water and treated with an excess of Ba(NO3<\/sub>)2<\/sub>, resulting in the precipitation of 0.6168 g of BaSO4<\/sub>.\r\n

[latex]{\\text{CaSO}}_{4}\\text{(}aq\\text{)}+\\text{Ba}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}\\rightarrow{\\text{BaSO}}_{4}\\text{(}s\\text{)}+\\text{Ca}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}[\/latex]<\/p>\r\nWhat is the concentration (percent) of CaSO4<\/sub> in the mixture?\r\n\r\n[reveal-answer q=\"25361\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"25361\"]\r\n\r\nThe plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of BaSO4<\/sub> and CaSO4<\/sub> through their stoichiometric factor. Once the mass of CaSO4<\/sub> is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.\r\n\r\n\"This\r\n\r\nThe mass of CaSO4<\/sub> that would yield the provided precipitate mass is\r\n

[latex]0.6168\\cancel{{\\text{g BaSO}}_{4}}\\times \\frac{1\\cancel{{\\text{mol BaSO}}_{4}}}{233.43\\cancel{{\\text{g BaSO}}_{4}}}\\times \\frac{1\\cancel{{\\text{mol CaSO}}_{4}}}{1\\cancel{{\\text{mol BaSO}}_{4}}}\\times \\frac{136.14{\\text{g CaSO}}_{4}}{1\\cancel{{\\text{mol CaSO}}_{4}}}=0.3597{\\text{g CaSO}}_{4}[\/latex]<\/p>\r\nThe concentration of CaSO4<\/sub> in the sample mixture is then calculated to be\r\n

[latex]\\begin{array}{l}\\\\ {\\text{percent CaSO}}_{4}=\\frac{{\\text{mass CaSO}}_{4}}{\\text{mass sample}}\\times 100\\%\\\\ \\frac{\\text{0.3597 g}}{0.4550 g}\\times 100\\%=79.05\\%\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n

Check Your Learning<\/strong><\/h4>\r\nWhat is the percent of chloride ion in a sample if 1.1324 g of the sample produces 1.0881 g of AgCl when treated with excess Ag+<\/sup>?\r\n

[latex]{\\text{Ag}}^{+}\\text{(}aq\\text{)}+{\\text{Cl}}^{-}\\text{(}aq\\text{)}\\rightarrow\\text{AgCl}\\text{(}s\\text{)}[\/latex]<\/p>\r\n[reveal-answer q=\"996093\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"996093\"]23.76%[\/hidden-answer]\r\n\r\n<\/div>\r\nThe elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis<\/strong>. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product (Figure\u00a03). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"879\"]\"This Figure\u00a03. This schematic diagram illustrates the basic components of a combustion analysis device for determining the carbon and hydrogen content of a sample.[\/caption]\r\n\r\n

\r\n

Example\u00a03:\u00a0Combustion Analysis<\/strong><\/h3>\r\nPolyethylene is a hydrocarbon polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a 0.00126-g sample of polyethylene yields 0.00394 g of CO2<\/sub> and 0.00161 g of H2<\/sub>O. What is the empirical formula of polyethylene?\r\n\r\n[reveal-answer q=\"511066\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"511066\"]\r\n\r\nThe primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:\r\n

[latex]{\\text{C}}_{\\text{x}}{\\text{H}}_{\\text{y}}\\text{(}s\\text{)}+{\\text{excess O}}_{2}\\text{(}g\\text{)}\\rightarrow x{\\text{CO}}_{2}\\text{(}g\\text{)}+y{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}[\/latex]<\/p>\r\nNote that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts x<\/em> and y<\/em> are needed.\r\n\r\nFirst, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:\r\n\r\n\"\r\n

[latex]\\begin{array}{l}\\\\ \\text{mol C}=0.00394{\\text{g CO}}_{2}\\times \\frac{1{\\text{mol CO}}_{2}}{\\text{44.01 g\/mol}}\\times \\frac{1\\text{mol C}}{1{\\text{mol CO}}_{2}}=8.95\\times {10}^{-5}\\text{mol C}\\\\ \\text{mol H}=0.00161{\\text{g H}}_{2}\\text{O}\\times \\frac{1{\\text{mol H}}_{2}\\text{O}}{18.02\\text{g\/mol}}\\times \\frac{2\\text{mol H}}{1{\\text{mol H}}_{2}\\text{O}}=1.79\\times {10}^{-4}\\text{mol H}\\end{array}[\/latex]<\/p>\r\nThe empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is\r\n

[latex]\\frac{\\text{mol H}}{\\text{mol C}}=\\frac{1.79\\times {10}^{-4}\\text{mol H}}{8.95\\times {10}^{-5}\\text{mol C}}=\\frac{2\\text{mol H}}{1\\text{mol C}}[\/latex]<\/p>\r\nand the empirical formula for polyethylene is CH2<\/sub>.\r\n\r\n[\/hidden-answer]\r\n

Check Your Learning<\/strong><\/h4>\r\nA 0.00215-g sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726 g of CO2<\/sub> and 0.00148 g of H2<\/sub>O in a combustion analysis. What is the empirical formula for polystyrene?\r\n\r\n[reveal-answer q=\"701960\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"701960\"]CH[\/hidden-answer]\r\n\r\n<\/div>\r\n

Glossary<\/h2>\r\ngravimetric analysis:\u00a0<\/strong>quantitative chemical analysis method involving the separation of an analyte from a sample by a physical or chemical process and subsequent mass measurements of the analyte, reaction product, and\/or sample\r\n\r\ncombustion analysis:\u00a0<\/strong>gravimetric technique used to determine the elemental composition of a compound via the collection and weighing of its gaseous combustion products","rendered":"

Gravimetric Analysis<\/h2>\n

A gravimetric analysis<\/strong> is one in which a sample is subjected to some treatment that causes a change in the physical state of the analyte that permits its separation from the other components of the sample. Mass measurements of the sample, the isolated analyte, or some other component of the analysis system, used along with the known stoichiometry of the compounds involved, permit calculation of the analyte concentration. Gravimetric methods were the first techniques used for quantitative chemical analysis, and they remain important tools in the modern chemistry laboratory.<\/p>\n

The required change of state in a gravimetric analysis may be achieved by various physical and chemical processes. For example, the moisture (water) content of a sample is routinely determined by measuring the mass of a sample before and after it is subjected to a controlled heating process that evaporates the water.<\/p>\n

\n

Example 2:\u00a0Gravimetric Analysis<\/strong><\/h3>\n

A 0.4550-g solid mixture containing CaSO4<\/sub> is dissolved in water and treated with an excess of Ba(NO3<\/sub>)2<\/sub>, resulting in the precipitation of 0.6168 g of BaSO4<\/sub>.<\/p>\n

[latex]{\\text{CaSO}}_{4}\\text{(}aq\\text{)}+\\text{Ba}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}\\rightarrow{\\text{BaSO}}_{4}\\text{(}s\\text{)}+\\text{Ca}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}\\text{(}aq\\text{)}[\/latex]<\/p>\n

What is the concentration (percent) of CaSO4<\/sub> in the mixture?<\/p>\n

Show Answer<\/span><\/p>\n
\n

The plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of BaSO4<\/sub> and CaSO4<\/sub> through their stoichiometric factor. Once the mass of CaSO4<\/sub> is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.<\/p>\n

\"This<\/p>\n

The mass of CaSO4<\/sub> that would yield the provided precipitate mass is<\/p>\n

[latex]0.6168\\cancel{{\\text{g BaSO}}_{4}}\\times \\frac{1\\cancel{{\\text{mol BaSO}}_{4}}}{233.43\\cancel{{\\text{g BaSO}}_{4}}}\\times \\frac{1\\cancel{{\\text{mol CaSO}}_{4}}}{1\\cancel{{\\text{mol BaSO}}_{4}}}\\times \\frac{136.14{\\text{g CaSO}}_{4}}{1\\cancel{{\\text{mol CaSO}}_{4}}}=0.3597{\\text{g CaSO}}_{4}[\/latex]<\/p>\n

The concentration of CaSO4<\/sub> in the sample mixture is then calculated to be<\/p>\n

[latex]\\begin{array}{l}\\\\ {\\text{percent CaSO}}_{4}=\\frac{{\\text{mass CaSO}}_{4}}{\\text{mass sample}}\\times 100\\%\\\\ \\frac{\\text{0.3597 g}}{0.4550 g}\\times 100\\%=79.05\\%\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

Check Your Learning<\/strong><\/h4>\n

What is the percent of chloride ion in a sample if 1.1324 g of the sample produces 1.0881 g of AgCl when treated with excess Ag+<\/sup>?<\/p>\n

[latex]{\\text{Ag}}^{+}\\text{(}aq\\text{)}+{\\text{Cl}}^{-}\\text{(}aq\\text{)}\\rightarrow\\text{AgCl}\\text{(}s\\text{)}[\/latex]<\/p>\n

Show Answer<\/span><\/p>\n
23.76%<\/div>\n<\/div>\n<\/div>\n

The elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis<\/strong>. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product (Figure\u00a03). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.<\/p>\n

\"This<\/p>\n

Figure\u00a03. This schematic diagram illustrates the basic components of a combustion analysis device for determining the carbon and hydrogen content of a sample.<\/p>\n<\/div>\n

\n

Example\u00a03:\u00a0Combustion Analysis<\/strong><\/h3>\n

Polyethylene is a hydrocarbon polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a 0.00126-g sample of polyethylene yields 0.00394 g of CO2<\/sub> and 0.00161 g of H2<\/sub>O. What is the empirical formula of polyethylene?<\/p>\n

Show Answer<\/span><\/p>\n
\n

The primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:<\/p>\n

[latex]{\\text{C}}_{\\text{x}}{\\text{H}}_{\\text{y}}\\text{(}s\\text{)}+{\\text{excess O}}_{2}\\text{(}g\\text{)}\\rightarrow x{\\text{CO}}_{2}\\text{(}g\\text{)}+y{\\text{H}}_{2}\\text{O}\\text{(}g\\text{)}[\/latex]<\/p>\n

Note that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts x<\/em> and y<\/em> are needed.<\/p>\n

First, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:<\/p>\n

\"This<\/p>\n

[latex]\\begin{array}{l}\\\\ \\text{mol C}=0.00394{\\text{g CO}}_{2}\\times \\frac{1{\\text{mol CO}}_{2}}{\\text{44.01 g\/mol}}\\times \\frac{1\\text{mol C}}{1{\\text{mol CO}}_{2}}=8.95\\times {10}^{-5}\\text{mol C}\\\\ \\text{mol H}=0.00161{\\text{g H}}_{2}\\text{O}\\times \\frac{1{\\text{mol H}}_{2}\\text{O}}{18.02\\text{g\/mol}}\\times \\frac{2\\text{mol H}}{1{\\text{mol H}}_{2}\\text{O}}=1.79\\times {10}^{-4}\\text{mol H}\\end{array}[\/latex]<\/p>\n

The empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H-to-C molar ratio is<\/p>\n

[latex]\\frac{\\text{mol H}}{\\text{mol C}}=\\frac{1.79\\times {10}^{-4}\\text{mol H}}{8.95\\times {10}^{-5}\\text{mol C}}=\\frac{2\\text{mol H}}{1\\text{mol C}}[\/latex]<\/p>\n

and the empirical formula for polyethylene is CH2<\/sub>.<\/p>\n<\/div>\n<\/div>\n

Check Your Learning<\/strong><\/h4>\n

A 0.00215-g sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726 g of CO2<\/sub> and 0.00148 g of H2<\/sub>O in a combustion analysis. What is the empirical formula for polystyrene?<\/p>\n

Show Answer<\/span><\/p>\n
CH<\/div>\n<\/div>\n<\/div>\n

Glossary<\/h2>\n

gravimetric analysis:\u00a0<\/strong>quantitative chemical analysis method involving the separation of an analyte from a sample by a physical or chemical process and subsequent mass measurements of the analyte, reaction product, and\/or sample<\/p>\n

combustion analysis:\u00a0<\/strong>gravimetric technique used to determine the elemental composition of a compound via the collection and weighing of its gaseous combustion products<\/p>\n","protected":false},"author":264249,"menu_order":12,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-6368","chapter","type-chapter","status-publish","hentry"],"part":3031,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6368","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/wp\/v2\/users\/264249"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6368\/revisions"}],"predecessor-version":[{"id":6375,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6368\/revisions\/6375"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/parts\/3031"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapters\/6368\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/wp\/v2\/media?parent=6368"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=6368"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/wp\/v2\/contributor?post=6368"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-binghamton-chemistry\/wp-json\/wp\/v2\/license?post=6368"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}