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Combustion Analysis

Combustion Analysis. The composition of a substance is often determined by using a specified reaction to break down the substance into known measurable compounds. Organic compounds are often analyzed by combustion. 1. Define combustion.

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Combustion Analysis

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  1. Combustion Analysis The composition of a substance is often determined by using a specified reaction to break down the substance into known measurable compounds. Organic compounds are often analyzed by combustion 1. Define combustion The reaction of a substance with molecular oxygen 2. Identify the products in combustion of a simple organic compound carbon dioxide and water

  2. For example, assume an unknown hydrocarbon is analyzed by combustion analysis. 1. Write the equation for the reaction. CxHy + O2 x CO2 + y H2O Note that all of the carbons are converted from the hydrocarbon to carbon dioxide. Measurement of the carbon dioxide provides a quantitative value for the amount of carbon in the substance Note that all of the hydrogens are converted from the hydrocarbon to water. Measurement of the water provides a quantitative value for the amount of hydrogen in the substance.

  3. Organic compounds are often analyzed by combustion The substance is heated to a high temperature with O2 to insure complete combustion. The resulting CO2 and H2O are absorbed using appropriate chemicals, and the amounts of CO2 and H2O are determined by measuring the increase in mass of the absorbents

  4. Assume a sample of an unknown hydrocarbon is completely burned in the presence of oxygen, and the resulting carbon dioxide and water are collected and measured. Results: 3.45 grams CO2 and 1.65 g H2O. Determine the empirical formula. Carbon 3.45 g CO2 1 mol CO2 1 mol C = .0784 mol C 44 g 1 mol CO2 Hydrogen 1.65 g H2O 1 mol H2O 2 mol H = 0.183 mol H 18 g 1 mol H2O

  5. Carbon = .0784 mol Hydrogen = 0.183 mol Carbon is the lowest value = .0784 mol Carbon .0784 = 1.00 Hydrogen = .183 mol = 2.33 mol .0784 .0784 Carbon: Hydrogen ratio is 1: 2.33 Hydrogen = 3 (2.33) = 7 Carbon = 3 (1) = 3 Empirical Formula = C3H7

  6. Complete combustion of a 0.523 g of unknown hydrocarbon produces 1.61 g of carbon dioxide and 0.743 g of water. In a separate experiment, it is determined that the molar mass of the compound is 114 g. What is the molecular formula? Carbon 1.61 g CO21 mol CO21 mol C = 0.0366 mol C 44 g 1 mol CO2 Hydrogen .743 g H2O 1 mol H2O2 mol H = .0826 mol H 18 g 1 mol H2O

  7. Complete combustion of a 0.523 g of unknown hydrocarbon produces 1.61 g of carbon dioxide and 0.743 g of water. In a separate experiment, it is determined that the molar mass of the compound is 114 g. What is the molecular formula? 0.0366 mol C and 0.0826 mol H 0.0826 = 2.26 0.0366 Carbon:Hydrogen ratio is 1: 2.26 Hydrogen 2.26 (4)  9 Carbon 1 (4) = 4 Empirical Formula is C4 H9

  8. Complete combustion of a 0.523 g of unknown hydrocarbon produces 1.61 g of carbon dioxide and 0.743 g of water. In a separate experiment, it is determined that the molar mass of the compound is 114 g. What is the molecular formula? Empirical Formula is C4 H9 = 57 g/mol Molecular Formula 114 g/mol = 2 57 g/mol Molecular Formula = 2 ( C4 H9 )= C8H18

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