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Molecular Formula Calculations Combustion vs. Weight Percent

Molecular Formula Calculations Combustion vs. Weight Percent. C x H y + ( x + y/4 ) O 2  x CO 2 + y/2 H 2 O. C x H y O z + ( x + y/ 4 - z ) O 2  x CO 2 + y/2 H 2 O. Combustion Analysis. C x H y + O 2  x CO 2 + y/2 H 2 O. y 4. y 2.

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Molecular Formula Calculations Combustion vs. Weight Percent

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  1. Molecular Formula CalculationsCombustion vs. Weight Percent • CxHy+(x+y/4)O2x CO2+y/2 H2O • CxHyOz+(x+y/4 -z)O2x CO2+y/2 H2O

  2. Combustion Analysis • CxHy+O2x CO2+y/2 H2O y 4 y 2 CxHy + ( x + ) O2 (g)x CO2(g) + H2O(g)

  3. Molecules with oxygen in their formula are more difficult to solve for Oz knowing the respective masses of CxHyOzsample, CO2 and H2O. • CxHyOz+(x + y/4 - z) O2xCO2+y/2 H2O Combustion Problem Erythrose is an important mono-saccharide that is used in chemical synthesis. It contains Carbon, Hydrogen and Oxygen. Problem: Combustion analysis of a 700.0 mg sample of erythrose yielded 1.027 g CO2 and0.4194 g H2O. (MM = 120.0 g/mol) Calculate the molecular formula.

  4. Erythrose Combustion Solution Mass fraction of C in CO2 = = = = 0.2729 g C / 1 g CO2 mol C xMM of C mass of 1 mol CO2 1 mol C x 12.01 g C/ 1 mol C 44.01 g CO2

  5. Erythrose Combustion Solution Mass fraction of C in CO2 = = = = 0.2729 g C / 1 g CO2 Mass fraction of H in H2O = = = = 0.1119 g H / 1 g H2O mol C xMM of C mass of 1 mol CO2 1 mol C x 12.01 g C/ 1 mol C 44.01 g CO2 mol H xMM of H mass of 1 mol H2O 2 mol H x 1.008 g H / 1 mol H 18.02 g H2O

  6. Erythrose Combustion Solution 0.2729 g C 1 g CO2 Mass (g) of C = 1.027 g CO2 x = 0.2803 g C Mass (g) of H = 0.4194 g H2O x = 0.04693 g H Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O Calculating moles of each element: C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O C0.02334H0.04656O0.02330 = CH2O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cmpd = C4H8O4 0.1119 g H 1 g H2O

  7. Molecules with oxygen in their formula are much easier to solve for Oz knowing the percent of C, H and O. • CxHyOz+(x + y/4 -z) O2xCO2+y/2 H2O Weight Percent Problem A sample is usually sent to a commercial laboratory for analysis. The analytical results provide the respective % of each of the elements in the sample, in the case of erythrose: Carbon, Hydrogen and Oxygen. Problem: A 1.2000 g sample of an unknown sugar that was thought to be erythrose was sent for analysis, the reported results were: C 40.00%; H 6.71%; O 53.29%; MM = 120.11 g/mol To solve: Let the analysis % values = mass in grams

  8. Erythrose Weight % Solution Let the weight % values = mass in grams of each element Mass (g) of C = 40.00 g Mass (g) of H = 6.71 gMass (g) of O = 53.29 g Calculating moles of each element: C = 40.00 g C / 12.01 g C/ mol C = 3.330 mol C H = 6.71 g H / 1.008 g H / mol H = 6.657 mol H O = 53.29 gO / 16.00 g O / mol O = 3.331 mol O C3.330H6.657O3.331= CH2O formula weight = 30.03 g / emp. formula (MM= 120.11 g/mol) / 30.03 g / formula = 4 formula units / cmpd= C4H8O4

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