1. 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

2. Combustion Analysis C x H y + ( x + ) O 2 (g)  x CO 2(g) + H 2 O (g) y 4 y 2 C x H y + O 2  x CO 2 + y/2 H 2 O

3. 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 CO 2 and 0.4194 g H 2 O. (MM = 120.0 g/mol) Calculate the molecular formula. Molecules with oxygen in their formula are more difficult to solve for O z knowing the respective masses of C x H y O z sample, CO 2 and H 2 O. C x H y O z + ( x + y/4 - z) O 2  x CO 2 + y/2 H 2 O

4. Mass fraction of C in CO 2 = = = = 0.2729 g C / 1 g CO 2 mol C x MM of C mass of 1 mol CO 2 1 mol C x 12.01 g C/ 1 mol C 44.01 g CO 2 Erythrose Combustion Solution

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

6. Mass (g) of C = 1.027 g CO 2 x = 0.2803 g C Mass (g) of H = 0.4194 g H 2 O 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 C 0.02334 H 0.04656 O 0.02330 = CH 2 O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cmpd = C 4 H 8 O 4 0.2729 g C 1 g CO 2 0.1119 g H 1 g H 2 O

7. 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 Molecules with oxygen in their formula are much easier to solve for O z knowing the percent of C, H and O. C x H y O z + ( x + y/4 -z) O 2  x CO 2 + y/2 H 2 O 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 g Mass (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 g O / 16.00 g O / mol O = 3.331 mol O C 3.330 H 6.657 O 3.331 = CH 2 O formula weight = 30.03 g / emp. formula (MM= 120.11 g /mol) / 30.03 g / formula = 4 formula units / cmpd = C 4 H 8 O 4