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 + ( 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

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 mg sample of erythrose yielded g CO 2 and g H 2 O. (MM = 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

Mass fraction of C in CO 2 = = = = g C / 1 g CO 2 mol C x MM of C mass of 1 mol CO 2 1 mol C x g C/ 1 mol C g CO 2 Erythrose Combustion Solution

Mass fraction of C in CO 2 = = = = g C / 1 g CO 2 Mass fraction of H in H 2 O = = = = g H / 1 g H 2 O mol C x MM of C mass of 1 mol CO 2 1 mol C x g C/ 1 mol C g CO 2 mol H x MM of H mass of 1 mol H 2 O 2 mol H x g H / 1 mol H g H 2 O Erythrose Combustion Solution

Mass (g) of C = g CO 2 x = g C Mass (g) of H = g H 2 O x = g H Calculating the mass of O: Mass (g) of O = Sample mass -( mass of C + mass of H ) = g g C g H = g O Calculating moles of each element: C = g C / g C/ mol C = mol C H = g H / g H / mol H = mol H O = g O / g O / mol O = mol O C H O = CH 2 O formula weight = 30 g / formula 120 g /mol / 30 g / formula = 4 formula units / cmpd = C 4 H 8 O g C 1 g CO g H 1 g H 2 O

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 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 = 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

Erythrose Weight % Solution Let the weight % values = mass in grams of each element Mass (g) of C = g Mass (g) of H = 6.71 g Mass (g) of O = g Calculating moles of each element: C = g C / g C/ mol C = mol C H = 6.71 g H / g H / mol H = mol H O = g O / g O / mol O = mol O C H O = CH 2 O formula weight = g / emp. formula (MM= g /mol) / g / formula = 4 formula units / cmpd = C 4 H 8 O 4