Molecular Formula Calculations

Combustion vs. Weight Percent• CxHy + (x + y/4) O2 x CO2 + y/2 H2O

• CxHyOz + (x + y/4 - z) O2 x CO2 + y/2 H2O

Combustion Analysis

CxHy + ( x + ) O2 (g) x CO2(g) + H2O(g)y 4

y 2

• CxHy + O2 x CO2 + y/2 H2O

Combustion ProblemErythrose 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 and 0.4194 g H2O. (MM = 120.0 g/mol) Calculate the molecular formula.

Molecules with oxygen in their formula are more difficult to solve for

Oz knowing the respective masses of CxHyOz sample, CO2 and H2O.

• CxHyOz + (x + y/4 - z) O2 x CO2 + y/2 H2O

Mass fraction of C in CO2 = =

= = 0.2729 g C / 1 g CO2

mol C x MM of Cmass of 1 mol CO2

1 mol C x 12.01 g C/ 1 mol C 44.01 g CO2

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 x MM of Cmass of 1 mol CO2

1 mol C x 12.01 g C/ 1 mol C 44.01 g CO2

mol H x MM of Hmass of 1 mol H2O

2 mol H x 1.008 g H / 1 mol H 18.02 g H2O

Erythrose Combustion Solution

Erythrose Combustion Solution

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 HCalculating 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 OCalculating 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 / formula120 g /mol / 30 g / formula = 4 formula units / cmpd = C4H8O4

0.2729 g C 1 g CO2

0.1119 g H 1 g H2O

Weight Percent ProblemA 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 Oz knowing the percent of C, H and O.

• CxHyOz + (x + y/4 -z) O2 x CO2 + y/2 H2O

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 = 40.00 gMass (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 g O / 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