Molecular Formula Calculations Combustion & Weight Percent C x H y + (x + y/4) O 2  x CO 2 + y/2 H 2 O C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O.

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Presentation transcript:

Molecular Formula Calculations Combustion & Weight Percent C x H y + (x + y/4) O 2  x CO 2 + y/2 H 2 O C 2 H 5 OH + 3 O 2  2 CO 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 Problem: Erythrose (MM = g/mol) is an important chemical compound in chemical synthesis. It contains Carbon, Hydrogen and Oxygen. Combustion analysis of a mg sample yielded g CO 2 and g H 2 O.

Erythrose 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 Calculating masses of C and H: Mass of Element = mass of compound x mass fraction of element 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 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

Empirical Formulas from Analyses

Empirical Formula Determination 1.Use percent analysis. Let 100 % = 100 grams of compound. 2.Determine the moles of each element. (Element % = grams of element.) 3.Divide each value of moles by the smallest of the mole values. 4.Multiply each number by an integer to obtain all whole numbers.

Empirical & Molecular Formula Determination Quinine: C 74.05%, H 7.46%, N 8.63%, O 9.86% 74.05/12.01, 7.46/1.008, 8.63/14.01, 9.86/16.00 C H 7.40 N O Empirical Formula: C 10 H 12 N 1 O 1 Empirical Formula Weight = ? Molecular Weight = Molecular Formula = 2x empirical formula Molecular Formula = C 20 H 24 N 2 O 2

A Carbon atom is at each angle. Each C has 4 bonds (lines + Hs ). Hs are not always drawn in & must be added. C = 20 H = 24 N = 2 O = 2 From the structures, determine the molecular formula of quinine. C 20 H 24 N 2 O 2