Question

0, 7.11, Elemental analysis of dry cell mass indicates that the cell is composed of 45% C, 25% 5% H, 12% N, 2.5% P, 0.5% S, and 10% minerals. a. By considering 1 g of cell dry weight, determine mole fractions of every element and the cell formula. b. If the cell formula is C4 Hs O22 N1.2 Poos S0,.02, determine the percent elemental com- position (i.e., mass fractions) of dry cell mass.

Answer 1

Given that analysis of dry cell mass indicates that the cell is compound of 45 % c. 25% D, 5% H, 12% N, 2.5% P, 0.5% s, and 10% minerals (a) Mole fraction of c = 45/45 + 25 + 5 + 12 + 12.5 + 0.5 + 10 c = 45/100 In 11g cell dry 5 C = 0.45 g similarly calculate every element %O = 0.25 g h = 0.05g n = 0.12 g P = 0.025g s = 0.005g mineral = 010 gm cell formula c_0.45 O_0.25 N_0.12 P_0.025 S_0.005 = c_g O_5 H_0.1 N_2.4 P_0.5 S_0.01 \r\n b) Given cell formula c_4 H_s O_2.2 N_1.2 P_0.08 S_0.02 mole fraction = 4/4 + 8 + 2.2 + 1.2 + 0.08 + 0.12 = 4/15.5 = 0.25 = 25% mass fraction of H = 8/15.5 = 0.516 = 51.6% O = 2.2/15.5 = 0.142 = 14.2% N = 1.2/15.5 = 0.017 = 7.7% P = 0.08/15.5 = 0.0056 = 0.5%