need help with a homework problem, very advanced

Question

Guanine is an essential component of DNA. It contains the elements carbon, hydrogen,

nitrogen and oxygen. When 615 mg of guanine is burned, 895 mg of carbon dioxide and 183 mg

of water are formed. In a separate analysis, a fresh 615 mg sample of guanine is boiled in excess

sulfuric acid, then neutralized with sodium hydroxide: this converts all of the nitrogen into NH 3 .

The NH 3 is then titrated with HCl: a total of 742 mg of HCl is required to consume all of the

NH 3 . The titration reaction is:

NH 3 + HCl NH 4 Cl

From this information, determine the empirical formula of guanine.

Answer 1

To determine the empirical formula, we need to find moles of C, H, O and N.

C, H and O are determined from the first analysis:

895 mg CO2 x 1 mmol CO2/44 mg x 1 mmol C/mmol CO2 = 20.3 mmoles C x 12 mg/mmol = 244 mg

183 mg H2O x 1 mmol H2O/18 mg x 2 mmol H/mmol H2O = 20.3 mmoles H x 1 mg/mmol = 20.3 mg

To find O, we must subtract mg of C, H and N (see below) from original mass, as follows:

mg O = 615 mg - 244 mg C - 20 mg H - 284 mg N = 67 mg O x 1 mmol/16 mg = 4.19 mmol O

N is determined in the second analysis:

742 mg HCl x 36.5 mg/mmol x 1 mmol N/mmol HCl = 20.3 mmoles N x 14 mg/mmol = 284 mg

SUMMARY:

moles C = 20.3

moles H = 20.3

moles N = 20.3

moles O = 4.19

Divide all by 4.19 to get whole numbers and we have

5 moles C

5 moles H

5 moles N

1 mole O

Empirical formula = C₅H₅N₅O