A compound contains only carbon, hydrogen, nitrogen and oxygen. Combustion of 0.

Question

A compound contains only carbon, hydrogen, nitrogen and oxygen. Combustion of 0.314 g of the compound produced 0.426 g CO2, and 0.0620 g H2O. In another experiment it is found that 0.103 g of the compound produces 0.0460 g of NH3. What is the empirical formula of the compound?: *
a. C14H10N6O12

b. C5H7N3O6

c. C7H5N3O6

d. C9H7N2O4

Explanation / Answer

Upon combustion, the CO2 has all the carbon originating from the CHNO compound 0.426 g CO2 contains 27.27% (from atomic wt of C/mol wt of CO2 = 12/44 = 0.2727) C => 0.1162 g of C originating from the CHNO compound => 0.1162/12 = 0.00968 moles , rounded to 0.010 moles of C in the CHNO compound. Similarly, upon combustion, all the H is from the CHNO compound: 0.0620 g of H2O has 11.11% H (from 2* atomic wt of H/mol wt of H2O = 2/18 = 0.111111) => 0.0068 g of H originating from the CHNO compound => 0.0068/1 = 0.0068, rounded to 0.006 moles of H in the CHNO compound Upon conversion to NH3, all the N originates from the CHNO compound: 0.0460 g NH3 contains 82.3% N (atom wt. of N/mol wt of NH3 = 14/17 = 0.823) =>0.103 g of CHNO compound contains 0.0460*0.823 = 0.03788 g of N hence, 0.314 g (to bring the N content in the CHNO compound to the same level for which the C and H levels were calculated above) = 0.314*0.03788/0.103 = 0.115 g of N => 0.115/14 = 0.008 moles of N Thus the CHN content of the 0.314 g of CHNO compound = 0.116 + 0.0068 + 0.115 = 0.2378 g, hence the oxygen content would be 0.314 - 0.2378 = 0.0762 g => 0.0762/16 = 0.00476 moles, rounded to 0.004 moles of O) Thus the CHNO compound contains the CHNO in the following ratio: C 0.010 H 0.006 N 0.008 O 0.004 The empirical formula would be: C5H3N4O2

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