Answer:
CHâ‚…N
Explanation:
In the combustion, all of the C in the compound was used to produce COâ‚‚ in a 1:1 ratio. Thus, the moles of COâ‚‚ (MW 44.01 g/mol) produced equals the moles of C in the compound:
(44.0 g)(mol/44.01g) = 0.99977 mol COâ‚‚ = 0.99977... mol C
Similarly, all of the H in the compound was used to produce Hâ‚‚O in a ratio of 2H:1Hâ‚‚O. The moles of Hâ‚‚O (MW 18.02 g/mol) produced was:
(45.0 g)(mol/18.02g) = 2.497...mol Hâ‚‚O
Moles of H is found using the molar ratio of 2H:1Hâ‚‚O:
(2.497...mol Hâ‚‚O)(2H/1Hâ‚‚O) = 4.994...mol H
The ratio of H to C in the compound is:
(4.994...mol H)/(0.99977... mol C) = 5 H:C
Some NOâ‚‚ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CHâ‚…N.
