Answer to Question #95571 in General Chemistry for Tea

Question

A compound contains only C , H, and N. Combustion of 35.0 mg of the compound produces 53 mg CO2 and 32.6 mg H20. What is the empirical formula of the compound?

Answer 1

w(C/CO₂) = 12g/mol / 44g/mol = 0.2727.

m(C) = w(C/CO₂) × m(CO2) = 0.2727 × 53 mg = 14.4545 mg.

w(H/H₂O) = 2g/mol / 18g/mol = 0.1111.

m(H) = w(H/H₂O) × m(H₂O) = 0.1111 × 32.6 mg = 3.6219 mg.

m(N) = m(comp) – m(C) – m(H) = 35 mg – 14.4545 mg – 3.6219 mg = 16.9236 mg.

C_XH_YN_Z

X : Y : Z = m(C)/M(C) : m(H)/M(H) : m(N)/M(N) = 14.4545mg/12mg/mmol : 3.6219mg/1mg/mmol : :16.9236mg/14mg/mmol = (1.2045 : 3.6219 : 1.2088)/1.2045 ≈ 1 : 3 : 1.

Therefore, the empirical formula of the compound is CH₃N.