Question

14. A compound contains C, H and N atoms. When burned in an air with excess oxygen, a 0.312 g sample produced 0.458 g CO2 and 0.374 g H2O. The nitrogen content of a 0.486 g sample is converted to 0.226 g N2. What is the empirical formula of the compound? -5 3.89 X10 mcle A) CHAN В) С2НаN C) CH4N2 D) C3H4N2 E) C3H4N 7.99 I65 mol4 1,50 molo

Answer 1

A) CH4N

Let's assume the compound is cxHyNz ne cyty Nz + O2(excess) XCO2 + 7 H₂O 0.458 9 co in moles les - 0.4589 - 0.4587 MMCO2 0.3549 H₂O in moles = 2 sy g = 0.01040909 1906 0.3749 o. MMHO x = 0.01040909 18 mod 0 020777777 Also, y = 20.020777777 = 0.041555555 -0.01040909, 0.04155555 оогоороо о. 0.01040909 1: 3.992237127 299 0:312 = 1200040909 +/X0041555555 + 14 x 2 z =0.010395383 "! 21:41 CHYN) t = 0.4869 02268.07 28g/mol x103 moi 0.226 g N in modes = 0.2269 Here sample weight = 0.486. Sample weight = 0.486g »r2 x 1 + moles ofo N in compound = 2x8.07 x 103 = 0.01614 mol when sample weight is 0.312 g moles of N 0.01614 0486X0312 = 0.01036148/ moles of c: moles of H: mol moles of N - 0.01040909 :0.041555555 0.010361481 - 0.01040909 0.onis55555 0.010361481 0.010361881 1.00: 4.01058 : 0.010361481 0.01036148) = . Empirical formula = CHUN)