Question

#1 the analysis of a compound shows it contains 5.4 mol C, 7.2 mol H, and 1.8 mol N.what is the empirical formual of the compound?

#2 A compound that contains only carbon, hydrogen, and oxygen is 58.8% C and 9.87 H by mass. What is the empirical Formuala of this substance ?

Answer 1

mol qc = 5.4 mol mol og M = 7.2 Mol mol G N = 1.8 mol Now divide each value of mol by 1.8' le smallest mole Value to get whole ratio son 3 1.8 H= 7.2 = 4 N = 18 = 1 C:H:n is in ratio of 3: 4:1 so Empirical formule = C3 H4N 1.8 so Answer 2. let Total many compound = loog mass of c= 58.89 moljc= mass molar masa moljca 58.8 g - 4.90 no port = 4.90 12.01 more

molo- Тој по- 90 онука - 3| 333 мо) до = 31:33 16. o Ymon1 = 1.96 mo) Now dirid de esch mol by 1196 to get whole ratio С- ч -4° - 25 ) :9 с н= - 4:21 1-94 1.9 - 4-48 - 1:44 o = - 1 1:46 mulripty C: и: o 2 : ч. 98 : 1 a lаlu E2 muhr С: :o S: до: 2 So formal = (541002 Answer