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A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas ? A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas ?
Number of Moles of H = 4.07/1 = 4 (almost)Number of Moles of C = 24.27/12 = 2(almost)Number of Moles of Cl = 71/35.5 = 2(almost)RatioC:H:Cl = 2:4:2 = 1:2:1Empirical Formula = CH2ClEmpirical Formula Mass =12+2+35 = 49n = 98.96/46 = 2(almost)Molecular formula = n x empirical formula = C2H5Cl2
In this question Gaurav answered right but in last steo i.e. [molecular formula = n x empirical formula = c(2)H(4)CL(2), thank u
He had divided it by 46. HE has to divided it by 49 n=98.96\49=2 appox. Molecular formula= n * emperical formula =C2hcl2
No: of moles of H=4.07÷2.01=2No: of moles of C=2.02÷2.01=1No: of moles of Cl=2.02÷2.01=1Therefore EF=CH2Cln=Mm÷EfMF=n(CH2Cl)Therefore MF=C2H4Cl2Therefore MF=n[CH2Cl)
Hello studentLet us assume 100 g of sample of given compound. Thus the percentage of each element will represent its mass content in sample.The number of moles ofHydrogen, H = 4.07/1 = 4 (approx)Carbon, C = 24.27/12 = 2 ( approx)Chlorine, Cl = 71/35.5 = 2 The ratio of moles and thus of elements in the given compound, C:H:Cl = 2:4:2 = 1:2:1Hence, Empirical formula = CH2ClThe molecular formula will be of the form, (CH2Cl)nEmpirical formula mass = 12 + 2 x 1 + 35.5 = 49.5 gNow, molar mass of Compound = 98.96 g Hence, n = 98.96/49.5 = 1.999 = 2Hence the molecular formula is : C2H4Cl2 Hope it helpsRegards,Kushagra
atomic mass of H C and cl are 1, 12 and 35.5 respectivelylet compound form is 100g, so the given percentages of each element represent its mass.so. theno. of moles of hydrogen= prcentage of hydrogen/atomic mass of hydrogen =4.07/1=4 (approx because AM of H is 1.008)no.of moles of H = 4 No. of moles of C is 24/12=2 [approx because carbon prcent is 24.7]No. of moles of C=2 no. of moles of Cl = 71\35.5=2no. of moles of Cl = 2 thus the ratio of given compound = C:H:Cl= 2:4:2=1:2:1so Emperical formula of this CH2Clemperical mass will be = mass of C + mass of H +mass of CL = 12+2*1+35.5=49.5 gso, the Emperical formula mass = 49.5 g molar mass of compound=98.96 g:.n= 68.96/49.5 = 1.99 =2 hence the molecular formula = C2H2*2Cl2=C2H4Cl2So the EF = CHCl MF= C2H4Cl2
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