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A 2.52- sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess O to yield 4.23g of CO2 and 1.01g of H2O. Another sample of the same compound, of mass 4.14g, yielded 2.11g of SO3 . A third sample, of mass 5.66g , yielded 2.27g of HNO3. Calculate the empirical formula of the compound.
here is the solution :
mass of CO2 = 4.23g or after dividing by the molar mass 44 =
0.0961moles of CO2 since there is 1 mole so C in CO2
=0.0961moles of C or
1.154g of Cwhich is the mass of C in 2.52 g of the compound
mass of H2O =1.01g or after dividing by the molar mass 18 =
0.0561moles of H2Oorsince there are 2 moles of H in 1 mole of H2O =
0.1122moles of H or
0.1131g of Hwhich is the mass of H in 2.52g of the compound
mass of compound burnt =2.5200g
mass of C+H =1.267g
mass of O + N + S =1.253g
mass of SO3 = 2.11g or after dividing by the molar mass 80 =0.0264moles of SO3 since there is 1 mole so S in SO3 =0.0264moles of S or
0.844g of Swhich is the mass of S in 4.14g of the compound
mass in 2.52 g of compound = .844 x 2.52 / 4.14 = 0.514 g or 0.0161 moles
mass of HNO3 = 2.27g or after dividing by the molar mass 60 =0.0378moles of HNO3since there is 1 mole so N in HNO3 =0.0378moles of N or
0.530g of Nwhich is the mass of N in 5.66 g of the compound so mass of N in 2.52 g = .530 x 2.52 / 5.66= 0.2360g or 0.01686 moles
from above we see that
mass of O + N + S =1.253
N + S in 2.52 g of sample = .2360 + .514
O = 1.253 -( .2360 + .514)
= 0.503 g or 0.031 moles
molar ratio of C : H : S : N : O = 0.0961:0.1122:0.0161:0.01686
:.031 or after dividing by the smallest we have
5.96: 6.96: 1: 1.047: 1.92
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