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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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