A 2.52-g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen

Question

A 2.52-g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess O to yield 4.36 g of CO2 and 0.892 g of H2O as the only carbon and hydrogen containing products respectively. Another sample of the same compound, of mass 4.14 g, yielded 2.60 g of SO3 as the only sulfur containing product. A third sample, of mass 5.66 g, was burned under different conditions to yield 2.80 g of HNO3 as the only nitrogen containing product. Calculate the empirical formula of the compound. Express your answer as a chemical formula. Enter the elements in the order: C, H, S, N, O.

Explanation / Answer

firstly C and H

# moles = mass / molar mass
molar mass of CO2 = 44 g/ mole so
4.36 g of CO2 has 4.360 /44 = 0.0991 moles of CO2
there is 1 mole of C in CO2 and all the C from the compound becomes CO2
so moles of C in the compound = 0.09909 moles
mass of C = 0.09909 x 12 = 1.1891 g

molar mass of H2O = 18 g/ mole
0.892 g of H2O has 0.892 / 18 = 0.04956 moles of H2O
there are 2 moles of H in H2O so moles of H in the compound = 0.09911 moles
mass of H = 0.09911 x 1.0079 = 0.100 g

mass of SO3 = 2.60 ; moles of SO3 = 2.60 / 80 = 0.0325 ;
moles of S in 4.14 g of compound = 0.0325
moles in 2.52 g of compound = .0325 x 2.52 / 4.14 = 0.01978 moles
mass of S = 0.01978 x 32 = 0.6330 g

mass of HNO3 = 2.80; moles of HNO3 = 2.80 / 63 = 0.04444 moles
moles of N in 5.66 g of sample = 0.04444
moles in 2.52 g of compound = 0.04444 x 2.52 / 5.66 = 0.01979 moles
mass of N = 0.01979 x 14 = 0.2770 g

mass C + H + S + N = 1.1891 + 0.100 + 0.6330 + 0.2770 = 2.1991
mass of O by difference = 2.52 - 2.1991 = 0.3209 g
moles of O in 2.52g = 0.3209 / 16 = 0.0201 moles

we now have alll the number of moles now we can do the empirical formula

molar ratio of C : H : S : N : O = 0.09909 : 0.09911 :0.01978 : 0.01979:0.0201
divide by the smallest number to simplify the ratio and we get
C : H : S : N : O = 5:5:1:1:1

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.