• A compound is 21.20%. Nitrogen, 6.06 percent of hydrogen, 24.30% sulpur and 48.45 percent oxygen. Write theEmpirical formula for the compound.

• A compound is 21.20%. Nitrogen, 6.06 percent of hydrogen, 24.30%

1.	• A compound is 21.20%. Nitrogen, 6.06 percent of hydrogen, 24.30% sulpur and 48.45 percent oxygen. Write theEmpirical formula for the compound.
Answer» n :-% N = 21.02 %H = 6.06 %S = 24.30 %O = 24.30 Relative no of atoms Divide the % composition of elements with their respective atomic masses Atomic MASS of N = 14Atomic mass of H = 1Atomic mass of S = 32Atomic mass of O = 16 ⇒ N = 21.20 / 14 = 1.5 ⇒ H = 6.06 / 1 = 6.06⇒ S = 24.3 / 32 = 0.75⇒ O = 48.45 /16 = 3.02 Simple Ratio Now divide all the values obtained above with the least no in order to GET the simple ratio ⇒ N = 1.5 /0.75= 2⇒ H = 6.06 /0.75= 8⇒ S = 0.75/ 0.75= 1⇒ O = 3.02/ 0.75= 4Empirical FORMULA = N2 H8 S O4 The EMPIRICAL formula of the compound is N2 H8 S O4