A gaseous compound is 30.4% N and 69.6% O by mass. A5.25-g sample of the gas occ

Question

A gaseous compound is 30.4% N and 69.6% O by mass. A5.25-g sample of the gas occupies a volume of 1.00L and exerts apressure of 1.26 atm at 4.0 degrees C. What is its molecularformula? Please explain how to determine this? A gaseous compound is 30.4% N and 69.6% O by mass. A5.25-g sample of the gas occupies a volume of 1.00L and exerts apressure of 1.26 atm at 4.0 degrees C. What is its molecularformula? Please explain how to determine this?

Explanation / Answer

30.4 g N *( 1mol/ 28 g) = 1.08571429 moles N 69.6 g O *( 1mol/ 16 g) = = 4.35 moles O Divide by 1.08 1 mol N => 4.35/1.08 = 4.02777778 moles O Thus Empirical formula is NO4 PV = nRT      => n = mass / molecularweight molecular weight = mass*R*T/PV = (5.25g *0.0821 L-atm/mol-K * 277K)/(1.00 L * 1.26 atm) molecular weight = 94.7570833 g/mol But empirical weight is 4*16 + 14 = 78 94.7/78 = 1.21 Thus formula is N1.2O4.8

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