Acompound has a molar mass of 92.02 grams/mole, and its percent composition is 30.4% nitrogen (n) and 69.6% oxygen (o). what is its molecular formula? a. n2o4 b. no2 c. n2o d. n4o2

The answer is: A. N2O4.

M(NₓOₙ) = 92.02 g/mol; molar mass of unknown compound.

If we take 100 grams of sample:

m(N) = 0.304 · 100 g.

m(N) = 30.4 g; mass of nitrogen.

n(N) = 30.4 g ÷ 14.01 g/mol.

n(N) = 2.17 mol; amount of nitrogen.

m(O) = 0.696 · 100 g.

m(O) = 69.6 g; mass of oxygen

n(O) = 69.6 g ÷ 16 g/mol.

n(O) = 4.35 mol; amount of oxygen.

n(N) : n(O) = 2.17 mol : 4.35 mol.

n(N) : n(O) = 1 : 2.

Answer is N₂O₄ or NO₂, so we must find molar mass.

M(N₂O₄) = 2Ar(N) + 4Ar(O) · g/mol.

M(N₂O₄) = 92.02 g/mol.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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