Question

An oxide of nitrogen contains 25.92% of nitrogen. Calculate the molecular formula if the V.D. of the compound is 54.

Answer 1

Answer:

Moles of nitrogen =  $\frac{25.9g\\2.}{14.01.g.mol}$

                               = 1.85 mol

Moles of oxygen   = $\frac{74.07.g}{15.999.g.mol}$

                               = 4.62 mol

And we divide through the individual molar quantities by the smallest such quantity, that of nitrogen

N = $\frac{1.85.mol}{1.85mol}$       O = $\frac{4.62mol}{1.85mol}$

   = 1                    = 2.50

Because, by definition, empirical formulae are the simplest whole number ratio definining constituent atoms in a species, we double this ratio to get $N_{2}$$O_{5}$ ( dinitrogen pentoxide ) .

Answer 2

Answer:

$hey \: mate \: here \: is \: ur \: ans$

Moles of nitrogen =

                               = 1.85 mol

Moles of oxygen   = \frac{74.07.g}{15.999.g.mol}15.999.g.mol74.07.g

                               = 4.62 mol

And we divide through the individual molar quantities by the smallest such quantity, that of nitrogen

N = \frac{1.85.mol}{1.85mol}1.85mol1.85.mol       O = \frac{4.62mol}{1.85mol}1.85mol4.62mol

   = 1                    = 2.50

Because, by definition, empirical formulae are the simplest whole number ratio definining constituent atoms in a species, we double this ratio to get N_{2}N2 O_{5}O5 ( dinitrogen pentoxide ) .

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