# Empirical and Molecular formula question?

Nitrogen forms more oxides than any other element. The percentage mass of nitrogen in one of the oxides is 36.85%.

(i) Determine the empirical formula of the compound

(ii) If the molecular weight of nitrogen oxide is 152.0g mol-1 , determine the molecular formula for this compound.

### 2 Answers

- davidLv 71 month agoFavorite Answer
1) percentage mass of nitrogen in one of the oxides is 36.85%.

N == 36.85 / 14 = 2.632

O == 63.15 / 16 = 3.947

N 2.632 / 2.632 = 1

O 3.947 / 2.632 = 1.5

N === 1 X 2 = 2

O === 1.5 X 2 = 3

Emp. Formula ... N2O3

2) Emp. weight = 76

152 / 76 = 2

(N2O3) X 2 = N4O6 <<< molec. formula

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- Dr WLv 71 month ago
there are a couple of different ways to do this.. we could

.. (1) assume 100g of compound

.. (2) convert that % N and 100g to mass N then mass O

.. (3) convert masses to moles

.. (4) simplify mole ratio to get empirical formula

.. (5) compare molecular mass to empirical unit mass to get

.. . . .# of empirical units per molecule and thus molecular formula

that's the typical way

but.... since we have molecular mass and %'s, we can just do this

.. (1) convert molecular mass * %N to mass N to moles N

.. (2) convert molecular mass * (100 - %N) to get mass O then moles O

.. (3) convert our molecular formula to empirical formula

*** method #1 ***

100g NxOy.. ... 36.85g N.. .. 1 mol N

----- ----- ----- x ----- ----- ---- x ---- ---- ---- = 2.6308 mol

... ... . 1.. ... ... .100g NxOy... . 14.007g N

100g NxOy.. ... 63.15g O.. ... 1 mol O

----- ----- ----- x ----- ----- ---- x ----- ---- ---- = 3.9471 mol

... ... . 1.. ... ... .100g NxOy... . 15.999g O

then we simplify by dividing both by the smallest

.. mole N = 2.6308 / 2.6308 = 1

.. mole O = 3.9471 / 3.9471 = 1.500

then we double to get rid of the fraction and our empirical formula is N2O3

with empirical unit mass = 2*14.007 + 3*15.999 = 76.011

then we divide molecular mass / empirical unit mass to get # empirical units / molecule

.. 152.0 / 76.011 = 2

then we write the molecular formula = 2 * N2O3 = N4O6

*** method #2 ***

nitrogen

.. 36.85g N.. ... 1 mol N.. .... 152.0g NxOy

----- ----- ----- x ----- ---- ---- x ----- ---- ---- --- = 4 mol N / mol NxOy

.. 100g NxOy.. .14.007g N... 1 mol NxOy

oxygen

.. 63.15g O.. ... 1 mol O.. .... 152.0g NxOy

----- ----- ----- x ----- ---- ---- x ----- ---- ---- --- = 6 mol O / mol NxOy

.. 100g NxOy.. .15.999g O... 1 mol NxOy

giving us a molecular formula = N4O6

and empirical formula N2O3

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