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Anonymous asked in Science & MathematicsChemistry · 1 month ago

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.

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  • david
    Lv 7
    1 month ago
    Favorite 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 W
    Lv 7
    1 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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