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A chemistry question about Stoichmetry?

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A sample containing NaCl, Na2SO4and NaNO3 , gives the following analysis in percent by mass : Na =32.08%, O=36.01% , Cl= 19.51%. calculate the percent by mass of each compound n the sample.

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  1. using molar masses, we can learn

    19.51 g Cl @ 58.44 g NaCl / 35.45 g Cl =

    "almost an answer":  32.16 grams NaCl

    =================================

    32.16 g NaCl - 19.51 =  12.65 g Na in the NaCl

    32.08 g Na - 12.65 g Na in NaCl = 19.43 g Na in the other two ...

    and  with "X" = moles  Na2SO4 , & "Y" = moles NaNO3:

    19.43 g Na = (X mol) (45.98g Na/ mol Na2SO4)  +  (Y mol)(22.99 g Na / mol NaNO3)

    19.43g = 45.98X + 22.99 Y   (aka equation A)

    ===========

    however with a total of 36.01 g of oxygen in the other two...

    and with "X" = moles  Na2SO4 , & "Y" = moles NaNO3:

    36.01 g O  = (X mol) (64.00 g O/ mol Na2SO4)  +  (Y mol)/(48.00 g O/ mol NaNO3

    36.01g = 64X  & 48Y

    therefor Y = (36.01 - 64 X) / 48  (aka equation B)

    ==============

    equation A :

    19.43g = 45.98X + 22.99 Y

    substituting equation B into equation A:

    19.43g = 45.98X + 22.99 (36.01 - 64 X) / 48

    multiplying all by 48:

    932.64 = 2207.04X + 22.99 (36.01 - 64 X)

    which gives:

    932.64 = 2207.04X + 827.87  - 1471.36 X)

    rearranging gives:

    104.77  = 735.68X

    solving:

    X = moles of Na2SO4 = 0.1424

    0.1424 moles Na2SO4 @ 142.04 g/mol =

    for another "almost an answer": 20.23 grams Na2SO4

    ====================

    now back to the finding the Na in another compound:

    0.1424 moles of Na2SO4 @ (2mol Na)(22.98 g/mol) Na / mole Na2SO4 =

    6.55 grams of Na are in the Na2SO4

    =========

    32.08 g Na total - 6.55 g Na in Na2SO4 - 12.65 g Na in NaCl =

    12.88 g Na are in NaNO3

    so lets use it & molar masses, to get the grams of NaNO3:

    12.88 g Na @ (84.99 g/mol) NaNO3 / 22.98 g Na=

    your last "almost an answer": 47.64g NaNO3

    ======================================...

    to summarize, we found:

    47.64 grams NaNO3

    20.23 grams Na2SO4

    32.16 grams NaCl

    which totals 100.0 grams of compounds

    so your final answers are:

    47.64 % NaNO3

    20.23 % Na2SO4

    32.16 % NaCl

    well there goes an hour, now back to putting my daughter's transmission back into her car

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