Suppose that a laboratory analysis of white powder showed 42.59% na, 12.02% c, and 44.99% oxygen. would

Suppose that a laboratory analysis of white powder showed 42.59% na, 12.02% c, and 44.99% oxygen. would you report that the compound is sodium oxalate or sodium carbonate? (use 43.38% na, 11.33% c, and 45.29% o for sodium carbonate, and 34.31% na, 17.93% c, and 47.76% o for sodium oxalate.)

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  1. e. 81.0%

    Explanation:

    Morphine has Molar mass = 285.34 g/mol

    Also, number of males of Morphine = Number of moles of HCl

    (At equivalence point )

    Now, number of moles of HCl = Molarity × volume

    =0.01 mol × 2.84×10^(-3) L

    = 2.84×10^(-5) mol

    therefore number of moles of morphine= 2.85×10^(-5)

    Mass of Morphine= No. of moles × molar mass

    = 2.84×10^(-5)×285.34 g/mol

    = 0.0081 g

    = 8.1 mg

    therefore

    percentage of morphine = mass of morphine/ mass of powder×100%

    = 8.1/10×100% = 81.0%

    hence option e is correct.

  2. 81.0% Calculate the number of moles of hcl used. 0.00284 l * 0.0100 = 0.0000284 mol = 2.84 x 10^-5 mol Now calculate the molar mass of C17H19NO3. First, lookup the atomic weights of the involved elements. Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight nitrogen = 14.0067 Atomic weight oxygen = 15.999 Molar mass = 17 * 12.0107 + 19 * 1.00794 + 14.0067 + 3 * 15.999 = 285.33646 g/mol Now determine the mass of 2.84 x 10^-5 moles of the substance. 2.84 x 10^-5 mol * 285.33646 g/mol = 0.008103555 g = 8.10 mg Finally, divide the now known mass of morphine by the mass of the sample, so 8.10 mg / 10.00 mg = 0.810 = 81.0%

  3. The white powder is sodium carbonate

    Explanation:

    If you supposed an ammount of 100 grams of the white powder,  it means there are 42.59 grams of Na, 12.02 grams of C and 44.99 grams of O. If you divided every compound by its molar mass, you will know the moles of every compound. (According to moles=mass (grams)/ molar mass (grams/mole)

    C=12.01 g/mole, Na=23 g/mole, O=16 g/mole

    So:

    C=12.02/12.01= 1.0 moles, Na=42.59/ 23= 1.85 moles, O= 44.99/16= 2.81 moles

    Knowing this, we can stimated that the white powder has 1 mol of C and approximately 2 and 3 moles of N and O, respectively

    Chemical formula for sodium carbonate is Na2CO3 and the formula for sodium oxalate is Na2C2O4. So the white powder is Na2CO3

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