If reproducibility is 60.5 mg, product mass might have been 0.133 9 to 0.134 9 g. Find wt% Al(C9H6NO)3 if the product we

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If reproducibility is 60.5 mg, product mass might have been
0.133 9 to 0.134 9 g. Find wt% Al(C9H6NO)3 if the product weighed
0.133 9 g. (Answer: 30.27%)

If Reproducibility Is 60 5 Mg Product Mass Might Have Been 0 133 9 To 0 134 9 G Find Wt Al C9h6no 3 If The Product We 1 (233.48 KiB) Viewed 63 times

EXAMPLE A Problem with Two Components A mixture of the 8-hydroxyquinoline complexes of Al and Mg weighed 1.084 3 g. When ignited in a furnace open to the air, the mixture decomposed, leaving a residue of A1,0; and MgO weighing 0.134 4 g. Find the weight percent of Al(C.H.NO); in the original mixture. heat Al + Mg Al2O3 + Mgo O ΑΙΟ, FM 459.43 MgQ FM 312.61 FM 101.96 FM 40.304 x + y Solution We will abbreviate the 8-hydroxyquinoline anion as Q. Letting the mass of AIQ: be x and the mass of MgQ2 be y, we can write 1.084 3 g Mass of Mass of AIO MgQ2 The moles of Al are x/459.43, and the moles of Mg are y/312.61. The moles of A1,03 are one-half of the total moles of Al, because it takes 2 mol of Al to make 1 mol of Al2O3. Moles of AlO3 = = () x 459.43 CHAPTER 27 Gravimetric and Combustion Analysis The moles of MgO will equal the moles of Mg = y/312.61. Now we can write Mass of AlO3 Mass of Mgo () .X (101.96) + 459.43 y (40.304) = 0.1344 g 312.61 mol AlO3 g ALLO mol ALO g MgO mol Mgo mol Mgo Substituting y = 1.084 3 – x into the preceding equation gives 1.084 3 (101.96) + (40.304) = 0.1344 g 459.43 312.61 from which we find x = 0.300 3 g, which is 27.70% of the original mixture. (X ). +) 40 TEST YOURSELF If reproducibility is +0.5 mg, product mass might have been 0.133 9 to 0.134 9 g. Find wt% Al(C,H.NO); if the product weighed 0.133 9 g. (Answer: 30.27%). Please appreciate the huge uncertainty: A 0.5-mg difference in mass of product gives a 9% difference in calculated composition of the mixture.