A 1.25-g sample contains some of the very reactive compound Al (C6H5) 3. On treating the compound with aqueous HCl, 0.951 g of C6H6 is obtained.

Al (C6H5) 3 (s) + 3HCl (aq) - - > AlCl3 (aq) + 3C6H6 (l)

Assuming that Al (C6H5) 3 was converted completely to products, what is the weight percent of Al (C6H5) 3 in original 1.25-g sample?

83.9% First, determine the molar masses of Al (C6H5) 3 and C6H6. Start by looking up the atomic weights of the involved elements. Atomic weight aluminum = 26.981539 Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Molar mass Al (C6H5) 3 = 26.981539 + 18 * 12.0107 + 15 * 1.00794 = 258.293239 g/mol Molar mass C6H6 = 6 * 12.0107 + 6 * 1.00794 = 78.11184 g/mol Now determine how many moles of C6H6 was produced Moles C6H6 = 0.951 g / 78.11184 g/mol = 0.012174851 mol Looking at the balanced equation, it indicates that 1 mole of Al (C6H5) 3 is required for every 3 moles of C6H6 produced. So given the number of moles of C6H6 you have, determine the number of moles of Al (C6H5) 3 that was required. 0.012174851 mol / 3 = 0.004058284 mol Then multiply by the molar mass to get the number of grams that was originally present. 0.004058284 mol * 258.293239 g/mol = 1.048227218 g Finally, the weight percent is simply the mass of the reactant divided by the total mass of the sample. So 1.048227218 g / 1.25 g = 0.838581775 = 83.8581775% And of course, round to 3 significant digits, giving 83.9%