Barrons SAT Subject Test Chemistry, 13th Edition

(Marvins-Underground-K-12) #1

  1. (F, F) The percent by mass of Ca in calcium citrate is about 24%. This is


estimated from the fraction , which is close to 25% (or ). The
numerator part of the fraction is found by recognizing that the molar mass of
calcium is 40 g and then tripling it due to the subscript of 3 on the calcium in
the formula given. The denominator part of the fraction is found by rounding
the molar mass of the compound, 498 g, to 500. By using similar thinking, the
percent by mass of calcium in calcium carbonate is found to be 40%. Since
the sample size is 1 gram for each compound, the mass of calcium in the
calcium citrate is only 0.24 g. In the calcium carbonate, though, the mass of
calcium is 0.40 g. These values make the first statement false. Additionally,
there are fewer Ca atoms in 1 mole of calcium carbonate (CaCO 3 ) than in
1 mole of calcium citrate (Ca 3 (C 6 H 5 O 7 ) 2 ) because each mole of calcium
citrate contains 3 times more Ca than each mole of calcium carbonate. This
makes statement II false as well.
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