Because the gas is 97.2% by mass chlorine, the fraction of chlorine in the gas can be
found as follows:%.
.(.)
whole %part
g total chlorineg gas(^100117)
0 802 0 972
##= 100 = 66.6% gas
fraction of chlorine in the solid product = 100.00% −66.6% = 33.4% solid
(c) Restatement: Formula of solid product.
You know from the question that the solid product contains Sb, Cl, and O atoms. The
weight of Sb can be found by taking the weight of antimony chloride (which has all of the
antimony atoms in it) and getting rid of the weight of chlorine atoms, which you have de-
termined to be 1.17 g. Therefore, 2.51 g SbCl 3 −1.17 grams of chlorine atoms = 1.34
grams of antimony.
The weight of chlorine in the solid product can be determined by taking the weight of
chlorine in the original compound (which has all of the chlorine atoms in it) and multiply-
ing it by the percent of chlorine found in the solid product. This becomes 1.17 grams of
chlorine ×0.334 = 0.391 g chlorine in the solid product.
The weight of oxygen in the solid product can be found by taking the total weight of the
solid product and subtracting the amount of antimony and chlorine previously determined.
This becomes 1.906 g solid product −1.34 g antimony −0.391 g chlorine = 0.175 g oxy-
gen atoms in the solid product.
Expressing these weights as moles yields.
1.34 g Sb →0.0110 mole Sb
0.175 g O →0.0109 mole O
0.391 g Cl →0.0110 mole Cl
Thus, they are in essentially a 1:1:1 molar ratio, which indicates the molecular formula
SbOCl.
(d) Given: Empirical formula is true formula.
Restatement: Balanced equation.
SbCl 3 (s) + H 2 O(,) →SbOCl(s) + 2 HCl(g)
Part IV: AP Chemistry Practice Test