Cliffs AP Chemistry, 3rd Edition

(c) Restatement: Compound F = CxHyGzor CxHyClz?

C H Clxy z+++O 2 __ggii"CO 2 H O 2 ^,h Cl 2 _gi

2 19.?. .?gg+++"3 88g0 80ggClg 2 _i

moles of carbon = moles of CO 2

.

.

.

g mole

gCO

44 01

388

• =

moles of hydrogen = 2 ×moles of H 2 O

.

.

.

g mole

gH O

2

18 02

080

#^21 0 088

:

• =

(2.19 grams of F)(1 mole F / 74.5 g F) = 0.0294 mole of compound F

This means that each mole of F contains 3 moles of C (0.0882 / 0.0294) and 3 moles of H

(0.088), or 39 grams of CH. This leaves 74.5 −39 = 36 grams, corresponding to 1 mole of

element G (Cl). Therefore, the empirical formula is C 3 H 3 Cl.

.

Gravimetrics