Page 172 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 10] STOICHIOMETRY 161

10.70. A sample of a hydrocarbon (a compound of carbon and hydrogen only) is burned, and 1.10 g CO 2 and 0.450 g H 2 O
are produced. What is the empirical formula of the hydrocarbon?
Ans. The empirical formula can be determined from the ratio of moles of carbon atoms to moles of hydrogen
atoms. From the masses of products, we can get the numbers of moles of products, from which we can get
the numbers of moles of C and H. The mole ratio of these two elements is the same in the products as in the
original compound.

1 mol CO 2
1.10gCO
2 = 0.0250 mol CO 2
44.0gCO 2

1 mol H 2 O
0.450gH O = 0.0250 mol H 2 O
2
18.0gH O
2
0.0250 mol CO 2 contains 0.0250 mol C
0.0250 mol H 2 O contains 0.500 mol H
The mole ratio is 1:2, and the empirical formula is CH 2 .
10.71. How many moles of each substance are present in solution after 0.100 mol of NaOH is added to 0.200 mol of
HC 2 H 3 O 2 ?
Ans. The balanced equation is
NaOH + HC 2 H 3 O 2 −→ NaC 2 H 3 O 2 + H 2 O
The limiting quantity is NaOH, and so 0.100 mol NaOH reacts with 0.100 mol HC 2 H 3 O 2 to produce
0.100 mol NaC 2 H 3 O 2 + 0.100 mol H 2 O. There is also 0.100 mol excess HC 2 H 3 O 2 left in the solution.
10.72. Combine Figs. 7-5 and 10-4 to show all the conversions learned so far.
Ans. The answer is shown on the factor-label conversion figure, page 348. It includes the boxes numbered 3 to 8,
11 to 14, 16 and 17, plus all the conversion factors on the arrows between them.