MOLE FRACTION

15.3. MOLE FRACTION

The other temperature-independent concentration unit introduced in this chapter is the mole fraction. The mole fraction of a substance in a solution is the ratio of the number of moles of that substance to the total number of moles in the solution. The symbol for mole fraction of A is usually X A , although some texts use the symbol N A . Thus, for a solution containing x mol of A, y mol of B, and z mol of C, the mole fraction of A is

x mol A

X A = x mol A + y mol B + z mol C

EXAMPLE 15.7. What is the mole fraction of CH 3 OH in a solution of 12.0 g CH 3 OH and 25.0 g H 2 O? Ans.

3 OH = 0.375 mol CH

12.0 g CH 3 OH

3 OH

32.0 g CH 3 OH

25.0 g H

1.39 mol H

18.0 g H

X CH 3 3 OH =

0.375 mol CH OH

0.375 mol CH 3 OH + 1.39 mol H 2 O

Since the mole fraction is a ratio of moles (of one substance) to moles (total), the units cancel and mole fraction has no units.

EXAMPLE 15.8. Show that the total of both mole fractions in a solution of two compounds is equal to 1.

Equivalents are measures of the quantity of a substance present, analogous to moles. The equivalent is defined in terms of a chemical reaction. It is defined in one of two different ways, depending on whether an oxidation-reduction reaction or an acid-base reaction is under discussion. For an oxidation-reduction reaction, one equivalent is the quantity of a substance that will react with or yield 1 mol of electrons. For an acid-base reaction, one equivalent is the quantity of a substance that will react with or yield 1 mol of hydrogen ions or hydroxide ions . Note that the equivalent is defined in terms of a reaction, not merely in terms of the formula of a compound. Thus, the same mass of the same compound undergoing different reactions can correspond to different numbers of equivalents. The ability to determine the number of equivalents per mole is the key to calculations in this section.

EXAMPLE 15.9. How many equivalents are there in 82.0 g of H 2 SO 3 , in each of the following reactions?

2 NaOH −→ Na 2 SO 3 + 2H 2 O Ans.

(a) H 2 SO 3 + NaOH −→ NaHSO 3 + H 2 O

(b)

H 2 SO 3 +

Since 82.0 g of H 2 SO 3 is 1.00 mol of H 2 SO 3 , we will concentrate on that quantity of H 2 SO 3 . Both of these reactions are acid-base reactions, and so we define the number of equivalents of H 2 SO 3 in terms of the number of moles of hydroxide ion with which it reacts. In the first equation, 1 mol of H 2 SO 3 reacts with 1 mol of OH − . By definition, that quantity of H 2 SO 3 is 1 equiv. For that equation, 1 equiv of H 2 SO 3 is equal to 1 mol of H 2 SO 3 . In the second equation, 1 mol of H 2 SO 3 reacts with 2 mol OH − . By definition, that quantity of H 2 SO 3 is equal to 2 equiv. Thus,

in that equation, 2 equiv of H 2 SO 3 is 1 mol of H 2 SO 3 .

(a) 1 equiv = 1 mol

(b) 2 equiv = 1 mol

We can use these equalities as factors to change moles to equivalents or equivalents to moles, especially to calculate normalities (Sec. 15.5) or equivalent masses (Sec. 15.6).

EXAMPLE 15.10. How many equivalents are there in 82.0 g of H 2 SO 3 in the following reaction?

6H + + H 2 SO 3 + 3 Zn −→ H 2 S + 3 Zn 2+ + 3H 2 O

Ans. This reaction is a redox reaction (Chap. 14), and so we define the number of equivalents of H 2 SO 3 in terms of the number of moles of electrons with which it reacts. Since no electrons appear explicitly in an overall equation, we

will write the half-reaction in which the H 2 SO 3 appears: 6H + + H 2 SO 3 + 6e − −→ H 2 S+3H 2 O

It is now apparent that 1 mol of H 2 SO 3 reacts with 6 mol e − , and by definition 6 equiv of H 2 SO 3 react with 6 mol e − . Thus, 6 equiv equal 1 mol in this reaction. Since 82.0 g H 2 SO 3 is 1 mol, there are 6 equiv in 82.0 g H 2 SO 3 .

Some instructors and texts ask for the number of equivalents per mole of an acid or base without specifying a particular reaction. In that case, merely assume that the substance undergoes an acid-base reaction as completely as possible. State that assumption in your answers on examinations.

EXAMPLE 15.11. What is the number of equivalents in 4.00 mol H 2 SO 3 ?

Ans. Assuming that the H 2 SO 3 will react with a base to replace both hydrogen atoms, we have reaction (b) from

Example 15.9, and there are 2 equiv per mole. In 4.00 mol H 2 SO 3 , 4.00 mol H 2 SO 3 =

8.00 equiv

1 mol

The major use of equivalents stems from its definition. Once you define the number of equivalents in a certain mass of a substance, you do not need to write the equation for its reaction. That equation has already been used in defining the number of equivalents. Thus, a chemist can calculate the number of equivalents in a certain mass of substance, and technicians can subsequently use that definition without knowing the details of the reaction.

One equivalent of one substance in a reaction always reacts with one equivalent of each of the other substances in that reaction .

EXAMPLE 15.12. How many equivalents of NaOH does the 82.0 g of H 2 SO 3 react with in reaction (b) of Example 15.9? Ans.

Since there are 2 equiv of H 2 SO 3 , they must react with 2 equiv of NaOH. Checking, we see that 2 equiv of NaOH liberate 2 mol of OH − and also react with 2 mol H + , and thus there are 2 equiv by definition also.