Hydrates

The most direct approach to conclude solving the problem would be to simply

express the given information as weight per volume and to evaluate the numerical data

as follows.

50.0 g

-------- = 20.0 g/dL or equivalently, 20.0

2.50 dL

Ratio and proportion could also be used to determine the concentration of the

solution obtaining the same results.

50.0 g

-------- = ----

2.50 dL

1 dL

(50.0 g)(1 dL) = (2.50 dL)(x g)

(50.0 g)(1 dL)

x g = ------------= 20.0 g

2.50 dL

Substituting for x in the original expression yields the desired results.

20.0 g

------ = -------- = 20.0 g/dL

1 dL

2-29. HYDRATES

Some salts come in several forms, such as the anhydrous form (no water) and in

the form of one or more hydrates. In a hydrate, a number of water molecules are

attached to each molecule of salt. The water that is attached to the salt contributes to

the molecular weight. Thus, if we were to weigh out equal amounts of a desired

chemical and one of its hydrates, the hydrate would not yield as much of the desired

chemical, per unit weight, as the anhydrous form, since some of the weight is

attributable to the water molecules. In some cases, the prescribed form of a salt may

not be available for the preparation of a solution. One must be able to determine how

much of the available form is equivalent to the quantity of the form prescribed. To do

this, we must first determine the amount of the prescribed form that is needed. Then,

using the molecular weights of both substances involved, use ratio and proportion to

determine the amount of available form needed.

a. Example 1. A procedure requires that 100 mL of a 10.0 solution be

prepared. Only CuSO4.H2O is available. How much of the hydrate is needed to

prepare this solution?

MD0837

2-11