Volume 2

Since the amount of the Ag

It might be tempting to choose the 50% KBr solution just because it is twice as concentrated as 25% LiCl. That would be a mistake. The mistake would be likely prompted by the deceiving mass percent concentration numbers. A solution that is more concentrated mass-wise might turn out to be less concentrated in terms of the number of molecules or ions it contains. Let us recall that the mass percent concentration refers to the gram amount of a solute in 100 g of solution. A simple calculation shows that 25 g of LiCl contains 1.4 times the number of molecules/ions as 50 g of KBr, because a molecule of KBr (MW = 119.0) is 2.8 times heavier than a molecule of LiCl (MW = 42.4).

In chemistry, it is often more important to know the number of particles such as molecules or ions in a given amount of solution than the mass of the solute. Of course, using molecular weights, we can always calculate which of two given solutions contains more molecules/ions, the way we just did for the 25% LiCl and 50% KBr. It would be much more convenient, though, to have solution concentration expressed in such a way that it specified the number of molecules of a solute rather than its mass.

Unsurprisingly, such a convenient measure of concentration has been developed and widely used for a long time. This type of concentration is

Moles? What moles? Certainly not those shown in Figure 2-118. In chemistry, the mole (abbreviated mol) is the unit of quantity of substance that contains 6.02 × 10

Why 6.02 × 10

In 1805,

Why are we talking grams? Why is the mole unit tied to the number of a.m.u.'s in 1 gram, not in 1 kilogram or 1 pound or 1 ton of a substance? Again, this is all about convenience. What is convenient and not convenient is determined by measurement capabilities and limitations of the instruments we have and use. Most laboratory balances are capable of accurately weighing masses ranging from a fraction of a gram to hundreds of grams. Ultimately, it is this mass range that dictates the choice of the mole as a unit tied to a mass on the order of a gram. For most substances, the mass of 1 mole varies from grams to hundreds of grams.

Chemistry is a practical science that is based on experimental studies. In chemical experiments, reagents need to be accurately measured for mixing in exact proportions. This is no different from other activities where some objects are used in a strict ratio, such as one nut per bolt in certain types of construction. We could certainly count out a dozen or so bolts and nuts one by one. What do we do however, if an accurate count is needed for hundreds, thousands, or even millions of identical objects? How did they count out the

Molecules and atoms are incomparably smaller than the tiniest bolt or nut ever made. Due to the miniscule size and mass of molecules, there is no practical way to count them out even by the millions or billions. Like in the bolt and nut example, we use the known mass of a single molecule (from the previously determined atomic masses) to weigh out

Picture a water tank in an area where water is scarce. The water in the tank is contaminated with 200 g of toxic BaCl

Both salts produced in this reaction are insoluble and will precipitate out. After filtration, the filtrate will be pure water. It is critical, however, that Ag

There are two different ways to solve this problem.

Knowing the quantity of BaCl

As follows from the equation, one molecule of BaCl

If 1 mol of BaCl

Note that both methods give the same result: 299.5 g of Ag

The convenience of the mole method is that once we know the amounts of reagents in moles, we can easily calculate not only their amounts in grams, but also the gram quantities of the products for a given reaction. Let us do that for the BaSO

0.9606 x 233.4 = 224.2 g for BaSO

and

0.9606 x 2 x 143.3 = 275.3 g for AgCl

As we are already aware of, it is important to know the limiting reagent in a reaction, the one that is totally consumed after the reaction has gone to completion. It is the limiting reagent that determines the quantity of reaction products formed. If the quantities of the reagents for a chemical reaction are given in a mass unit, such as gram, a calculation is needed to identify the one that is limiting. In contrast, it takes just a quick glance at the same quantities expressed in moles to recognize the limiting reagent.

For example, what is the limiting reagent in the reaction of 135 g of NaOH with 110 g of HCl? From the mass quantities and molecular weights (Figure 2-121, top), it is not evident, without a calculation, which of the two is in excess and which is limiting. With the amounts expressed in moles, however, it is immediately clear that the limiting reagent is HCl (3.0 mol) and that NaOH (3.4 mol) is used in excess (Figure 2-121, bottom).

There are certain advantages in doing chemical calculations in moles rather than in grams or another mass unit. Ultimately, however, it does not matter if one prefers to use the molecular mass method or the mole method for calculations, so long as the result is correct.

Many chemical reactions occur in solution. Such reactions are conveniently performed by mixing reagents in the form of pre-made solutions of known concentrations. First, let us refresh our memory on mass percent concentration, which refers to the amount of a solute in grams per 100 grams of solution.

In this formula, we have *grams* (marked in red) in both the numerator and denominator. By cancelling out the gram units we can see that mass percent concentration is expressed in percentage (%).

As mentioned above, molarity (depicted M) refers to the number of moles of a solute in one liter (L) of solution and is measured in mol/L, as is clear from the formula below.

As mentioned above, molarity (depicted M) refers to the number of moles of a solute in one liter (L) of solution and is measured in mol/L, as is clear from the formula below.

The superior convenience of using solutions of a known molar concentration is illustrated by the following simple example. Suppose we need to make batches of Cu(OH)_{2} in various quantities on a regular basis, several times a day. For the preparation, a solution of CuCl_{2} is treated with a solution of NaOH and the blue precipitate of insoluble Cu(OH)_{2} formed is separated by filtration, washed with water, and dried.

**CuCl**_{2} + 2 NaOH = Cu(OH)_{2}↓ + 2 NaCl

One way to deal with the assignment is each time we need to make a batch of Cu(OH)_{2}, we calculate the amounts of CuCl_{2} and NaOH needed for making the required quantity of the product, weigh out the reagents, and do the reaction.

There is, however, another, much more efficient and easy way of dealing with the task. We prepare two*stock solutions*, one of NaOH and the other of CuCl_{2}. As follows from the equation, per each molecule (mole) of CuCl_{2} we need two molecules (moles) of NaOH for the reaction. It would be handy to have a 1 molar (1M) solution of CuCl_{2} and a 2 molar (2M) solution of NaOH. The prepared solutions can be stored on the shelf and used at any moment to make Cu(OH)_{2} by simply mixing equal volumes of each. Moreover, it is very easy to calculate what volume of each solution is needed to make a particular quantity of Cu(OH)_{2}. For example, if 10 g of Cu(OH)_{2} is needed to be made, we convert the mass amount to the molar amount by dividing 10 g by 97.6 g, the mass of 1 mole of Cu(OH)_{2}. The result, 10/97.6 = 0.102, is the number of moles in 10 g of Cu(OH)_{2}. Consequently, 0.102 mol of CuCl_{2} and 0.204 mol of NaOH are needed for the reaction.

Now that we know the number of moles of each reagent, we can calculate the volumes of the stock solutions to be mixed together to make the required amount of the product. Again, we think proportions. The molarity of our CuCl_{2} stock solution is 1M, meaning that 1 L of the solution contains 1 mole of CuCl_{2}. We need the fraction volume *x* of this solution containing 0.102 mol of CuCl_{2}.

1 mol – contained in 1 L

0.102 mol – contained in*x* L

x = (1 x 0.102)/1 = 0.102 L = 102 mL. This is the volume of the CuCl_{2} stock solution to be used for the reaction.

Similarly, we come up with a proportion to find out the volume*y* of the 2M NaOH stock solution.

2 mol – contained in 1 L

0.204 mol – contained in*y* L

*y* = (1 x 0.204)/2 = 0.102 L = 102 mL. It is not surprising that the same volume of the NaOH stock solution is needed because its molarity (2M) is twice that of the CuCl_{2} solution (1M).

If a different gram quantity of Cu(OH)_{2} should be made, we just convert it to mol and calculate the volumes of the stock solutions to be used.

It is worth to reemphasize that molarity expresses the number of moles of a solute per__volume__ of solution. Why per volume, not mass, like for mass percent concentration? Because liquids are more convenient to measure by volume than by mass. The ease of accurately measuring volumes of solutions is the foundation for the broadly used important analytical method, called *volumetric analysis*.

One way to deal with the assignment is each time we need to make a batch of Cu(OH)

There is, however, another, much more efficient and easy way of dealing with the task. We prepare two

Now that we know the number of moles of each reagent, we can calculate the volumes of the stock solutions to be mixed together to make the required amount of the product. Again, we think proportions. The molarity of our CuCl

1 mol – contained in 1 L

0.102 mol – contained in

x

Similarly, we come up with a proportion to find out the volume

2 mol – contained in 1 L

0.204 mol – contained in

If a different gram quantity of Cu(OH)

It is worth to reemphasize that molarity expresses the number of moles of a solute per

Now let us practice a bit. There is 250 mL of an aqueous solution containing 20 g of NaOH. What is the molar concentration of this solution? First, we need to convert the mass amount of NaOH to moles. As the molecular weight of NaOH is 40 a.m.u., 1 mole of NaOH is 40 g. By denoting the number of moles in 20 g of NaOH as *x*, we come up with the following proportion.

40 g of NaOH – 1 mol

20 g of NaOH –*x* mol

*x* = (20 x 1)/40 = 0.5 mol

This is the number of moles in the given amount of NaOH (20 g). To calculate the molarity, we need to divide this number of moles by the volume of our solution. However, we need to express the volume in liters because molarity is the number of moles of a solute per__liter__ of solution. The given volume is 250 mL (milliliters). As "milli" denotes a factor of one thousandth, we convert the milliliters to liters by dividing 250 by 1,000. The volume in liters is therefore 250/1,000 = 0.25 L. Finally, we divide the number of moles of NaOH in the given solution by the volume of the solution in liters: 0.5 mol/0.25 L = 2 mol/L = 2M. The molarity of the given NaOH solution is 2M.

It is useful to learn how solutions of a certain molarity are prepared. For example, we need to make 1 L of a 1M aqueous solution of CuCl_{2}. The desired molarity suggests that one liter of the solution should contain 1 mole of CuCl_{2}, which is 134.5 g. Should we just weigh out 134.5 g of CuCl_{2} and dissolve it in 1 L of water? Not a very good idea. Why? Because molarity is the number of moles of a solute in 1 L of solution, __not__ in 1 L of solvent (water in this case). What is the difference, you might ask? Without going into details, the volume of water (or another solvent) sometimes contracts and sometimes expands on dissolution of a substance in it. In many cases, the change in volume is insignificant and can be neglected. But to be on the safe side and make sure that the molarity of our solution to be prepared is accurate, we need to use a special vessel called a **volumetric flask**. On the neck of every volumetric flask there is a mark the flask should be filled up to in order to accurately measure out the specified volume of a liquid. Since we need to prepare 1 L of a solution, we need a 1-liter volumetric flask.

After placing 134.5 g of CuCl_{2} in a 1-L volumetric flask, most of the water is added in portions with occasional gentle swirling to have the salt dissolve. Finally, water is slowly and carefully added to the mark. Watch this **video** to see how this is done.

In general, the standard procedure for making up a solution of a certain molarity is pretty straightforward. However, there is a trick involved when using CuCl_{2} as the solute. This salt exists in two forms, anhydrous (CuCl_{2}) and dihydrate (CuCl_{2}•2H_{2}O). The anhydrous salt is brown in color, whereas the dihydrate is blue (Figure 2-122). On contact with water, brown anhydrous CuCl_{2} is instantaneously converted to blue CuCl_{2}•2H_{2}O.

40 g of NaOH – 1 mol

20 g of NaOH –

This is the number of moles in the given amount of NaOH (20 g). To calculate the molarity, we need to divide this number of moles by the volume of our solution. However, we need to express the volume in liters because molarity is the number of moles of a solute per

It is useful to learn how solutions of a certain molarity are prepared. For example, we need to make 1 L of a 1M aqueous solution of CuCl

After placing 134.5 g of CuCl

In general, the standard procedure for making up a solution of a certain molarity is pretty straightforward. However, there is a trick involved when using CuCl

How can we prepare a 1M solution of CuCl

The same number of molecules in a given volume of

Imagine a developer building houses on one square mile of land with the goal of having as many houses built as possible. It is clear that to achieve the goal, houses with a smaller rather than larger footprint should be built. However, if there is a regulation stipulating that houses in that area must be separated by a distance of say at least 0.6 miles, the land will fit the same number of houses regardless of their footprint size (Figure 2-123).

The situation with molecules in gases is similar. The "regulation" set up by Mother Nature requires that the distance separating molecules in gases be much larger than the size of the molecules (Figure 2-124). It is this long distance between the molecules that controls their number per gas volume, not their size. Although, unlike houses on land, molecules in a gas are constantly and quickly moving in 3-dimensional space, the long distances between them are maintained.

If the same number of molecules of any gas occupies the same volume at the same temperature and under the same pressure, then

Those familiar with the

Until 1982, 0

Knowing the volume of one mole of a gas under certain conditions has important benefits, including simpler chemical calculations and analytical procedures. Here is an example. Quicklime (CaO) is used in large quantities to make steel and concrete. The production of quicklime is based on the high-temperature decomposition of limestone, naturally occurring calcium carbonate.

As quicklime is produced on a very large scale, it is important to know the purity of limestone in the deposit before mining it. One way of determining the percentage of CaCO

1. We write a balanced chemical equation for the decomposition of CaCO

2. We calculate the number of moles in 50 g of pure CaCO

3. Next, we note that 1 mol of CaCO

4. Given that 10.2 L of CO

At the end of this subsection, we will calculate both the volume and mass amount of oxygen a modern gasoline car consumes. Chances are that you will find the result of the calculation quite

We drive the distance of 365 km (227 miles) from New York City to Washington, DC. Our car's fuel economy is 7.84 L of gasoline per 100 km (~30 miles per gallon). The total amount of gas needed for this trip is: (7.84 L x 365 km)/100 km = 28.6 L. Knowing the density of gasoline (0.7 kg/L), we can calculate its mass: 28.6 L x 0.7 kg/L = 20 kg.

To calculate how much oxygen is needed to burn 20 kg of gasoline, we need a balanced equation for the combustion reaction. For that, the formula of gasoline is required. Although gasoline is a mixture of hydrocarbons (Volume 4), it is appropriate to use the formula of octane, C

The number of moles of gasoline consumed in the combustion is 175.7, as calculated by dividing its mass (20 kg = 20,000 g) by the molecular weight of C

For comparison, the passenger volume of a mid-size car is roughly 1/100 that of the air needed to get enough oxygen for the drive. Even more surprising might be the mass of O

Without these calculations, would we have ever guessed that as we drive, we consume that much atmospheric oxygen?

1. Calculate the number of moles in (a) 16 g of O

2. Calculate mass amounts in grams for (a) 1.5 mol of CaCO

3. Calculate the number of moles at 0

4. Calculate the density of xenon in g/L and g/cm

5. Determine the molarity of (a) 500 mL of solution containing 112.2 g of KOH; (b) 250 mL of solution containing 98.1 g of H

6. To 50 g of a 20% solution of AgNO

7. How would you prepare 0.5 L of a 2M solution of H

8. Do you think molar concentration could be converted to mass percent concentration and vice versa? Answer

9. How much CuSO

10. A chunk of lithium metal was thrown into water (excess). The volume of H