STOICHIOMETRY:REAL WORLD REACTION

Summary (or Welcome to the Real World)

We don't live in an ideal world. Humans aren't able to measure exact quantities of substances; not even specially designed machines are completely reliable. In any reaction, there will always be one reactant that you should have added more of. However, even if you were a superhuman and could measure perfectly, you still would be unsure of the results of your reaction.

Chemical reactions never completely consume reactants. You will always only get a percentage of the product you think you will. Chemical reactions are not just equations that you write down on paper; they are living works of nature that produce heat and energy.

This lesson provides you with all the real-life details you need to know about Stoichiometry. You will learn about limiting reagents, percent yield,heat of reaction, and energy changes.

Terms

Actual yield - The amount of product that is actually produced in a chemical reaction.

Endothermic reaction - A reaction that requires heat energy to be carried to completion.

Enthalpy - The amount of heat a substance has at a given temperature and pressure.

Excess reagent - The reactant that is in excess in a chemical reaction. There is more than enough of it to react with the other reactant(s).

Exothermic reaction - A reaction that releases energy in the form of heat.

Heat of reaction - The change in enthalpy in a chemical reaction.

Limiting reagent - The reactant that limits or determines the amount of product that can be formed in a chemical reaction.

Percent yield - The decimal percentage resulting from dividing actual yield by theoretical yield.

Standard heat of formation - Unique to every substance, it is a measure of the change in enthalpy in the reaction that produces 1 mole of the substance from its constituent elements.

Theoretical yield - The calculated or expected amount of product to be formed in a chemical reaction.

Thermochemical reaction - Chemical reactions that include the amount of heat energy produced or absorbed.

Formulas

Formula for Percent Yield

Percent Yield =

δH	δH = δH f (products) - δH f (reactants)

Conversion Factor between kJ and kcal

STOICHIOMETRY

Summary

This first, short introductory SparkNote to Stoichiometry is meant to give you a very basic and general understanding of what stoichiometry is. It also introduces the most important concept of stoichiometry: the mole.

Terms

Avogadro's number - 6.02×10²³ ; the number of particles that make up the measure of 1 mole.

Chemically engineered - Produced by chemical means.

Mole - A measure of amount. 1 mole equals 6.02×10²³ items of something. The number is so massive because moles are used to measure the amount of molecules, atoms, and other incredibly miniscule particles.

Stoichiometry - The calculation of quantities in chemical equations.

Overview

What is Stoichiometry?

Stoichiometry is at the heart of the production of many things you use in your daily life. Soap, tires, fertilizer, gasoline, deodorant, and chocolate bars are just a few commodities you use that are chemically engineered, or produced through chemical reactions. Chemically engineered commodities all rely on stoichiometry for their production.

But what is stoichiometry? Stoichiometry is the calculation of quantities in chemical equations. Given a chemical reaction, stoichiometry tells us what quantity of each reactant we need in order to get enough of our desired product. Because of its real-life applications in chemical engineering as well as research, stoichiometry is one of the most important and fundamental topics in chemistry.

Introduction to the Mole

Which weighs more, 100 pounds of feathers or 100 pounds of bowling balls? You've probably heard this riddle before. Obviously they both weigh the same since I told you I have 100 pounds of each. But if I have 100 pounds of bowling balls and 100 pounds of feathers, do I have more feathers or more bowling balls? The quantities of feathers and bowling balls would not be equal. An individual feather weighs a lot less than a bowling ball. It would take only about 10 bowling balls to get 100 pounds whereas it would take a LOT more feathers.

When you measure quantities in moles, however, the situation is exactly opposite from when you measure according to weight. A mole is defined as the amount of a substance. More specifically, there are 6.02×10²³particles in a mole of substance. Therefore, if you had 1 mole of feathers and 1 mole of bowling balls, you would have 6.02×10²³ feathers and6.02×10²³ bowling balls. Now suppose you were asked the question, "Which weighs more, 100 moles of feathers or 100 moles of bowling balls?" The answer this time would be the bowling balls. Although there is an equal number of both feathers and bowling balls, an individual bowling ball weighs much more than an individual feather, and so an equal number of bowling balls would weigh more than an equal number of feathers.

Now, let's return to the number 6.02×10²³ . This number is known as Avogadro's number and you should definitely commit it to memory. You are probably wondering why it's so large, and it does indeed look intimidating without the exponential notation:

6.02×1023 = 60, 200, 000, 000, 000, 000, 000, 000, 000

Although you will never have a mole of bowling balls, you will soon be calculating moles of compounds, molecules, atoms, and ions. These representative particles are extremely and incredibly small. That is why there are so many particles in a mole of substance. When you appreciate just how small these particles are, 6.02×10²³ stops seeming like such a crazy number.

Calculation

Summary

In this lesson you will learn all about conversion factors and how to use them. You will use them to solve general problems and then to solve your first stoichiometry problems. Most importantly, you will learn the four steps necessary to solve ANY stochastic problem:

1. Balance the equation
2. Convert units of given substance to moles
3. Find moles of wanted substance using mole ratio
4. Convert moles of wanted substance to desired units

Only move on to the next lesson after you:

1. Completely understand the steps involved in stoichiometric problems.
2. Memorize the three all-important conversion factors in the formulas section of the lesson.

Terms

Conversion Factor - A ratio (or fraction) that represents the relationship between two different units.

Formula Unit - The representative particle of a substance. It is the smallest unit of a substance that still retains that substance's properties and is the simplest way to write the formula of the substance without coefficients.

Gram Formula Mass - The mass of one mole of compound.

Gram Molecular Mass - The mass of one mole of a molecular substance.

Mole - A measure of amount. 1 mole equals 6.02×10²³ items of something. The number is so massive because moles are used to measure the amount of molecules, atoms, and other incredibly miniscule particles.

Mole ratio - The ratio between any two constituents in chemical equation

Standard Temperature and Pressure - The arbitrarily decided standard conditions at which experiments are done: 1 atm and 273 degrees Kelvin (0 degrees Celsius).

Formulae

General Conversion Factor Equation -

Conversion Factor =

Converting from Formula Units to Moles -

Moles =

Converting from Volume of a Gas to Moles -

moles =

Converting from Mass to Moles -

Moles =

Converting from Molality of a Solution to Moles - Moles = molality × kilograms of solution

Converting from Molarity of a Solution to Moles - Moles = molarity × liters of solution

Conversion Factors

The easiest way to do stoichiometric calculations involves using conversion factors. A conversion factor is a ratio (or fraction) which represents the relationship between two different units. A conversion factor is ALWAYS equal to 1. Here are some examples of conversion factors:

All these conversion factors are equal to 1. If it doesn't seem obvious at first, think about it for a second. Realize that 1 minute is equivalent to 60 seconds. Simply replace 1 minute in the fraction with its equivalent 60 seconds and it becomes clear that 60 seconds / 60 seconds = 1. Numerator and denominator are equivalent; they are just expressed differently.

As you can see it is extremely important to keep track of your units when using conversion factors. Without units, the first fraction would be 1 / 60. This is not equal to 1 and could very easily lead to wrong answers.

Furthermore, when you use units, you make it very easy to check your work. For example, perhaps you are trying to find out how many dozen eggs you have to buy to make three cakes. If you're getting an answer of 12 dozen eggs you might want to check your work. Could you even fit 12 of those cartons in your refrigerator? If you look back on your calculations you may immediately see the incorrect conversion factor: 1 egg / 12 dozen. It is easy to see that this is where the error occurred since this does NOT equal 1.

How do you use Conversion Factors?

We all know from elementary school math that if you multiply any quantity by 1 you get the same quantity back. You can do this as many times as you want. For example, 2×1 = 2 , and 18×1×1×1 = 18 .

Multiplication by 1 is what you do whenever you do a problem involving conversion factors. The best way to explain how to solve using conversion factors is to work through some simple examples.

Problem: How many days are there in 3 years? (Assume none of these years are leap years)

Solution: Here we basically want to convert years to days. Our conversion factor is:

Since this is equivalent to 1, multiplication of this ratio with our original value will only change its units and not its magnitude. Therefore:

3 years× = 1, 095 days

Notice that years is on the bottom of the conversion factor. This is VERY important. You always want to have the units of what you currently have on the bottom of the conversion factor and the units you want on the top.

Conversion Factor =

In this case we are multiplying our conversion factor by years. We therefore put years on the bottom of the conversion factor. When multiplied together, the resulting fraction has years in both numerator AND denominator. These units can now "cancel each other out". How? You might want to think about it like this. When you see the fraction 2 / 2, you cancel the 2s in both numerator and denominator. You can do the same thing with units.

When doing any type of problem involving conversion factors, feel free to draw a line through any unit you see on the top and bottom of the fraction to make it visually obvious that the units cancel.

3 years× = 1, 095 days

Canceling units in this way makes it much easier to check your work. The units you want in your answer should be the only unit not to cancel. If your calculations yield other units, which cannot be cancelled, you have made a mistake likely resulting from a missing conversion factor or an upside down conversion factor that needs to be flipped.

Stoichiometric Calculations

Applying Conversion Factors to Stoichiometry

Now you're ready to use what you know about conversion factors to solve some stoichiometric problems in chemistry. Almost all stoichiometric problems can be solved in just four simple steps:

1. Balance the equation.
2. Convert units of a given substance to moles.
3. Using the mole ratio, calculate the moles of substance yielded by the reaction.
4. Convert moles of wanted substance to desired units.

These "simple" steps probably look complicated at first glance, but relax, they will all become clear.

Let's begin our tour of stoichiometry by looking at the equation for how iron rusts:

Fe + O2→Fe2O3

Step 1. Balancing the Equation

The constituent parts of a chemical equation are never destroyed or lost: the yield of a reaction must exactly correspond to the original reagents. This fact holds not just for the type of elements in the yield, but also the number. Given our unbalanced equation:

Fe + O2→Fe2O3

This equation states that 1 iron (Fe) atom will react with two oxygen (O) atoms to yield 2 iron atoms and 3 oxygen atoms. (The subscript number, such as the two inO₂ describe how many atoms of an element are in a molecule.) This unbalanced reaction can't possibly represent a real reaction because it describes a reaction in which one Fe atom magically becomes two Fe atoms.

Therefore, we must balance the equation by placing coefficients before the various molecules and atoms to ensure that the number of atoms on the left side of the arrow corresponds exactly to the number of elements on the right.

4Fe +3O2→2Fe2O3

Let's count up the atoms in this new, balanced version of the reaction. On the left of the arrow we have 4 atoms of iron and 6 atoms of oxygen (since 3×2 = 6 ). On the right we also have 4 iron (since 2×2 = 4 ) and 6 oxygen ( 2×3 = 6 ). The atoms on both sides of the equation match.

The process of balancing an equation is basically trial and error. It gets easier and easier with practice. You will likely start to balance equations almost automatically in your mind.

Step 2. Converting Given Units of a Substance to Moles

The process of converting given units into moles involves conversion factors. Below we will provide the most common and important conversion factors to convert between moles and grams, moles and volumes of gases, moles and molecules, and moles and solutions. These conversion factors function in the same way as those discussed in the previous section Note also that though these conversion factors focus on converting from some other unit to moles, they can also be turned around, allowing you to convert from moles to some other unit.

Converting from Grams to Moles

The gram formula mass of a compound (or element) can be defined as the mass of one mole of the compound. As the definition suggests, it is measured in grams/mole and is found by summing the atomic weights of every atom in the compound. Atomic weights on the periodic table are given in terms of amu (atomic mass units), but, by design, amu correspond to the gram formula mass. In other words, a mole of a 12 amu carbon atom will weigh 12 grams.

The gram formula mass can be used as a conversion factor in stoichiometric calculations through the following equation:

Moles =

Gram formula mass is also known as GFM. You may also see the term gram molecular mass, abbreviated GMM. This term is often used instead of GFM when the substance is molecular and not ionic. However, only the terminology is different, GMM is used in the same way as GFM. Therefore, I will use the catch- all term GFM in this study guide.

Converting between Volume of a Gas and Moles

The Ideal Gas law, discussed at length in the Sparknote on Gases, provides a handy means of converting between moles and a gas, provided you know certain qualities of that gas. The Ideal Gas Law is PV = nRT , withn representing the number of moles. If we rearrange the equation to solve for n , we get:

n =

with P representing pressure in atm, V representing volume in liters, T representing temperature in Kelvins, and R the gas constant, which equals .0821 L-atm/mol-K. Given P , V , and T , you can calculate the number of moles of substance in a gas.

In those instances when a problem specifies that the calculations are to be made at STP (Standard Temperature and Pressure; P = 1 atm, T = 273 K)), the problem becomes even simpler. At STP, a mole of gas will always occupy 22.4 L of volume. If you are given a volume of a gas at STP, you can calculate the moles in that gas by calculating the volume you are given as a fraction of 22.4 L. At STP, 11.2 L of a gas will be .5 moles; 89.6 L of gas will be 4 moles.

Converting between Individual Particles and Moles

Avogadro's Number provides the conversion factor for moving from number of particles to moles. There are 6.02×10²³ formula units of particles in every mole of substance, with formula unit describing the substance we are looking at, whether it is a compound, molecule, atom, or ion. A formula unit is the smallest unit of a substance that still retains that substance's properties and is the simplest way to write the formula of the substance without coefficients. Some representative formula units are listed below.

• Compounds: Cu₂S , NaCl
• Molecules: N₂ , H₂
• Atoms: Fe, Na
• Ions: Na⁺(aq) , Cl^-(aq)

Since 1 mole = 6.02×10²³ formula units, the conversion from formula units to moles is simple:

Moles =

Converting between Solutions and Moles

Solutions are discussed in much greater detail in the series of Solutions SparkNotes. But it is possible, and fairly easy to convert between the measures of solution (molarity and molality) and moles.

Molarity is defined as the number of moles of solute divided by the number of liters of solvent. Rearranging the equation to solve for moles yields:

Moles = molarity × liters of solution

MolaLity is defined as the number of moles of solute divided by the number of kilograms of solvent. Rearranging the equation to solve for moles yields:

Moles = molality × kilograms of solution

Using the Mole Ratio to Calulate Yield

Before demonstrating how to calculate how much yield a reaction will produce, we must first explain what the mole ratio is.

The Mole Ratio

Let's look once again at our balanced demonstration reaction:

4Fe +3O2→2Fe2O3

The coefficients in front of iron, oxygen, and iron (III) oxide are ratios that govern the reaction; in other words, these numbers do not demand that the reaction can only take place with the presence of exactly 4 moles of iron and 3 moles of oxygen, producing 2 moles of iron (III) oxide. Instead, these numbers state the ratio of the reaction: the amount of iron and oxygen reaction together will follow a ratio of 4 to 3. The mole ratio describes exactly what its name suggests, the molar ratio at which a reaction will proceed. For example, 2 moles of Fe will react with 1.5 moles of O₂ to yield 1 mole of Fe₂O₃ . Alternatively, 20 moles of Fe will react with 15 moles ofO₂ to yield 10 moles of Fe₂O₃ . Each of these examples of the reaction follow the 4:3:2 ratio described by the coefficients.

Now, with a balanced equation, the given units converted to moles, and our understanding of the mole ration, which will allow us to see the ratio of reactants to each other and to their product, we can calculate the yield of a reaction in moles. Step 4 demands that we be able to convert from moles to back to the units requested in a specific problem, but that only involves turning backwards the specific converstion factors described above.

Sample Problems

Problem: Given the following equation at STP:

N2(g) + H2(g)→NH3(g)

Determine what volume of H₂(g) is needed to produce 224 L of NH₃(g).

Solution:
br> Step 1: Balance the equation.

N2(g) + 3H2(g)→2NH3(g)

Step 2: Convert the given quantity to moles. Note in this step, 22.4 L is on the denominator of the conversion factors since we want to convert from liters to moles. Remember your conversion factors must always be arranged so that the units cancel.

= 10 moles of NH3(g)

Step 3: mole ratio.

= 15 moles H2(g)

Step 4, convert to desired units:

= 336 L H2(g)

Now for a more challenging problem:

Given the following reaction:

2H2S(g) + O2(g)→SO2(g) + 2H2O(s)

How many atoms of oxygen do I need in order to get 18 g of ice?

Solution

Step 1. The equation is partially balanced already, but let's finish the job.

2H2S(g) +3O2(g)→2SO2(g) + 2H2O(s)

Step 2, convert to moles:

1 formula unit of H₂O has 2 atoms of H and 1 atom of O
The atomic mass of H is 1 gram/mole
Atomic mass of O = 16 grams/mole

GFM of H2O(s) = + = 18 grams / mole

×1 mole = 1 mole of H2O(s)

Step 3, mole ratio:

×3 moles O2(g) = 1.5 moles O2(g)

Step 4, convert to desired units:

= 9.03×1023 molecules O2(g)

Is this the answer? No. The question asks for ATOMS of oxygen. There are two atoms of oxygen in each molecule of O₂(g).

×2 atoms O = 1.806×1024 atoms O

Now we're done. Note how important it was to write out not only your units, but what substance you're currently working with throughout the problem. Only a brief check was needed to ascertain if we were really answering the given question. Always check to make sure you have answered the correct question.

SOLUBILITY OF SOLUTION

Introduction and Summary

Have you ever wondered why oil and water don't mix while sugar and water do? The answer to that question, and many others, comes from an analysis of the factors affecting solubility--the propensity of a solute to form a solution with a given solvent. A thermodynamic cycle constructed for the formation of a solution shows that a solution will form when the strength of the interaction (from dipole-dipole, van der Waals, hydrogen bonding, etc.) between solvent and solute is greater than the energies of the separated solute and solvent. The strength of that solute-solvent interaction is largely determined by the structures of the solvent and the solute. Like dissolves like--only solutes with similar properties or structural features in common with the solvent will be dissolved.

The solubility of solids in aqueous solution, generally increases with the temperature of the solution. There are a few exceptions to this rule, such as Na₂SO₄ and all gasses. That temperature effect is governed by the entropy of solution. Solids become less ordered when dissolved in water giving a positive ΔS_soln. Because ΔG = ΔH - TΔS, a positive entropy term will lead to an increase in solubility with increasing temperature. Gasses become more ordered when dissolved in water (due to the large negative change in volume for the gas) so they have negative entropies of solvation Therefore, gasses have a decreasing solubility with increasing temperature.

However, it is possible to increase the solubility of a gas in a given solvent by increasing the pressure of the gas above the solvent. That effect reduces the size of the negative entropy of solvation for gasses by increasing the order of the gas in the gas phase. The equation that describes this effect is called Henry's law.

Terms

Concentration - The amount of solute dissolved in a given amount of solvent.

Dissolve - To become a part of a solution.

Enthalpy of Solution - The amount of heat absorbed upon the formation of a solution by a solute and a solvent.

Entropy of Solution - The amount of disorder created by the formation of a solution by a solute and a solvent.

Henry's Law - The solubility of a gas in a solvent is proportional to the pressure of the gas above the solvent.

Le Chatelier's Principle - A system at equilibrium will react to a change by opposing that change.

Non-polar - A molecule with a low net dipole.

Polar - A molecule with a large net dipole.

Solubility - The amount of a particular solute that can dissolve in a given amount of a particular solvent. Solubilities are generally listed in g / L.

Solute - A minor component of a solution.

Solution - A homogeneous mixture.

Solvent - The major component of a solution.

Solubility

Solution Formation

To understand why things dissolve at all, we will look at the solution formation process from a thermodynamic point of view. shows a thermodynamic cycle that represents the formation of a solution from the isolated solute and solvent. From Hess's law we know that we can add the energies of each step in the cycle to determine the energy of the overall process. Therefore, the energy of solution formation, the enthalpy of solution, equals the sum of the three steps--ΔH_soln = ΔH₁ + ΔH₂ + ΔH₃.

Figure %: The Solution Formation Process

ΔH₁ and ΔH₂are both positive because it requires energy to pull molecules away from each other. That energy cost is due to the intermolecular forces present within any solute or solvent. The forces acting between molecules such as CH₃Cl are largely van der Waals and dipole-dipole interactions. Some molecules that contain O-H, N-H, or F-H bonds can form hydrogen bonds that are relatively strong intermolecular forces. Ions of opposite charge, such as in a crystal of NaCl, are attracted to each other because of electrostatic forces. Each of those forces increase with decreasing distance. Therefore, it should make sense that it costs energy to pull molecules and ions away from each other. When the expanded form of the solvent and the solute are combined to form a solution, energy is released, causing ΔH₃ to be negative. This makes sense because the solute and solvent can interact through the various types of intermolecular forces.

What determines the enthalpy of solution is, therefore, the difference between the energy required to separate the solvent and solute and the energy released when the separated solvent and solute form a solution. To restate that in simpler terms, solutions will form only when the energy of interaction between the solvent and solute is greater than the sum of the solvent-solvent and solute-solute interactions. That situation can only occur when the solvent and solute have similar properties. For example, if a non-polar molecule, such as oil, is mixed with a polar molecule like water, no solution forms. Water's solvent-solvent intermolecular interactions are mostly hydrogen bonds and dipole-dipole while oil has only van der Waals. Water can satisfy its hydrogen bonds and become stabilized by dipole-dipole interactions only when near other water molecules. Therefore, water is destabilized when it forms a solution with oil. That is why such a solution will never form between oil and water. Therefore, the primary rule of solubility is that like dissolves like. Only when the solute and solvent molecules have several common structural features such as their polarities will a solution form.

Pressure and Temperature Effects

If you have not yet studied thermodynamics or you do not know what ΔG or ΔS stands for, then please skip to the next heading in which the following discussion on temperature and pressure effects on solubility is summarized without the thermodynamics. The following discussion is a slightly more advanced treatment of the same phenomena.

The creation of disorder during the solution formation process is its essential driving force. In fact, most compounds that are soluble in water have positive enthalpies of solution. The only reason why those solutions form is due to the positive entropy of solution, ΔS_soln. As shows, both the solvent and the (solid or liquid) solute become less ordered upon solution formation. Therefore, from the equation ΔG = ΔH - TΔS we should predict that the solubility of every compound should increase with increasing temperature. That prediction turns out to be correct for nearly every solvent and solute. However, there are some exceptions, such as sodium sulfate in water that actually become less soluble at higher temperatures. That is usually due to their negative entropies of solution. The reason why some solutions have a negative entropy of solution is beyond the scope of this SparkNote. If you wish to pursue this topic further, search for the hydrophobic effect.

Using the idea of the entropy of solution, we can predict other properties of solutions. For example, we should predict that all gasses should bcome less soluble in water with increasing temperature because they have a negative entropy of solution. Gasses have a negative entropy of solution in water because they are confined to a smaller volume when dissolved as compared to their volumes as gasses. As we should predict, all gasses become less soluble in water with increasing temperature.

To make gasses more soluble in water, we could think about trying to decrease the magnitude of their negative entropies of solution. One way to accomplish that is to make the gas above the solvent more ordered by increasing the pressure of the gas. In fact, William Henry discovered this property of gaseous solutes around the turn of the nineteenth century. Because he discovered that behavior first, the law that describes the increase in the solubility of a gas with increasing pressure is called Henry's law. Henry's law is given below in , C stands for the concentration of the dissolved gas and P represents the partial pressure of the gas above the solution. The units on the constant, k, are adjusted to suit the pressure and concentration units used.

Figure %: Henry's Law

Summary of Factors Affecting Solubility

Normally, solutes become more soluble in a given solvent at higher temperatures. One way to predict that trend is to use Le Chatelier's principle. Because ΔH_soln is positive for most solutions, the solution formation reaction is usually endothermic. Therefore, when the temperature is increased, the solubility of the solute should also increase. However, there are solutes that do not follow the normal trend of increasing solubility with increasing temperature. One class of solutes that becomes less soluble with increasing temperature is the gasses. Nearly every gas becomes less soluble with increasing temperature.

Another property of gaseous solutes in summarized by Henry's law (see ) which predicts that gasses become more soluble when their pressures above a liquid solution are increased. That property of gaseous solutes can be rationalized by using Le Chatelier's principle. Imagine that you have a glass of water inside of a sealed container filler with nitrogen gas. If the size of that container were suddenly halved, the pressure of nitrogen would suddenly double. To decrease the pressure of nitrogen above the solution (as is required by Le Chatelier's principle), more nitrogen gas becomes dissolved in the glass of water.

COLLIGATIVE PROPERTIES OF SOLUTION

Introduction and Summary

Solutions, especially liquid solutions, generally have markedly different properties than either the pure solvent or the solute. For example, a solution of sugar in water is neither crystalline like sugar nor tasteless like water. Some of the properties unique to solutions depend only on the number of dissolved particles and not their identity. Such properties are called colligative properties. The colligative properties we will consider in this SparkNote are vapor pressure lowering, freezing point depression, boiling point elevation, and osmotic pressure.

When a nonvolatile solute is dissolved in a solvent, the vapor pressure of the resulting solution is lower than that of the pure solvent. The amount of the vapor pressure lowering is proportional to the amount of solute and not its identity. Therefore, vapor pressure lowering is a colligative property. The equation that describes that phenomenon is called Raoult's law.

Boiling point elevation is a colligative property related to vapor pressure lowering. The boiling point is defined as the temperature at which the vapor pressure of a liquid equals the atmospheric pressure. Due to vapor pressure lowering, a solution will require a higher temperature to reach its boiling point than the pure solvent.

Every liquid has a freezing point--the temperature at which a liquid undergoes a phase change from liquid to solid. When solutes are added to a liquid, forming a solution, the solute molecules disrupt the formation of crystals of the solvent. That disruption in the freezing process results in a depression of the freezing point for the solution relative to the pure solvent.

When a solution is separated from a volume of pure solvent by a semi-permeable membrane that allows only the passage of solvent molecules, the height of the solution begins to rise. The value of the height difference between the two compartments reflects a property called the osmotic pressure of a solution. As you know, if you add more solvent to a solution, the two mix together to form a more dilute solution. The same forces allowing that mixing serve to force solvent molecules from the pure solvent compartment across the membrane into the solution compartment causing the change in volume. The amount of osmotic pressure is directly related to the concentration of the solute. That is because more concentrated the solutions have greater potentials for dilution.

Terms

Colligative Property - A property that depends only on the amount of solute in a solution and not the identity of the solute.

Concentrated - A solution of high concentration.

Concentration - The amount of solute dissolved in a given amount of solvent.

Dilute - A solution of low concentration or the process of lowering the concentration of a solution by the addition of solvent.

Dilution - The decrease of the concentration of a solution by the addition of more solvent.

Dissolve - To become a part of a solution.

Nonvolatile - A substance that does not have an appreciable vapor pressure.

Raoult's Law - The vapor pressure of a solution is the product of the mole fraction of the solvent and the vapor pressure of the pure solvent.

Solubility - The amount of a particular solute that can dissolve in a given amount of a particular solvent. Solubilities are generally listed in g / L.

Solute - A minor component of a solution.

Solution - A homogeneous mixture.

Solvent - The major component of a solution.

Colligative Properties

What are Colligative Properties?

A we have discussed, solutions have different properties than either the solutes or the solvent used to make the solution. Those properties can be divided into two main groups--colligative and non-colligative properties. Colligative properties depend only on the number of dissolved particles in solution and not on their identity. Non-colligative properties depend on the identity of the dissolved species and the solvent.

To explain the difference between the two sets of solution properties, we will compare the properties of a 1.0 M aqueous sugar solution to a 0.5 Msolution of table salt (NaCl) in water. Despite the concentration of sodium chloride being half of the sucrose concentration, both solutions have precisely the same number of dissolved particles because each sodium chloride unit creates two particles upon dissolution--a sodium ion, Na⁺, and a chloride ion, Cl^-. Therefore, any difference in the properties of those two solutions is due to a non-colligative property. Both solutions have the same freezing point, boiling point, vapor pressure, and osmotic pressure because those colligative properties of a solution only depend on the number of dissolved particles. The taste of the two solutions, however, is markedly different. The sugar solution is sweet and the salt solution tastes salty. Therefore, the taste of the solution is not a colligative property. Another non-colligative property is the color of a solution. A 0.5 M solution of CuSO₄ is bright blue in contrast to the colorless salt and sugar solutions. Other non-colligative properties include viscosity, surface tension, and solubility.

Raoult's Law and Vapor Pressure Lowering

When a nonvolatile solute is added to a liquid to form a solution, the vapor pressure above that solution decreases. To understand why that might occur, let's analyze the vaporization process of the pure solvent then do the same for a solution. Liquid molecules at the surface of a liquid can escape to the gas phase when they have a sufficient amount of energy to break free of the liquid's intermolecular forces. That vaporization process is reversible. Gaseous molecules coming into contact with the surface of a liquid can be trapped by intermolecular forces in the liquid. Eventually the rate of escape will equal the rate of capture to establish a constant, equilibrium vapor pressure above the pure liquid.

If we add a nonvolatile solute to that liquid, the amount of surface area available for the escaping solvent molecules is reduced because some of that area is occupied by solute particles. Therefore, the solvent molecules will have a lower probability to escape the solution than the pure solvent. That fact is reflected in the lower vapor pressure for a solution relative to the pure solvent. That statement is only true if the solvent is nonvolatile. If the solute has its own vapor pressure, then the vapor pressure of the solution may be greater than the vapor pressure of the solvent.

Note that we did not need to identify the nature of the solvent or the solute (except for its lack of volatility) to derive that the vapor pressure should be lower for a solution relative to the pure solvent. That is what makes vapor pressure lowering a colligative property--it only depends on the number of dissolved solute particles.

summarizes our discussion so far. On the surface of the pure solvent (shown on the left) there are more solvent molecules at the surface than in the right-hand solution flask. Therefore, it is more likely that solvent molecules escape into the gas phase on the left than on the right. Therefore, the solution should have a lower vapor pressure than the pure solvent.

Figure %: The Vapor Pressure of a Solution is Lower than that of the Pure Solvent

The French chemist Francois Raoult discovered the law that mathematically describes the vapor pressure lowering phenomenon. Raoult's law is given in :

Figure %: Raoult's Law Describes the Mathematics of Vapor Pressure Lowering

Raoult's law states that the vapor pressure of a solution, P, equals the mole fraction of the solvent, c _solvent, multiplied by the vapor pressure of the pure solvent, P^o. While that "law" is approximately obeyed by most solutions, some show deviations from the expected behavior. Deviations from Raoult's law can either be positive or negative. A positive deviation means that there is a higher than expected vapor pressure above the solution. A negative deviation, conversely, means that we find a lower than expected vapor pressure for the solution. The reason for the deviation stems from a flaw in our consideration of the vapor pressure lowering event--we assumed that the solute did not interact with the solvent at all. That, of course, is not true most of the time. If the solute is strongly held by the solvent, then the solution will show a negative deviation from Raoult's law because the solvent will find it more difficult to escape from solution. If the solute and solvent are not as tightly bound to each other as they are to themselves, then the solution will show a positive deviation from Raoult's law because the solvent molecules will find it easier to escape from solution into the gas phase.

Solutions that obey Raoult's law are called ideal solutions because they behave exactly as we would predict. Solutions that show a deviation from Raoult's law are called non-ideal solutions because they deviate from the expected behavior. Very few solutions actually approach ideality, but Raoult's law for the ideal solution is a good enough approximation for the non- ideal solutions that we will continue to use Raoult's law. Raoult's law is the starting point for most of our discussions about the rest of the colligative properties, as we shall see in the following section.

Boiling Point Elevation

One consequence of Raoult's law is that the boiling point of a solution made of a liquid solvent with a nonvolatile solute is greater than the boiling point of the pure solvent. The boiling point of a liquid or is defined as the temperature at which the vapor pressure of that liquid equals the atmospheric pressure. For a solution, the vapor pressure of the solvent is lower at any given temperature. Therefore, a higher temperature is required to boil the solution than the pure solvent. is a phase diagram for both a pure solvent and a solution of that solvent and a nonvolatile solute that explains that point graphically.

Figure %: Phase Diagram for a Solvent and its Solution with a Nonvolatile Solute

As you can see in the the vapor pressure of the solution is lower than that of the pure solvent. Because both pure solvent and solution need to reach the same pressure to boil, the solution requires a higher temperature to boil. If we represent the difference in boiling point between the pure solvent and a solution as ΔT_b, we can calculate that change in boiling point from the :

In the we use the units molality, m, for the concentration, m, because molality is temperature independent. The term K_b is a boiling point elevation constant that depends on the particular solvent being used. The term i in the above equation is called the van't Hoff factor and represents the number of dissociated moles of particles per mole of solute. The van't Hoff factor is 1 for all non-electrolyte solutes and equals the total number of ions released for electrolytes. Therefore, the value of i for Na₂SO₄ is 3 because that salt releases three moles of ions per mole of the salt.

Freezing Point Depression

As you may have noticed when we looked at the , the freezing point is depressed due to the vapor pressure lowering phenomenon. The points out that fact:

Figure %: Phase Diagram for a Solution and the Pure Solvent Indicating the Freezing Point Depression

In analogy to the boiling point elevation, we can calculate the amount of the freezing point depression with the :

Note that the sign of the change in freezing point is negative because the freezing point of the solution is less than that of the pure solvent. Just as we did for boiling point elevation, we use molality to measure the concentration of the solute because it is temperature independent. Do not forget about the van't Hoff factor, i, in your freezing point calculations.

One way to rationalize the freezing point depression phenomenon without talking about Raoult's law is to consider the freezing process. In order for a liquid to freeze it must achieve a very ordered state that results in the formation of a crystal. If there are impurities in the liquid, i.e. solutes, the liquid is inherently less ordered. Therefore, a solution is more difficult to freeze than the pure solvent so a lower temperature is required to freeze the liquid.

Osmotic Pressure

Osmosis refers to the flow of solvent molecules past a semipermeable membrane that stops the flow of solute molecules only. When a solution and the pure solvent used in making that solution are placed on either side of a semipermeable membrane, it is found that more solvent molecules flow out of the pure solvent side of the membrane than solvent flows into the pure solvent from the solution side of the membrane. That flow of solvent from the pure solvent side makes the volume of the solution rise. When the height difference between the two sides becomes large enough, the net flow through the membrane ceases due to the extra pressure exerted by the excess height of the solution chamber. Converting that height of solvent into units of pressure (by using the ) gives a measure of the osmotic pressure exerted on the solution by the pure solvent. P stands for pressure, r is the density of the solution, and h is the height of the solution.

shows a typical setup for measuring the osmotic pressure of a solution.

Figure %: Setup for Measuring the Osmotic Pressure of a Solution

You can understand why more molecules flow from the solvent chamber to the solution chamber in analogy to our discussion of Raoult's law. More solvent molecules are at the membrane interface on the solvent side of the membrane than on the solution side. Therefore, it is more likely that a solvent molecule will pass from the solvent side to the solution side than vice versa. That difference in flow rate causes the solution volume to rise. As the solution rises, by the pressure depth equation, it exerts a larger pressure on the membrane's surface. As that pressure rises, it forces more solvent molecules to flow from the solution side to the solvent side. When the flow from both sides of the membrane are equal, the solution height stops rising and remains at a height reflecting the osmotic pressure of the solution.

The equation relating the osmotic pressure of a solution to its concentration has a form quite similar to the ideal gas law:

Although the above equation may be more simple to remember, the is more useful. This form of the equation has been derived by realizing that n / V gives the concentration of the solute in units of molarity, M.

P=iMRT