


CHEMICAL CALCULATIONS

USER MANUAL

VERSION 1.5

















SPARTAN SOFTWARE

PROVIDER OF SCIENTIFIC SOFTWARE FOR EDUCATION































INTRODUCTION

Chemical Calculations was designed to help high school and college students solve problems in an introductory chemistry class. The program is easy to learn and run. If you have used other Windows programs before, Chemical Calculations will not cause you any problems. You can use your mouse or the keyboard. The menu functions operate the same as the mouse functions. 

Students can use Chemical Calculations as a homework helper, a study guide, or as a review for a test. One of the strong points about Chemical Calculations is that it shows the work needed to get the answer to the problem. The answer alone would not help much. With the work shown, students can see how a problem is addressed and use that in solving other similar problems.

Teachers can use Chemical Calculations as an instructional tool in the classroom. With a computer hooked up to an overhead projector, a teacher can demonstrate many types of chemical principles and problems. You can experiment with a particular variable in order to illustrate a particular concept. For example, you can change the temperature and show the effect on a volume of a gas. Then have the students see if they can identify the relationship that exists. The possibilities are limited only by one's imagination.

In the media center, groups of students or an individual can receive extra assistance with a particular problem or concept. Students can use Chemical Calculations to study for a quiz or a test. It can be used in  group competitions in the classroom. 

Chemical Calculations is a tool that can be used by both students and teachers. 





























TUTORIAL

In this tutorial, you will receive a brief introduction to all sections of Chemical Calculations. Hopefully, this will help you in your understanding of chemistry and make using the program easier.

One word of advice is necessary. Chemistry problems are not all solved exactly the same way. By this is meant that some problems can be solved using the Celsius scale and some using the Kelvin scale. The value of R in gas problems will vary depending upon the units used in the problem. Pressure may be assigned mm of Hg, or a number of atmospheres or kiloPascals. In each section of the program, there are two sources of additional help. First you have the help button, illustrated by the question mark. Press that button and you are given a brief explanation of the concept or problem type. Secondly, by placing the cursor over certain data entry boxes, you will be given an additional prompt or reminder. 

Many steps have been taken to prevent the entry of invalid data. If you attempt to enter invalid data or fail to enter data where it is required, a message box will notifying you of the error. You can then enter the correct data and the program will continue normally.



































General

The General section contains problems that are found throughout chemistry. It does not refer to a particular chemical concept but rather it addresses some of the math that is used throughout the program and throughout chemistry.

1.  Significant Figures. This causes students a lot of confusion that is really not necessary once you understand what it is all about. Calculations in chemistry require numbers to be very accurate. There is no room for numbers that are guesses or just estimates. Each number that is used must be an accurate, dependable number. So rules have been established to guarantee that the numbers used are just that - dependable and reliable. You may have been introduced to these rules and have noticed that several of the rules are concerned with zeroes.

Example 1: 123456789 contains 9 significant figures. Each non-zero number is assumed to be an accurate, reliable number.

Example 2: 100 has only 1 significant figure. The zeroes are there only to place the 1.

Example 3: 100.00 has 5 significant figures. Zeroes between a non-zero number and the decimal point are significant as well as zeroes after the decimal point.

Example 4: If you add the numbers 10.65 and 12.3, the final answer has only three significant digits. An answer can have only as many significant digits as the number with the least number of significant digits. 12.3 has three and 10.65 has four. So the final answer can only have three. What number comes after the 3 in 12.3? It could be almost anything from 0 to 9. So we cannot take a chance by guessing and therefore limit the answer to three significant figures.

2. Time Conversions  Occasionally you need to convert from one time interval to another. You might need to know the number of moles produced per second and are given minutes. Whatever time unit you are given you sometimes need to change it. To change the time unit, select Time Conversion from the General Choice screen. Enter the numerical value of your time then click on the button corresponding to the time unit you want.

Example 1:  Convert 30 minutes into hours. Enter 30 into the data entry box. Since you want to convert minutes to seconds, click on the button labeled 'Minutes - hours'.  The answer appears in the answer box at the bottom. Time conversion is relatively simple but once in a while you get a time unit that might not be easy. This section is provided for your convenience.

3. Temperature Conversion. Temperatures in chemistry cannot be in the Fahrenheit scale. They must be in the Celsius or Kelvin scale. So you must be able to convert from one scale to another. Gas laws usually are in the Kelvin scale. This section of the program does the conversion for you. Enter the temperature you want to convert, select the scale that it is presently in, select the scale you want to convert it to and press Ok. The new temperature is given along with the work to get that answer.

Example 1: Convert 32 F to C.  Enter 32 in the entry box. Select F in the From box and C in the To box. Press Ok. The new temperature is 0 Celsius.

Example 2: Convert 100 C to K. Enter 100 in the entry box. Select C in the From box and K in the To box and press Ok. The new temperature is 373 Kelvin.

4. Density. Density refers to the thickness of a material. The thicker a substance is the higher the density. The thinner a substance is the lower the density. Air is a mixture of gases and is very thin. Therefore, it has a very low density. There is not much matter in any volume of air. Liquids are a little thicker than gases. So the density is higher than that for gases. You can move easily through a gas and not quite so easily in a liquid like water. Solids are very thick and so have a much higher density. You can?t move through a solid. Density is mass / volume. You will notice that you can select to solve for density, mass or volume. Check the appropriate option, enter the requested data and press Ok. 

Example 1: Determine the density of an object that has a mass of 1.5 grams and occupies a volume of 50 ml. Select density. Where prompted, enter 1.5 for mass and 50 for volume and press Ok. The density is 0.03. 

Example 2: An object has a density of 2.7 and has a mass of 10. What is the volume? Select volume and where prompted enter 2.7 for density and 10 for mass. The volume has to be 3.704.

5. Metric Conversions. Sometimes you have to convert from one metric unit to another. Basically, any metric conversion consists of multiplying or dividing by 10 or a multiple of 10. The problem arises when you convert from a large number to a real small one or vice versa. Select the option - mass, distance or volume. Enter your value. Select the From unit and the To unit and press Ok.

Example 1: Convert 2500 grams to kilograms. Press mass. Enter 2500 in the entry box. Select grams in the From box, kilograms in the To box and press Ok. Your answer is 2.5 kilograms.

Example 2: Convert 1000000 milliliters to liters. Select volume. Enter 1000000 in the entry box. Select milliliters in the From box and liters in the To box. Press Ok. You have 1000 liters.
6. Volumes. Occasionally, you have to know the volume of a container. The formulas for volumes you will most likely encounter are given. Select the volume you want and enter the data where prompted. The program will prompt you for different measurements depending upon the type of volume you select. You will see volume in density problems, gas problems, mole calculations plus a few others. Just remember that volume refers to the amount of space that an object occupies or contains.

Example 1: What is the volume of a cylinder with a radius of 1.50 millimeters and a height of 15 millimeters? Select cylinder in the option box. Enter 1.5 for the radius and 15 for the height. Then press Ok. The volume is 106.029.

7. English - Metric Conversions.  All measurements in chemistry must be metric. If you are given an English quantity, it  must first be converted into a metric equivalent. This section is provided to give you that capability. However, you might also want your answer in  the English system. This section also gives you that ability. After entering your quantity, select one of two options - English to metric or metric to English. Then click on the correct button. The Pounds - Kilograms button will make a conversion two ways depending upon which option you selected.

Example 1: How many kilograms in 5 pounds? Enter 5 in the data entry box then select the 'English To Metric' option. Finally click on the 'Pounds - Kilograms' button. The final answer is 2,270 kilograms.

Example 2: How many pounds in 10,000 kilograms? Enter 10000 into the data entry box. Select the 'Metric To English' option. Click the 'Pound - Kilogram' button. The answer is 22.026 pounds.

8. Mass And Weight.  One area that causes confusion is that of mass and weight. The two terms are used interchangeably but are not the same. Mass refers to the amount of 'stuff' that makes up an object. Weight is a measure of the force of gravity that pulls objects downward to the center of the Earth. In chemistry, we need to use mass, measured in grams or kilograms, not weight. This section allows you to convert weight into mass and, for convenience, mass into weight. 

Example 1: What is the weight of 1 kilogram? Enter 1 into the data entry box and click the 'Kilograms To Newtons' button. One kilogram weighs 9.810 Newtons. Be careful when entering grams as they are kilograms divided by 1000.

Example 2: What is the weight of 25 grams? First convert 25 grams into kilograms by dividing 25 by 1000. Enter  0.025 into the data entry box and click the 'Kilograms To Newtons' button. The answer is 0.245 Newtons.

A more common conversion is from Newtons to kilograms. Similarly, enter the weight in Newtons and press the 'Newtons To Kilograms' button. If you need other metric conversions or conversion from English to metric, there are other sections in the program that will do that for you.

Example 3: What is the mass of an object that weights 2.5 Newtons? Enter 2.5 into the data entry box and click on the 'Newtons To Kilograms' button. The answer is 0.255 kilograms.



























Periodic Tables

You will notice that there are seven versions of the periodic table. Versions two through six are present to illustrate the periodic nature of the table by examining several characteristics of elements like density, boiling point and electronegativity. The first one is the most important. When you select the regular periodic table, you will see what appears to be a regular periodic table. However, each element is really a button that you can click. When you click on a button, a data form for that particular element appears showing you all the important data for that element. The table is color coded as well. All information you will probably need to find is present on this table.

Example 1: What is the boiling point of hydrogen? Press the H button on the table and H?s data form will appear. As you can see, the boiling point of hydrogen is -252.76  degrees Celsius. Remember that temperatures in chemistry cannot be Fahrenheit.

Example 2: What happens to the radius of elements as you go through the periodic table? Select the Atomic Radius version of the periodic table and press Ok. As you can see, there is a gradual increase in radius. The plotted line goes up and down but generally the trend is for the line to go upward. Each up and down cycle corresponds to an energy level.



































Moles

This section is one of the main workhorses of the program. In many problems that you will encounter, you must be able to calculate grams, moles or the number of atoms. The mole is the major unit of quantity in chemistry. Consider it to be the chemical equivalent of a dozen. The following equation is absolutely essential to remember.

1 mole = 1 gram atomic weight = 6.02E23 atoms. No matter which element you are working with, the above relationship is true.

There are three subsections to the   Mole category:  Mole-Elements, Average Atomic Mass and Mass - Energy.  The first one, Mole - Elements is extremely critical to the study of chemistry and so the student must be sure that they understand completely the concept of the mole.

1.  Moles - Elements   The ability to calculate the mass of an element, the number of moles and the number of atoms is one of the  main skills a student must master.

Example 1: How many grams are there in 1 mole of Carbon. Enter a 1 in the 'Enter an amount'  box at the top of the form. Select moles in the From section and grams in the To section. You then scroll down the 'Select an element' list box until you find Carbon and click on it. When you select an element, the atomic weight of that element appears directly beneath the list box. Press Ok and you will receive the answer. There are 12.011 grams in 1 mole of Carbon. If you wish to continue with Carbon and see how many atoms are in 1 mole, simply select atoms in the To section and press Ok. You do not have to clear the form. The number of atoms is 6.02E23. 

Example 2: How many moles are there in 100 grams of Iron (Fe)? Enter 100 in the 'Enter an amount' box, select grams in the From section and moles in the To section. Scroll down the 'Select and element' list box until you come to Iron and click on it. The atomic weight of Iron will then appear. Click Ok and you will receive your answer. 100 grams of Iron is equal to 1.791 moles. To determine how many atoms of Iron you have, simply select atoms in the To section and press Ok. There are 1.08E24 atoms in 100 grams of Iron.

Example 3:  If you have 1.2E24 atoms of Gold, how many grams do you have? Enter 1.2E24 in the 'Enter an amount' box, select atoms from the From section and select grams in the To section. Scroll down to Gold and click on it. Press Ok and you will see that you have 3.92E2 or 392 grams of Gold. How many moles? Just select moles in the To section and press Ok. There are 1.992 moles of Gold. 

Become familiar with this portion of the program. It will be extremely useful in many types of calculations.

2.  Average Atomic Mass  When you look up the atomic mass of an element, it is rarely a whole number. Instead it is a decimal number. The reason for this is that there are several different versions, or isotopes, for most atoms. Another reason is that these different version occur at different rates. So when you consider that there are different isotopes for most elements and that these have different rates of occurrence, you can see why you do not get a whole number.

Example 1.  What is the average atomic mass for magnesium? Magnesium has three isotopes. Isotope 1 have a mass of 23.985 and has a fractional abundance of 0.787. Isotope 2 has a mass of 24.986 and has a fractional abundance of 0.1013. Finally, isotope three has a mass of 25.983 and an abundance of 0.1117. Click on the 'Average Atomic Mass' button on the Mole choice form. As you can see, you may enter up to six different isotopes with their respective occurrences. Enter 23.985 for the weight of isotope 1 and 0.787 for its occurrence. Enter 24.986 for the weight of isotope 2 and 0.1013 for its occurrence. Lastly enter 25.983 for the weight of isotope 3 and 0.1117 for its occurrence. Press OK and the average atomic mass is 24.31 which agrees with the mass given on the periodic table.

3.  Mass - Energy  When moles of two substances react together, heat is either gained or lost. Which of the two happens depends upon the substances reacting. If the product is more stable by releasing energy then that is what will happen. If stability is greater when energy is absorbed that is also what will happen. 

Example 1.  If 16 grams of Fe2O3 reacts with Al, how much energy is produced? First you need to have a balanced reaction. The reaction is as follows:

Fe2O3 + 2Al => Al2O3 + 2Fe + 852 kJ

Select Mass - Energy from the Mole choice form. We are concerned with Fe2O3. Enter a 1 for the coefficient for material in question in data box one. Enter 16 for the grams of material. The molecular weight of Fe2O3 is 160 so enter that in box 3. The reaction released 852 kJ of energy so enter 852 for the kJ gained/released. Press OK. The heat energy is 85.2 kJ.


































Weights

Weight is another extremely important function in chemistry. The ability to determine the weight of a molecule is critical to a majority of chemical calculations. To determine the weight of a molecule, you must determine how many of each type of atom there is in the molecule. The program does this for you. In order for you to fully utilize this part of the program, you must become familiar with the correct entry of the formula. Efforts have been made to help you with this. 

Example 1: In the formula entry box at the top of the form enter the following molecule exactly as written.   2Mn(NO3)2.10H2O    As you enter the formula, pay attention to capital letters and small letters. You do not have to worry about whether a number is a coefficient or a subscript. It is taken care of for you. Once you have entered the formula, click on Ok. The molecule has a weight of 718.14. You will notice that this value is given to you twice. The first time is  just the weight. The second is an additional reminder of what a mole is. One mole of this molecule has a weight of 718.14 grams. 

Example 2: Enter the following molecules to determine their weights and see if you get the same answer as the one given.

H2O              18.015
C6H12O6   180.15
H2SO4          98.079
Ca(OH)2       74.09
Cu(NO3)2   187.55

Example 3:  What if you have a molecule that is a hydrate?  This means it contains H2O at the end. With these kinds of molecules, one of the things you might want to know is the percent of the molecule that is H2O (water).  This is referred to as the % Hydrate.  Go back to example 1 and re-enter the molecule given. You notice that it has .10H2O at the end. Don't forget the period. That  means that each  molecule contains 10 molecules of water. But what is the % hydrate? After you have entered the molecule, check the "Check for % Hydrate' box. When you check it, a check mark appears. Now all you have to do is click Ok. The weight of the water is 360.260 and the percent of the total molecule that is water (H2O) is 50.165%.  To check the percent hydrate, divide the weight of the water by the weight of the molecule and multiply by 100%.

Example 4:  Lets continue using the same molecule to look at another powerful analysis. This is the percent composition. This determines what percent of the total molecule is made up by each type of atom in the molecule. After you have entered the molecule, check the percent composition check box and click Ok. At the bottom of the form, there are six data boxes. In these boxes you will see the elements listed, the total weight contributed by each element, and the percent of the total molecule that is made up of each element.

In this case, you will see:   Mn   109.876   15.3%
                              N     56.027    7.8%
                              O    511.981   71.3%
                              H     40.318    5.6%

 This means that there are 109.876 grams of Manganese in the molecule and that they make up 15.3% of the total weight. Nitrogen contributes 56.027 grams and makes up 7.8% of the molecule. Oxygen's 511.981 grams makes up the bulk of the molecule comprising 71.3 percent. Hydrogen, the lightest element, comes in at 40.318 grams and 5.6%.

Mastery of this calculation is mandatory for the vast majority of all calculations in chemistry.




















































Reactions

In reactions, you have a choice of two options. The first is to balance a chemical reaction. The second is to take a balanced reaction and calculate certain quantities from it. In order to make predictions about how much of a reactant is needed to make a certain quantity of a product, you need a balanced reaction. The same is true if you want to know how much of a product you can make from a known quantity of a reactant.

1. Balancing Reactions.  Chemical Calculations will not balance the reaction for you but it will help you balance it. The first step is to correctly enter the equation. There is an example at the top of the form and if you place the cursor over the entry box, there is an additional reminder about correct entry.

When you enter the formula, be sure to correctly use capital letters and small letters. Add numbers where you normally would enter them. Enter a + sign between molecules on the left side and between molecules on the right side. It is important that you use the => combination to  separate the reactants on the left from the products on the right. The program looks for the = sign and the > sign. 

Example 1: Enter the following reaction exactly as shown.

		CO2 + H2O => C6H12O6 + O2

Press Ok. What you will see is a list of elements and the number of each element on the reactants side. The same is true for the products on the right side. You will see a list of elements and their numbers. On the left you will see that there is 1 C, 3 O, and 2 H atoms. On the right side you will see that there are 6 C, 12 H, and 8 O atoms. By simple inspection, you can see that the elements on both sides of the => are not equal. What you have to do is provide the correct coefficients only to the correct molecules. Begin by placing a 6 in front of the C on the left side and press Ok. As you can see, all the elements are not equal so you must continue. Place a 6 in front of the H on the left side and press Ok. We are getting closer but they are still not equal. Now place a 6 in front of the O2 on the right side and press Ok. At this point a message box appears and tells you that the total atoms on the left are equal to the total atoms on the right. This does not mean that the reaction is balanced. It is up to you to check each element on the left and right side to see if the numbers of each kind of atom is the same.

The final, balanced equation is: 6CO2 + 6H2O => C6H12O6 + 6O2.

2. Mass - Mass. Now that we have a complete, balance reaction, lets make some predictions using this equation.

6CO2 + 6H2O => C6H12O6 + 6O2

How many grams of C6H12O6 are made from 50 grams of CO2? Entering data into the correct boxes is important. We want to find a quantity of a product and are given a quantity of a reactant. In the first box at the top, enter a 6 which corresponds to the coefficient of the given reactant. Next enter 50 in the second box at the top. This is the actual mass in grams of the reactant that you were given. Finally, at the top, enter the molecular weight of the CO2 which is 44. So you have the coefficient, the mass and the molecular weight for the CO2 which is the given reactant. On the bottom in the first box, enter a 1 since this is the coefficient for the C6H12O6. Remember that if no number is given, it is assumed to be a 1. Leave the second box empty since this is the quantity that you want to find. Lastly, enter 180.15 for the molecular weight of the C6H12O6. Press Ok. The answer is 34.119. This means that, if you are given 50 grams of CO2, you will produce 34.119 grams of C6H12O6. Likewise, we could have entered the 34.119 and left the 50 out. Your answer would have been the 50 grams of CO2. This capability is a very powerful technique in chemistry. But it all depends upon a balanced chemical reaction!



















































Solutions

The solutions section is composed of eight sub-sections. Solutions are made when one substance is dissolved in another. You can have solutions of gases, liquids and even solids. Once we have a solution, we must have the means to study them. Of the various means of studying solutions, Molarity and Molality are the most common. 

1. Molarity / Molality.  Molarity and Molality are the two most common means of measuring the concentration of a solution. This is another one of the critical skills that you will need to master in chemistry. Molarity refers to the moles per liter of solution. If a solution has a molarity of 2, that means that there are 2 moles of a solute in the liter of solution. It you have a 0.5 M solution, there is 0.5 mole dissolved into a liter of solution. Molality is a variation of molarity. Molality refers to the number of moles per kilogram of solution

Example 1: What is the molarity of a solution in which 3 moles of a solute are dissolved into a liter of solution. First select molarity at the top of the form. Where prompted, enter the number of moles and then the liters of solution. Press Ok and the molarity is given along with the calculations used to get the answer.  The molarity is 3.000. 

Example 2: What is the molality of a solution where 0.25 moles of a solute are dissolved into 500 grams of solution. First convert the 500 grams to kilograms.  500 grams is equal to 0.5 kilogram. Select molality at the top.  Enter 0.25 for the number of moles and .500  for the kilograms. Press Ok and you will see a molality of .500.

2. Henry's Law.  Another important calculation in chemistry and one that you are indirectly familiar with  is Henry's Law.  This can be called the 'carbonated beverage' law. In making carbonated beverages, CO2 is dissolved under pressure into the beverage. CO2 is used as a preservative.  In order to determine solubility using Henry's Law, two pieces of information are required.  One of these is the pressure used to force the gas into solution.  Pressure in chemistry can be given in several forms and you need to know the difference between them. In this particular case, the pressure is in atmospheres. Sometimes pressure is given in millimeters of Mercury. When you listen to the weather, you are given the atmospheric pressure or barometric pressure in mm of Hg. Another means of measuring pressure is in units called kiloPascals. The three forms are equivalent.  

1 atmosphere = 760 mm of Hg = 101.3 kiloPascals.   Remember these numbers!

Solubility refers to the amount of a solute that will dissolve in a certain amount of solvent at a certain temperature. Different substances dissolve at different rates depending upon the temperature. Some substances do not dissolve hardly at all in certain solvents.

Example 1: What is the solubility of 450 ml of H2S dissolved in 200 grams of water at 10 atmospheres? First you need to calculate the molality which is moles per kilogram of solvent. 450 ml is equal to 0.450 liters. As you will see later, one mole of a gas occupies 22.4 liters. So 0.45 liters divided by 22.4 liters per mole gives you .02 moles. Next you need the number of kilograms. 200 grams of water is equal to .2 kilograms. Therefore the molality is .02 moles divided by .2 kilograms. The result is 0.1 molality.  In the Henry's Law form, enter the .1 in the first data box that asks you for molality. In the previous  section, you  saw how to have Chemical Calculations solve for molality for you.  Next you enter the number of atmospheres (10) into the second data box and press Ok. The solubility is 1 for 10 atmospheres.

3. Freezing Point / Boiling Point.  Another useful calculation involving solutions is the change in freezing point and boiling point. Think of the antifreeze/coolant you put into your car. When you add antifreeze, you change the freezing point of water by lowering it. Likewise, when you add coolant, you are increasing the boiling point. The amount you change the freezing point or boiling point depends upon two things: molality and the freezing point depression or boiling point elevation for the solvent. In the freezing point / boiling point changes section, you will see a list of some common solvents with their freezing point, boiling point, freezing point depression, and boiling point elevation. Locate water and notice the freezing point depression and the boiling point elevation. The freezing point of water is 0 degrees Celsius. The boiling point of water is 100 degrees Celsius. For every 1 molal change to the solution, the freezing point of water will drop 1.86 degrees Celsius. Also, the boiling point will rise by 0.51 degrees Celsius. For a 1 molal change to a water solution, the new freezing point will drop to -1.86 degrees and the boiling point will rise to 100.51 degrees.

Example 1: What is the boiling point and freezing point of a 1.5 molal benzene solution ? Select the boiling point option. Enter 2.53 for the boiling point elevation in the first data box. The normal boiling point of benzene is 80.10 degrees.  Enter 80.10 in the second data box for the original temperature. Enter 1.5 in the third data box. A 1.5 molal solution increases the boiling point by 1.5 times the boiling point elevation of 2.53 degrees. Press Ok. The new boiling point is 83.895 degrees.  To calculate the new freezing point, select freezing, enter 5.12 as the freezing point depression in the first data box, 5.53 as the freezing point in the second data box, and 1.5 for the molality in the third data box. The new freezing point is -2.15 degrees.

4. Solubility Product Constant.  The solubility product constant is a means of determining if a precipitate (solid) will occur when two solutions are mixed.  In order to calculate this, you must have a balanced equation and know the numbers of each ion produced. The concentrations of each product must be in moles per liter. Once you have those, look at the ions produced  and in particular their coefficients. In Chemical Calculations, it is assumed that you have only two products. You raise the concentration of each product to the power of their coefficient. Then you multiply the concentrations together. The result is the solubility product constant. If the number is greater than the experimental value for the solubility product constant, a precipitate will occur. If it is less than or equal to the solubility product constant, no precipitation will occur.

Example 1:  What is the solubility product constant of AgI if its concentration is 2.88E-6  grams per liter. First change the grams per liter to moles per liter. The molecular weight of AgI is 234.768. The 2.88E-6  grams divided by the 234.768 grams per mole yields 1.2E-8 moles per liter. Secondly, AgI breaks down into two ions as shown.  

AgI => Ag + I 

The concentration of Ag is 1.2E-8 moles per liter. The same is true with I. The coefficients of both ions is 1. The equation then is 1.2E-8 raised to the first power times 1.2E-8 raised to the first power. Enter 1.2E-8 into the first data box on the solubility product constant form. Enter a 1 in the second data box as the coefficient. Enter the same numbers into the third and fourth data boxes and press Ok. The resulting solubility product constant is 1.44E-16. If it is greater than the experimental value, which is available in a data table, a precipitate will form.

Example 2:  What is the solubility product constant of CaF2 if its concentration is .017 grams per liter? Change the grams to moles per liter. .017 grams divided by 78.075 grams per mole yields 0.000217739 moles. Next write out the ionic equation.

CaF2 => Ca + 2F

The concentration of Ca is 0.000217739. Since there are two F ions, the concentration of F is 2 times 0.000217739 or 0.000435478. Enter 0.000217739 in the first data box and a 1 in the second data box. Enter 0.000435478 in the third data box and a 2 in the last data box. Press Ok. The solubility product constant is 4.14E-11.

5. Equilibrium Constant.  Another constant expression is the equilibrium constant. This is used to determine if the formation of products is favored or if the formation of reactants is favored. To start, you need a balanced equation. Secondly, you need the concentrations of materials in moles per liter. To determine the equilibrium constant, divide the product of the concentrations of the products by the product of the reactants. Each concentration is raised to a power equal to its coefficient. If the equilibrium constant is greater than 1, the formation of products is favored. If it less than 1, the formation of reactants is favored.

Example 1: What is the equilibrium constant for the equation   N2 + O2  <=> 2NO? The concentration of each reactant is 0.72 M and the concentration of the product is 4.9E-16 M.

Starting with reactants, enter a 1 in the first reactant coefficient box. Next enter 0.72 into the first reactant concentration box. Enter a 1 in the second reactant coefficient box and 0.72 in the second reactant concentration box. For the products, enter a 2 in the first product coefficient box and 4.9E-16 in the first product concentration box. Press Ok. The equilibrium constant is 4.63E-31.  Since this value is less than 1, the reaction tends to favor the formation of reactants.

6. Titration.  Titration is the process of determining the concentration or volume of an unknown solution. Given the volume and concentration of a known solution and the volume of the unknown solution, you can determine the concentration of the unknown. This is used a lot in determining the concentration of an acid or a base. The basic equation is the molarity of the known times the volume of the known equals the molarity of the unknown times the volume of the unknown.

Example 1: If 100 ml of .5 M NaOH is needed to neutralize 200 ml of HCl, what is the molarity of the acid? In the titration form, there are four data boxes towards the top. These correspond to the concentrations and volumes of the known and unknown. You may determine any of these four values simply by filling in the other three and leaving the one you want to find blank. You must also fill in the ratio of base molecules to acid molecules. This requires a balanced equation.

Since you want to find out the molarity of the acid, leave the first data box blank. Put .200 in the second box where you are asked the volume. The volume must be in liters. Put .5 in the third box for the molarity of the base and .100 in the fourth box for the volume of the base. In the balanced equation, all the coefficients are 1 so put a 1 in  the ratio box and press Ok. The molarity of the acid is 0.250.

7. Ionization Constant.  The last solution problem type is the ionization constant. This deals with ionic molecules, particularly acids and bases, in water. The ionization constant is found in a manner similar to the other constants. The product of the product concentrations is divided by the product of the reactant concentrations. Each concentration is raised to a power equal to their coefficients.

Example 1: What is the ionization constant of a .1 M CH3COOH solution mixed with enough water to make 1 liter of solution. The H3O+ concentration is 0.00135. You must have a balanced ionic equation. 

CH3COOH + H2O <=> H3O+  +  CH3COO-

The ionization constant is equal to   [0.00135][0.00135] / [.1 - 0.00135]. The concentrations of the products are the same since the coefficients are both 1. The concentration of the reactant is .1 minus the amount that ionizes or breaks down into the two product ions. In the first  box on the ionziation constant form enter .1 for the molality and enter a 1 for the coefficient in the second box. Enter 0.00135 for both concentrations for the products and 1 for the coefficients. Press Ok and the ionization constant is 1.85E-5.

8.  Mole Fraction/Percent  Another way to express solution concentration is through the mole fraction. The Mole Fraction is the moles of solute divided by the sum of the moles of solute and moles of solvent. In a solution, the total of the mole fractions must equal 1. 

Example 1.  7.59 grams of glucose are dissolved in 125 grams of water. What is the mole fraction and mole percent of the glucose? Enter 7.89 in data box 1 for the mass of the solute.  The molecular weight of glucose is 180. Enter 180 in data box two. Enter 125 in data box three for the mass of the solvent (water) and 18 in  data box four for the molecular weight of the solvent. Press OK. The mole fraction is 0.006 which means that less than 1 percent of the moles in the solution are glucose. The mole percent is calculated by  multiplying the mole fraction by 100. Therefore, the mole fraction for glucose is 0.627 %. 


















Gas Laws

1. Charles's Law.  Charles's Law deals with a direct relationship between gas temperature and volume. The higher the temperature, the higher the volume. The lower the temperature, the lower the volume. All temperatures must be Kelvin. In this section of the program, you have the option of finding either the volume or the temperature.

Example 1:  What volume will a sample of a gas occupy at 28 Celsius if it presently occupies 457 ml at 0 degrees Celsius? On the Charles's Law form, check the volume option. You might notice that the prompts change depending upon whether you chose temperature or volume. The initial temperature is 0 Celsius so it must be converted to Kelvin. The Kelvin temperature is 273 (0 + 273) degrees. The initial volume must be entered in liters. 457 ml is equal to .457 liters. The final temperature of 28 Celsius must be converted to 301 Kelvin. Enter 273 in the first box, .457 into the second and 301 into the third. Press Ok and the new volume is .504 liters. Since the temperature increased, the volume had to increase also.

Example 2: If a gas occupies .733 liters at a temperature of 10 degrees Celsius, at what temperature will it occupy a volume of 1.225 liters? Check the temperature option. Enter 283 in the first box since 10 degrees C is equal to 283 degrees K. Enter .733 in the second box for the volume and 1.225 in the final box for the new volume. Press Ok. The new temperature is 472.954 degrees Kelvin. Since the volume increased, and other factors are held constant, the temperature had to have increased.

2. Boyle's Law.  Boyle's Law deals with the effect of pressure on the volume of a gas. This is an inverse relationship in that when the pressure increases, the volume decreases. When the pressure decreases, the volume increases. Think of applying pressure to a cylinder. The more pressure you apply, the smaller the volume of the cylinder. In the Boyle's Law form, you have the option of finding pressure or volume.

Example 1: If a sample of oxygen occupies a volume of 425 ml when under a pressure of 745 mm Hg, what will its volume be if the pressure is changed to 760 mm Hg? Select the volume option at the top of the form. Enter 745 in the first data box. Enter .425 in the second box for the volume in liters. Finally enter 760 in the third data box for the final pressure and press Ok. The new volume is .417 liter. Since the pressure went up, the volume had to come down.

Example 2:  A gas occupies a volume of 125 ml at a pressure of 760 mm Hg. What pressure is required for the gas to occupy a volume of 120 ml?  Convert the volumes to liters. 125 ml is .125 liter and 120 ml is .120 liter. Enter the initial volume of .125 in data box one. Enter 760 in the second data box for the initial pressure and .120 in the third data box for the final volume. Press Ok. The new volume is 791.667 mm Hg. The volume went down  so the pressure had to increase.

3. Ideal Gas Equation.  The ideal gas equation is one that combines four variables into one equation. The ideal gas equation considers pressure, temperature, volume, and the number of particles. The law is stated as:  PV = nRT. R is a gas constant and has the value of 0.83. n is the number of molecules. P is pressure, V is volume, and T is Kelvin temperature. By knowing the value of three of the variables plus the value of R, you can calculate the value of the fourth variable.

Example 1: What pressure is exerted by 4.5 moles of a gas with a volume of 198 liters at a temperature of 8 degrees Celsius? Select the pressure option and enter 198 in the second box for the volume. Enter 4.5 for the number of moles and 281 for the Kelvin temperature in the last box. Press enter and the pressure is 0.53 atmospheres.

4. Molecular Mass.  Once you have the ideal gas equation mastered, there are a couple of other values that you can determine. One of these is the molecular mass (weight) of the gas. The new form of the equation for molecular mass is 

M  = mRT / PV  where M is the molecular mass (weight) and m is mass in grams.

Example 1: Calculate the molecular mass of a gas with a volume of .273 liters, mass of 0.75 grams, pressure of .959 atmospheres, and at a temperature of 61 degrees Celsius.

In the molecular mass form, enter .959 in the first data box. Enter 0.75 in the second data box, .273 in the volume box, and 333 degrees Kelvin in the temperature box. Press enter and the  molecular mass of this sample is 79.415 grams per mole.

5. Gas Density.  The density of a gas is another value that can be derived from the ideal gas equation. The revised formula is:

D = PM / RT

Example 1: What is the density of oxygen that has  a Kelvin temperature of 273 degrees, and .977 atmospheres? The molecular weight of Oxygen, a diatomic gas, is 32. Enter 32 for the molecular weight in box one, .977 atmospheres for the pressure in box two, and 273 degrees Kelvin for the temperature. Press Ok and the density is 1.397.

6. Dalton's Law.  When a gas is collected from a chemical reaction, many times it is collected by a process called water displacement. If we want to know the pressure exerted by the gas alone, we must remove the pressure exerted by the water. The water pressure is a function of its temperature. Dalton's Law states that the total pressure in a container is the sum of the partial pressures of all the gases in the container. So we must remove the pressure of the water vapor. We can also determine the volume of a dry gas by removing the effect of water.

Example 1: What is the partial pressure of Oxygen collected over water if the temperature is 28 degrees Celsius and the total pressure is 98.74 kiloPascals?  On the Dalton's form, select the pressure option. Enter the 28 in the first data box. Do not change it to Kelvin. The reason is that Chemical Calculations needs to determine the vapor pressure of water based upon the Celsius scale. This is done for you. Next enter the 98.74 in the second data box and press Ok. The dry pressure of the Oxygen is 94.94 kiloPascals.

Example 2: A quantity of a gas is collected over water at a temperature of 10 degrees Celsius and at a volume of .125 liters. The pressure is 98 kiloPascals. What will the volume be at a standard pressure? A standard pressure is 101.3 kiloPascals. Enter the 10 in the first data box, .125 in the second, 98 in the third data box, and 101.3 in the last box. Press Ok and the new volume is .119 liters.

7. Combined Gas Law.  The Combined Gas Law literally combines Boyle's law, Charles's law and Dalton's law all into one. Seldom will any one of the laws operate alone.

Example 1: 400 ml of a gas are collected over water at 20 degrees Celsius and at an atmospheric pressure of 97 kiloPascals. What will be the new volume at STP? STP refers to standard temperature (0 degrees Celsius) and pressure (101.3 kiloPascals). Since the gas is collected over water, check the 'collected over water' box at the top of the Combined Gas Law form. Enter .400 in the first box, 97 in the second, 293 (20 + 273) in the third, 101.3 in the fourth box, and 273 (273 + 0) in the last  box. Press Ok and the new volume is 0.35 liters. Chemical Calculations converts the Kelvin temperature back to Celsius in order to determine the water vapor pressure. Otherwise, Kelvin temperatures must be used.

8. Molar - Volume.  According to Avogadro's Principle, 1 mole of a gas occupies 22.4 liters and contains 6.02E23 molecules. The weight of the gas will vary depending on which gas it is. The kind of gas does not matter. 1 mole of any gas will occupy 22.4 liters and contain 6.02E23 molecules.

Example 1: How many grams of CO2 will occupy a volume of 0.5 liters? First you must have the molecular weight of the gas. In this case it is 44. On the Molar Volume form select the mass option. Enter 44 in the first data box as the molecular weight. Enter .5 as the volume in liters in the second box and press Ok. The answer is 0.982 grams.

Example 2: What volume will 200 grams of H2S occupy?  Select the volume option. Enter 34.066 in box 1 as the molecular weight of H2S. Enter 200 in the second data box and press Ok. The volume is 131.5 liters. 

9. Mass - Volume.  If you know the mass of reactant and the complete, balanced chemical equation, you can calculate the volume of the gas produced.

Example 1:  If you have 10 grams of KClO3, what will be the volume of the O2 produced? First you must have the balanced equation.

  2KClO3 => 2 KCl + 3 O2
(10 grams)                 ? liters

In the mass - volume form, select the volume option. Enter 10 in the first data box for the mass of the reactant. Enter the molecular weight (123) of the reactant in box 2. Next the ratio of product to reactant must be entered. These numbers are the coefficients from the  balanced reaction. Enter a 3 in the first ratio box and 2 in the  second. Press Ok and the volume is 2.732 liters.

Example 2: A quantity of Mg reacts with HCl to form MgCl2 and H2.

Mg + 2HCl => MgCl2 + H2

The temperature is 20 degrees Celsius, the volume produced is .042 liters,  and the pressure is 99.3 kiloPascals. What is the mass of the Mg? In the mass - volume form select the mass option. Enter .042 in box 1 as the volume, 293 as the temperature in degrees Kelvin, 1 for both ratio numbers, 24.3 as the molecular weight for Mg, and 99.3 kiloPascals as the temperature. When you press Ok, the mass is .041 grams. 

10. Volume - Volume.  If you know the volume of a gas in a reaction, you can calculate the volume of another gas in the same reaction. 

Example 1: If 6 liters of O2 are used in a reaction with CS2 to produce CO2 and SO2, what is the volume of the CO2 produced? You must have a balanced chemical reaction.

CS2 + 3 O2 => CO2 + 2SO2

In the volume - volume form, enter 6 in the first data box as the original volume. Enter the 3 as the coefficient of the given gas and 1 as the coefficient of the gas you are trying to find. Press Ok. The volume of the CO2 is 2 liters. 

11. Gay - Lussac's Law.  According to Gay-Lussac's Law, if you have a constant volume, the pressure exerted by the gas is directly proportional to the Kelvin temperature.

Example 1: The contents of an aerosol can are at a pressure of 3.00 atmospheres and at a temperature of 25 degrees Celsius. If the temperature was raised to 52 degrees Celsius, what would the pressure be?

Select the pressure option in Gay-Lussac's form. The temperatures must be converted to degrees Kelvin. Enter 3 in the first data box for pressure, 298 for the initial temperature, and 325 for the final temperature. Press Ok and the new pressure is 3.272 atmospheres.

Example 2: The pressure in a tire was 1.8 atmospheres at a temperature of 20 degrees Celsius. If, after a long drive, the pressure now reads 1.9 atmospheres, what is the new temperature?

Select the temperature option in the Gay-Lussac's form. Enter 293 for the initial temperature, 1.8 for the initial pressure and 1.9 for the final pressure. Pressing Ok will give you a new temperature of 309.278 Kelvin or 36.278 degrees Celsius.

12. Raoult's Law.  Raoult's Law states that the vapor pressure of a solution varies directly with the mole fraction of solvent. In other words, the more solute you have in a solution, the lower the vapor pressure of the solvent.

Example 1: 200 grams of C12H22O11 are dissolved in 100 grams of water. The temperature is 70 degrees Celsius. What is the vapor pressure of the water?  To solve for the vapor pressure, you must have the molecular weights of the solvent and the solute. The molecular weight of water is 18 and for C12H22O11 it is 342. In the Raoult's Law form, enter 70 in the first box for the temperature, 200 for the mass of the solute, 342 for the molecular weight of the solute, 100 for the mass of the solvent, and 18 for the molecular weight of the solvent. When you press Ok, the vapor pressure of the water is 28.229.

13. Graham's Law.  Graham's Law deals with rates of diffusion of gases. When two gases are introduced into a container, the rates at which they diffuse varies inversely as the square root of their molecular weights.

Example 1: Compare the rates of diffusion of Hydrogen and Oxygen at the same pressure and temperature.  All you need are the molecular weights for the two gases. Since Hydrogen was mentioned first in the problem, enter 2 into the first data box in Graham's form. Enter 32 in the second data box. Remember, you are dealing with diatomic molecules and not elements. Press Ok. The rate of diffusion of  Hydrogen is 4 times that of Oxygen.
























































Acids / Bases

The Acid/Base section allows you to calculate the pH given either the H3O+ concentration or the OH- concentration. Also, you can calculate the concentrations of OH- and H3O+ from the pH.

Example 1: What is the pH if the H3O+ concentration is 1E-4? In the Acid/Base form select the first option. Enter 1E-4 in the data box and press Ok. The pH is 4.00.

Example 2: What is the pH if the OH- concentration is 1.8E-2? Select the second option and enter 1.8E-2. Press Ok. The pH is 12.255.

Example 3: What is the H3O+ concentration if the pH is 3.72? Select option 3. Enter 3.72 into the data box and press Ok. The H3O+ concentration is 1.91E-4.

Example 4: What is the OH- concentration if the pOH is 11? Select option 4 and enter 11 in the data box. Press Ok and the OH- concentration is 1E-11.








































Energy Calculations

1. Energy Exchange / Transfer.  The last major section is Energy Calculations. No chemical reaction will occur without at least some degree of energy change. Energy will be released or taken in. 

Whenever a substance undergoes a physical change there is a change of heat energy. The amount of heat energy gained or lost depends upon the mass of the substance, the amount of temperature change and the specific heat of the substance. Not all substances gain or lose the same amount of heat energy.

Example 1: How much heat is lost when a piece of solid Aluminum with a mass of 4000 grams cools from 660 degrees Celsius to 25 degrees Celsius? Choose the energy exchange/transfer option. Enter 4000 in the first data box for the mass. Enter the temperature change of 635 (660 - 25) in box two. The specific heat is unique to each substance. For  Aluminum it is 0.903.  Press Ok and the heat energy change is 2,293,620. 

2. Thermodynamics.  Thermodynamics deals with energy interactions in a chemical reaction. When you select thermodynamics from the Energy Calculations section, you will see three options: enthalpy, entropy, and Gibb's free energy. All three deal with the  energy interactions between reactants and products in a chemical reaction.

When you select enthalpy from the list of options, you will see that you need to have the enthalpy totals for the reactants and the enthalpy totals for the products. These values you can look up in the table at the bottom of the main screen or in your text. From a balanced equation, you will also need the coefficients of each substance.

Example 1:  What is the enthalpy change for the following chemical equation?

Cl2 + 2HBr => 2HCl + Br2

The enthalpy  values for elements are zero. From the data table, the enthalpy for HCl is -92.3 kiloJoules. The value for HBr is -36.4 kiloJoules. From the balanced equation, the ratio is 2 to 2. In the first data box enter -36.4 as the enthalpy for the reactant HBr. In the second data box, enter -92.3 as the enthalpy  value for the product HCl. Enter a 2 in both ratio boxes and press Ok. The enthalpy change is -111.8 kiloJoules.  What this means is that the products have less energy than the reactants so the reaction must be exothermic (gives off energy).

Entropy refers to the amount of disorder in a reaction. Solids going to a liquid or a liquid going to a gas are common examples of increased disorder or entropy.

Example 2:  What is the change in entropy for the following chemical equation.

Ca + 2H2O => Ca(OH)2 + H2

You need to calculate the sum of the entropies of the products and the sum of the entropies of the reactants. You can look up these values in the data table provided or in your text. The entropy  values for the products are 75.9 for Ca(OH)2 and 131 for H2. The entropy values for the products are 41.4 for Ca and 69.9 for H2O. The product total is 206.9. The reactant total is 111.3.  After selecting entropy, enter 206.9 in the first data box and 181.2 in the second data box. Remember to multiply by any coefficients. The overall entropy change is 25.7.

Gibb's free energy refers to the tendency of a reaction to be spontaneous.  It requires the enthalpy change, the entropy change, and the Kelvin temperature. If the free energy value is negative then the reaction is spontaneous.

Example 3: Is the following reaction spontaneous? The enthalpy change is -424 kiloJoules, the entropy change is -209 J/K, and the Celsius temperature is 25 degrees. Select the Gibb's free energy option. Enter -424 in the first data box. Enter 298 (25 + 273) in the second box and enter -209 in the third data box. Press Ok and the Gibb's free energy value is -361.718. Since the value is negative, the reaction is spontaneous.

3. Faraday's Law.  Faraday's Law deals with the relationship between chemical reactions, chemical change and electric energy. The following equality is important in all calculations using Faraday's Law.

1 mole e- = 96500 coulombs = 1 ampere-second

An atom of Al tends to lose 3 moles of e-.   Al => Al+  + 3e-
An atom H tends to lose 1 mole of e-.          H => H+  + 1e-
An atom of Cu tends to lose 2 moles of e-.   Cu => Cu++  + 2e-

Example 1: How many grams of Al will be deposited if 96500 coulombs of electricity pass through an aluminum nitrate solution? Select the mass option in the Faraday's form. Enter 96500 in the first data box. Enter the molecular weight of Al (26.98) in the second data box. Enter 3 in the moles of electrons box since Al tends to lose 3 moles of electrons. Since no seconds have been given, leave the last box blank. If a time quantity had been given, you would convert to seconds and enter that value into that box. Press Ok and the number of grams is 8.993.

Example 2:  If 10 amperes of current flow for 20 seconds through a copper(II) nitrate solution, how many moles of copper are deposited? Select option two in the Faraday's Law form. Enter 10 in the first data box, 20 in the second data box for time (do not convert to seconds), and 2 in the last box for the number of moles of electrons released. Press Ok. There are 0.062 moles of copper released.

Example 3:  How many liters of Cl2 are released when 8.12 amperes of current pass for 2 hours through molten magnesium chloride? Select option 3 on the Faraday's law form.  Enter 8.12 in the first data box, 120 minutes (2 hours = 120 minutes) in the second box and 2 in the third box for the number of moles of  electrons released. Press Ok. There are 6.785 liters of chlorine gas released.

4. Energy and Change of State.  In the Energy and Change of State form, you deal with the energies needed to melt or boil a substance. You need the mass and the molecular weight of the substance in question. You also need the enthalpy of fusion or vaporization. These numbers are specific for different materials. Different substances have different capacities for heat. Some can hold more than others. This data is available in the Enthalpies data chart at the bottom of the main screen. They are also available in a textbook.

Example 1:  How much heat is required to melt 5.67 grams of iron(II) oxide? Then enthalpy of fusion is 32.2 kiloJoules/mole. Select the mass option. Enter 5.67 for the mass of the substance, 32.2 for the enthalpy of fusion, and 71.8 for the molecular weight. Press Ok. The energy required is 2.543.

Example 2: How much energy is required to boil 53.5 grams of Benzene (C6H6)?  Select the boiling option. Enter 53.5 in the first data box for the mass. Enter 30.8 as the heat of vaporization in the second box and the molecular weight of 78 in the third box. The energy needed to boil is 21.126.

5. Half Lives.  Half Life calculations deal with radioactive materials. It is possible to determine the amount of time a material will be radioactive or how long it takes to reach a certain level of radioactivity. You can also determine the quantity of radioactive material remaining after a certain period of time.

Example 1: The half life of Co is 5.26 years. If you start out with 4.516E8 atoms, how long will in order to get to 1.764E6 atoms? Select the time option on the Half Life form. Enter 5.26 as the half life of Co. Enter 4.516E8 as the initial quantity and 1.764E6 as the final quantity. The time is 42.08 years.

Example 2: Strontium 90 has a half life of 28.1 years. If you start out with 1.94E17 atoms, how much will be left after 140.5 years?  Enter 28.1 as the half life. Enter 1.94E17 as the initial quantity and 140.5 as the time. The number of atoms left is 6.0625E15.




































Calculator

If you need a calculator, simply press the Calculator button. You will see the Windows calculator appear. This is the calculator built into the Windows operating system. There are two versions that you can select. 

Data Tables

There are numerous data tables available for you to use. The majority of data that you will probably need is available. If the value you need is not given, consult your textbook's data tables.

























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