Why do we need mols?
Percents by weight.
Basic stoichiometry.
Density times volume of a pure material.
Atoms or molecules to mols.
Concentration times volume of a solution.
Stoichiometry roadmap.
Mol and percent worksheet.
Stoichiometry problems with mass and gas at STP.
Stoichiometry problems on concentration and density.
Stoichiometry problems using complete roadmap.



Every chemist has dreamed that atoms were large enough to see and manipulate one at a time. The same chemist realizes after considering it, that if individual molecules were available for manipulation, it would take far too long to get anything done. The view from the atom is very different from the view of trillions and trillions of atoms. The mass action of the atoms that we see on our "macro" view of the world is the result of the action of an incredibly large number of atoms averaged in their actions. The most usual way we count the atoms is by weighing them. The mass of material as weighed on a balance and the atomic weight of the material being weighed is the way we have of knowing how many atoms or molecules we are working with. Instead of counting eggs, we can count cartons of eggs, each carton of which has a given number, a dozen. Instead of counting B-B's, we can count liters of B-B's and find out how many B-B's are in a liter. Instead of counting rice grains, we buy kilograms or pounds of rice and have an idea of how many rice grains are in the container.

There are less than one hundred naturally occurring elements. Each element has a characteristic atomic weight. Most Periodic Charts include the atomic weight of an element in the box with the element. The atomic weight is usually not an integer because it is close to being the number of protons plus the average number of neutrons of an element. Let's use the atomic weight as a number of grams. This will give us the same number of any atom we choose. If we weigh out 1.008 grams of hydrogen and 35.45 grams of chlorine and 24.3 grams of magnesium, we will have the same number of atoms of each one of these elements. The neat trick with this system is that we can weigh the atoms on a grand scale of number of atoms and get a count of them. This number of atoms that is the atomic weight expressed in grams is Avogadro's number, 6.022 E 23. The name for Avogadro's number of ANYTHING is a mole or mol. A mol of aluminum is 27.0 grams of aluminum atoms. Aluminum is a metal element, so the particles of aluminum are atoms. There are Avogadro's number of aluminum atoms in 27.0 grams of it. But 1.008 grams of hydrogen is NOT a mol of hydrogen! Why not? Remember that hydrogen is one of the diatomic gases. There is really no such thing as loose hydrogen atoms. The total mass of a single hydrogen diatomic molecule (H2) is 2. 016 AMU. A mol of hydrogen gas has a mass of 2.016 grams. In that 2.016 gram mass is Avogadro's number of H2 molecules because that is the way hydrogen comes. A mol of water is 18.016 grams because each water molecule has two hydrogen atoms and one oxygen atom. A mol of water has in it Avogadro's number of water molecules. Another way to view the same thing is that a formula weight is the total mass of a formula in AMU expressed with units of grams per mol.

So Avogadro's number is just a number, like dozen or score or gross or million or billion, but it is a very large number. You could consider a mol of sand grains or a mol of stars. We are more likely to speak of a mol of some chemical, for which we can find the mass of a mol of the material by adding the atomic weights of all the atoms in a formula of the chemical. The unit of atomic weight or formula weight is grams/mol.

The chemical formula of a material should tell you; (a) which elements are in the material, (b) how many atoms of each element are in the formula, (c) the total formula weight, and (d) how the elements are attached to each other. The symbols of the elements tell you which elements are in the material. The numbers to the right of each symbol tells how many atoms of that element are in the formula. The type of atoms and their arrangement in the formula will tell how the elements are attached to each other. A metal and a nonmetal or negative polyatomic ion shows an ionic compound. A pair of non-metals are bonded by covalent bonds. Some crystals have water of hydration loosely attached in the crystal. This is indicated by the dot such as in blue vitriol, Cu(SO4) · 5H2O, showing five molecules of water of hydration to one formula of cupric sulfate.

The unit of the formula weight or molecular weight or atomic weight is "grams per mol," so it provides a relationship between mass in grams and mols of material.

nFw = m

'n' is the number of mols, 'Fw is the formula weight, and 'm' is the mass.


Back to the beginning of Mols, Percents, and Stoichiometry.



All men weigh 200 pounds. All women weigh 125 pounds. What is the percent by weight of woman in married couples? A married couple is one man and one woman. (No political implications intended.) The total weight is 325 pounds. The formula for percent is:

In this case the woman is the target, so the weight of the woman goes on top, and the total weight goes on the bottom of the fraction.

125 # 325 # x100%=38.461%= 38.5%

Notice that the units of pound cancel to make the percent a pure number of comparison.

The weights of atoms are the atomic weights. What is the percentage of chloride in potassium chloride? The atomic weight of potassium is 39.10 g/mol. The atomic weight of chlorine is 35.45 g/mol. So the formula weight of potassium chloride is 74.55 g/mol. The chloride is the target and the potassium chloride is the total. 35.45 g/mol / x 100% = 47.55198 % or 47.6 % to three significant figures.

35.45g/mol 74.55g/mol x100%=47.5520%=47.6%



You can do that with any part of a compound. What is the percentage of sulfate in beryllium sulfate tetrahydrate?

Notice that the examples here are done to two decimal points of the atomic weights. The problems in the practice bunch at the end of this chapter are done to one decimal point of the atomic weight.


Back to the beginning of Mols, Percents, and Stoichiometry.



Pronounce stoichiometry as "stoy-kee-ah-met-tree," if you want to sound like you know what you are talking about, or "stoyk," if you want to sound like a real geek. Stoichiometry is just a five dollar idea dressed up in a fifty dollar name. You can compare the amounts of any materials in the same chemical equation using the formula weights and the coefficients of the materials in the equation. Let's consider the equation for the Haber reaction, the combination of nitrogen gas and hydrogen gas to make ammonia.

N2 + 3 H2 —> 2 NH3
The formula for nitrogen is N2 and the formula for hydrogen is H2. They are both diatomic gases. The formula for ammonia is NH3. The balanced equation requires one nitrogen molecule and three hydrogen molecules to make two ammonia molecules, meaning that one nitrogen molecule reacts with three hydrogen molecules to make two ammonia molecules or one MOL of nitrogen and three MOLS of hydrogen make two MOLS of ammonia. Now we are getting somewhere. The real way we measure amounts is by weight (actually, mass), so 28 grams (14 g/mol times two atoms of nitrogen per molecule) of nitrogen and 6 grams of hydrogen (1 g/mol times two atoms of hydrogen per molecule times three mols) make 34 grams of ammonia. Notice that no mass is lost or gained, since the formula weight for ammonia is 17 (one nitrogen at 14 and three hydrogens at one g/mol) and there are two mols of ammonia made. Once you have the mass proportions, any mass-mass stoichiometry can be done by good old proportionation. What is the likelihood you will get just a simple mass-mass stoich problem on your test? You should live so long. Well, you should get ONE.

Rather than thinking in terms of proportions, think in mols and mol ratios, a much more general and therefore more useful type of thinking. A mol ratio is just the ratio of one material in a chemical equation to another material in the same equation. The mol ratio uses the coefficients of the materials as they appear in the balanced chemical equation. What is the mol ratio of hydrogen to ammonia in the Haber equation? 3 mols of hydrogen to 2 mols of ammonia. Easy. In the standard stoichiometry calculations you should know, ALL ROADS LEAD TO MOLS. You can change any amount of any measurement of any material in the same equation with any other material in any measurement in the same equation. That is powerful. The setup is similar to Dimensional Analysis, and the calculations can include portions of DA.

1. Start with what you know (GIVEN), expressing it as a fraction.

2. Use definitions or other information to change what you know to mols of that material.

3. Use the mol ratio to exchange mols of the material given to the mols of material you want to find.

4. Change the mols of material you are finding to whatever other measurement you need.

How many grams of ammonia can you make with 25 grams of hydrogen? (Practice your mol math rather than doing this by proportion. Check it by proportion in problems that permit it.)

You are given the mass of 25 grams of hydrogen. Start there.

Visual for solving stoichiometry problems in a Dimensional Analysis style.

25 g H2/1 Change to mols of hydrogen by the formula weight of hydrogen 1 mol of H2 = 2.0 g. (The 2.0 g goes in the denominator to cancel with the gram units in the material given.) Change mols of hydrogen to mols of ammonia by the mol ratio. 3 mols of hydrogen = 2 mols of ammonia. (The mols of hydrogen go in the denominator to cancel with the mols of hydrogen. You are now in units of mols of ammonia.) Convert the mols of ammonia to grams of ammonia by the formula weight of ammonia, 1 mol of ammonia = 17 g. (Now the mols go in the denominator to cancel with the mols of ammonia.) Cancel the units as you go.

The math on the calculator should be the last thing you do. 2 5 ÷ 2 . 0 x 2 ÷ 3 x 1 7 = and the number you get (141.66667) will be a number of grams of ammonia as the units in your calculations show. Round it to the number of significant digits your instructor requires (often three sig. figs.) and put into scientific notation if required. Most professors suggest that scientific notation be used if the answer is over one thousand or less than a thousandth. The answer is 142 grams of ammonia.

The calculator technique in the preceding paragraph illustrates a straightforward way to do the math. If you include all the numbers in order as they appear, you will have less chance of making an error. Many times students have been observed gathering all the numbers in the numerator, gathering all the numbers in the denominator, presenting a new fraction of the collected numbers,