Chapter 3

Atoms are very small, but we need to know the mass of different atoms to compare them. To do this, we define a unit, called the atomic mass unit (amu). and then measure the mass of all atoms in terms of amu’s. Thus the atomic mass of H is amu and that of O is 16 amu. We rarely need to know the actual mass of any atom, just its mass compared to others.

The atomic mass of an element as listed in the Periodic Table is the weighted average of all the naturally occurring isotopes for that element. See example 3.1 on page 79.

Mole – x 1023 units of anything (atoms, molecules, ions etc) # is called Avogadro’s # (named after chemist who discovered concept) Molar Mass– Mass of Avogadro’s units of a substance (16.00g of O atoms / mole or g of C / mole). These are conversion factors between grams and moles

Mole calculations g  moles  # of atoms and vice versa Let’s do some examples.(3.2, 3.3, and 3.4 on pages 81-83)

Molecular Mass (Molecular Weight [MW]) is the same, only referring to molecules rather than atoms. A general term referring to both is Formula Weight (FW), calculated by adding all atomic masses for every atom in the molecule. The same mole calculations are done with MW’s as atoms.

The Per Cent composition by mass is the % by mass of each element in a compound.

Determination of % Composition: From formula – Calculate the formula weight. % is always =. The formula weight is the whole. Multiply the atomic weight of each element by the # of atoms of that element. That value is the part due to that element. Doing that calculation will produce the % of that element in the compound. Repeat this for each element.

From experimental data – You will have obtained the total mass of the sample of substance and the masses of each element in the substance from the experiment. It is not important right now to know how this data was obtained. In the % formula above, the whole is the total mass of the sample of substance and the parts are the individual element masses.

One can determine, experimentally the empirical formula (that is why it is called an empirical formula) for a compound. Calculating Empirical Formulas STEP 1: Obtain the masses of each element in your sample. If your data is % composition, just use the % values (essentially you are assuming you have a 100 g sample) STEP 2: Divide the mass of each element by its Atomic weight (this gives the moles of each element)

STEP 3: Divide each value from step 2 by the smallest of those values. (This gives the mole ratio of the elements in the substance) STEP 4: If all values in Step 3 are whole #’s, each value is the subscript for that element in the empirical value. If all these values are not whole #’s, then find a whole # to multiply the non-whole # value from step 3 that converts it to a whole #. Multiply all the values from Step 3 by that multiplier. Now you have the subscripts for each element in the empirical formula.

Example: Substance with the following composition: 43.40% Na, 11.32% C and 45.28% O Let’s do this calculation together.

If you know the approximate molecular weight and the empirical formula, you can determine the molecular formula, and from that the exact molecular weight.

For example: Suppose you had a substance with an approximate molecular weight of 32.0 and an empirical formula of HO. Determine the formula weight of the empirical formula (in this case HO, which equals 17.0) Divide the approximate molecular weight by this formula weight and round off your answer to the nearest whole number. In our case: 32.0 / 17.0 = 1.88, round off to 2. Now multiply all subscripts in the empirical formula by this whole #. HO becomes H 2 O 2

Chemical Reactions -> The manipulation of atoms &/or molecules to form different atoms &/or molecules. VERY IMPORTANT – In all chemical reactions, atoms just move around. No new atoms are ever made. Molecules are made but not atoms.

Remember the Law of Conservation of Mass – In its most general form, this law says that the total mass of all substances in the Universe never changes. On a more practical level, applied to chemical reactions, the law means that the total mass of all substances that are present at the beginning of a reaction must equal the total mass of all substances present at the end of the reaction.

Chemical Equations – A shorthand format for stating what takes place during a chemical reaction. It is called an equation because everything that is present at the start is on the left side (called the reactants) and everything present at the end is on the right (called the products). Instead of an equal sign between, an arrow is used to show that the reactants are changing into the products. The total number and type of all atoms must be equal on both sides, in order for the equation to obey the Law of Conservation of Mass. This is called a Balanced Equation.

Balancing: You may only change coefficients, but NEVER subscripts. 1. Balance elements only found in one compound on each side first 2. Treat polyatomic ions as one unit if they appear on both sides unchanged. 3. Don’t unbalance something already balanced unless no choice 4. Use fractions, if necessary

Let’s do some examples together: 1.BCl 3 + H 2 O -> H 3 BO 3 + HCl

2. K + Al 2 O 3 -> Al + K 2 O

3. C 6 H 14 + O 2 -> CO 2 + H 2 O

4. Al 2 (SO 4 ) 3 + Ba(NO 3 ) 2 -> Al(NO 3 ) 3 + BaSO 4

5. NaNO 3 -> NaNO 2 + O 2

Stoichiometry: - The process of determining theoretical amounts of reactants and/or products using the balanced chemical equation for a reaction. As long as we know the amount (either in g, or moles) of any substance involved in a particular chemical reaction and we know the balanced equation, then we can calculate the theoretical amounts of any reactants necessary for that reaction and the theoretical maximum amounts of all products produced in that reaction. This can all be done using one mathematical equation, modifying it as needed for any specific problem.

A is the substance we know information about. The first conversion factor changes grams into moles of A. The second conversion factor, simply a ratio of coefficients from the balanced equation, converts moles A to moles B. The final conversion factor changes moles B into grams of B.

Remember: We can only convert A to B using the coefficients from the equation, and only from moles to moles. If you already know the moles of A then skip the first conversion factor and if you only want to know moles of B, then skip the third conversion factor.

Example: Suppose we have the following balanced equation: BCl 3 + 3H 2 O  H 3 BO 3 + 3HCl How many grams of H 3 BO 3 can be produced if g of H 2 O are completely reacted?

In our general formula, for this problem, A is H 2 O and B is H 3 BO 3.

We could also have determined the amount of HCl that could have been produced. These calculations give the theoretical maximum amount of product (or usually stated as the theoretical yield). We could also have calculated the amount of BCl 3 needed to react with that amount of H 2 O. This calculation is called stoichiometry.

Another important calculation based on chemical equations and stoichiometry is % yield. In the previous case, if we actually obtained 8.56 grams of H 3 BO 3 then the % yield is: 8.56/12.08 x 100 = 70.9 %

Usually, when a reaction is really done, the reactants are not added in exact stoichiometric amounts. Thus, one reactant will be used up before the others. This one controls the actual amounts of product and is called the limiting reagent.

As an example, think of a cake recipe that calls for 2 cups of flour, 1 cup of sugar and I cup of water for each cake. Suppose you have 12 cups of flour, 10 cups of water and 1 cup of sugar. You can only make one cake, even though you have lots of extra flour and water. They are in excess. The sugar is the limiting ingredient and controls how many cakes you can make.

Once you have the balanced equation, calculate the moles of each reactant present. Then determine how many moles of product Q would be formed from the number of moles of each reactant that you have. The reactant that produces the least amount of Q is the limiting reagent. Do all stoichiometric calculations using the amount of the limiting reagent you have.

Let’s do example 3.15 on page 102.