Chapter 5

• To accurately and concisely represent chemical reactions we use the symbols of the elements and

compounds in the reaction.

• This is done with a chemical equation.

• But the equation must be balanced (because according to the Law of Conservation of Mass, matter is

never created or destroyed and during a chemical reaction atoms are not created nor destroyed, only

rearranged).

• Therefore all equations must contain the same number of each particular atom on both sides of

the equation.

• This is accomplished using coefficients.

1. C + O2 CO

• We will practice balancing a few simple equations: NOTE: When balancing you can

only change coefficients. (coefficients are the numbers in front of the formulas of the

elements or compounds)

2. C + O2 CO2

3. H2 + Cl2 HCl

4. Ba(NO3)2 + H2SO4 BaSO4 + HNO3

(Hint: treat polyatomic ions as one item)

Let’s do one more example and look at the balanced equation as a picture.

05_02.JPG

05_01.JPG

• Earlier we talked about atomic weights(AW) for the elements.

• A new value we need to understand now is molecular weight (MW) or Molecular Mass for

compounds (a more general term is formula mass). This is determined by simply adding the AW’s of

every atom in the molecule.

• For example, 3 of the compounds we predicted formulas for earlier have the following MW’s

NaCl 22.99 + 35.45 = 58.44

CaCl2 40.08 x1 + 35.45 x 2 = 110.98

MgSO4 24.31 + 32.06 + 16.00 x 4 = 120.37

Let’s look at what a balanced equation tells us in more detail.

First, a new concept.

• An Italian chemist by the name of Amedeo Avogadro, in the nineteenth century, realized a very

important concept;

• Equal volumes of pure substance gases have equal numbers of atoms or molecules.

• Amount equal to the atomic weight(for a monatomic element) or the molecular weight for a

compound, will have the same volume in the gas phase.

• Therefore this # becomes very important and is known as Avogadro’s #.

• Several experiments were developed to determine the value of this #.

• As of today, we have extremely accurate instrumental ways of determining this # and our best estimate

today is that it’s value is 6.02 X 1023

.

• If we have a sample of any substance containing this # of molecules (for a compound) or atoms (for a

monatomic element) we call that sample a MOLE, abbreviated mol.

6.02 X 1023

items = 1 mole

• Thus a mole is a specific # of particles. But this # of particles will have a specific mass.

For NaCl it would be 58.44 grams, for CaCl2 it would be 110.98 g and for MgSO4 it

would be 120.37 g.

• Use the molecular weight such as 58.44 for NaCl with the proper units Grams/mol. This is essentially

equal to 1 because 58.44 grams of NaCl = 1 mole of NaCl, therefore dividing 58.44g by 1 mole or

dividing 1 mole by 58.44 g for NaCl = 1.00.

• Now I can convert moles of NaCl to grams or vice versa by multiplying by the appropriate value of 1.00

(multiplying by 1 does not change the value), either 58.44 grams/mol or 1 mol/58.44 grams.

• Important to be able to calculate grams from moles and vice versa. Chemists

use # of molecules (too small to be seen) but measure grams (large enough to

see or handle).

Example: Convert 1.65 moles of NaCl to equal value in grams and then convert back

to moles:

NaCl of g 96.426 NaCl of mol 1

NaCl of g 58.44 x NaCl of moles 1.65

moles 1.65 g 58.44

mol 1 x g 96.426

Let’s return to what we can learn from a balanced chemical equation. Such as:

Cu2S + 2 Cu2O 6 Cu + SO2

1 mole 2moles 6 moles 1 mole

0.1 moles 0.2 moles 0.6 moles 0.1 moles

5 moles 10 moles 30 moles 5 moles

• Literally, this says that 1 molecule of Cu2S plus 2 molecules of Cu2O react together to form 6

atoms of Cu plus 1 molecules of SO2.

• But this also means that 1 mole of Cu2S plus 2 moles of Cu2O react together to form 6 moles of Cu

plus 1 mole of SO2, or any equivalent ratio, such as 0.10 mole of Cu2S plus 0.20 moles of Cu2O

react together to form 0.60 moles of Cu plus 0.10 mole of SO2.

• But if we know moles, we can also determine grams, which is what we need to know in a lab,

because we measure grams, not moles or molecules in a typical lab experiment.

• Earlier we discussed homogeneous and heterogeneous mixtures and pointed out that

homogeneous mixtures are all called solutions.

• In solutions there is a solvent, which enables the solution to form (the dissolver) and one or

more solutes, (those substances that are broken down to molecular level and mixed in between

the molecules of the solvent (the dissolvees)

• The maximum amount of solute that can dissolve in a particular solution (depends on the solute, the

solvent, and the temperature and sometimes the pressure (especially when gases are dissolved in

liquids) is called its Solubility.

• We frequently use the terms soluble or insoluble when talking about a solid solute dissolving in a

liquid solvent. We use the terms miscible or immiscible when talking about the solubility of a liquid

solute in a liquid solvent

Concentrations of solutes in a water based solution are measured in many ways. One of

the most common is a term called Molarity. It measures how many moles of solute are

present in every liter of solution.

solution of liters

solute of moles (M)Molarity

Let’s practice a calculation using this equation. See example 5.9 on page 141

WORKED EXAMPLE 5.9 Solution Concentration: Molarity and Moles

Calculate the molarity of a solution made by dissolving 3.50 mol of NaCl in enough water to produce 2.00 L of solution.

Calculate the molarity of a solution that has 0.0500 mol of NH3 in 5.75 L of solution.

Solution

Exercise 5.9A

We read 1.75 M NaCl as “1.75 molar NaCl.”

Calculate the molarity of a solution made by dissolving 0.750 mol of H 3PO4 in enough water to produce 775 mL of solution.

Exercise 5.9B

Another common concentration term is % concentration. This can be % by volume or % by

weight or mass (more common)

100 x solution of Volume

solute of Volume volumeby %

This is used for a solution between 2 liquids

100 x solution of mass

solute of mass massby %

This is used for any solutions, most commonly for a solid solute in a liquid solvent. Let’s

do example 5.14 on page 144.

WORKED EXAMPLE 5.14 Percent by Mass

What is the percent by mass of a solution of 25.5 g of NaCl dissolved in 425 g (425 mL) of water?

Hydrogen peroxide solutions for home use are 3.0% by mass solutions of H 2O2 in water. What is the percent by mass of a solution of 9.40 g of H2O2 dissolved in 335 g (335 mL) of water?

Solution

Exercise 5.14A

Sodium hydroxide (NaOH, lye) is used to make soap and is very soluble in water. What is the percent by mass of a solution that contains 1.00 kg of NaOH dissolved in 950 mL of water?

Exercise 5.14B

Use these values in the above percent-by-mass equation: