Chapter 5 To accurately and concisely represent chemical

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.

• 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) 1. C + O 2 CO

2. C + O 2 CO 2

3. H 2 + Cl 2 HCl 4. Ba(NO 3)2 + H 2 SO 4 Ba. SO 4 + HNO 3 (Hint: treat polyatomic ions as one item)

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

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• 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 Na. Cl 22. 99 + 35. 45 = 58. 44 Ca. Cl 2 40. 08 x 1 + 35. 45 x 2 = 110. 98 Mg. SO 4 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 Na. Cl it would be 58. 44 grams, for Ca. Cl 2 it would be 110. 98 g and for Mg. SO 4 it would be 120. 37 g.

• 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). • Use the molecular weight such as 58. 44 for Na. Cl with the proper units Grams/mol. This is essentially equal to 1 because 58. 44 grams of Na. Cl = 1 mole of Na. Cl, therefore dividing 58. 44 g by 1 mole or dividing 1 mole by 58. 44 g for Na. Cl = 1. 00. • Now I can convert moles of Na. Cl 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.

Example: Convert 1. 65 moles of Na. Cl to equal value in grams and then convert back to moles:

Let’s return to what we can learn from a balanced chemical equation. Such as: Cu 2 S + 2 Cu 2 O 6 Cu + SO 2 1 mole 2 moles 0. 1 moles 0. 2 moles 5 moles 10 moles 6 moles 1 mole 0. 6 moles 0. 1 moles 30 moles 5 moles • Literally, this says that 1 molecule of Cu 2 S plus 2 molecules of Cu 2 O react together to form 6 atoms of Cu plus 1 molecules of SO 2.

• But this also means that 1 mole of Cu 2 S plus 2 moles of Cu 2 O react together to form 6 moles of Cu plus 1 mole of SO 2, or any equivalent ratio, such as 0. 10 mole of Cu 2 S plus 0. 20 moles of Cu 2 O react together to form 0. 60 moles of Cu plus 0. 10 mole of SO 2. • 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.

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 Na. Cl in enough water to produce 2. 00 L of solution. Solution We read 1. 75 M Na. Cl as “ 1. 75 molar Na. Cl. ” Exercise 5. 9 A Calculate the molarity of a solution that has 0. 0500 mol of NH 3 in 5. 75 L of solution. Exercise 5. 9 B Calculate the molarity of a solution made by dissolving 0. 750 mol of H 3 PO 4 in enough water to produce 775 m. L of solution.

Another common concentration term is % concentration. This can be % by volume or % by weight or mass (more common) This is used for a solution between 2 liquids

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 Na. Cl dissolved in 425 g (425 m. L) of water? Solution Use these values in the above percent-by-mass equation: Exercise 5. 14 A Hydrogen peroxide solutions for home use are 3. 0% by mass solutions of H 2 O 2 in water. What is the percent by mass of a solution of 9. 40 g of H 2 O 2 dissolved in 335 g (335 m. L) of water? Exercise 5. 14 B Sodium hydroxide (Na. OH, 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 Na. OH dissolved in 950 m. L of water?