Unit 8. The Mathematics Of Chemical Equations

Transcription

1 Unit 8 The Mathematics Of Chemical Equations

2 Stoichiometry Knowing the amounts of substances that enter into a chemical reaction as well as the amounts of products that result is crucial. In this unit, you will find that you can also apply the mole concept to relate quantities of reactants and products in chemical reactions.

3 What is Stoichiometry? Stoichiometry is the study of the quantitative, or measurable, relationships that exist in chemical formulas and chemical reactions. One aspect of stoichiometry comprises converting among moles, particles, mass, and volume. Another aspect of stoichiometry is concerned with chemical reactions and involves the relationships between reactants and products in a chemical reaction.

4 Interpreting Balanced Chemical Equations The coefficients in a balanced equation can directly indicate any of the following: they can indicate the # of particles of each substance taking part in the reaction. they can indicate the # of moles of each substance taking part in the reaction. they can indicate the volume of each substance taking part in the reaction (@STP) but they can never, ever, ever indicate the mass of each substance taking part

5 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O Now look at the above equation. There are several quantities that you can relate to the coefficients in a balanced equation...moles, particles, and volume...but never grams, because each substance has its own molar mass.

6 Let s look at moles 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 2 moles plus 7 moles make 4 moles and 6 moles The coefficients can represent the # of moles of each substance 6 moles plus 21 moles make 12 moles and 18 moles If I triple one of the coefficients, I must triple the others 1 mole plus 3.5 moles make 2 moles and 3 moles If I cut one of the coefficients in half, I must cut the others in half

7 Let s look at particles 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 25 mlcls plus 175 mlcls make 100 mlclsand 150 mlcls The coefficients can represent the # of particles of each substance, but they must keep the same relationship as the coefficients. 2 A# plus 7 A# make 4 A# and 6 A# Each coefficient can represent a multiple of Avogadro s number (remember 1 mole = 6.02 x particles)

8 If STP, we can compare volumes 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 2 L plus 7 L make 4 L and 6 L The coefficients can represent the # of L of each substance 14 gal plus 49 gal make 28 gal and 42 gal It does not matter what volume unit is used, I multiply one of the coefficients by seven, I must multiply the others by seven 60 cm 3 plus 210 cm 3 make 120 cm 3 and 180 cm 3 Again, it does not matter what volume unit is used what does matter is the relationship among the coefficients.

9 Never, ever relate grams Each substance has its own mass. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O Using a periodic table to determine the mass involved. 60 g plus 224 g make 176 g and 108 g Now totaling the reactant and the product masses. 284 g make 284 g This proves the Law of Conservation of Mass.

10 Stoichiometric Problems Before solving these problems, you must be able to do the following. write correct chemical formulas (know the difference among ionic compounds, molecular compounds, and acids) predict products (know the five types of equations) balance chemical equations know the three definitions of the mole and how to convert among them

11 Steps for Solving Box Problems Step 1 Write a complete balanced equation (predicting products when needed) Step 2 Mark the equation with the given and the wanted (extremely important) Step 3 Arrow down -(always divide) convert the given to moles; if the given is moles, move it to the bottom of the arrow Step 4 Arrow across -divide by the coefficient of the given; then multiply by the coefficient of the wanted Step 5 Arrow up -(always multiply) convert the wanted moles to the desired unit; if the desired unit is moles, rewrite the answer in the correct number of significant digits and use scientific notation Step 6 Compute the answer to the correct number of significant digits and use scientific notation Note: Only do step 4, if you are going from moles to moles, volume to volume, or particles to particles.

12 Example 1: Determine the number of molecules of water produced from the reactions of 87.4 L oxygen with excess hydrogen. Step 1 1 O H 2 2 H 2 O know who is diatomic Step L don t need the H 2? mlcls mark the given and the wanted Step L/mole Step 5 x 6.02 x mlcls/mole convert the given to moles convert moles of wanted to molecules divide by coefficient of given, multiply by coefficient of wanted 3.90 mol 1, x mol Step 4 Step 6 -answer is x (cut it off for 3 sig digs) -answer becomes 4.70 x mlclsh 2 O

13 Example 2: How many grams of nitric acid are produced from the reaction of 94.5 grams of aluminum nitrate with excess sulfuric acid? Step 1 2 Al(NO 3 ) H 2 SO 4 6 HNO Al 2 (SO 4 ) 3 Step g don t need the H 2 SO 4? g don t need the Al 2 (SO 4 ) 3 Step g/mole Step 5 x g/mole convert the given to moles convert moles of wanted to grams 1 Al N O g divide by coefficient of given, multiply by coefficient of wanted.444 mol 1.33 mol 2, x 6 Step 4 1 H N O g Please note: do not use the coefficient when determining the mass of a substance Step 6 -answer is grams (cut it off for 3 sig digs) -answer becomes 8.38 x 10 1 g HNO 3

14 Limiting Reactants When quantities of reactants are available in the exact ratio described by the balanced equation, chemists say that the reactants are in stoichiometric proportions. All of the reactants take part in the reaction. There will be no reactants left over. Most often, however, reactants are available in other than stoichiometric proportions. There is an excess of one of the reactants. The reactants are said to be in nonstoichiometric proportions.

15 Let s look at a ham sandwich!! To make a ham sandwich we need.. 2 pieces + 3 pieces + 1 piece 1 ham of bread of ham of cheese sandwich When we look for supplies, we find that we have 20 pieces 15 pieces 16 pieces how many sandwiches possible? Only 5 sandwiches possible because we will run out of ham before running out of bread or cheese. So the ham is the limiting reactant, the other two reactants are in excess.

16 Identifying Limiting Reactants When chemicals are combined in nonstoichiometric proportions, 1 reactant will limit the amount of product formed. The reactant that limits the amount of product formed is called the limiting reactant. The reactant that is left over is called the reactant in excess. The quantities of products formed in a reaction are always determined by the quantity of the limiting reactant. In order to determine which reactant is limiting and which is in excess, you must compare the moles of each reactant to the mole relationship determined by the balanced equation.

17 Examples 1) 2 Na + 1 I 2 2 NaI If I start with 100 atoms of Na, how many molecules of I 2 do I need? 100 atoms 100 mlcls If I start with 100 atoms of Na, divide by 2 and multiply by 1, tells me that I need 50 molecules of the iodine. This means that Na is limiting and the iodine is in excess by or 50 molecules I need 50 mlclsof iodine 2) 2 Na + 1 I 2 2 NaI If I start with.28 moles of Na, how many moles of I 2 do I need?.28 moles.14 moles If I start with.28 moles of Na, divide by 2 and multiply by 1, tells me that I need.14 moles of the iodine. This means that I have a stoichiometric mixture. I need.14 moles of iodine

18 % Yield The amount of a product that should be produced based on calculations is called the expected yield (could be thought of as the theoretical yield). The amount of a product that is really obtained from a chemical reaction is called the actual yield (could be thought of as the experimental yield). There are many reasons why the actual yield may differ from the expected yield. Some portion of the reactants may simply not react. Some of the reactants may take part in side reactions. Some of the product may be lost during the process of recovering it or transferring it from one container to another.

19 % Yield Formula % x 100% Example: Using our box example #2.instead of making 83.8 g HNO 3, we only make 75.6 g HNO 3, what is the % yield? %.. x 100 % = 90.2 % yield Use some common sense -% yield will not be over 100 % nor should the % yield be under 75 % (for any problems that come from me).