1 IB Topic 1: Quantitative Chemistry 1.4: Mass Relationships in Chemical Reactions  Calculate theoretical yields from chemical equations.  Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given.  Solve problems involving theoretical, experimental and percentage yield.  Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas.  Solve problems using the ideal gas equation, PV = nRT  Apply Avogadro’s law to calculate reacting volumes of gases  Apply the concept of molar volume at standard temperature and pressure in calculations  Analyze graphs relating to the ideal gas equation

2 Review: Identify the mole ratio of any two species in a chemical reaction. How do you calculate the quantity of reactants and products in chemical reactions? Use the chemical equation to predict ratio of reactants and products N 2 +3H 2  2NH 3 1 molecule N 2 reacts with 3 molecules H 2 producing 2 molecules NH 3 1 mole N 2 reacts with 3 moles H 2 producing 2 moles NH 3 The balanced equations gives us the ratio in particles or moles (not mass) of the chemicals involved in the reaction. The ratio of N 2 to NH 3 is 1:2 The ratio of NH 3 to H 2 is 2:3 The ratio of H 2 to N 2 is 3:1

3 Review: Identify the mole ratio of any two species in a chemical reaction. Consider the reaction: 2C 8 H O 2  16CO H 2 O What is the ratio of O 2 to CO 2 ? 25:16 What is the ratio of H 2 O to C 8 H 18 ? 18:2 What is the ratio of CO 2 to H 2 O? What is the ratio of O 2 to C 8 H 18 ? Remember these ratios are in particles or moles. We will be using moles.

4 Calculate theoretical yields from chemical equations. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. In essence, theoretical yield is what you calculate We use a process called stoichiometry: The calculation of quantities in chemical reactions

5 Mole-Mole Conversions  How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl 2  2 NaCl 2.6 moles Cl 2 2 mol NaCl 1 mol Cl 2 = 5.2 moles NaCl

6 Mole-Mass Conversions  Most of the time in chemistry, the amounts are given in grams instead of moles  We still go through moles and use the mole ratio, but now we also use molar mass to get to grams  Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2  2 NaCl 5.00 moles Na 1 mol Cl g Cl 2 2 mol Na 1 mol Cl 2 = 177g Cl 2

7 Practice  Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. 2 Al + 3 I 2  2 AlI moles Al 3 moles I g I 2 2 moles Al 1 mole I 2 = g I 2

8 Mass-Mole  We can also start with mass and convert to moles of product or another reactant  We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water 2 C 2 H O 2  4 CO H g H 2 O 1 mol H 2 O 2 mol C 2 H g H 2 O 6 mol H 2 0 = mol C 2 H 6

9 Practice  Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide  4 Al + 3 O 2  2 Al 2 O g Al 2 O 3 1 mol Al 2 O 3 3 mol O g Al 2 O 3 2 mol Al 2 O g Al 2 O 3 2 mol Al 2 O 3 = mol O 2

10 Mass-Mass Conversions  Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!)  Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

11 Mass-Mass Conversion  Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.  N H 2  2 NH g N 2 1 mol N 2 2 mol NH g NH g N 2 1 mol N 2 1 mol NH 3 = 2.4 g NH 3

Calculate theoretical yields from chemical equations. How much oxygen does it take to burn grams of octane?

Calculate theoretical yields from chemical equations. How much oxygen does it take to burn grams of octane? Write the balanced chemical equation: 2C 8 H O 2  16CO H 2 O

Calculate theoretical yields from chemical equations. How much oxygen does it take to burn grams of octane? 2C 8 H O 2  16CO H 2 O Mass of O 2 required: mol (32.00 g/mol) =351.0 g O 2

Calculate theoretical yields from chemical equations. In a spectacular reaction called the thermite reaction, iron(III) oxide reacts with aluminum producing iron and aluminum oxide. How many grams of iron will be produced from 43.7 grams of aluminum?

Calculate theoretical yields from chemical equations. In a spectacular reaction called the thermite reaction, iron(III) oxide reacts with aluminum producing iron and aluminum oxide. How many grams of iron will be produced from 43.7 grams of aluminum? Write the balanced chemical equation: Fe 2 O 3 + 2Al  2Fe + Al 2 O 3

Calculate theoretical yields from chemical equations. In a spectacular reaction called the thermite reaction, iron(III) oxide reacts with aluminum producing iron and aluminum oxide. How many grams of iron will be produced from 43.7 grams of aluminum? Write the balanced chemical equation: Fe 2 O 3 + 2Al  2Fe + Al 2 O 3 Mass of Fe produced: 1.62 mol (55.8 g/mol) =90.4 g Fe

18 Limiting Reactant: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given. Limiting Reactants You are given amounts for two reactants and one reactant will run out first. This is called the limiting reactant. The reactant that is left over is called the excess reactant. For example: A strip of zinc metal weighing 2.00 g is placed in a solution containing 2.50 g of silver nitrate causing the following reaction to occur: Zn + 2AgNO 3  2Ag + Zn(NO 3 ) 2 How many grams of Ag will be produced?

20 Limiting Reactant  Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant.  That reactant is said to be in excess (there is too much).  The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

21 Limiting Reactant  To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one.  The lower amount of a product is the correct answer.  The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it!  Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

22 Limiting Reactant: Example  10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2  2 AlCl 3  Start with Al:  Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl g AlCl g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl g Cl 2 1 mol Cl 2 2 mol AlCl g AlCl g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3 Limiting Reactant

23 LR Example Continued  We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete.

24 Limiting Reactant Practice  15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

25 Limiting Reactant Practice 2 K + I 2  2 KI  Potassium:  Iodine:  Iodine is the limiting reactant and we get 19.6 g of potassium iodide 15.0g K 1 mol K 2 mol KI 166 g KI 39.1g K 2 mol K 1 mol KI = 63.7g KI 15.0 g I 2 1 mol I 2 2 mol KI 166 g KI 254 g I 2 1 mol I 2 1 mol KI = 19.6 g KI

26 Finding the Amount of Excess  By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.  Can we find the amount of excess potassium in the previous problem?

27 Finding Excess Practice  15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI  We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

28 Limiting Reactant: Recap 1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2. Convert ALL of the reactants to the SAME product (pick any product you choose.) 3. The lowest answer is the correct answer. 4. The reactant that gave you the lowest answer is the LIMITING REACTANT. 5. The other reactant(s) are in EXCESS. 6. To find the amount of excess, subtract the amount used from the given amount. 7. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

29 Practice  How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?  3 Ca + N 2  Ca 3 N 2 Ca 3 N g Ca 3 N g Ca 1 mol Ca 1 mol Ca 3 N g Ca 3 N g Ca 3 mol Ca 1 mol Ca 3 N g Ca 3 mol Ca 1 mol Ca 3 N 2 Ca 3 N 2 = 2.47 g Ca 3 N 2

Solve problems involving theoretical, experimental and percentage yield.

Solve problems involving theoretical, experimental and percentage yield.  Theoretical Yield is the amount of product that would result if all the limiting reagent reacted (i.e. the calculated amount)  Actual Yield is the amount of product actually obtained from a reaction in lab

Solve problems involving theoretical, experimental and percentage yield. The theoretical yield from a chemical reaction is the yield calculated by assuming the reaction goes to completion. In practice we often do not obtain as much product from a reaction mixture as theoretically possible due to a number of factors:  Many reactions do not go to completion  Some reactants may undergo two or more reactions simultaneously forming undesired products  Not all of the desired product can be separated from the rest of the products The amount of a specified pure product actually obtained from a given reaction is the experimental or actual yield.

Solve problems involving theoretical, experimental and percentage yield. Percentage yield indicates how much of a desired product is obtained from a reaction % Yield = Actual Yield Theoretical Yield x 100

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Solve problems involving theoretical, experimental and percentage yield. Consider the reaction: P 4 O H 2 O  4H 3 PO 4 If 558 g of P 4 O 10 reacts, the experimental yield of H 3 PO 4 is 746 g. What is the percentage yield of H 3 PO 4 ? Theoretical amt:  558 g/284 g/mol = mol P 4 O 10  X H 3 PO 4 = 4 H 3 PO 4 = 7.86 mol H 3 PO P 4 O 10 1 P 4 O 10  7.86 mol H 3 PO 4 x 98.0 = 770. g H 3 PO 4 Percentage yield:  746 g x 100% = 96.9% 770 g

Solve problems involving theoretical, experimental and percentage yield. Consider the reaction: P 4 + 5O 2  P 4 O 10 If 272 g of phosphorus reacts, the percentage yield of tetraphosphorus decoxide is 89.5%. What mass of P 4 O 10 is obtained? Theoretical amt:  272 g/124.0 g/mol = mol P 4  X P 4 O 10 = 1 P 4 O 10 = mol P 4 O P 4 1 P 4  mol P 4 O 10 x = 623 g Experimental amt:  623 g x.895 = 558 g

Solve problems involving theoretical, experimental and percentage yield. Consider the reaction: Fe 2 O 3 + 3CO  2Fe + 3CO 2 When 84.8 g iron(III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the percentage yield of this reaction?

Solve problems involving theoretical, experimental and percentage yield. Consider the reaction: Fe 2 O 3 + 3CO  2Fe + 3CO 2 When 84.8 g iron(III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the percentage yield of this reaction? Theoretical amt:  84.8 g/159.2 g/mol = mol Fe 2 O 3  X Fe = 2 Fe= mol Fe mol Fe 2 O 3 1 Fe 2 O 3  mol Fe x 55.8 g.mol = 59.4 g Fe Percentage yield:  54.3 g x 100% = 91.4% 59.4 g

Solve problems involving theoretical, experimental and percentage yield. If 50.0 g silicon dioxide is heated with an excess of carbon, the percentage yield of silicon carbide 76.0%. What mass of silicon carbide will actually be produced? SiO 2 + 3C  SiC + 2CO

Solve problems involving theoretical, experimental and percentage yield. If 50.0 g silicon dioxide is heated with an excess of carbon, the percentage yield of silicon carbide 76.0%. What mass of silicon carbide will actually be produced? SiO 2 + 3C  SiC + 2CO Theoretical amt:  50.0 g/60.1 g/mol = mol SiO 2  X SiC = 1 SiC= mol SiC mol SiO 2 1 SiO 2  mol SiC x 40.1 g.mol = 33.4 g SiC Experimental yield:  33.4 g x.760% = 25.4 g SiC Read pg pg. 259: 30,32

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Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Variables That Describe a Gas  Pressure (P) measured in kPa, mm Hg, atm kPa = 760 mm Hg = 1.00 atm  Volume (V) measured in dm 3 or L 1 dm 3 = 1000 cm 3 = 1 L = 1000 mL  Temperature (T) measured in K (Kelvin) K = o C  Amount of matter (n) measured in moles

43 Gaseous Volume Relationships in Chemical Reactions Kinetic Theory: Tiny particles in all forms of matter are in constant motion Application to Gases 1) A gas is composed of particles that are considered to be small, hard spheres that have insignificant volume and are relatively far apart from one another. Between the particles there is empty space. No attractive or repulsive forces exist between the particles. 2) The particles in a gas move rapidly in constant random motion. They travel in straight paths and move independently of each other. They change direction only after a collision with one another or other objects. 3) All collisions are perfectly elastic. Total kinetic energy remains constant.

Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Boyle’s Law Relates Pressure-Volume  As pressure increases, volume decreases if temperature and amount remain constant.  Spaces between particles so particles can move close closer together  P 1 x V 1 = P 2 x V 2  See pg 335: Sample problem Do practice problems pg 335: 10,11

Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Charles’s Law Relates Temperature-Volume  As temperature increases, volume increases if pressure and amount remain constant  Particles gain kinetic energy, move farther apart  V 1 /T 1 = V 2 /T 2 ;  T has to be in Kelvin ; K= o C  See pg 337: Sample problem Do practice problems pg 337: 12,13

Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Gay-Lussac’s Law Relates Temperature-Pressure  As temperature increases, pressure increases if volume and amount remain constant.  Particles gain kinetic energy so they move faster and have more collisions  P 1 /T 1 = P 2 /T 2 ; T has to be in Kelvin ; K = o C  See pg 338: Sample problem Do practice problems pg : 14,15

Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas. Gas Laws: Combined Gas Law Relates Temperature-Pressure-Volume  P 1 x V 1 /T 1 = P 2 x V 2 /T 2 T has to be in Kelvin  See pg 340: Sample problem Do practice problems pg 340: 16,17

Solve problems using the ideal gas equation, PV = nRT Gas Laws: Ideal Gas Law Relates Temperature-Pressure-Volume-Amount PV = nRT  P = pressure in kPa  V = volume in dm 3 or L  n = moles  R = Gas constant (8.31)  T = temperature in K Ideal gas: particles have no volume and are not attracted to each other Pg 342: Sample Practice 22,23 Pg 343: Sample Practice 24,25

Apply Avogadro’s law to calculate reacting volumes of gases Avogadro’s Hypothesis: Equal volumes of gases at the same temperature and pressure contain equal numbers of particles (moles). At STP (standard temperature & pressure: 273 K and kPa) 1 mole of any gas occupies a volume of 22.4 dm 3 (L). Read pg Pg : Practice 31-36

Apply Avogadro’s law to calculate reacting volumes of gases Assuming STP, how many dm 3 of oxygen are needed to produce 19.8 dm 3 SO 3 according to: 2SO 2 (g) + O 2 (g)  2SO 3 (g) Since equal volumes of gases contain the same number of moles, we can use the equation coefficients with the volumes. X dm 3 O 2 = 1 O 2 = 9.90 dm 3 O dm 3 SO 2 2 SO 2

Apply Avogadro’s law to calculate reacting volumes of gases Nitrogen monoxide and oxygen combine to form the brown gas nitrogen dioxide. How many cm 3 of nitrogen dioxide are produced when 3.4 cm 3 of oxygen reacts with an excess of nitrogen monoxide? Assume conditions of STP.

Apply Avogadro’s law to calculate reacting volumes of gases Nitrogen monoxide and oxygen combine to form the brown gas nitrogen dioxide. How many cm 3 of nitrogen dioxide are produced when 3.4 cm 3 of oxygen reacts with and excess of nitrogen monoxide? Assume conditions of STP. 2NO + O 2  2NO 2 X cm 3 NO 2 = 2 NO 2 = 6.8 cm 3 O cm 3 O 2 1 O 2 Pg. 249: Pg. 250: 17,18

Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) a) How many dm 3 of HF are needed to produce 9.40 dm 3 H 2 at STP? b) How many grams of Sn are needed to react with 20.0 dm 3 HF at STP? c) What volume of H 2 is produced from 37.4 g Sn?

Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) a) How many dm 3 of HF are needed to produce 9.40 dm 3 H 2 at STP? x HF = 2 HF = 18.8 dm 3 HF 9.40 dm 3 H 2 1 H 2

Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) b) How many grams of Sn are needed to react with 20.0 dm 3 of HF at STP? 20.0 dm 3 HF / 22.4 dm 3 mol -1 =.893 mol HF x Sn = 1 Sn =.447 mol Sn.893 HF 2 HF.893 mol Sn x = 53.0 g

Apply the concept of molar volume at standard temperature and pressure in calculations Tin(II) fluoride, formerly found in many kinds of toothpaste, is formed by this reaction: Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) c) What volume of H 2 at STP is produced from 37.4 g Sn? 37.4 g / gmol -1 =.315 mol Sn x H 2 = 1 H 2 =.315 mol H Sn 1 Sn.315 mol H 2 x 22.4 dm 3 = 7.06 dm 3

Analyze graphs relating to the ideal gas equation

Analyze graphs relating to the ideal gas equation Real gases deviate from ideal behavior at low and high pressures and temperatures.  Gas molecules do have some attraction for each other  Gas molecules have a volume