# Chapter 4 Chemistry’s Accounting: Reaction Stoichiometry Life as an astronaut depends on knowing how much fuel to load for reaching orbit and for survival.

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Chapter 4 Chemistry’s Accounting: Reaction Stoichiometry Life as an astronaut depends on knowing how much fuel to load for reaching orbit and for survival in the hostile environment of space. These matters of life and death depend on calculations like those in this chapter. We see how to predict how much oxygen needs to be carried to react with a given amount of hydrogen to power the space shuttle and to generate electricity. Back on Earth, much of industry depends on similar calculations. Quantitative information on the reaction (how much of one reactant combines with a given amount of another).

How to Use Reaction Stoichiometry Mole-to-mole calculation 2H 2 (g)+O 2 (g)  2H 2 O(l)  1 mol O 2 ~ 2 mol H 2 O (stoichiometric relation) “chemically equivalent to” How many moles of nitrogen are needed to produce 5.0 mol of ammonia by reaction with H 2 ? N 2 (g)+3H 2 (g)  2NH 3 (g) Mass-to-mass calculation How many grams of carbon dioxide are produced by burning 100.0 g of propane? C 3 H 8 (g)+5O 2 (g)  3CO 2 (g)+4H 2 O(l) 1 mol 3 mol 1 mol 3mol 2 mol ? ? 5.0 mol ?= mol 44.09 g 132 g 100.0 g ? g ? = g

Gravimetric Analysis Solid waste 25.40 g, silver known to be present Concentrated nitric acid added  Ag ion Hydrochloric acid added  precipitate AgCl 16.1 g Mass percentage of silver in the waste? 1 mol 1 mol x g 16.1 g 107.87 g 143.32 g x=12.1 g Mass percentage of silver = 12.1 g/25.4 g=47.7%

Case Study 4 (a) Concentration of carbon dioxide in the atmosphere over the past 1000 years as determined from ice core measurements. The blue line in the inset shows the increase in the rate of emission of CO 2 since 1850. Greenhouse gas

Case Study 4 (b) The average surface temperature of the Earth from 1860 to 1990. Greenhouse effect: greenhouse gases Trap radiation and raise earth temperature

Case Study 4 (c) The extent to which glaciers and ice caps have receded is shown by these pictures of a boulder in the Andes. The pictures, which were taken in 1978 (a) and 1995 (b), show extensive local warming. (a)(b) Greenhouse effect in the making!

Figure 4.6 (a) When an octane molecule undergoes complete combustion, it forms carbon dioxide and water: one CO 2 molecule is formed for each carbon atom present (yellow arrows). (b) However, in a limited supply of oxygen, some of the carbon atoms end up as carbon monoxide molecules, CO, so the yield of carbon dioxide is reduced (blue arrows).

Figure 4.7 (a) The yield of a product would be 100% if no competing reactions were taking place. (b) However, if a reactant can take part in more than one reaction at the same time, then the yield of a particular product will be less than 100% because other products will also form. (b) (a) The Limits of Reaction

2C 8 H 18 (l)+25O 2 (g)  16CO 2 (g)+18H 2 O(l) 1.0 L(702g) ? g 2*114.2 g 16*44.2 g ?=2160 g (theoretical yield) When oxygen is limited, following reaction also occurs: 2C 8 H 8 (l)+17O 2 (g)  16CO(g)+18H 2 O(l) Suppose we gathered 1.14 kg of CO 2, then Percentage yield = 1.14/2.16=52.8% The Limits of Reaction: An Example

Figure 4.8 Limiting reactant How to decide which is the limiting reactant. (a) The gold and green boxes depict the relative amounts of each reactant that are required by the stoichiometric relation. (b) If the amount of reactant B is less than that required for all A to react, then B is the limiting reactant. (c) If the amount of A is less than that required for all B to react, then A is the limiting reactant. 100 g of water reacts with 100.0 g of calcium carbide. Which is limiting reactant? CaC 2 (s)+2H 2 O(g)  Ca(OH) 2 (aq)+C 2 H 2 (g) 1 mol 2 mol 64.10 g 2*18.02 g=36.04 g 100 g 100 g 1.56 mol 5.55 mol Water is too much! Only 3.12 mol is needed! Excess=5.55 mol –3.12 mol =2.43 mol

Combustion Analysis 1.621g newly synthesized compound (known to contain C,H,O) after combustion analysis yields 1.902 g of water and 3.095 g of carbon dioxide. Determine the empirical molecular formula. After combustion: C  CO 2  +NaOH(s)  NaHCO3 (s) H  H 2 O  +P 4 O 10 (s)  H 3 PO 4 (l) Rest  total - C - H

Combustion Analysis 1.902 g H 2 O  1.902/18.02 mol H 2 O  0.2111 mol H 3.095 g CO 2  3.095/44.01 mol CO 2  0.07032 mol C Oxygen = 1.621g – 0.211*1.079g – 0.07032*12.01g = 0.564 g = 0.0352 mol C:H:O = 0.07032 : 0.2111 : 0.0352 = 2:6:1 C 2 H 6 O (C 2n H 6n O n )

Figure 4.9 A pipet is an accurate means of transferring a fixed volume of solution. Here, a solution containing a reactant is being added to a reaction vessel. 1.345 g KNO3 == 1.345 g/(101.11 g/mol) KNO3 1.345 g of potassium nitrate dissolved In water to produce 25.0 ml of solution Molarity = (1.345/101.11)mol/0.025 L =0.5321 M

Figure 4.10 The steps involved in making up a solution of known concentration (here, a solution of potassium permanganate, KMnO 4 ). (a) A known mass of the compound is dispensed into a volumetric flask. (b) Some water is added to dissolve it. (c) Finally, water is added up to the mark. The bottom of the solution’s meniscus, the curved top surface, should be level with the mark.

Figure 4.11 A schematic summary of how to use molarity to convert volume of solution to moles of solute (left) and the amount of solute to the volume of solution that contains that amount of solute (right).

Figure 4.12 The steps involved in dilution. A small sample of the original solution is transferred to a volumetric flask, and then water is added up to the mark. The diluted solution (right) has fewer solute molecules per given volume than the concentrated solution.

Figure 4.13 When a solution is diluted, the same number of solute molecules occupy a larger volume, so there is a smaller number of molecules in a given volume (indicated by the small square).

Figure 4.14 The apparatus typically used for a titration: magnetic stirrer; flask containing the analyte; clamp and buret containing the titrant.

Figure 4.15 An acid-base titration in progress. The indicator is phenolphthalein.

Figure 4.16 The schematic procedure for volume-to-volume conversions.

Figure 4.17 The schematic procedure for determining a concentration or amount from a titration.

Oxalic acid

Titration: An Example 25.0 ml of solution of oxalic acid is titrated with 0.5 M NaOH (aq) and the stoichiometeric point is reached when 38.0 ml of solution of base is added. Molarity of oxalic acid? NaOH added = 0.038 * 0.5 mol =0.019 mol H 2 C 2 O 4 (aq)+2NaOH  Na 2 C 2 O 4 (aq)+2H 2 O(l) 1 mol 2 mol x mol 0.019 mol  x= 0.0095 mol Molarity of oxalic acid = 0.0095 mol / 0.025 L = 0.38 M

Figure 4.18 In this redox titration, the titrant is the oxidizing agent potassium permanganate, KMnO 4, and the analyte contains iron(II) ions. The purple color of the permanganate ion disappears as it reacts with iron(II). However, at the stoichiometric point, the purple color persists, showing that all the iron(II) has been oxidized to iron(III).

Connection 1 (a) The vitamin supplements on drugstore shelves are the end product of years of discovery, synthesis, and testing.

Connection 1 (b) This field biologist is collecting specimens of plants that may contain compounds with medicinal value.

Assignment for Chapter 4 19, 27, 41, 53

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