Laboratory Chemistry Emergency Plan Days 6-10

Transcription

1 Laboratory Chemistry Emergency Plan Days 6-10 Day 4: Day 5: Reference: Skill Building Topic 9 A Quantitative Understanding of Formulas Skill Building Topic 10 Understanding Mass Relationships in Chemical Reactions Putting It all Together You Are the Chemist - The Solvay Process Periodic Table

2 DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS Skill Building Topic 9 A QUANTITATIVE UNDERSTANDING OF FORMULAS Calculating Percent Composition from a Formula The percent by mass of each material found in an item is called the element s percent composition. To find the percent composition of an element in a compound, first calculate the compound s molar mass. Then divide the total mass of the element in the compound by the molar mass of the compound and multiply the result by 100%. Example Calculate the molar mass of sucrose, C 1 H O mole of C (1.00g/1mole) = gram mole of H (1.01g/mole) =.0 grams 11 mole of O (16.00g/mole) = grams Molar mass of C 1 H O 11 = 34. grams Find the mass % of each element in sucrose mass C/mol g C/mol % C = 100 = 100% = 4.08% mass sucrose/mol 34. g sucrose/mol C % H = mass H/mol mass sucrose/mol 100 =.18 gh/mol 34. g sucrose/mol 100% = 6.48% H % O = mass O/mol mass sucrose/mol x 100 = g O/mol 34. g sucrose/mol 100% = 51.43% O Calculating Percent Composition from Mass In the laboratory, it is possible to determine the percentage composition for many compounds. Example In the laboratory, g magnesium ribbon is burned in the presence of oxygen to form magnesium oxide. The product has a mass of g. What is the percentage composition of the compound? Mass of Magnesium: g Mass of magnesium oxide: g g Mg % Mg = 100 = 60.3% Mg g compound % O = 100% 60.3% = 39.7% O

3 DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS Formulas from Percentage Composition: Empirical Formulas An empirical formula gives the relative numbers of different elements in a substance, using the smallest whole numbers for subscripts. The empirical formula of the compound can be calculated from percent composition data or from the number of grams of each element in the sample. A formula compares the relative numbers of moles of each element. If percents are used, they must be converted to grams, and grams to moles. It is easiest to assume there is 100 g of the compound. Then the percentages are equal to the number of grams of each element. Example 1 A hydrocarbon (the only elements in the compound are hydrogen and carbon) consists of 85.7% carbon and 14.3% hydrogen. What is its empirical formula? Step 1: Assume that you have 100 g of a compound, and that 85.7 g of it are carbon and 14.3 g are hydrogen. Step : Use dimensional analysis to find the number of moles of each element in the compound mol C 85.7 g C = 7.14 mol C 1.0 g C 1.00 mol H 14.3 g H = 14. mol H 1.01g H Step 3: Determine the smallest whole number ratio of moles of elements by dividing all mole values by the smallest mol C = 1mol C mol H = 1.99 mol H 7.14 The ratio of moles of C to moles of H = 1: and the empirical formula for the compound is CH. Example The percentage composition of one of the oxides of nitrogen is 74.07% oxygen and 5.93% nitrogen. What is the empirical formula? Step g of compound consists of g oxygen and 5.93 g nitrogen. Step : Step 3: 1.00 mol O g O = 4.6 mol O g O 1.00 mol N 5.93 g N = 1.85 mol N 14.01g N 4.6 mol O =.5 mol O mol N = 1.0 mol N 1.85

4 DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS Step 4: This ratio does not consist only of whole numbers, but formulas are not expressed with decimals: NO.5. Multiply all the numbers in the ratio by a number that converts the decimal to a whole number. In this case the number is, and the empirical formula becomes N O 5.

5 DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS Skill Building Topic 9 A QUANTITATIVE UNDERSTANDING OF FORMULAS Practice Exercises Complete the following exercises. 1. Find the percent carbon and percent hydrogen in C H 4.. Find the percent carbon and the percent hydrogen in C 3 H What do the answers to problems 1 and have in common? Explain your answer. 4. Find the percent of each element in the following compounds. a. MgSO 4 b. MgSO 4 7 H O c. (CH 3 ) SO d. C 5 H 5 N

6 DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS e. Ca 3 (PO 4 ) 5. A compound consists of g Fe and 5.01 g O. What is the percent of each element in the compound? 6. A sample of adrenaline, an important hormone, is made up of g carbon, g hydrogen, 0.6 g oxygen, and g nitrogen. What is the percent of each element in the compound? 7. Copper (II) sulfate commonly appears as a hydrated salt. In the lab, a sample of hydrated salt was heated until all the water was driven off. The original sample had a mass of 4.50 g. The dried sample has a mass of g. What is the percent of water in this compound? 8. A compound is 8.3% nitrogen and 17.6% hydrogen. What is the empirical formula? 9. Another compound is 5.% C, 13.0% H, and 34.8% O. What is the empirical formula? 10. Nicotine is 74.03% C, 8.70% H, and 17.7% N. What is the empirical formula?

7 DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS 11. A hydrocarbon is found to be 8.66% carbon. What is the empirical formula? 1. A sample of a compound contains 0.5 g titanium and g chlorine. Determine the empirical formula of this compound.

8 DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS Skill Building Topic 10 UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS If the number of atoms is conserved in a chemical reaction, the mass must also be conserved as expected from the Law of Conservation of Mass. In the equation for the formation of water H (g) + O (g) H O(l) molecules of hydrogen and 1 molecule of oxygen combine to form molecules of water. We could also say that moles of hydrogen react with 1 mole of oxygen to form moles of water. Using the number of grams in a mole of each substance, the mass relationships in the table can be determined. The ratio of moles of hydrogen to moles of oxygen in forming water will be :1. If 10 moles of hydrogen are available, 5 moles of oxygen are required. H (g) + O (g) H O(l) molecules 1 molecule molecules mol 1 mol mol moles x (.0g/mol) 1 mole x (3.00g/mol) moles x (18.0g/mol) 4.04 g 3.00 g g Solving problems involving the masses of products and or reactants is conveniently accomplished by dimensional analysis. All numerical problems involving chemical reactions begin with a balanced equation. Example Find the mass of water produced when 10.0 grams hydrogen react with plenty of oxygen. H (g) + O (g) H O(l) Step 1: We know the amount of oxygen is in excess, so we will focus on hydrogen. Find the moles of hydrogen represented by 10.0 g by using the molar mass of H. 1 mol H 10.0 g H = 4.95 mol H.0 g H Step : Find the number of moles of H O produced by 4.95 mole H. From the balanced equation, we know that for every mol H we produce mol H O. mol H O 4.95 mol H = 4.95 mol H O mol H Step 3: Find the mass of H O that contains 4.95 mol H O by using the molar mass of water g HO 4.95 mol H O = 89.1 g HO 1mol HO Most chemistry students find it is more convenient to set up all three steps in one problem. Make sure that all labels cancel except the g H O, an appropriate unit to express the mass of H O as required in the problem. 1 mol H mol HO 18.0 g HO 10.0 g H = 89.1 g HO.0 g H mol H 1 mol H O Molar mass Coefficients Molar Mass

9 DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS of H in equation of H O Understanding Limiting Reactants In Unit 4, Section B of your ChemCom textbook you will find a conceptual atom-counting method for finding the limiting reactant of an equation. Refer to that discussion before you begin this presentation, which presents two additional methods for finding limiting reactants. If a combustion problem states that excess oxygen is available, we need not concern ourselves with the oxygen-to-water ratio. The mass of water produced is predicted from (and limited by) the mass of hydrogen available. Of course, in the laboratory we often deal with specified masses of each reactant, but that requires an enhanced problem-solving method. Suppose a family wants to make several chile rellenos casseroles to serve at a neighborhood party. The recipe lists required ingredients, which are itemized in the left column of the table in Figure 3. The right column represents a survey of pantry and refrigerator contents. Required Ingredients Available Ingredients Possible Casseroles 1 7-oz can whole green 7-oz cans whole green chiles chiles 1 pound Monterrey Jack 3 pounds Monterrey Jack 3 cheese Cheese 1 pound cheddar cheese 3 pounds cheddar cheese 3 3 eggs 1 dozen eggs 4 3 Tablespoons flour 5 pounds flour Many 5 oz canned evaporated milk 4 cans evaporated milk 4 Figure 3: Required Ingredients Table How many casseroles can be made? Although four casseroles can be made from the available eggs or milk, there are only enough cans of green chiles for two casseroles. In other words, the number of cans of green chiles can be called the limiting factor. After the two casseroles are prepared, cheese, eggs, flour, and milk will remain, but all the green chiles will be used. Therefore, no more casseroles can be made. In this example, the green chiles are the limiting reactant. The limiting reactant is the reactant that is consumed first and limits the amount of product that can be made. The same principle applies in determining the quantity of product that can be produced in a chemical reaction. Let s take another look at the reaction of hydrogen and oxygen to produce water, then consider what would happen if.00 mol hydrogen and.00 mol oxygen were available. How many moles of water can be produced? What is the limiting reactant? Which reactant will be in excess and by how much? H (g) + O (g) H O(l) The balanced chemical equation states that.00 mol hydrogen react with 1.00 mol oxygen. When the reaction is complete,.00 mol water are produced and 1.00 mol oxygen remains unreacted. This problem is easy to solve by inspection. A more systematic way to solve the problem is to create an SRF table, as shown in Figure 4. The table is composed of the following lines: Line 1: The balanced chemical equation is listed. Line : The Starting number of moles of each substance is listed. This would be what is available, the same as the ingredients for the casseroles. Line 3: The Reacting ratio determined from the coefficients in the balanced equation is multiplied by x, the basic amount of moles that will react. The reactants are being consumed so a minus sign is placed in front of them. The products are increasing so a plus sign is placed in front of them.

10 DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS Line 4: This line contains what will be left and what will be formed when the reaction is complete. The values of this line are obtained by adding lines and 3. This line will provide all the answers, in Final moles, to the problems. Line 1 H + O H O Line Starting moles 0 Line 3 Reacting moles -x -1x +x Line 4 Final moles -x -1x 0+x Figure 4 Sample SRF Table When the reaction is completed, either the hydrogen will be consumed or the oxygen will be consumed. That means that either x = 0 (hydrogen) or - 1x = 0 (oxygen) If we solve for x in both equations the smallest x value will be the limiting reactant. The larger value will represent an amount in excess of what is possible with the given ingredients. So for this example, x = 1 (hydrogen) or x = (oxygen). Since x = 1 is smallest, it is the value we choose. This step also identifies the substance that will run out first, the limiting reactant. In this case the hydrogen is the limiting reactant. Using the value of x = 1, we can determine the final amounts. The amount of product is x = (1) = moles of H O. The reactant in excess is O and it is excess by 1x = 1(1) = 1 mole of O. This method of working the problem gives the same result as the visual inspection hydrogen is the limiting reactant, and.00 mol H O are produced. The advantage of learning this method is that it works even when the coefficients become difficult to use with the visual method. Example Aluminum chloride, AlCl 3, has many uses including in deodorants and antiperspirants. It is synthesized from aluminum and chlorine. What mass of AlCl 3 can be produced if 100 g of each reactant are available? What is the limiting reactant? How many grams of the excess reactant remain? Step 1: Write the balanced equation. Al(s) + 3 Cl (g) AlCl 3 (s) Step : Set up the SRF Table. Since only moles can go into the table, the grams of each reactant will first need to be converted to moles. See Figure 4. 1 mol Al 100 g Al = 3.70 moles Al 7.0 g Al 1 mol Cl 100 g Cl = moles Cl 71.0 gcl Al(s) + 3 Cl (g) AlCl 3 (s) Starting moles Reacting moles -x -3x +x Final moles x x x Step 3: Set the Final moles equal to zero and solve for x. Choose the smallest value of x x = 0 or x = 0

11 DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS x = 1.85 moles or x = 0.47 moles Since x = 0.47 mol is smaller, Cl will be consumed first; it is the limiting reactant. Step 4: Determine the amount of product formed and the amount of Al remaining by substituting x = 0.47 into the corresponding equations on the Final moles line. Amount of product (AlCl 3 ): x = (0.47 mole) = 0.94 moles AlCl 3 Amount of Al remaining: 3.70 (0.47 mole) =.76 moles of Al Step 5: Convert the moles back to grams. 134 grams 0.94 moles AlCl3 = 16 grams AlCl3 formed 1 mole.76 moles Al 7.0 grams 1 mole = 74.5 grams Al left

12 DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS Skill Building Topic 10 UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS Practice Exercises 1. Acetylene burns in air to form carbon dioxide and water: 5 C H (g) + O (g) 4 CO (g) + H O(l) How many moles of CO are formed from 5.0 moles C H?. If insufficient oxygen is available, carbon monoxide can be a product of the combustion of butane: 9 C 4 H 10 (l) + O (g) 8 CO(g) + 10 H O(l). What mass of CO could be produced from 5.0 g butane? g NaNH is required for an experiment. Using the following reaction, what mass of sodium is required for reaction. Na(s) + NH 3 (g) NaNH (s) + H (g)? 4. Ethanol and acetic acid react to produce ethyl acetate according to the reaction C H 5 OH + CH 3 COOH CH 3 COOC H 5. If the reaction is only 35% efficient at the conditions used, what mass of CH 3 COOH will be necessary to produce 100 g CH 3 COOC H 5? Assume that sufficient ethanol is available. 5. Heating CaCO 3 yields CaO and CO. Write the balanced equation. Calculate the mass of CaCO 3 consumed when 4.65 g of CaO forms.

13 DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS 6. In the synthesis of sodium amide (NaNH ), what is the maximum mass of NaNH possible if 50.0 g of Na and 50.0 g NH 3 were used? Na(l) + NH 3 (g) NaNH (s) + H (g) 7. The fuel methanol, CH 3 OH, can be made directly from carbon monoxide (CO) and hydrogen (H ). a. Write a balanced equation for the reaction. b. Calculate the maximum mass of methanol if one starts with 5.75 g CO and 10.0 g H. c. Which reactant is the limiting reactant? d. How much of the excess reactant remains? 8. Aspirin (C 9 H 8 O 4 ) is synthesized in the laboratory from salicylic acid (C 7 H 6 O 3 ) and acetic anhydride (C 4 H 6 O 3 ): C 7 H 6 O 3 (s) + C 4 H 6 O 3 (l) C 9 H 8 O 4 (s) + CH 3 COOH(l) a. What is the theoretical yield of aspirin if you started with 15.0 g salicylic acid and 15.0 g acetic anhydride? b. Which reactant is the limiting reactant? c. What mass of the excess reactant remains?

14 DAY FIVE: YOU ARE THE CHEMIST Putting it all Together You are the Chemist Introduction As an Undergraduate Senior in a Chemical Science, you have been hired by the SPC Corporation to work at their Green River Wyoming Plant as an Engineering Summer Intern. You will be working with the Plant Engineer on the Solvay Process. You have done an internet search on this process and have found the following information: Solvay Process for the Production of Sodium Carbonate History Two Belgian brothers, Ernest and Alfred Solvay, perfected a new procedure for producing soda ash (sodium carbonate) in They licensed their patent for this procedure in several countries. The American patents were held by the Solvay Process Company, incorporated in New York in The company's first plant was near Syracuse, New York, near a limestone quarry that provided the essential raw materials. Other plants were built in Hutchinson, Kansas; Detroit; and Milwaukee. The manufacture of soda ash proved extremely profitable, as it was sold for a wide variety of industrial uses. Solvay eventually expanded into the production of other alkali products, including caustic soda (sodium hydroxide), salts, calcium chloride and baking soda (sodium bicarbonate). A second plant was opened in Detroit, Michigan in The Solvay Process Company was absorbed in 191 by the Allied Chemical & Dye Corporation. Process Summary Sodium carbonate, Na CO 3, soda ash, has a number of uses but its most common use is in the production of glass. Since the 1860's, sodium carbonate has been produced using the Solvay Process. The Solvay Process is a continuous process using limestone (CaCO 3 ) to produce carbon dioxide (CO ) which reacts with ammonia (NH 3 ) dissolved in brine (concentrated NaCl (aq) ) to produce sodium carbonate (Na CO 3 ). The steps in the Solvay Process are: 1. Brine Purification (Sodium Chloride Purification). Sodium Hydrogen Carbonate Formation 3. Sodium Carbonate Formation 4. Ammonia Recovery Sodium carbonate, Na CO 3, dissolves in water to form an alkaline solution. Used as a base, sodium carbonate is cheaper and safer than sodium hydroxide.

15 DAY FIVE: YOU ARE THE CHEMIST Uses of Sodium Carbonate Use Process Notes Glass Making A mixture of Na CO 3, CaCO 3 and SiO (silicon dioxide sand) is used for window or bottle glass. Water Softening Agent CO 3 - from dissolved Na CO 3 can precipitate Mg + and Ca + ions from hard water as the insoluble carbonates, preventing them from forming a precipitate with soap resulting in scum. For this reason, sodium carbonate is also known as washing soda. Paper Making Na CO 3 is used to produce the NaHSO 3 necessary for the sulfite method of separating lignin from cellulose. Baking Soda Production Baking soda (or sodium hydrogen carbonate or sodium bicarbonate), NaHCO 3, is used in food preparation and in fire extinguishers. Sodium Hydroxide Production for Soaps and Detergents Na CO 3 is reacted with a Ca(OH), slaked lime, suspension. Wool Processing Na CO 3 removes grease from wool and neutralises acidic solutions. Power Generation Na CO 3 is used to remove SO (g) from flue gases in power stations.

16 DAY FIVE: YOU ARE THE CHEMIST Solvay Process The Solvay Process for the production of sodium carbonate is summarised in the flowchart below: brine ammonia ----->ammoniated brine<----- NaCl (aq) NH 3 /\ NaCl limestone H CaCO 3 O NH 3 NH 3 \/ lime kiln \/ CO -----> carbonating tower H O CaO \/ filter NH 4 Cl \/ \/ > lime Ca(OH) slaker > \/ product NaHCO o C \/ product Na CO 3 ammonia recovery \/ by-product CaCl

17 DAY FIVE: YOU ARE THE CHEMIST Process Steps Brine Purification Brine (Sodium Chloride solution) is concentrated by evaporation to at least 30% Impurities such as calcium, magnesium and iron are removed by precipitation as insoluble salts: Ca + (aq) + CO 3 - (aq) -----> CaCO 3(s) Mg + (aq) + OH - (aq) -----> Mg(OH) (s) Fe 3+ (aq) + 3OH - (aq) -----> Fe(OH) 3(s) Brine solution is then filtered and passed through an ammonia tower to dissolve ammonia. This process is exothermic, releases energy, so the ammonia tower is cooled. Sodium Hydrogen Carbonate Formation Carbon dioixide is produced by the thermal decomposition of limestone, CaCO 3(s), in the lime kiln: CaCO 3(s) -----> CO (g) + CaO (s) Carbon dioxide is bubbled through the ammoniated brine solution in the carbonating tower. The carbon dioxide dissolves to form a weak acid: CO (g) + H O (l) HCO 3 - (aq) + H + (aq) The ammonia in the brine reacts with H + to form ammonium ions: NH 3(aq) + H + (aq) NH 4 + (aq) The HCO 3 - then reacts with the Na + to form a suspension of sodium hydrogen carbonate: HCO 3 - (aq) + Na + (aq) NaHCO 3(s) NaHCO 3 precipitates because of the large excess of Na + present in the brine which forces the equilibrium position to shift to the right by Le Chatelier's Principle (NaHCO 3 is quite soluble in water). The overall molecular equation for the formation of sodium hydrogen carbonate in the carbonating tower is: NH 3(aq) + CO (g) + NaCl (aq) + H O (l) -----> NaHCO 3(s) + NH 4 Cl (aq) The net ionic equation for the formation of sodium hydrogen carbonate: NH 3(aq) + CO (g) + Na + (aq) + H O (l) -----> NaHCO 3(s) + NH 4 + (aq) where Cl - is a spectator ion (it does not participate in the reaction).

18 DAY FIVE: YOU ARE THE CHEMIST Sodium Carbonate Formation Suspended sodium hydrogen carbonate is removed from the carbonating tower and heated at 300 o C to produce sodium carbonate: NaHCO 3(s) -----> Na CO 3(s) + CO (g) + H O (g) The carbon dioxide produced is recycled back into the carbonating tower. Ammonia Recovery CaO is formed as a by-product of the thermal decomposition of limestone in the lime kiln. This CaO enters a lime slaker to react with water to form calcium hydroxide: CaO (s) + H O (l) -----> Ca(OH) (aq) The calcium hydroxide produced here is reacted with the ammonium chloride separated out of the carbonating tower by filtration: Ca(OH) (aq) + NH 4 Cl (aq) -----> CaCl (aq) + H O (l) + NH 3(g) The ammonia is recycled back into the process to form ammoniated brine. Calcium chloride is formed as a by-product of the Solvay Process. Environmental Issues 1. Solid Wastes Calcium chloride, CaCl, is a by-product of the Solvay Process. There are a limited number of uses for CaCl as a drying agent in industry, de-icing roads, an additive in soil treatment, and an additive in concrete. The rest must be disposed of by evaporating to dryness and disposing of the solid. CaCl can not be pumped into the Green River because it will raise the concentration of chloride ion to unacceptable levels. Other solid wastes include unreacted calcium carbonate, sand and clay from the kiln. These are used to make bricks, landfill, or road base.. Air Pollution Some ammonia is lost to the atmosphere during the Solvay Process. Ammonia is a toxic atmospheric pollutant. Ammonia losses are minimized by recycling the gas to reduce plant operation costs. Carbon dioxide gas is also recycled. 3. Thermal Pollution Some of the processes involved in the Solvay Process are exothermic, they release heat. The Green River Plant has to cool the water first before releasing in order to prevent disruption to aquatic organisms.

19 DAY FIVE: YOU ARE THE CHEMIST Based on the manufacturing information in this packet and in the skills that you learned in the Skill Building assignments answer the following questions: 1. If SPC s Green River Plant produces 35,000,000 Kg per year of sodium carbonate. How many moles of calcium carbonate are needed per year to produce this? Take the overall equation as, CaCO 3 (s) + NaCl (aq) Na CO 3 (aq) + CaCl (aq) Would this problem have been easier for you to do the calculation if the scientific notation value of 3.5 x 10 8 Kg had been used?. Evaporative basins at Dry Creek produce an average of 650,000,000 Kg per year of salt. This is purified, then dissolved to form a saturated brine solution that is pumped to the Solvay plant. Ammonia is dissolved in the brine solution and then the ammoniated brine is reacted with carbon dioxide. A. Write an equation for this reaction. B. If 50% of the original salt is sodium chloride, what mass of ammonia will be needed to react with it? C. Since this is a continuous process and almost all of the ammonia used is continuously recycled and reused many times on a daily basis, is the amount calculated in B realistic? 3. How much carbon dioxide must be made per year by heating calcium carbonate? (Hint: one mole of carbon dioxide is produced for each mole of calcium carbonate used.) 4. Can all of the waste carbon dioxide from the last step in the process be reused in the carbonating tower? Since this is a continuous process, is this amount of carbon dioxide the same amount as calculated in problem 4 or is it very much less? Explain your answer in terms of the Sodium Carbonate formation reaction and the nature of recycling materials in a continuous process. NaHCO 3(s) -----> Na CO 3(s) + CO (g) + H O (g)

20 Periodic Table of the Elements Chemistry Reference Sheet California Standards Test Sodium.99 Na 11 Atomic number Element symbol Average atomic mass* Element name Hydrogen 1.01 H 1 Lithium 6.94 Li 3 Sodium.99 Na 11 Potassium K Nickel Ni 8 Rubidium Rb 37 Rutherfordium (61) Rf 104 Molybdenum Mo 4 Germanium 7.61 Ge 3 1 1A A B 11 1B 1 B 13 3A 16 6A Key B A 15 5A 17 7A 18 8A 3 3B 4 4B 5 5B 6 6B Copper Cu 9 Cobalt Co 7 Helium 4.00 He Boron B 5 Carbon 1.01 C 6 Nitrogen N 7 Oxygen O 8 Fluorine F 9 Neon 0.18 Ne 10 Aluminum 6.98 Al 13 Silicon 8.09 Si 14 Phosphorus P 15 Sulfur 3.07 S 16 Chlorine Cl 17 Argon Ar 18 Calcium Ca 0 Scandium Sc 1 Titanium Ti Chromium 5.00 Cr 4 Iron Fe 6 Zinc Zn 30 Gallium 69.7 Ga 31 Arsenic 74.9 As 33 Selenium Se 34 Bromine Br 35 Krypton Kr 36 Strontium 87.6 Sr 38 Yttrium Y 39 Zirconium 91. Zr 40 Niobium 9.91 Nb 41 Technetium (98) Tc 43 Ruthenium Ru 44 Rhodium Rh 45 Palladium Silver Ag 47 Cadmium Cd 48 Indium In 49 Tin Sn 50 Antimony Sb 51 Tellurium Te 5 Iodine I 53 Xenon Xe 54 Cesium Cs 55 Barium Ba 56 Lanthanum La 57 Hafnium Hf 7 Tantalum Ta 73 Tungsten W 74 Rhenium Re 75 Osmium Os 76 Iridium 19. Ir 77 Platinum Pt 78 Gold Au 79 Mercury Hg 80 Thallium Tl 81 Lead 07. Pb 8 Bismuth Bi 83 Polonium (09) Po 84 Astatine (10) At 85 Pd Radon () Rn 86 Francium (3) Fr 87 Radium (6) Ra 88 Actinium (7) Ac 89 Dubnium (6) Db 105 Seaborgium (66) Sg 106 Bohrium (64) Bh 107 Hassium (69) Hs 108 Meitnerium (68) Mt 109 Magnesium 4.31 Mg 1 Beryllium 9.01 Be 4 Vanadium V 3 Manganese Mn 5 * If this number is in parentheses, then it refers to the atomic mass of the most stable isotope. Praseodymium Pr 59 Mendelevium (58) Md 101 Cerium Ce 58 Neodymium Nd 60 Promethium (145) Pm 61 Samarium Sm 6 Europium Eu 63 Gadolinium Gd 64 Terbium Tb 65 Dysprosium Dy 66 Holmium Ho 67 Erbium Er 68 Thulium Tm 69 Ytterbium Yb 70 Lutetium Lu 71 Thorium 3.04 Th 90 Protactinium Pa 91 Uranium U 9 Neptunium (37) Np 93 Plutonium (44) Pu 94 Americium (43) Am 95 Curium (47) Cm 96 Berkelium (47) Bk 97 Californium (51) Cf 98 Einsteinium (5) Es 99 Fermium (57) Fm 100 Nobelium (59) No 10 Lawrencium (6) Lr 103 Copyright 008 California Department of Education

21 Formulas, Constants, and Unit Conversions Chemistry Reference Sheet California Standards Test Formulas Ideal Gas Law: PV = nrt Calorimetric Formulas P 1 V 1 P V Combined Gas Law: = No Phase Change: Q = m(δt)c p T1 T Pressure Formula: P= F A Latent Heat of Fusion: Q = mδh fus Mass-Energy Formula: E = mc Latent Heat of Vaporization: Q = mδh vap Constants L Volume of Ideal Gas at STP:.4 mol Speed of Light in a Vacuum: c = m s Specific Heat of Water: C p (H O) = 1.00 cal J (g C) = 4.18 (g C) Latent Heat of Fusion of Water: ΔH cal J fus (H O)= 80 g = 334 g Latent Heat of Vaporization of Water: ΔH cal J vap (H O)= 540 g = 60 g Calorie-Joule Conversion: 1 cal = J Unit Conversions Absolute Temperature Conversion: K = C + 73 lbs. Pressure Conversions: 1 atm = 760 mm Hg = 760 Torr = kpa = 14.7 in. = 9.9 in. Hg Copyright 008 California Department of Education

22 Day Four Honors Chemistry Addendum In a laboratory after the proper mass measurements had been made, a sample of pure nickel was heated in crucible, the nickel reacted with oxygen. Once the reaction was complete the product formed was nickel oxide, the mass was taken and the following data was collected Mass of crucible = 30.0 Mass of Nickel and Crucible = Mass of Nickel Oxide and crucible = Determine the following information based on the data given Mass of nickel = Mass of nickel oxide = Mass of oxygen = Based on these calculations what is the empirical formula for the nickel oxide? 1. Write the chemical formula for the reactants in the above reaction.. Write chemical formula for the product in the above reaction. 3. Describe what happened in the above reaction.

23 Day Five Honors Chemistry Addendum 1. Look up if you do not remember the equation for respiration and fill in the blanks + +. Make sure the equation is balanced if you have not done so already. 3. The average person consumes 0.84kg of oxygen per day, how many liters of Carbon dioxide will be produced by the average person in one day? 4. If there is 00grams of glucose and 840g of Oxygen available what is the limiting reagent? 5. Using the limiting reagent from problem 4 calculate the amount of Carbon dioxide produced and the amount of water produced. 6. Using problems 4 and 5 as evidence, explain why the intake of food is necessary.