Stoichiometry. Can You Make 2.00 Grams of a Compound?

Introduction Stoichiometry... Can You Make 2.00 Grams of a Compound? Catalog No. AP4554 Publication No. 4554 Use your skills of predicting chemical reactions, balancing equations, and calculating molar mass to solve a complex stoichiometry problem. Then test your laboratory techniques by mixing the reactants and isolating exactly 2.00 g of a compound. Concepts Balancing equations Stoichiometry Molar mass Solubility rules Background Stoichiometry is the branch of chemistry that deals with the numerical relationships and mathematical proportions of reactants and products in chemical reactions. For any chemical equation the mole, mass, and atom ratios must remain constant. The coefficients from a balanced chemical equation provide definite ratios between the number of moles of reactants and products involved (equation 1). Equation 1: N 2 (g) + 3H 2 (g) 2N (g) ne mole of any chemical substance contains 6.02 10 23 particles (molecules, atoms, etc.) and has a mass equal to the sum of atomic weights of all elements in the formula. Therefore, the mole ratio of a balanced chemical equation can be converted into a mass ratio or number of particles ratio. These ratios can be used as conversion factors to determine the exact amounts of reactants that combine and products that form during a chemical change (equation 2). Equation 2: Balanced equation: N 2 (g) + 3H 2 (g) 2N (g) Molar mass: 28.0 2.0 17.0 Mass: 28.0 g + 6.0 g = 34.0 g When two ionic compounds are placed in solution, a double replacement reaction can occur. Double replacement reactions are also called double displacement or exchange reactions and follow the pattern: A + X + B + Y B + X + A + Y. Normally, the reactants are chosen so one product will precipitate out of solution while the other ions stay in solution. In this way, one of the products can be easily isolated by filtration. Equation 3 is an example of a double replacement reaction. Equation 3: Pb(N 3 (aq) + KI(aq) PbI 2 (s) + KN 3 (aq) When the reacting ionic compounds are dissolved in water, they separate into the four different ions that are free to move around and then re-combine to form the new products. If one of the combination of ions is insoluble in water, it will precipitate out of solution. In equation 3, lead iodide (PbI 2 ) precipitates out of solution. Some ionic solids also contain water molecules that are trapped within their crystal structure. These are called hydrated compounds. An example of a hydrated compound would be copper(ii) sulfate pentahydrate, Cu(S 4 ) 5H 2. During the dissolving process, the water molecules are released into the solution. In order to produce the 2.00 grams of desired product in the laboratory using a double replacement reaction, a balanced equation is required and the product that will precipitate must be identified. The next step is to determine the molar mass of each reactant and product, and finally use dimensional analysis to determine the quantities of reactants required. The following four steps describe the process in detail. CHEM-FAX...makes science teaching easier. IN4554 041509

Step 1. Write a balanced formula equation for the reaction. Preparing a balanced chemical formula equation is the first step in understanding a reaction. Start the process by determining what reaction is occurring and write out the names of all reactants and products. Equation 4: Reaction: nitrogen + hydrogen = ammonia Next, prepare an unbalanced equation by writing out the chemical formula for each reactant and product. Equation 5: Unbalanced equation: N 2 + H 2 N Finally, the chemical equation must be balanced. The key principle in balancing equations is that atoms are conserved in a chemical reaction. Therefore, the same number and type of atoms must be found among the reactants and products of a reaction. For example, if you start with two nitrogen atoms in the reactants, you must have two nitrogen atoms in the products. Equation 6: Balanced equation: N 2 + 3H 2 2N Most chemical equations can be balanced by trial and error. This is done by adding up the quantity of each atom present on the reactants side of the equation and on the product side of the equation. Identify which atoms are not balanced and determine what coefficients are necessary in front of one of the reactants or products to balance the quantity of that atom on both sides of the equation. Step 2. Predict which compound will form a precipitate using general solubility rules. When working with aqueous solutions, it is helpful to understand a few rules concerning which substances are soluble in water and which will form precipitates. The more common solubility rules are listed below: 1. All common salts of the Group IA (Li, Na, K, etc.) elements and the ammonium ion are soluble. 2. All common acetates and nitrates are soluble. 3. All binary (two element) compounds of Group VIIA elements (other than F) with metals are soluble, except those of silver, mercury(i), and lead. 4. All sulfates are soluble except those of barium, strontium, lead, calcium, silver, and mercury(i). 5. Except for those in Rule #1, carbonates, hydroxides, oxides, and phosphates are insoluble. Using these solubility rules, predict if any of the possible combination of reactant ions can form insoluble compounds. If so, the product will precipitate out of solution. Step 3. Calculate the mass of reactants needed to produce the 2.00 g of precipitate. To calculate the mass of reactants needed to produce 2.00 g of a precipitated product requires the determination of the molar mass of each reactant and product. Molar mass (also called molecular weight) is the mass in grams of one mole of a substance. Molar mass is determined by adding the atomic masses of all of the atoms in the chemical formula. For example, the chemical formula CuS 4 5H 2 shows that in 1 mole of this compound, there are 1 mole of Cu ions, 1 mole of S 4 ions, and 5 moles of H 2 which gives a total of 1 mole of Cu, 1 mole of S, 10 moles of H and 9 moles of atoms. The total mass of each element in the compound is equal to the number of moles of atoms of the element multiplied by its atomic mass. The molar mass of copper(ii) sulfate pentahydrate is equal to the sum of the total masses of each element: number of moles of atoms atomic mass = total mass of element 1 mole of Cu 63.55 g Cu/mole = 63.55 g Cu 1 mole of S 32.07 g S/mole = 32.07 g S 9 moles of 16.00 g /mole = 144.00 g 10 moles of H 1.01 g H/mole = 10.10 g H Molar Mass of CuS 4 5H 2 249.72 g

nce the molar mass of each reactant and product is determined, the next step is to determine the number moles in 2.00 g of precipitate using equation 5. Equation 5: moles = grams/molar mass For example, if 2.00 g of copper(ii) sulfate pentahydrate is desired, the number of moles will be: 2.00 g 249.72 g/mol or 0.00801 moles. Using dimensional analysis, the equations would be: 200 g mole 249.72 g = 0.00801 mole Lastly, use the coefficients in the balanced equation to determine the number of moles of each of the reactants required to produce the desired number of moles of precipitate. nce the number of moles is obtained, use the following equation to calculate the number of grams of each reactant. Equation 6: grams = moles molar mass Step 4. Use the calculated masses of reactants to make and recover the 2.00 grams of precipitate. After the chemical equation is balanced, the precipitated product is determined and the mass of each reactant required to produce 2.00 g of product is calculated, the reaction can be performed in lab. Both of the reactants need to be dissolved in separate beakers using 25 ml of distilled water for each reactant. Then the two solutions need to be mixed together to form the precipitate. The precipitate will be recovered by filtration using a pre-massed piece of filter paper. The precipitate will be dried under a heat lamp or in a drying oven, and then the mass of the solid determined. The precipitate will be turned in to your teacher and the percent error calculated for the experiment. Student Tips When writing chemical formulas for the reactants and products, remember that the sum of all of the positive and negative charges in the compound must equal zero. Use subscripts to indicate the number of ions needed to equalize the charges. Parentheses are required to enclose polyatomic ions when there is more than one in the formula, such as the two nitrate ions in Ba(N 3. After the formulas for individual compounds are correctly written, only the coefficients can be changed to balance the equation and to obtain the same number of atoms of each element in both the reactants and products. Before measuring out the reactants, do the calculated mass values seem reasonable? According to the Law of Conser vation of Mass, the total mass of all reactants must equal the total mass of all products. Not all solids precipitate out of solution immediately. Some precipitates, such as calcium sulfate, require more than an hour to fully precipitate out of solution. Patience is required. Safety Information The compounds selected for this experiment have low to moderate toxicity, but avoid breathing the dust or getting any in yours eyes or on your skin. Zinc sulfate is a skin and mucous membrane irritant and is mildly toxic. Sodium carbonate and potassium carbonate may be skin irritants. Magnesium sulfate irritates eyes and the respiratory tract. Calcium chloride is moderately toxic. All solutions may be flushed down the drain and solids may be thrown in the trash. Wear chemical splash goggles, chemicalresistant gloves, and chemical-resistant aprons.

Background Teacher Notes Stoichiometry... Can You Make 2.00 Grams of a Compound? This experiment is designed to be a culminating activity to provide the teacher with feedback on how well students have mastered several fundamental skills in chemistry. It is assumed that the students already have an understanding of the concepts of formula writing, balancing chemical equations, determining molar mass values, using dimensional analysis for problem solving, and applying stoichiometric relationships to determine experimental proportions. The students should also be proficient in basic lab techniques, such as filtration, the proper handling of chemicals and using mechanical or electronic balances to measure mass. Perhaps they may have already used solubility rules or charts to study precipitate reactions between ions in solution. In order to successfully complete this experiment, students will need to apply all of these skills and plan the steps of an experiment the way a chemist does in industry. Materials Included in Kit Additional Materials Needed for the Lab Calcium acetate monohydrate, Ca(C 2 2 H 2, 65 g Balance to measure to 0.01g Calcium chloride dihydrate, CaCl 2 2H 2, 60 g Filtration setup (see Figure 1a or 1b) Magnesium sulfate heptahydrate, Mg(S 4 7H 2, 100 g Heat lamp or drying oven Potassium carbonate, K 2 C 3, 60 g Filter paper, 1 per lab group Sodium carbonate, Na 2 C 3, 50 g Beakers, 100-mL, 2 per lab group Zinc Sulfate heptahydrate, ZnS 4 7H 2, 100 g Student and teacher instructions Figure 1a: How to Set Up a Gravity Filtration Place filter paper in funnel and wet slightly with distilled water. Figure 1b: How to Set Up a Vacuum Filtration Aspirator screws onto faucet. funnel support filter paper long stem pipet beaker (or flask) Pour solution slowly into funnel with the aid of a stirring rod. Using a wash bottle, wash any solid remaining in the beaker into the filter paper. Safety Precautions The six chemicals provided in this kit were selected because of their relatively low toxicity levels and ease of disposal. Zinc sulfate is a skin and mucous membrane irritant and mildly toxic. Sodium carbonate and potassium carbonate may be skin irritants. Magnesium sulfate irritates eyes and respiratory tract. Calcium chloride is moderately toxic. Please review all MSDS sheets before starting lab. Wear chemical splash goggles, chemical-resistant gloves, and chemical-resistant aprons. 8 The flow of water through the aspirator creates a partial vacuum in the flask.

Pre-Lab Teacher Tips The pre-lab is essentially a dry lab and should highlight any deficiencies or problem areas before a student attempts to actually isolate 2.00 g of product. Typical mistakes made by the students include incorrectly written chemical formulas, errors in balancing the equations, wrong molar mass calculations or mistakes in the dimensional analysis using the stoichiometric conversion factors. Ideally, the pre-lab should be corrected and handed back to the students before they begin the actual laboratory procedure. The pre-lab also makes an excellent guide for the experiment. Results for Pre-Lab Calculations Sample Problem: How would you prepare 2.00 grams of precipitate by reacting barium chloride dihydrate with silver nitrate? Directions: Complete the following equations/problems showing ALL WRK, including the necessary units and labels. 1. Write a complete word equation listing the names of the reactants and the predicted products. A: B: C: D: barium chloride dihydrate + silver nitrate silver chloride + barium nitrate + H 2 2. Write a balanced formula equation for your double displacement reaction. Predict which compound will form a precipitate and indicate this in the equation using the symbol, (s), following the chemical formula. A: B: C: D: BaCl 2 2H 2 + 2AgN 3 2AgCl(s) + Ba(N 3 + 2 H 2 3. Calculate the molar mass of each reactant and product and put the answers in the blank spaces below. A: B: C: D: H 2 244.28 + 169.88 143.32 + 261.36 + 18.02 molar mass reactant A: molar mass reactant B: molar mass product C: molar mass product D: Work Space 1 (137.34 g Ba/mol) = 137.34 g/mol 2 (35.45 g Cl/mol) = 70.90 g/mol 4 (1.01 g H/mol) = 4.04 g/mol 2 (16.00 g /mol) = 32.00 g/mol 1 (107.87 g Ag/mol) = 107.87 g/mol 1 (14.01 g N/mol) = 14.01 g/mol 3 (16.00 g /mol) = 48.00 g/mol 1 (107.87 g Ag/mol) = 107.87 g/mol 1 (35.45 g Cl/mol) = 35.45 g/mol 1 (137.34 g Ba/mol) = 137.34 g/mol 2 (14.01 g N/mol) = 28.02 g/mol 6 (16.00 g /mol) = 96.00 g/mol molar mass H 2 : 2 (1.01 g H/mol) = 2.02 g/mol 1 (16.00 g /mol) = 16.00 g/mol 9

4. Calculate the number of moles in 2.00 g of the precipitated product. Molar mass of precipitate: 143.32 # of moles of precipitate: 0.0140 5. Use dimensional analysis to determine the mass of each reactant required to make 2.00 grams of your precipitate. First, determine the number of moles of each reactant and product and then calculate the mass of all reactants and product(s). Put the answers in the blank spaces below. Does the total mass of all reactants equal the total mass of all products? If not, check your work. Moles: A: B: C: D: H 2 0.00700 + 0.0140 0.0140 + 0.0070 + 0.0140 Grams: A: B: C: D: H 2 1.71 + 2.38 2.01 + 1.83 + 0.250 mass of reactant A needed: mass of reactant B needed: mass of product C that should form: mass of product D that should form: mass of H 2 that should form: Work Space 0.00700 mole 244.28 g/mol = 1.71 g 0.0140 mole 169.88 g/mol = 2.38 g 0.0140 mole 143.32 g/mole = 2.01 g 0.00700 mole 261.36 g/mol = 1.83 g 0.0140 mol 18.02 g/mol = 0.252 g 6. Complete the data chart below to determine the mass of precipitate formed in the sample experiment: Mass of dry filter paper + precipitate 2.97 g Mass of dry filter paper 1.02 g = Mass of precipitate 1.95 g 7. Calculate your percent error based on the relationship below: % error = 1.95 g 2.00 g 100 = 2.5% 2.00 g 10

Teacher Tips for Experiment This laboratory procedure is an open-ended lab and not a cookbook lab. Therefore, the students must develop a detailed procedure and the laboratory portion of this activity should not be started until all chemical formula equations, calculations, and procedures have been checked by the teacher. Any mistakes can be corrected at this point or you can simply let them find the error of their ways for themselves. There are enough chemicals included in this kit for five sets of experiments. There are 12 different combinations that will produce a precipitate reaction, so each lab group has a unique problem to solve. Cut out the possible pairs of reactants found on page 13 and assign them to lab groups. Experimental errors can result if the students do not dissolve the reactants completely in distilled water before mixing. Some of the solid filtered out may be the undissolved reactant(s). Tap water should not be used because the impurities in it, such as calcium and magnesium ions in hard water, can also precipitate out and interfere with the desired reaction. The filter paper and precipitate must be completely dried and any water of hydration must be removed by the drying process. Usually, the most accurate method to ensure complete dehydration is by heating until a constant mass is achieved. Most results fall within a 5 10% error range. For any group that produces CaS 4 (s), a 1.7% error is caused by the very slight solubility of this precipitate (0.033 g CaS 4 dissolves in 50 ml of the original solution). It is not unusual for some groups to obtain less than a 5% error through careful measurements and techniques. A few groups may obtain greater than a 25% error, which can usually be traced to an incorrect formula or calculation resulting in the wrong mass of the starting reactant(s). Some of the solids, such as calcium sulfate, do not immediately precipitate out of solution. Patience and additional time are required. Disposal All solids can be buried in a landfill site approved for the disposal of chemical and hazardous wastes, following the Flinn Disposal Method #26a. Solutions can be simply poured down the drain if your school is connected to a sanitary sewer system with a water treatment plant according to Flinn Disposal Method #26b. Acknowledgments Special thanks to Mark A. Case from Emmaus High School, Emmaus, PA for developing this laboratory procedure. Materials for Stoichiometry... Can You Make 2.00 g of a Compound? Kit are available from Flinn Scientific, Inc. Catalog No. AP4554 AP1055 AP1284 Description Stoichiometry... Can You Make 2.00 Grams of a Compound? Kit Laboratory oven Büchner Funnel, Porcelain, 100-mm AP1521 Stopper for Büchner funnel, no. 6 GP4072 Filtering Flask, 250-mL Consult your Flinn Scientific Catalog/Reference Manual for current prices. 11

Possible Pairs of Reactants Directions: Use a scissors to cut apart the possible pairs of assigned reactants. Have each lab group draw a card to determine which pair they will use for their experiment. #1 #2 A. zinc sulfate heptahydrate A. zinc sulfate heptahydrate B. calcium acetate monohydrate B. sodium carbonate #3 #4 A. zinc sulfate heptahydrate A. zinc sulfate heptahydrate B. calcium chloride dihydrate B. potassium carbonate #5 #6 A. magnesium sulfate heptahydrate A. magnesium sulfate heptahydrate B. calcium acetate monohydrate B. sodium carbonate #7 #8 A. magnesium sulfate heptahydrate A. magnesium sulfate heptahydrate B. calcium chloride dihydrate B. potassium carbonate #9 #10 A. calcium acetate monohydrate A. calcium acetate monohydrate B. sodium carbonate B. potassium carbonate #11 #12 A. sodium carbonate A. calcium chloride dihydrate B. calcium chloride dihydrate B. potassium carbonate 12

Theoretical Results For each pair of reactants below there is a complete word equation, balanced formula equation, molar mass values and grams of reactants needed and products formed. Reactant Pair #1 zinc sulfate heptahydrate + calcium acetate monohydrate calcium sulfate + zinc acetate + water ZnS 4 + Ca(C 2 2 H 2 CaS 4 (s) + Zn(C 2 2 + 8H 2 287.56 g/mole 176.19 g/mole 136.15 g/mole 183.48 g/mole 18.02 g/mole 4.22 g 2.59 g 2.00 g 2.70 g 2.12 g Reactant Pair #2 zinc sulfate heptahydrate + sodium carbonate zinc carbonate + sodium sulfate + water ZnS 4 + Na 2 C 3 ZnC 3 (s) + Na 2 S 4 + 7H 2 287.56 g/mole 105.99 g/mole 125.38 g/mole 142.02 g/mole 18.02 g/mole 4.59 g 1.69 g 2.00 g 2.27 g 2.01 g Reactant Pair #3 zinc sulfate heptahydrate + calcium chloride dihydrate calcium sulfate + zinc chloride + water ZnS 4 + CaCl 2 2H 2 CaS 4 (s) + ZnCl 2 + 9H 2 287.56 g/mole 147.02 g/mole 136.15 g/mole 136.29 g/mole 18.02 g/mole 4.22 g 2.16 g 2.00 g 2.00 g 2.38 g Reactant Pair #4 zinc sulfate heptahydrate + potassium carbonate zinc carbonate + potassium sulfate + water ZnS 4 + K 2 C 3 ZnC 3 (s) + K 2 S 4 + 7H 2 287.56 g/mole 138.21 g/mole 125.38 g/mole 174.27 g/mole 18.02 g/mole 4.59 g 2.20 g 2.00 g 2.78 g 2.01 g Reactant Pair #5 magnesium sulfate heptahydrate + calcium acetate monohydrate calcium sulfate + magnesium acetate + water MgS 4 + Ca(C 2 2 H 2 CaS 4 (s) + Mg(C 2 2 + 8H 2 246.50 g/mole 176.19 g/mole 136.15 g/mole 142.38 g/mole 18.02 g/mole 3.62 g 2.59 g 2.00 g 2.09 g 2.12 g Reactant Pair #6 magnesium sulfate heptahydrate + sodium carbonate magnesium carbonate + sodium sulfate + water MgS 4 + Na 2 C 3 MgC 3 (s) + Na 2 S 4 + 7H 2 246.50 g/mole 105.99 g/mole 84.31 g/mole 142.02 g/mole 18.02 g/mole 5.85 g 2.51 g 2.00 g 3.37 g 2.99 g 13

Reactant Pair #7 magnesium sulfate heptahydrate + calcium chloride dihydrate calcium sulfate + magnesium chloride + water MgS 4 + CaCl 2 2H 2 CaS 4 (s) + MgCl 2 + 9H 2 246.50 g/mole 147.02 g/mole 136.15 g/mole 95.76 g/mole 18.02 g/mole 3.62 g 2.16 g 2.00 g 1.41 g 2.38 g Reactant Pair #8 magnesium sulfate heptahydrate + potassium carbonate magnesium carbonate + potassium sulfate + water MgS 4 + K 2 C 3 MgC 3 (s) + K 2 S 4 + 7H 2 246.50 g/mole 138.21 g/mole 84.31g/mole 174.27 g/mole 18.02 g/mole 5.85 g 3.28 g 2.00 g 4.13 g 2.99 g Reactant Pair #9 calcium acetate monohydrate + sodium carbonate calcium carbonate + sodium acetate + water Ca(C 2 2 H 2 + Na 2 C 3 CaC 3 (s) + 2NaC 2 2 + H 2 176.19 g/mole 105.99 g/mole 100.09 g/mole 82.03 g/mole 18.02 g/mole 3.52 g 2.12 g 2.00 g 3.28 g 0.36 g Reactant Pair #10 calcium acetate monohydrate + potassium carbonate calcium carbonate + potassium acetate + water Ca(C 2 2 H 2 + K 2 C 3 CaC 3 (s) + 2KC 2 2 + H 2 176.19 g/mole 138.21 g/mole 100.09 g/mole 98.15 g/mole 18.02 g/mole 3.52 g 2.76 g 2.00 g 3.92 g 0.36 g Reactant Pair #11 sodium carbonate + calcium chloride dihydrate calcium carbonate + sodium chloride Na 2 C 3 + CaCl 2 2H 2 CaC 3 (s) + 2NaCl + 2H 2 105.99 g/mole 147.02 g/mole 100.09 g/mole 58.45 g/mole 18.02 g/mole 2.12 g 2.94 g 2.00 g 2.34 g 0.72 Reactant Pair #12 calcium chloride + potassium carbonate calcium carbonate + potassium chloride CaCl 2 2H 2 + K 2 C 3 CaC 3 (s) + 2KCl + 2H 2 147.02 g/mole 138.21 g/mole 100.09 g/mole 74.56 g/mole 18.02 g/mole 2.94 g 2.76 g 2.00 g 2.98 g 0.72 The steps used to determine the above values are the same as those described in the background section of the experiment. The total masses of reactants needed for a class with 12 lab groups to complete the experiment are 17.62 g ZnS 4 ; 18.94 g MgS 4 ; 12.22 g Ca(C 2 2 H 2 ; 8.44 g Na 2 C 3 ; 10.2; and 11.00 g K 2 C 3. There are enough chemicals included in this kit for five sets of experiments. They are also common laboratory chemicals that have applications in many other experiments. 14