The Mathematics of Chemical Equations Chapter 11 STOICHIOMETRY aka USING THE REACTION EQUATION LIKE A RECIPE SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds (Tools>Options>Print>UNcheck "Background Printing")!

USING EQUATIONS Nearly everything we use is manufactured from chemicals. Soaps, shampoos, conditioners, cd’s, cosmetics, medications, clothes, etc. For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them. Chemical processes carried out in industry must be economical, this is where balanced equations help.

USING EQUATIONS Equations are a chemist’s recipe. Equations tell chemists what amounts of reactants to mix and what amounts of products to expect. When you know the quantity of one substance in a rxn, you can calculate the quantity of any other substance consumed or created in the rxn. Quantity meaning the amount of a substance in grams, liters, molecules, or moles.

N2 + 3H2  2NH3 1 mol N2 + 3 mol N2  2 mol NH3 28 g N2 + 3 (2 g H2)  2 (17 g NH3) 34 g reactants  34 g products +  22.4 L 22.4 L 22.4 L 22.4 L N2 67.2 L H2  44.8 L NH3

USING EQUATIONS The calculation of quantities in chemical reactions is called Stoichiometry. When you bake cookies you probably use a recipe. A cookie recipe tells you the amounts of ingredients to mix together to make a certain number of cookies. If you need more cookies than the yield of the recipe, the amounts of ingredients can be doubled or tripled.

Chocolate Chip 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 How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white sugar?

Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them chemical equations Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds & elements as ingredients

Chemistry Recipes An equation tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start! Example: 2 Na + Cl2  2 NaCl This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

Conversion Factors from a Chemical Equation 2 C6H6 (l) + 15 O2 (g)  12 CO2 (g) + 6 H2O (g) In this equation there are 2 molecules of benzene reacting with 15 molecules of oxygen to produce 12 molecules of carbon dioxide and 6 molecules of water. This equation could also be read as 2 moles of benzene reacts with 15 moles of oxygen to produce 12 moles of carbon dioxide and 6 moles of water.

Conversion Factors from a Chemical Equation Since 6.02 x 1023 is the common factor between the relationship between the actual number of molecules and the number of moles present we can use the ratio of the # moles of one substance in the equation to find the # of moles another substance in the equation. This ratio, # moles of unknown # moles of given is known as the MOLE RATIO

Conversion Factors from a Chemical Equation 2 C6H6 (l) + 15 O2 (g)  12 CO2 (g) + 6 H2O (g) The MOLE RATIO, from the balanced equation, for oxygen and carbon dioxide is 15 : 12. It can be written as: 12 moles CO2 or as 15 moles of O2 15 moles O2 12 moles of CO2 NOTE: The MOLE RATIO is used for converting moles of one substance into moles of another substance. The numbers that go in front of “moles of given” & “moles of unknown” are the coefficients from the balanced equation! IT IS ALWAYS MOLES OVER MOLES! Moles of unknown Moles of Given

What was that again?! # Moles Unknown # Moles Given The #’s in the numerator & denominator MUST come from the balanced chemical equation

I Y Stoichiometry !!! Consider: 4NH3 + 5O2  6H2O + 4NO Is this equation balanced? Yes!! You must always start stoich. problems with a balanced equation! Write all of the possible mole ratios that can be formed from this equation. YES! NOW!!!

4 moles NH3 5 moles O2 5 moles O2 4 moles NH3 5 moles O2 6 moles H2O 6 moles H2O 5 moles O2 4 moles NH3 6 moles H2O 6 moles H2O 4 moles NH3 6 moles H2O 4 moles NO 4 moles NO 6 moles H2O ETC.

I Y Stoichiometry !!! Using the coefficients of balanced equations and our knowledge of mole conversions we can perform powerful calculations! A.K.A. stoichiometry. A balanced equation is essential for all calculations involving amounts of reactants and products. If you know the number of moles of 1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn equation.

Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H2 + O2  2 H2O How many moles of reactants are needed? What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?

Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl2  2 NaCl 5 moles Na 1 mol Cl2 2 mol Na = 2.5 moles Cl2

MOLE – MOLE EXAMPLE The following rxn shows the synthesis of aluminum oxide. 3O2(g) + 4Al(s)  2Al2O3(s) 3O2(g) + 4Al(s)  2Al2O3(s) How many moles of aluminum are needed to form 3.7 mol Al2O3? Given: 3.7 moles of Al2O3 Uknown: ____ moles of Al

MOLE – MOLE EXAMPLE Solve for the unknown: 3O2(g) + 4Al(s)  2Al2O3(s) 4 mol Al 3.7 mol Al2O3 2 mol Al2O3 = 7.4 mol Al Mole Ratio

Mole - Mole Problems Consider : 4NH3 + 5O2  6H2O + 4NO How many moles of H2O are produced if 0.176 mol of O2 are used? How many moles of NO are produced in the reaction if 17 mol of H2O are also produced? 6 mol H2O 5 mol O2 x 0.211 mol H2O = 0.176 mol O2 # mol H2O= 4 mol NO 6 mol H2O x 11 mol NO = 17 mol H2O # mol NO=

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?

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 + Cl2  2 NaCl 5.00 moles Na 1 mol Cl2 71.0g Cl2 2 mol Na 1 mol Cl2 = 178g Cl2

Mole-Mass Problems Consider : 4NH3 + 5O2  6H2O + 4NO How many grams of H2O are produced if 1.90 mol of NH3 are combined with excess oxygen? How many grams of O2 are required to produce 0.3 mol of H2O? 6 mol H2O 4 mol NH3 x 18.0 g H2O 1 mol H2O x 51.3 g H2O = 1.90 mol NH3 32.0 g O2 1 mol O2 x 5 mol O2 6 mol H2O x 8 g O2 = 0.3 mol H2O

Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.

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 (C2H6) needed to produce 10.0 g of water 2 C2H6 + 7 O2  4 CO2 + 6 H20 10.0 g H2O 1 mol H2O 2 mol C2H6 18.0 g H2O 6 mol H20 = 0.185 mol C2H6

Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide

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

Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N2 + 3 H2  2 NH3 2.00g N2 1 mol N2 2 mol NH3 17.0g NH3 28.0g N2 1 mol N2 1 mol NH3 = 2.43 g NH3

Mass – Mass Problems Consider : 4NH3 + 5O2  6H2O + 4NO How many grams of NO is produced if 12 g of O2 is combined with excess ammonia? 1 mol O2 32.0 g O2 x 4 mol NO 5 mol O2 x 30.0 g NO 1 mol NO x 12 g O2 9.0 g NO =

Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

Volume – Volume Problems Ex. What volume of hydrogen gas will be needed to produce 256.7 L of ammonia gas (NH3) at STP? N2 + 3 H2 2 NH3 256.7 L NH3 1 mole NH3 mole H2 22.4 L H2 3 1 mole H2 2 mole NH3 22.4 L NH3 = 385.1 L H2 Mole Ratio

Caution: this stuff is difficult to follow at first. Limiting Reagents Caution: this stuff is difficult to follow at first. Be patient.

Limiting Reactants Available Ingredients Limiting Reactant 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly Limiting Reactant bread Excess Reactants peanut butter and jelly

Grilled Cheese Sandwich Bread + Cheese  ‘Cheese Melt’ 2 B + C  B2C 100 bread 30 slices ? sandwiches

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?

Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up another 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.

Limiting Reactants Limiting Reactant Excess Reactant used up in a reaction determines the amount of product Excess Reactant More than enough to react with the limiting reagent – some left over! added to ensure that the other reactant is completely used up cheaper & easier to recycle Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: limiting reactant, and amount of product formed

Real-World Stoichiometry: Limiting Reactants Ideal Stoichiometry Limiting Reactants LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 366

Real-World Stoichiometry: Limiting Reactants Fe + S FeS Ideal Stoichiometry Limiting Reactants S = Fe = LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 366

Limiting Reactants aluminum + chlorine gas  aluminum chloride Al(s) + Cl2(g)  AlCl3 2 Al(s) + 3 Cl2(g)  2 AlCl3 100 g 100 g ? g A. 200 g B. 125 g C. 667 g D. ???

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 reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Remember! You can’t compare to see which is greater and which is lower unless the product is the same!

Limiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting 2 Al + 3 Cl2  2 AlCl3 Start with Al: Now Cl2: Limiting Reactant 10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3 27.0 g Al 2 mol Al 1 mol AlCl3 = 49.4g AlCl3 35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3 71.0 g Cl2 3 mol Cl2 1 mol AlCl3 = 43.9g AlCl3

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 .

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.

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?

Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2  2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I2 1 mol I2 2 mol K 39.1 g K 254 g I2 1 mol I2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 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!

Excess Reactant 2 Na + Cl2  2 NaCl 50 g 50 g x g 81.9 g NaCl / 23 g/mol / 71 g/mol x 58.5 g/mol 1 : 2 1.40 mol “Have” 2.17 mol 0.70 mol coefficients “Need” 1.40 mol EXCESS LIMITING

Excess Reactant (continued) 2 Na + Cl2  2 NaCl 50 g 50 g x g 81.9 g NaCl All the chlorine is used up… 81.9 g NaCl 50.0 g Cl2 31.9 g Na is consumed in reaction. How much Na is unreacted? 50.0 g - 31.9 g = 18.1 g Na total used “excess”

Conservation of Mass is Obeyed 2 Na + Cl2  2 NaCl 50 g 50 g x g 81.9 g NaCl 2 Na + Cl2  2 NaCl + Na 50 g 50 g x g 81.9 g NaCl 18.1 g 31.9 g + 18.1 g 100 g product 100 g reactant

Limiting Reactant: Recap You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. Convert ALL of the reactants to the SAME product (pick any product you choose.) The reactant that gave you the lowest answer is the LIMITING REACTANT. The other reactant is in EXCESS. To find the amount of excess left over, subtract the amount used from the given amount. 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!

Percentage Yield

Percent Yield measured in lab calculated on paper

% yield Experimental yield Theoretical yield X 100 = % yield

Sample problem 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Step 1: write the balanced chemical equation 2H2 + O2  2H2O Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 16 g H2 143 g = Step 3: Calculate % yield actual theoretical 138 g H2O 143 g H2O = % yield = x 100% x 100% 96.7% =

Practice problem Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2? Step 1: write the balanced chemical equation N2 + 3H2  2NH3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 2 mol NH3 3 mol H2 x 17.04 g NH3 1 mol NH3 x # g NH3= 20.0 mol H2 227 g = Step 3: Calculate % yield actual theoretical 40.5 g NH3 227 g NH3 = % yield = x 100% x 100% 17.8% =

Percent Yield When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g Theoretical Yield: 45.8 g K2CO3 1 mol K2CO3 138.21 g 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl = 49.4 g KCl

Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.4 g actual: 46.3 g Theoretical Yield = 49.4 g KCl 46.3 g 49.4 g % Yield =  100 = 93.7%

Challenging question 2H2 + O2  2H2O What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2? Hint: determine limiting reagent first 1 mol O2 32 g O2 x 2 mol H2O 1 mol O2 x 18.02 g H2O 1 mol H2O x # g H2O= 60 g O2 68 g = 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 7.0 g H2 62.4 g = actual theoretical 58 g H2O 62.4 g H2O = % yield = x 100% x 100% 92.9% =

PARTICLES B MOLE 6.02X10 A 22.4 L 23 MASS VOLUME MOLE MAP MOLAR MASS RATIO

6 Na(s) + Fe2O3(s)  3 Na2O(s) + 2 Fe (s) Air Bag Design Exact quantity of nitrogen gas must be produced in an instant. Use a catalyst to speed up the reaction 2 NaN3(s)  2 Na(s) + 3 N2(g) 6 Na(s) + Fe2O3(s)  3 Na2O(s) + 2 Fe (s)

6 Na(s) + Fe2O3(s)  3 Na2O(s) + 2 Fe(s) 2 NaN3(s)  2 Na(s) + 3 N2(g) 6 Na(s) + Fe2O3(s)  3 Na2O(s) + 2 Fe(s) Airbag Design Assume that 65.1 L of N2 gas are needed to inflate an air bag to the proper size. How many grams of NaN3 must be included in the gas generant to generate this amount of N2? (Hint: The density of N2 gas at this temperature is about 0.916 g/L). 65.1 L N2 x 0.916 g/L N2 X g NaN3 = 59.6 g N2 (1 mol N2) (2 mol NaN3) (65 g NaN3) (28 g N2) (3 mol N2) (1 mol NaN3) 59.6 g N2 X = 92.2 g NaN3 How much Fe2O3 must be added to the gas generant for this amount of NaN3? X g Fe2O3 = 92.2 g NaN3 (1 mol NaN3) (2 mol Na) (1 mol Fe2O3) (159.6 g Fe2O3) (65 g NaN3) (2 mol NaN3) (6 mol Na) (1 mol Fe2O3) X = 37.7 g Fe2O3

Water from a Camel Camels store the fat tristearin (C57H110O6) in the hump. As well as being a source of energy, the fat is a source of water, because when it is used the reaction takes place. What mass of water can be made from 1.0 kg of fat? 2 C57H110O6(s) + 163 O2(g)  114 CO2(g) + 110 H2O(l) X g H2O = 1 kg ‘fat” (1000 g ‘fat’) (1 mol “fat”) (110 mol H2O) (18 g H2O) (1 kg ‘fat’) (890 g ‘fat’) (2 mol ‘fat’) (1 mol H2O) X = 1112 g H2O or 1.112 liters water

Rocket Fuel B2H6 + O2  B2O3 + H2O B2H6 + 3 O2  B2O3 + 3 H2O The compound diborane (B2H6) was at one time considered for use as a rocket fuel. How many grams of liquid oxygen would a rocket have to carry to burn 10 kg of diborane completely? (The products are B2O3 and H2O). B2H6 + O2  B2O3 + H2O Chemical equation Balanced chemical equation B2H6 + 3 O2  B2O3 + 3 H2O 10 kg X g X g O2 = 10 kg B2H6 (1000 g B2H6) (1 mol B2H6) (3 mol O2) (32 g O2) (1 kg B2H6) (28 g B2H6) (1 mol B2H6) (1 mol O2) X = 34,286 g O2

Water in Space In the space shuttle, the CO2 that the crew exhales is removed from the air by a reaction within canisters of lithium hydroxide. On average, each astronaut exhales about 20.0 mol of CO2 daily. What volume of water will be produced when this amount of CO2 reacts with an excess of LiOH? (Hint: The density of water is about 1.00 g/mL.) CO2(g) + 2 LiOH(s)  Li2CO3(aq) + H2O(l) 20.0 mol excess x g X mL H2O = 20.0 mol CO2 (1 mol H2O) (18 g H2O) (1 mL H2O) (1 mol CO2) (1 mol H2O) (1 g H2O) X = 360 mL H2O

Real Life Problem Solving Determine the amount of LiOH required for a seven-day mission in space for three astronauts and one ‘happy’ chimpanzee. Assume each passenger expels 20 mol of CO2 per day. Note: The lithium hydroxide scrubbers are only 85% efficient. (4 passengers) x (10 days) x (20 mol/day) = 800 mol CO2 Plan for a delay CO2(g) + 2 LiOH(s)  Li2CO3(aq) + H2O(l) 800 mol X g

Note: The lithium hydroxide scrubbers are only 85% efficient. CO2(g) + 2 LiOH(s)  Li2CO3(aq) + H2O(l) 800 mol X g x 23.9 g/mol 1:2 800 mol 1600 mol Needed (actual yield) X g LiOH = 800 mol CO2 (2 mol LiOH) (1 mol CO2 ) (23.9 g LiOH) ( 1 mol LiOH) = 38,240 g LiOH Note: The lithium hydroxide scrubbers are only 85% efficient. Actual Yield Theoretical Yield 38,240 g LiOH X g LiOH % Yield = 0.85 = Amount of LiOH to be taken into space X = 44,988 g LiOH

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 79.1 g Zn 1 mol Zn 65.39 g Zn 1 mol H2 Zn 22.4 L H2 1 mol = 27.1 L H2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L 0.90 2.5 mol HCl 1 L 1 mol H2 2 mol HCl 22.4 L H2 1 mol H2 = 25 L H2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H2 left over zinc Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

% error Experimental yield – Theoretical yield Theoretical yield X 100 = % error What you predicted you would get (use stoichiometry) What YOU got in the experiment

Review: Molar Mass of Compounds The molar mass (MM)- add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl2 Avg. Atomic mass of Calcium = 40.1g Avg. Atomic mass of Chlorine = 35.5g Molar Mass of calcium chloride = 40.1 g/mol Ca + (2 X 35. 5) g/mol Cl  111.1 g/mol CaCl2 20 Ca  40.08 17 Cl 35.45

Review: Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 1023 Multiply by 6.02 X 1023 Moles Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table Mass (grams)