1. PACKET #6 Moles/Stoichiometry Textbook: Chapters 6 & 9 Reference Table: T & PT www.regentsprep.org

2. Let’s start with Vocabulary!! Mole (mol): The SI base unit that measures an amount of a substance; 1 mole of molecules has a mass equal to the molecular weight in grams. Mole (mol): The SI base unit that measures an amount of a substance; 1 mole of molecules has a mass equal to the molecular weight in grams. 1 mole = 6.02x10 23 atoms (molecules) 1 mole = 6.02x10 23 atoms (molecules) When looking at a chemical reaction, the number of moles are indicated by the coefficients in front of the formula’s. When looking at a chemical reaction, the number of moles are indicated by the coefficients in front of the formula’s. Avogadro’s Number: The number of molecules in one mole for any substance; Avogadro’s Number: The number of molecules in one mole for any substance; 1mole of particles = 6.02 x 10 23 2 moles of particles = 12.04 x 10 23 3 moles of particles = 18.06 x 10 23 etc...

3. Molecular Formula (MF): indicates the total number of atoms of each element needed to form the molecule (actual ratio of atoms). Ex. C 4 H 10 (4 carbon atoms and 10 hydrogen atoms) Molecular Formula (MF): indicates the total number of atoms of each element needed to form the molecule (actual ratio of atoms). Ex. C 4 H 10 (4 carbon atoms and 10 hydrogen atoms) Empirical Formula (EF): the simplest ratio in which atoms combine to form a compound. Ex. C 4 H 10 can be divided by 2 to give you C 2 H 5 Empirical Formula (EF): the simplest ratio in which atoms combine to form a compound. Ex. C 4 H 10 can be divided by 2 to give you C 2 H 5 Molecular Mass: the total mass of the molecular formula Molecular Mass: the total mass of the molecular formula Empirical Mass: the total mass of the empirical formula Empirical Mass: the total mass of the empirical formula

4. Stoichiometry: The quantitative relationships between the amounts of reactants and products formed during a chemical reaction. Stoichiometry: The quantitative relationships between the amounts of reactants and products formed during a chemical reaction. 2H 2 + O 2  2H 2 O Notice the ratio between all of the substances in this reaction. Stoichiometry implies that no matter how you change the quantity of any of these substances, the others will have to change in order for the reaction to remain the same. Notice the ratio between all of the substances in this reaction. Stoichiometry implies that no matter how you change the quantity of any of these substances, the others will have to change in order for the reaction to remain the same.

5. Calculating Molecular Formula When supplied with the Empirical Formula, and the Molecular Mass, we are capable of calculating the Molecular Formula with the following equation: When supplied with the Empirical Formula, and the Molecular Mass, we are capable of calculating the Molecular Formula with the following equation: Mass MF Mass EF Example: The empirical formula is CH and the molecular mass is 26, what is the molecular formula?

6. Gram Formula Mass (GFM) The GFM is the total mass of any substance (element or compound) The GFM is the total mass of any substance (element or compound) Example: K 2 CO 3 (2 K atoms, 1 C atom, and 3 O atoms) Mass K = 39g Mass C = 12g Mass O = 16g (39x2) + (12x1) + (16x3) = 138 grams The gram formula mass of K 2 CO 3 = 138g

7. WHAT YOU KNOW!!!! GFM = 1 mole = 6.02 x 10 23 molecules = 22.4 L This is the most important information for you to memorize in order to master this unit. This is the most important information for you to memorize in order to master this unit. The above means that the GFM of any substance (element or compound) is the same thing as saying 1 mole of that substance, which is the same as saying there are 6.02 x 10 23 molecules of that substance, which is the same as saying there is 22.4 L of that same substance!! The above means that the GFM of any substance (element or compound) is the same thing as saying 1 mole of that substance, which is the same as saying there are 6.02 x 10 23 molecules of that substance, which is the same as saying there is 22.4 L of that same substance!!

8. WHAT YOU KNOW!!!! GFM of K 2 CO 3 = 138 grams GFM of K 2 CO 3 = 138 grams 138 g K 2 CO 3 = 1 mole 138 g K 2 CO 3 = 1 mole 1 mole K 2 CO 3 = 6.02 x 10 23 atoms of K 2 CO 3 1 mole K 2 CO 3 = 6.02 x 10 23 atoms of K 2 CO 3 6.02 x 10 23 atoms of K 2 CO 3 = 22.4 L K 2 CO 3 6.02 x 10 23 atoms of K 2 CO 3 = 22.4 L K 2 CO 3 With “what you know”, you can set up a proportion to solve for any problem. With “what you know”, you can set up a proportion to solve for any problem.

9. Converting (ANYTHING) Example: How many molecules are there if you have 4.2 moles of NaCl? 1. Start with what you know!! 2. Set up a proportion based on what you know, and what you are given. 1 mole = 6.02 x 10 23 4.2 moles x 4.2 moles x

10. Converting (ANYTHING) Example: How many liters are there in 3.8 moles of MgF 2 ? Example: How many molecules in 104.2 L of ZnO?

11. Mole  Mass / Mass  Mole Converstions Table T: mole = given mass Table T: mole = given mass GFM GFM Example: If you have 54 grams of LiF, how many moles do you have? Example: If you have 2.3 moles of CuCl 2, how many grams do you have?

12. Stoichiometry Mole-Mole Problems: Answers how many moles of one element or compounds react with a given number of moles of another element or compound. Answers how many moles of one element or compounds react with a given number of moles of another element or compound. Example: How many moles of Ca are needed to react completely with 6 moles of H 2 O in the following reaction: Ca + 2H 2 O  Ca(OH) 2 + H 2 1. Cross out anything in the equation that the problem is not considering. 2. Write the coefficient of each substance remaining under the element or compound. 3. On top of the element or compound write the number of moles given in the problem. 4. Set up a proportion:

13. Mass-Mass Problems: Answers how many grams of one element or compounds react with a given number of grams of another element or compound. Answers how many grams of one element or compounds react with a given number of grams of another element or compound. Example: How many grams of calcium are needed to react completely with 108 grams of H 2 O in the following equation : Ca + 2H 2 O  Ca(OH) 2 + H 2

14. Volume-Volume Problems: Answers how many grams of one element or compounds react with a given number of grams of another element or compound. Answers how many grams of one element or compounds react with a given number of grams of another element or compound. Example: How many liters of calcium are needed to react completely with 179.2 L of H 2 O in the following equation : Ca + 2H 2 O  Ca(OH) 2 + H 2

15. Limiting Reactant (Reagant) The reactant that is completely consumed when a reaction is run to completion. The reactant that is completely consumed when a reaction is run to completion. Let’s assume the following reaction: Let’s assume the following reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g) If we have 28 grams of N 2 and 6 grams of H 2, then the reaction would be considered a stoichiometric mixture because the relative amounts of reactants matches the numbers in the balanced equation. If we have 28 grams of N 2 and 6 grams of H 2, then the reaction would be considered a stoichiometric mixture because the relative amounts of reactants matches the numbers in the balanced equation.

16. Limiting Reactant (Reagant) Now, let’s consider the following balanced reaction and question. Now, let’s consider the following balanced reaction and question. What mass of water is required to react exactly with 249 grams of methane? CH 4 (g) + H 2 O(g)  3H 2 (g) + CO(g) NOW, if 249 grams of methane is mixed with 300 grams of water, which reactant will be considered the limiting reactant?

17. PRACTICE PROBLEMS: For each of the following balanced chemical equations, suppose that exactly 50.0 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of the product in boldface is expected. (Assume that the limiting reactant is completely consumed). 2NH 3 (g) + 2Na(s)  2NaNH 2 (s) + H 2 (g) 2NH 3 (g) + 2Na(s)  2NaNH 2 (s) + H 2 (g) 2Al(s) + 3H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (s) + 3H 2 (g) 2Al(s) + 3H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (s) + 3H 2 (g)

18. MORE MORE 2Al(s) + 6HCl(aq)  2AlCl 3 (aq) + 3H 2 (g) 2Al(s) + 6HCl(aq)  2AlCl 3 (aq) + 3H 2 (g) 2K(s) + I 2 (s)  2KI(s) 2K(s) + I 2 (s)  2KI(s)

19. Given the balanced equation: Cu + 4HNO 3  Cu(NO 3 ) 2 + 2H 2 O + 2NO 2 15g of copper reacts with 28g of HNO 3. a) Determine the limiting reactant. b) How much NO 2 can be made from the conditions above (in grams!!) c) What mass of the excess reactant will be left over at the end of the reaction?

20. Percent Composition The percentage by mass of each of the elements in compound The percentage by mass of each of the elements in compound Table T: Table T: % composition = mass part x 100 mass whole mass whole Example: What is the percent by mass of magnesium in magnesium oxide (MgO)?

21. Percent Composition  Hydrate: a compound that incorporates water molecules into its fundamental solid structure. The compound has a dot after it followed by the number of water molecules attached.  Ex: CaSO 4 2H 2 O (SOLVE THE GFM)  The dot in the hydrate equates to an addition sign when calculating the GFM of a hydrate.

22.  To calculate the percent of water in a hydrate, you must determine the mass of the whole, and the mass of the part. Example: Determine the percent composition of water in CaSO 4 2H 2 O. GFM CaSO 4 2H 2 O = whole GFM CaSO 4 2H 2 O = whole GFM 2H 2 O = part GFM 2H 2 O = part

23. In lab, we will be performing an experiment called “Water in a Hydrate” in which we will take a hydrate and remove as much water from a given amount by heating it. In lab, we will be performing an experiment called “Water in a Hydrate” in which we will take a hydrate and remove as much water from a given amount by heating it. The purpose of removing the water is to turn the hydrate into what is called an anhydrous, which is a hydrated compound minus the bonded water molecules. The purpose of removing the water is to turn the hydrate into what is called an anhydrous, which is a hydrated compound minus the bonded water molecules. By calculating the amount of water in the sample of hydrate we can determine the “part” in the equation, and therefore calculate the percent composition of water in the hydrate. By calculating the amount of water in the sample of hydrate we can determine the “part” in the equation, and therefore calculate the percent composition of water in the hydrate.

24. Calculating Empirical Formula from Percent Composition If you know the percent of each element in a compound, you can find the empirical formula for of the compound. If you know the percent of each element in a compound, you can find the empirical formula for of the compound. 1. Assume that you have 100 grams. If there is 86% C, then there are 86 grams of carbon. If there is 14% hydrogen then there is 14 grams of hydrogen. 2. Find number of moles of carbon and hydrogen by using the mole equation (Table T). 3. The number of moles become the subscripts for each of the elements. 4. Divide everything out by the smallest subscript to get the ratio among the elements, and therefore giving you the empirical formula.

25. Calculating Empirical Formula from Percent Composition  Ex: A compound is 86% C and 14% H by mass. What is the empirical formula for this compound?

26. Review Questions 1) What is the total number of moles of hydrogen gas contained in 9.03 x 10 23 molecules? A) 6.02 moles B) 2.00 moles C) 9.03 moles D) 1.50 moles 2) What is the gram formula mass of Na 2 CO 3 10H 2 O? A) 266 g B) 286 g C) 100 g D) 142 g

27. 3) The total number of molecules in 34.0 grams of NH 3 is equal to A) 1.00 x 6.02 x 10 23 B) 1.00 x 22.4 C) 2.00 x 22.4 D) 2.00 x 6.02 x 10 23 4) An 8.24-gram sample of a hydrated salt is heated until it has a constant mass of 6.20 grams. What was the percent by mass of water contained in the original sample? A) 75.2% B) 14.1% C) 24.8% D) 32.9% 5) A compound with an empirical formula of CH 2 has a molecular mass of 70. What is the molecular formula? A) C 2 H 4 B) CH 2 C) C 4 H 8 D) C 5 H 10

28. 6) What is the percent by mass of oxygen in magnesium oxide, MgO? A) 50% B) 60% C) 40% D) 20% 7) Given the reaction: PbCl 2 (aq) + Na 2 CrO 4 (aq)  ‚ PbCrO 4 (s) + 2NaCl(aq) What is the total number of moles of NaCl formed when 2 moles of Na 2 CrO 4 react completely? A) 1 mole B) 2 moles C) 3 moles D) 4 moles 8) What is the percent composition by mass of aluminum in Al 2 (SO 4 ) 3 (gram-formula mass = 342 grams/mole)? A) 15.8% B) 7.89% C) 36.0% D) 20.8%

29. 9) What is the mass in grams of 3.0 x 1023 molecules of CO 2 ? A) 66 g B) 88 g C) 44 g D) 22 g 10) Which compound has the empirical formula CH? A) C 3 H 8 B) CH 4 C) C 6 H 6 D) C 2 H 4 11) The gram formula mass of NH 4 Cl is A) 22.4 g/mole B) 28.0 g/mole C) 95.5 g/mole D) 53.5 g/mole

30. 12. In the lab, a student performed an experiment to determine the percent by mass of water in a hydrate. The following data was recorded: Mass of crucible + cover = 11.70g Mass of crucible + cover + hydrated salt before heating = 14.90g Mass of crucible + cover + anhydrous salt after heating = 14.53g Determine the approximate percent by mass of water in the hydrated salt (show all work).