Agenda 2/18/ A Day Slip Quiz - timed The Mole is a number – and so much more! Study Guide – self-check and grade Homework worksheets Equations and Mole Ratios Work to do at home

5A Day Slip quiz Write the correct formulas (including states) for the following substances – use your Naming flow chart – 4 mins. then turn in – I will grade again today. 1.sulfuric acid (you have to know) 2.Hydrochloric acid (you have to know) 3.Silicon tetrachloride (liquid) 4.Ammonia (liquid) 5.Hydrogen chloride (gas)

Chapter 11. Study Guide for Content Mastery Section 11.4 Empirical and Molecular Formulas 1. The percent composition of a compound is the percent by mass of each of the elements in a compound.

Chapter 11. Study Guide for Content Mastery Section 11.4 Empirical and Molecular Formulas 2. To find the percent composition of a compound divide the mass of each element in the sample by the mass of the sample, than multiply each quotient by 100.

Chapter 11. Study Guide for Content Mastery Section 11.4 Empirical and Molecular Formulas 3. Which information about a compound can you use to begin to determine the empirical and molecular formulas of the compound? C. Percent composition of the compound.

Chapter 11. Study Guide for Content Mastery Section 11.4 Empirical and Molecular Formulas 4. You have determined that a compound is composed of moles of carbon and moles of oxygen. What must you do to determine the mole ratio of the elements in the empirical formula of the compound? C. Divide each mole value by mol (the smallest value).

Chapter 11. Study Guide for Content Mastery Section 11.4 Empirical and Molecular Formulas 5. The mole ratio fo carbon to hydrogen to oxygen in a compounds is 1 mol C: 2 mol H: 1 mol O. What is the empirical formula of this compound? B. CH 2 O

Chapter 11. Study Guide for Content Mastery Section 11.4 Empirical and Molecular Formulas 6. You calculate the mole ratio of oxygen to aluminum in a compound to be 1.5mol O : 1 mol Al. What should you do to determine the mole ratio in the empirical formula of the compound? B. Multiply each mole value by 2 (to get whole number ratios 3 mol O : 2 mol Al)

Chapter 11. Study Guide for Content Mastery Section 11.4 Empirical and Molecular Formulas 7. What is the relationship between the molecular formula and the empirical formula of a compound? C. Molecular formula = (empirical formula)n

Chapter 11. Study Guide for Content Mastery Section 11.4 Empirical and Molecular Formulas 8. You know that the empirical formula of a compound has a molar mass of 30.0g/mol. The experimental molar mass of this compound is 60.0g/mol. What must you do to determine the value of n in the relationship between molecular formula an the empirical formula? c)Divide 60.0g/mol by 30.0 g/mol (n = 2)

Chapter 11. Study Guide for Content Mastery Section 11.4 Empirical and Molecular Formulas 9. You know that the experimental molar mass of a compound is three times the molar mass of its empirical formula. If the compound’s empirical formula is NO 2, what is its molecular formula? (empirical formula)n = molecular formula 3(NO 2 ) = N 3 O 6 D. 10 pts (I’ll look at qu 10)

Chapter 11. Study Guide for Content Mastery Section 11.5 The formula for a hydrate An hydrate is a compound that has a specific number of water molecules bound to its atoms. Molecules of water that become part of a hydrate are called waters of hydration. In the formula for a hydrate, the number of water molecules associated with each formula unit of the compound are written following a dot.

Chapter 11. Study Guide for Content Mastery Section 11.5 The formula for a hydrate The substance remaining after a hydrate has been heated and its waters of hydration released is called anhydrous. The ratio of the number of moles of Water of hydration to one mole of the anhydrous compound indicates the coefficient of H 2 O that follows the dot in the formula of the hydrate. Because the anhydrous form of the hydrate can absorb water into its crystal structure, hydrates are used as desiccants, which are drying agents.

Chapter 11. Study Guide for Content Mastery Section 11.5 The formula for a hydrate Chemical FormulaName CdSO 4 Cadmium sulfate, anhydrous CdSO 4  H 2 OCadmium sulfate monohydrate CdSO 4  4H 2 OCadmium sulfate tetrahydrate 10 pts (I’ll look at qu 11)

Turn in your Study Guide Packet – make it clear how many out of 20 you got for Sections 11.4 and 11.5 Peer review your Mole as a Number of Particles Practice problems – check each other’s answers and make sure the work shown explains what you did to find the answers. Grade A (superior understanding and execution) B (good understanding) C Average

Mole Ratios How can the coefficients in a chemical equation be interpreted? Student to read Why?

Model 1 – A Chemical Reaction nitrogenhydrogenammonia 1. Consider the reaction in Model 1. a) What are the coefficients for each of the following substances in the reaction? N 2 H 2 NH 3 N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

Model 1 – A Chemical Reaction nitrogenhydrogenammonia 1. Consider the reaction in Model 1. a) What are the coefficients for each of the following substances in the reaction? 1 N 2 3 H 2 2 NH 3 N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

Model 1 – A Chemical Reaction nitrogenhydrogenammonia b) Draw particle models below to illustrate the reaction in Model 1. N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

Model 1 – A Chemical Reaction nitrogenhydrogenammonia 2. a) Calculate the amount of reactants consumed and products made. b) Record the ratio of nitrogen to hydrogen to ammonia. Reduce the ratio to the lowest whole numbers possible. N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

Model 1 – A Chemical Reaction nitrogenhydrogenammonia N 2 (g)+ 3 H 2 (g)  2NH 3 (g) N 2 consumed H 2 consumed NH 3 produced Ratio N 2 : H 2 : NH 3 For a single reaction, how many molecules of each substance would be consumed or produced? 1 molecule3 molecules 2 molecules1:3:2 If the reaction occurred 100 times, how many molecules of each substance would be consumed or produced? If the reaction occurred 538 times, how many molecules of each substance would be consumed or produced?

Model 1 – A Chemical Reaction nitrogenhydrogenammonia N 2 (g)+ 3 H 2 (g)  2NH 3 (g) N 2 consumed H 2 consumed NH 3 produced Ratio N 2 : H 2 : NH 3 For a single reaction, how many molecules of each substance would be consumed or produced? 1 molecule3 molecules 2 molecules1:3:2 If the reaction occurred 100 times, how many molecules of each substance would be consumed or produced? 100 molecules 300 molecules 200 molecules 1:3:2 If the reaction occurred 538 times, how many molecules of each substance would be consumed or produced?

Model 1 – A Chemical Reaction nitrogenhydrogenammonia N 2 (g)+ 3 H 2 (g)  2NH 3 (g) N 2 consumed H 2 consumed NH 3 produced Ratio N 2 : H 2 : NH 3 For a single reaction, how many molecules of each substance would be consumed or produced? 1 molecule3 molecules 2 molecules1:3:2 If the reaction occurred 100 times, how many molecules of each substance would be consumed or produced? 100 molecules 300 molecules 200 molecules 1:3:2 If the reaction occurred 538 times, how many molecules of each substance would be consumed or produced? 538 molecules 1614 molecules 1076 molecules 1:3:2

3. Refer to the data table in question 2. a)How do the reduced ratios in the last column compare to the coefficients in the reaction shown in Model 1? The reduced ratio is the same as the coefficients in the reaction equation. b) Each coefficient in the chemical equation was multiplied by the same number, so the reduced ratio is still the same. N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

4. Even 538 is a small number of molecules to use in a reaction. Typically chemists us much larger numbers of molecules. (Recall that one mole is equal to 6.02 x particles.) Consider each situation below as it relates to the reaction in Model 1. N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

Model 1 – A Chemical Reaction nitrogenhydrogenammonia N 2 (g)+ 3 H 2 (g)  2NH 3 (g) N 2 consumed H 2 consumed NH 3 produced Ratio N 2 : H 2 : NH 3 1 molecule3 molecules 2 molecules1:3:2 If the reaction occurred 6.02 x times, how many molecules of each substance would be consumed or produced? 6.02 x molecules x molecules x :3:2 How many moles of each substance would be consumed or produced in the previous situation? 1 mole3 moles2 moles1:3:2

5. a) How do the reduced ratios in the last column compare to the coefficients in the reaction in Model 1? The reduced ratio is always the same as the coefficients in the reaction. b) Each part of the ratio of coefficients for the chemical reaction was multiplied by the same number, so the reduced ratio is the same. N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

6. The ratio obtained from the coefficients in a balanced chemical equation is called the mole ratio. a)The mole ratio for the reaction in Model 1 is 1 N 2 : 3 H 2 : 2 NH 3 b)This ratio is called the mole ratio because it is ratio of the number of moles of reactants and products that you would need or make if the reaction occurred a mole number of times (6.02 x times). N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

7. a) How many moles of nitrogen would be needed to make 10.0 moles of ammonia? x moles N 2 = 1 mole N moles NH 3 2 moleNH 3 x moles N 2 = 10 moles NH 3 1mole N 2 2 mole NH 3 x = 5 Therefore 5 moles Nitrogen would be needed to make 10.0moles ammonia. N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

7. b) How many moles of ammonia could be made by completely reacting 9.00 moles of hydrogen? x moles NH 3 = 2 moles NH moles H 2 3 moles H 2 x moles NH 3 = 2 moles NH 3 (9.00 moles H 2 ) 3 moles H 2 x = 6.00 Therefore 6.00 moles ammonia could be made by reacting 9.00 moles hydrogen. N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

7. c) How many moles of hydrogen would be needed to completely react with 7.41 moles of nitrogen? 8., 9., and 10. Work to end of page 4 before end of class – if done carry on to the extension questions. I will collect to take a look at what you have done today. N 2 (g)+ 3 H 2 (g)  2NH 3 (g)

Work to do at home Written work due on: 20 Feb, Friday 7am Find out why ammonia is a very important chemical and how it is produced on a large (industrial) scale. 2 paragraphs MLA formatting through out, works cited at end (not wikis or blogs) Post to Google Classroom