Chemical Reactions and the Mole…. 1 Fe (s) + 1 S (s) 1 FeS (s)1 Fe (s) + 1 S (s) 1 FeS (s) This chemical reaction means one atom of iron reacts with one atom of sulfur to form one molecule of iron (II) sulfideThis chemical reaction means one atom of iron reacts with one atom of sulfur to form one molecule of iron (II) sulfide Can we do this chemical reaction?Can we do this chemical reaction? No - atoms are too small to see -No - atoms are too small to see - We must perform the reaction many times larger to see it!We must perform the reaction many times larger to see it! We always perform the reaction 6.02 x times larger, or 1 mole times largerWe always perform the reaction 6.02 x times larger, or 1 mole times larger

1 Fe (s) + 1 S (s) 1 FeS (s) If we do this reaction 6.02 x times bigger, it would be written: If we do this reaction 6.02 x times bigger, it would be written: 6.02 x Fe (s) x S (s) 6.02 x FeS (s) Since 6.02 x is 1 mole, we could also say: Since 6.02 x is 1 mole, we could also say: 1 mole Fe (s) + 1 mole S (s) 1 mole FeS (s) 1 mole Fe (s) + 1 mole S (s) 1 mole FeS (s) So we could view the 1 coefficient as either 1 atom, or 1 mole, depending on whether we were looking at the reaction on the atomic scale, or the large, human, molar scale! So we could view the 1 coefficient as either 1 atom, or 1 mole, depending on whether we were looking at the reaction on the atomic scale, or the large, human, molar scale!

1 Fe (s) + 1 S (s) 1 FeS (s) Here’s the problem - If you have 10 grams of Fe, how many grams of S do you need? Most would say 10 grams! It is a 1:1 ratio! But the units in the chemical reaction aren’t grams! The 1 coefficient does not stand for grams - it stands for moles! If you are going to Mexico, your dollars have to be converted into pesos before you go! When you look at the ratio in a chemical reaction, you must be in moles! 1 Fe (s) + 1 S (s) 1 FeS (s) Here’s the problem - If you have 10 grams of Fe, how many grams of S do you need? Most would say 10 grams! It is a 1:1 ratio! But the units in the chemical reaction aren’t grams! The 1 coefficient does not stand for grams - it stands for moles! If you are going to Mexico, your dollars have to be converted into pesos before you go! When you look at the ratio in a chemical reaction, you must be in moles!

Let’s look at a sample problem… Number one in your homework packet under Easy Stoichiometry Problems… Carbon dioxide can be commercially prepared by heating chalk, or calcium carbonate. How many moles of carbon dioxide can be produced when 3.05 moles of calcium carbonate are heated? How do we approach this problem? There are four steps we follow in any problem we do with chemical reactions and amounts… Let’s look at a sample problem… Number one in your homework packet under Easy Stoichiometry Problems… Carbon dioxide can be commercially prepared by heating chalk, or calcium carbonate. How many moles of carbon dioxide can be produced when 3.05 moles of calcium carbonate are heated? How do we approach this problem? There are four steps we follow in any problem we do with chemical reactions and amounts…

1.Write a balanced chemical reaction 2.Make sure you are in the unit of moles (mole map) 3.Set up a ratio from the chemical reaction, putting what you want to solve for on top, and what you want to cancel on bottom 4.Convert out of the unit of moles, if necessary (mole map) 1.Write a balanced chemical reaction 2.Make sure you are in the unit of moles (mole map) 3.Set up a ratio from the chemical reaction, putting what you want to solve for on top, and what you want to cancel on bottom 4.Convert out of the unit of moles, if necessary (mole map)

Volume of A (liquid) Volume of B (liquid) Volume of A (gas) Volume of B (gas) Grams AGrams B Moles AMoles B Use Molar Mass Use Molar Mass STP Use 22.4 L 1 mole STP Use 22.4 L 1 mole Use Molarity moles Liter moles Liter USE YOUR “CHEMICAL” RECIPE HERE SET UP A RATIO FROM YOUR BALANCED EQUATION X moles you want moles you have

Step 1: Write a balanced chemical equation: Step 1: Write a balanced chemical equation: 1 CaCO 3 (s) 1 CaO (s) + 1 CO 2 (g) 1 CaCO 3 (s) 1 CaO (s) + 1 CO 2 (g) Step 2: Convert into the unit of moles: Step 2: Convert into the unit of moles: Done! We already have 3.05 moles CaCO 3 Done! We already have 3.05 moles CaCO 3 Step 3: Set up a ratio: Step 3: Set up a ratio: The reaction says that I can create 1 mole of CO 2 for every 1 mole of CaCO 3 I start with The reaction says that I can create 1 mole of CO 2 for every 1 mole of CaCO 3 I start with It is just like a recipe! It is just like a recipe!

1 CaCO 3 (s) 1 CaO (s) + 1 CO 2 (g) 3.05 moles CaCO 3 x 1 mole CO 2 = 3.05 moles CaCO 3 x 1 mole CO 2 = 1 mole CaCO 3 1 mole CaCO moles CaCO moles CaCO 3 No step 4 is required, because the problem asks for the answer in moles! No step 4 is required, because the problem asks for the answer in moles!

Let’s look at problem #10…. Group 1 metals are explosive when they come into contact with water. If 20.0 grams of potassium were to explode with excess water, how many moles of hydrogen would be produced? Let ’ s write a balanced chemical equation - step 1! 2 K (s) + 2 HOH (l) 2 KOH (aq) + 1 H 2 (g) Writing reactions will be CRITICAL on the test!!! Let’s look at problem #10…. Group 1 metals are explosive when they come into contact with water. If 20.0 grams of potassium were to explode with excess water, how many moles of hydrogen would be produced? Let ’ s write a balanced chemical equation - step 1! 2 K (s) + 2 HOH (l) 2 KOH (aq) + 1 H 2 (g) Writing reactions will be CRITICAL on the test!!!

2 K (s) + 2 HOH (l) 2 KOH (aq) + 1 H 2 (g) Step 2: Convert into moles: 20.0 g K x 1 mole =.51 moles K 39.1 grams I MUST be in the unit of the mole to be able to look at the ratios or amounts from the chemical reaction! Step 3: Set up a ratio from the reaction:.51 moles K x 1 mole H 2 =.255 moles H 2 2 mole K No step 4 is required because the problem asks for moles of hydrogen 2 K (s) + 2 HOH (l) 2 KOH (aq) + 1 H 2 (g) Step 2: Convert into moles: 20.0 g K x 1 mole =.51 moles K 39.1 grams I MUST be in the unit of the mole to be able to look at the ratios or amounts from the chemical reaction! Step 3: Set up a ratio from the reaction:.51 moles K x 1 mole H 2 =.255 moles H 2 2 mole K No step 4 is required because the problem asks for moles of hydrogen

Let’s look at problem #29…. Cigarette lighters use butane, or C 4 H 10, as their fuel. If 120 grams of butane burn with only 55 liters of oxygen gas at STP, how many liters of carbon dioxide are produced? Which reactant is your limiting reactant? Let ’ s write a balanced chemical equation - step 1! 2 C 4 H 10 (g) + 13 O 2 (g) 10 HOH (g) + 8 CO 2 (g) There is a problem here…. One of our starting chemicals is going to run out! Let’s look at problem #29…. Cigarette lighters use butane, or C 4 H 10, as their fuel. If 120 grams of butane burn with only 55 liters of oxygen gas at STP, how many liters of carbon dioxide are produced? Which reactant is your limiting reactant? Let ’ s write a balanced chemical equation - step 1! 2 C 4 H 10 (g) + 13 O 2 (g) 10 HOH (g) + 8 CO 2 (g) There is a problem here…. One of our starting chemicals is going to run out!

2 C 4 H 10 (g) + 13 O 2 (g) 10 HOH (g) + 8 CO 2 (g) 2 C 4 H 10 (g) + 13 O 2 (g) 10 HOH (g) + 8 CO 2 (g) To figure out which starting material runs out, called our limiting reactant, we must first convert both into moles: To figure out which starting material runs out, called our limiting reactant, we must first convert both into moles: g C 4 H 10 x 1 mole = 2.07 moles C 4 H g C 4 H 10 x 1 mole = 2.07 moles C 4 H grams 58 grams 55 L O 2 x 1 mole = 2.46 moles O 2 55 L O 2 x 1 mole = 2.46 moles O liters 22.4 liters Which one runs out? Is it always the one we have less of? What if I was making sandwiches, and used two pieces of bread, and one piece of bologna, per sandwich? With 8 pieces of bologna, and 10 pieces of bread, what runs out….? Which one runs out? Is it always the one we have less of? What if I was making sandwiches, and used two pieces of bread, and one piece of bologna, per sandwich? With 8 pieces of bologna, and 10 pieces of bread, what runs out….?

The bread!The bread! I need twice as much bread as bologna to make a sandwich….I need twice as much bread as bologna to make a sandwich…. In this problem, what runs out?In this problem, what runs out?

2 C 4 H 10 (g) + 13 O 2 (g) 10 HOH (g) + 8 CO 2 (g) g C 4 H 10 x 1 mole = 2.07 moles C 4 H g C 4 H L O 2 x 1 mole O 2 = 2.46 moles O L O 2 To see which one runs out, we divide each number by the number needed in the reaction… 2.07/2 for the C 4 H 10, and 2.46/13 for the O 2 Which of these numbers is less? That is the one that runs out… The O 2 is less, so it runs out! We finish the problem with the amount of O 2, and ignore the C 4 H 10 All problems are done this way - finish the problem with the chemical that runs out, and ignore the one you have extra of! 2 C 4 H 10 (g) + 13 O 2 (g) 10 HOH (g) + 8 CO 2 (g) g C 4 H 10 x 1 mole = 2.07 moles C 4 H g C 4 H L O 2 x 1 mole O 2 = 2.46 moles O L O 2 To see which one runs out, we divide each number by the number needed in the reaction… 2.07/2 for the C 4 H 10, and 2.46/13 for the O 2 Which of these numbers is less? That is the one that runs out… The O 2 is less, so it runs out! We finish the problem with the amount of O 2, and ignore the C 4 H 10 All problems are done this way - finish the problem with the chemical that runs out, and ignore the one you have extra of!

2 C 4 H 10 (g) + 13 O 2 (g) 10 HOH (g) + 8 CO 2 (g) Step 3: Set up a ratio from the reaction: 2.46 moles O 2 x 8 moles CO 2 = 1.51 moles CO 2 13 moles O 2 Step 4: Convert out of the mole: 1.51 moles CO 2 x 22.4 L = L CO 2 1 mole CO 2 Now try some problems on your own! 2 C 4 H 10 (g) + 13 O 2 (g) 10 HOH (g) + 8 CO 2 (g) Step 3: Set up a ratio from the reaction: 2.46 moles O 2 x 8 moles CO 2 = 1.51 moles CO 2 13 moles O 2 Step 4: Convert out of the mole: 1.51 moles CO 2 x 22.4 L = L CO 2 1 mole CO 2 Now try some problems on your own!