1. Stoichiometry Continued True Stoichiometry So far, we have talked about changing from units of one thing (atoms, compounds, ions, etc) to different units of the same thing. So far, we have talked about changing from units of one thing (atoms, compounds, ions, etc) to different units of the same thing. But stoichiometry is not limited to this. Its true purpose is to change units of one thing into units of a different thing. But stoichiometry is not limited to this. Its true purpose is to change units of one thing into units of a different thing. In other words, grams of NaCl into grams of NaOH. In other words, grams of NaCl into grams of NaOH.

2. The most important thing you need to remember is you will always go through the mole. The mole is the key

3. We have already touched on the mole and … Mass (use the molar mass) Mass (use the molar mass) Density (convert it to mass first then pick it up from there) Density (convert it to mass first then pick it up from there) Atoms (Avogadro’s Number) Atoms (Avogadro’s Number) Molecules (Avogadro’s Number) Molecules (Avogadro’s Number) Ions (Avogadro’s Number) Ions (Avogadro’s Number) Formula Units (Avogadro’s Number) Formula Units (Avogadro’s Number)

4. So, what is molarity and how do volumes of gases fit into this?

5. What is Molarity? Molarity (M) is the concentration of a solution expressed as moles of solute dissolved per Liter of solution. Molarity (M) is the concentration of a solution expressed as moles of solute dissolved per Liter of solution. In other words, it is moles/Liter In other words, it is moles/Liter Its formula is M=moles/volume (in liters) Its formula is M=moles/volume (in liters)

6. How do we use this information? In an experiment, we use a certain volume of the solution. In an experiment, we use a certain volume of the solution. By using the molarity, we will know how many moles are being used. By using the molarity, we will know how many moles are being used. Remember, we usually use volumes in mL in our labs. So, we need to convert mL to L for these problems. Remember, we usually use volumes in mL in our labs. So, we need to convert mL to L for these problems. For example, we are going to use 50 mL of 3 M HCl in a lab, how many moles of HCl are we going to use? For example, we are going to use 50 mL of 3 M HCl in a lab, how many moles of HCl are we going to use?

7. Examples (Remember M=mol/V) How many moles are in 25 mL of a 15 M NaCl solution? How many moles are in 25 mL of a 15 M NaCl solution? We have a 50 mL solution of HCl that contains 0.52 moles of HCl. What is the concentration of this solution? (In other words, what is the molarity?) We have a 50 mL solution of HCl that contains 0.52 moles of HCl. What is the concentration of this solution? (In other words, what is the molarity?) What volume does 0.34 moles of 12 M HCl occupy? What volume does 0.34 moles of 12 M HCl occupy?

8. What about volumes of gases? At normal conditions, one mole of any gas will occupy 22.4 liters. At normal conditions, one mole of any gas will occupy 22.4 liters. In other words, a gas will have a value of 22.4 L/mol at normal conditions. In other words, a gas will have a value of 22.4 L/mol at normal conditions. If it isn’t normal conditions, a density will be given so solve for mass. If it isn’t normal conditions, a density will be given so solve for mass. Examples (both at normal conditions): How many moles of H 2 (g) occupy 89.76 L. How many moles of H 2 (g) occupy 89.76 L. What is the volume occupied by 2.5 moles of O 2 gas? What is the volume occupied by 2.5 moles of O 2 gas?

9. Now, let us practice getting various things into the mole. 1. 100 mL of 5.6 M of HCl 2. 68.5 L of O 2 (g) at normal conditions 3. 89 g of AgCl 4. 9.65 x 10 22 ions of Cl - 5. 75 mL of nitrogen (l) whose density is 0.808 g/mL

10. So, that is the first step. Get it into moles. After that, we need to change moles to moles.

11. How do we do that? That is where a balanced equation comes into play. That is where a balanced equation comes into play. We can change moles to moles of any two substances in the chemical equation by using the coefficients in the balanced equation. We can change moles to moles of any two substances in the chemical equation by using the coefficients in the balanced equation. The conversion factor (also known as the mole ratio) will be: The conversion factor (also known as the mole ratio) will be: moles solving for /moles given moles solving for /moles given

12. For Example For example, let’s say we have the balanced equation: 2NaCl  2 Na + Cl 2 If we have 5 moles of NaCl, how many moles a Na and Cl 2 will we have after the reaction? 5 moles NaCl (given) x ______________= 5 moles Na 5 moles NaCl (given) x ______________=2.5 moles Cl 2 2 moles NaCl 2 moles Na 1 moles Cl 2 2 moles NaCl Coefficients

13. Then after you have the moles, you can change it back to whatever you need Convert the following: Convert the following: 5.5 moles of NaOH into grams 5.5 moles of NaOH into grams 10 moles of copper into atoms 10 moles of copper into atoms 0.8 moles of neon gas at normal conditions to Liters 0.8 moles of neon gas at normal conditions to Liters 3.5 moles of HCl dissolved in 100 mL of water is what molarity? 3.5 moles of HCl dissolved in 100 mL of water is what molarity? The volume of 15 moles of gasoline, if the density of gasoline is 0.77 g/mL The volume of 15 moles of gasoline, if the density of gasoline is 0.77 g/mL

14. In other words… 1. Is the number they gave me in moles? If not, how can I change it into moles? 2. If I am changing compounds, I need to change moles to moles. Use the coefficients from the balanced equation (the coefficient of the substance you were given goes on the bottom). 3. Do they want the answer in moles? If not, how can I change the moles into that unit?

15. Mass In grams Molarity (moles of solute per liter of solution) Use the formula: M = mol/L Volume (Liters of a gas at STP) Atoms (or molecules) MOLE ÷ ÷ ÷ x x x 22.4 NANA Molar mass 6.02 x 10 23 atoms (molecules) per mole Summary Chart

16. Practice Problems 2.00 g H 2 O (g) = ____________ L H 2 O (g) 2.00 g H 2 O (g) = ____________ L H 2 O (g) 8.00 x 10 22 molecules C 6 H 12 O 6 = _________ g C 6 H 12 O 6 8.00 x 10 22 molecules C 6 H 12 O 6 = _________ g C 6 H 12 O 6 1.00 x 10 24 atoms Ar = ________ liters Ar 1.00 x 10 24 atoms Ar = ________ liters Ar

17. More Practice Problems Use the following balanced equation to complete the following: Use the following balanced equation to complete the following: CaC 2 (s) + 2H 2 O (l)  C 2 H 2 (g) + Ca(OH) 2 How many grams of water are needed to react with 485 g calcium carbide? How many grams of water are needed to react with 485 g calcium carbide? How many grams of CaC 2 are needed to make 23.6 g C 2 H 2 ? How many grams of CaC 2 are needed to make 23.6 g C 2 H 2 ? If 55.3 g Ca(OH) 2 are formed, how many grams of water reacted? If 55.3 g Ca(OH) 2 are formed, how many grams of water reacted?

18. Homework Page 329 and 330 – 9, 22, 25, 28, 30 Page 329 and 330 – 9, 22, 25, 28, 30