1. Moles and Molecules Moles and Chemical Reactions Moles, Chemical Reactions, and Molarity All done as UNIT CONVERSIONS!!! …and practice, practice, practice Chemical Calculations CHM 1010 - Sinex

2. Mass of substance in grams Moles of substance Number of particles of substance (atoms, molecules, or ions) Molar mass of substance - g/mole Avogrado’s number - 6.02 x 10 23 particles/mole Mass-moles-particles conversions Conversion factors Molar mass – sum of atomic masses from periodic table 2

3. 3 How many atoms of carbon? atoms C = 9 atoms/molecule * 3.3 x 10 21 molecules = 3.0 x 10 22 atoms Now find molecules: molecules = 0.0055 moles * 6.02x10 23 molecules/mole = 3.3 x 10 21 molecules First find moles: moles = 1000 mg *1g/1000mg * 1mole/180g = 0.0055 moles How many molecules of aspirin (C 9 H 8 O 4 ) in a dose of 1000 mg? Find MM of aspirin: MM = 9*12+8*1+4*16 = 180 g/mole Conversion factors in boxes from the formula

4. 4 For 12.6g NaHCO 3, how many moles do you have? Mass of substance AMoles of substance A MM How many CO 2 molecules in 0.17 moles CO 2 ? Moles of substance AMolecules of substance A 6.02 x 10 23

5. 5 Another way to find moles… Molarity = moles of solute/volume of solution in liters (moles/L) A unit of concentration – amount solute per amount solution or solvent Rearranging the definition of molarity: moles = M x V L For a solution (homogeneous mixture), there are two components: solute (substance being dissolved) and solvent (dissolving medium, usually water) Another way to find moles! How many moles NaCl in 50 mL of 0.28 M NaCl? 50 mL x 1L/1000 mL x 0.28 mole/L = 0.014 mole NaCl

6. 6 What is the molarity if 35g KCl are dissolved to make 350 mL of solution? How many moles are in 300 mL of a 0.23 M NaOH solution? Moles of substance AVolume of substance A Molarity

7. 7 2C 4 H 10 + 13O 2  8CO 2 + 10H 2 O 213810 4261620 13080100 16.545 6 1.3 32 50 1.37 X 13/2X 8/2X 10/2 1.378.915.486.85 Mole ratios The ratio of moles in a reaction Click for answers

8. 8 1 mole CaCO 3 x 2 mole HCl/1 mole CaCO 3 = 2 mole HCl Moles of substance AMoles of substance B Mole ratio This comes from the balanced chemical reaction! …and it relates substances in the reaction! Mole-to-mole conversions 2HCl + CaCO 3  CaCl 2 + H 2 O How many moles of HCl react with a mole of CaCO 3 ? mole ratio

9. 9 Moles of substance AMoles of substance B Mole ratio Mass of substance A Molar mass of substance A Mass of substance B Molar mass of substance B Volume of substance AVolume of substance B Molarity of solution AMolarity of solution B Stoichiometry Molarity (M) = moles of solute/volume in liters of solution – moles/L Conversion factors

10. 10 Moles of substance AMoles of substance B Mole ratio Mass of substance A Molar mass of substance A Mass of substance B Molar mass of substance B Volume of substance AVolume of substance B Molarity of solution AMolarity of solution B How many grams of O 2 are required to reaction with 10g C 3 H 8 ? 10g C 3 H 8 x 1 mole/44g = 0.23 moles C 3 H 8 0.23 moles C 3 H 8 x 5 moles O 2 /1mole C 3 H 8 = 1.15 moles O 2 1.15 moles O 2 x 32.0g/mole = 37 g O 2 C 3 H 8 + 5O 2  3CO 2 + 4H 2 O This is a classic stoichiometry problem! Conversion factors in boxes

11. 11 Moles of substance AMoles of substance B Mole ratio Mass of substance A Molar mass of substance A Mass of substance B Molar mass of substance B Volume of substance AVolume of substance B Molarity of solution AMolarity of solution B How many grams of O 2 are required to reaction with 10g C 3 H 8 ? 10g C 3 H 8 x 1 mole/44g = 0.23 moles C 3 H 8 0.23 moles C 3 H 8 x 5 moles O 2 /1mole C 3 H 8 = 1.15 moles O 2 1.15 moles O 2 x 32.0g/mole = 37 g O 2 C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 1 2 3 Step 1 Step 3 Step 2 Another view of solving

12. 12 Moles of substance AMoles of substance B Mole ratio Mass of substance A Molar mass of substance A Mass of substance B Molar mass of substance B Volume of substance AVolume of substance B Molarity of solution AMolarity of solution B C 3 H 8 + 5O 2  3CO 2 + 4H 2 O If 74g O 2 react, how many grams of CO 2 will be produced?

13. 13 Moles of substance AMoles of substance B Mole ratio Mass of substance A Molar mass of substance A Mass of substance B Molar mass of substance B Volume of substance AVolume of substance B Molarity of solution AMolarity of solution B 2HCl + CaCO 3  CaCl 2 + H 2 O What volume of stomach acid, 0.16 M HCl, reacts with 1.0 g CaCO 3 ? 0.010 mole CaCO 3 x 2 mole HCl/1 mole CaCO 3 = 0.020 mole HCl 1.0 g CaCO 3 x 1 mole/100g = 0.010 mole CaCO 3 0.020 mole HCl x 1L/0.16 moles = 0.13 L = 130 mL stomach acid

14. 14 Moles of substance AMoles of substance B Mole ratio Mass of substance A Molar mass of substance A Mass of substance B Molar mass of substance B Volume of substance AVolume of substance B Molarity of solution AMolarity of solution B This is a classic titration problem! CH 3 COOH + NaOH  CH 3 COONa + H 2 O What volume of 0.25 M CH 3 COOH will reaction with 25.0 mL of 0.37M NaOH? 25.0 mL x 1L/1000 mL x 0.37 mole/L = 0.0093 moles NaOH 0.0093 moles NaOH x 1 mole CH 3 COOH/1 mole NaOH = 0.0093 mole CH3COOH 0.0093 mole x 1L/0.25 moles = 0.037 L = 37 mL CH 3 COOH

15. 15 Could you find one of the conversion factors? Moles of substance AMoles of substance B Mole ratio Mass of substance A Molar mass of substance A Mass of substance B Molar mass of substance B Volume of substance AVolume of substance B Molarity of solution AMolarity of solution B 2HCl + CaCO 3  CaCl 2 + H 2 O Suppose 2.0g CaCO 3 reacted with 25.0 mL of HCl. What is the molarity of the HCl? 2.0 g CaCO 3 x 1 mole/100g = 0.020mole CaCO 3 0.020 mole CaCO 3 x 2 mole HCl/1 mole CaCO 3 = 0.040 mole HCl Molarity = moles/ V L = 0.040 mole HCl / (25.0 mL x 1L/1000mL) = 1.6 M HCl Moles of substance B Volume of substance B

16. 16 Moles of substance AMoles of substance B Mole ratio Mass of substance A Molar mass of substance A Mass of substance B Molar mass of substance B Volume of substance AVolume of substance B Molarity of solution AMolarity of solution B Al(OH) 3 + 3HCl  AlCl 3 + 3H2O How many grams of Al(OH) 3 will react with 100 mL of 0.15 M HCl?

17. 17 2Al 2 O 3 (l) + 3C (s)  4Al (l) + 3CO 2 (g) For a 1.0 kg of Al 2 O 3, how much aluminum metal, in grams, can be produced? How many atoms of carbon are consumed for the problem above?

18. 18 CH 3 COOH + NaOH  CH 3 COONa + H 2 O How many grams of CH 3 COOH are needed to react with 27.3 mL of 0.21 M NaOH? HX + NaOH  NaX + H 2 O If 0.121g HX react with 26.2 mL of 0.229 M NaOH, what is the molar mass of HX? What is element X?

19. 19 What if two reagents are added, how much product do you get? 2Na + Cl 2  2NaCl Suppose 10 g Na and 10 g Cl 2 are mixed together. How much NaCl can be produced? First find the moles of each reactant: 10 g Cl 2 x 1 mole/71.0 g = 0.14 mole Cl 2 10 g Na x 1 mole/23.0 g = 0.43 mole Na Find the moles of NaCl produced for each reactant: 0.14 mole Cl 2 x 2 mole NaCl/1 mole Cl 2 = 0.28 mole NaCl 0.43 mole Na x 2 mole NaCl/2 mole Na = 0.43 mole BrCl  This is what can be produced! Cl 2 is limiting reagent. 21 Have 0.43 Need 0.22 Need 0.28 Have 0.14 2Na + 1Cl 2

20. 20 A chemical analysis problem A chemist decides to check the molarity of H 2 SO 4 in her car battery. A 10.0 mL sample is reacted with 31.03 mL of 2.73M NaOH. What is the molarity of H 2 SO 4 ? H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O

21. 21 Answers to problems by slide number: 4- 0.15 mole NaHCO 3 ; 1.0 x 10 23 molecules CO 2 6- 1.3 M KCl; 0.069 mole NaOH 12- 61g CO 2 16- 0.39g Al(OH) 3 17- 5.3 x 10 2 g Al; 8.9 x 10 24 atoms C 18- 0.34g CH 3 COOH; MM = 20.2 g/mole; X = F 20- 4.2M H 2 SO 4