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Chemical Calculations

Chemical Calculations. Moles and Molecules Moles and Chemical Reactions Moles, Chemical Reactions, and Molarity All done as UNIT CONVERSIONS!!! …and practice, practice, practice. CHM 1010 - Sinex. Mass-moles-particles conversions. Mass of substance in grams.

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Chemical Calculations

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

  2. Mass-moles-particles conversions Mass of substance in grams Molar mass of substance - g/mole Moles of substance Conversion factors Avogrado’s number - 6.02 x 1023 particles/mole Number of particles of substance (atoms, molecules, or ions) Molar mass – sum of atomic masses from periodic table

  3. How many molecules of aspirin (C9H8O4) in a dose of 1000 mg? Find MM of aspirin: MM = 9*12+8*1+4*16 = 180 g/mole First find moles: moles = 1000 mg *1g/1000mg * 1mole/180g = 0.0055 moles Conversion factors in boxes Now find molecules: molecules = 0.0055 moles * 6.02x1023 molecules/mole = 3.3 x 1021 molecules How many atoms of carbon? atoms C = 9 atoms/molecule * 3.3 x 1021 molecules = 3.0 x 1022 atoms from the formula

  4. 6.02 x 1023 MM Mass of substance A Moles of substance A Moles of substance A Molecules of substance A For 12.6g NaHCO3, how many moles do you have? How many CO2 molecules in 0.17 moles CO2?

  5. Another way to find moles… For a solution (homogeneous mixture), there are two components: solute (substance being dissolved) and solvent (dissolving medium, usually water) A unit of concentration – amount solute per amount solution or solvent Molarity = moles of solute/volume of solution in liters (moles/L) Rearranging the definition of molarity: moles = M x VL 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. Molarity Moles of substance A Volume of substance A 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?

  7. The ratio of moles in a reaction 2C4H10 + 13O2 8CO2 + 10H2O Mole ratios Click for answers

  8. Mole-to-mole conversions This comes from the balanced chemical reaction! …and it relates substances in the reaction! Mole ratio Moles of substance A Moles of substance B 2HCl + CaCO3 CaCl2 + H2O How many moles of HCl react with a mole of CaCO3? 1 mole CaCO3 x 2 mole HCl/1 mole CaCO3 = 2 mole HCl mole ratio

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

  10. Mass of substance A Mass of substance B Molar mass of substance B Molar mass of substance A Mole ratio Moles of substance A Moles of substance B Molarity of solution A Molarity of solution B Volume of substance A Volume of substance B How many grams of O2 are required to reaction with 10g C3H8? C3H8 + 5O2 3CO2 + 4H2O 10g C3H8 x 1 mole/44g = 0.23 moles C3H8 Conversion factors in boxes 0.23 moles C3H8 x 5 moles O2/1mole C3H8 = 1.15 moles O2 1.15 moles O2 x 32.0g/mole = 37 g O2 This is a classic stoichiometry problem!

  11. Mass of substance A Mass of substance B Molar mass of substance B Molar mass of substance A Mole ratio Moles of substance A Moles of substance B Molarity of solution A Molarity of solution B Volume of substance A Volume of substance B 1 3 How many grams of O2 are required to reaction with 10g C3H8? 2 Another view of solving C3H8 + 5O2 3CO2 + 4H2O Step 1 10g C3H8 x 1 mole/44g = 0.23 moles C3H8 Step 2 0.23 moles C3H8 x 5 moles O2/1mole C3H8 = 1.15 moles O2 Step 3 1.15 moles O2 x 32.0g/mole = 37 g O2

  12. Mass of substance A Mass of substance B Molar mass of substance B Molar mass of substance A Mole ratio Moles of substance A Moles of substance B Molarity of solution A Molarity of solution B Volume of substance A Volume of substance B If 74g O2 react, how many grams of CO2 will be produced? C3H8 + 5O2 3CO2 + 4H2O

  13. Mass of substance A Mass of substance B Molar mass of substance B Molar mass of substance A Mole ratio Moles of substance A Moles of substance B Molarity of solution A Molarity of solution B Volume of substance A Volume of substance B What volume of stomach acid, 0.16 M HCl, reacts with 1.0 g CaCO3? 2HCl + CaCO3 CaCl2 + H2O 1.0 g CaCO3 x 1 mole/100g = 0.010 mole CaCO3 0.010 mole CaCO3 x 2 mole HCl/1 mole CaCO3 = 0.020 mole HCl 0.020 mole HCl x 1L/0.16 moles = 0.13 L = 130 mL stomach acid

  14. Mass of substance A Mass of substance B Molar mass of substance B Molar mass of substance A Mole ratio Moles of substance A Moles of substance B Molarity of solution A Molarity of solution B Volume of substance A Volume of substance B What volume of 0.25 M CH3COOH will reaction with 25.0 mL of 0.37M NaOH? CH3COOH + NaOH  CH3COONa + H2O 25.0 mL x 1L/1000 mL x 0.37 mole/L = 0.0093 moles NaOH 0.0093 moles NaOH x 1 mole CH3COOH/1 mole NaOH = 0.0093 mole CH3COOH 0.0093 mole x 1L/0.25 moles = 0.037 L = 37 mL CH3COOH This is a classic titration problem!

  15. Mass of substance A Mass of substance B Molar mass of substance B Molar mass of substance A Mole ratio Moles of substance A Moles of substance B Molarity of solution A Molarity of solution B Volume of substance A Volume of substance B Could you find one of the conversion factors? Suppose 2.0g CaCO3 reacted with 25.0 mL of HCl. What is the molarity of the HCl? 2HCl + CaCO3 CaCl2 + H2O 2.0 g CaCO3 x 1 mole/100g = 0.020mole CaCO3 0.020 mole CaCO3 x 2 mole HCl/1 mole CaCO3 = 0.040 mole HCl Molarity = moles/ VL = 0.040 mole HCl / (25.0 mL x 1L/1000mL) = 1.6 M HCl Moles of substance B Volume of substance B

  16. Mass of substance A Mass of substance B Molar mass of substance B Molar mass of substance A Mole ratio Moles of substance A Moles of substance B Molarity of solution A Molarity of solution B Volume of substance A Volume of substance B How many grams of Al(OH)3 will react with 100 mL of 0.15 M HCl? Al(OH)3 + 3HCl  AlCl3 + 3H2O

  17. For a 1.0 kg of Al2O3, how much aluminum metal, in grams, can be produced? 2Al2O3 (l) + 3C (s)  4Al (l) + 3CO2 (g) How many atoms of carbon are consumed for the problem above?

  18. How many grams of CH3COOH are needed to react with 27.3 mL of 0.21 M NaOH? CH3COOH + NaOH  CH3COONa + H2O If 0.121g HX react with 26.2 mL of 0.229 M NaOH, what is the molar mass of HX? HX + NaOH  NaX + H2O What is element X?

  19. What if two reagents are added, how much product do you get? Suppose 10 g Na and 10 g Cl2 are mixed together. How much NaCl can be produced? 2Na + Cl2 2NaCl 2Na + 1Cl2 First find the moles of each reactant: 10 g Na x 1 mole/23.0 g = 0.43 mole Na 10 g Cl2 x 1 mole/71.0 g = 0.14 mole Cl2 Find the moles of NaCl produced for each reactant: 0.43 mole Na x 2 mole NaCl/2 mole Na = 0.43 mole BrCl 0.14 mole Cl2 x 2 mole NaCl/1 mole Cl2 = 0.28 mole NaCl • This is what can be produced! Cl2 is limiting reagent.

  20. A chemical analysis problem A chemist decides to check the molarity of H2SO4 in her car battery. A 10.0 mL sample is reacted with 31.03 mL of 2.73M NaOH. What is the molarity of H2SO4? H2SO4 + 2NaOH  Na2SO4 + 2H2O

  21. Answers to problems by slide number: 4- 0.15 mole NaHCO3; 1.0 x 1023 molecules CO2 6- 1.3 M KCl; 0.069 mole NaOH 12- 61g CO2 16- 0.39g Al(OH)3 17- 5.3 x 102g Al; 8.9 x 1024 atoms C 18- 0.34g CH3COOH; MM = 20.2 g/mole; X = F 20- 4.2M H2SO4

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