1. Molar Relations

2. Stoichiometry The mathematics of chemical formulas and chemical equations. Chemists use a mole to “count” atoms.

3. The Mole A mole is defined as 6.022 x 10 23 of anything. In this sense it is like a “dozen”.

4. For any substance (an element or compound): 1 mole = 6.022 x 10 23 atoms, formula unit, or molecules 1 mole = 1 gram formula mass 1 mole = 22.4 L for any GAS at STP STP – Standard Temperature and Pressure 0 C & 1 atm 273 K & 101.3 kPa 32 F & 760 mmHg

5. For any substance (an element or compound): 1 mole = 6.022 x 10 23 atoms, formula unit, or molecules 1 mole = 1 gram formula mass

6. 14.8 mole of copper = ?? g 7.98 x 10 23 molecules CO 2 = ?? mole 8.907 x 10 23 formula units Cl 2 = ?? L 123.8 g AgNO 3 = ?? Formula units 6.022 x 10 23 atoms, molecules, or formula units 1 mole gram formula mass (P.table) 22.4 L(dm 3 ) For a gas at STP

7. (c) 2006, Mark Rosengarten Grams Moles How many grams will 3.00 moles of NaOH (40.0 g/mol) weigh? 3.00 moles X 40.0 g/mol = 120. g How many moles of NaOH (40.0 g/mol) are represented by 10.0 grams? (10.0 g) / (40.0 g/mol) = 0.250 mol

8. Percent composition Gives the percent by mass of each element in the compound. % Element = Mass element X 100 Mass compound What is the percent composition of (NH 4 ) 2 CO 3 ?

9. Empirical Formulas The formula with the simplest ratio of atoms in a compound. NaCl C 6 H 12 O 6 K 2 SO 4 H 2 O 2 The chemical formula for most ionic compounds is usually the empirical formula. This is not always the case with covalent compounds.

10. Molecular formulas are always a multiple of the empirical formula. Empirical Formula X 2X 6 CH C2H2C2H2 C6H6C6H6

11. What is the molecular formula of a compound that has an empirical formula of CH 2 and a molecular mass of 84 g/mol?

12. What is the empirical formula of a compound that is 32.79 % Na, 13.02 % Al, and 54.19 % F?

13. The coefficients of a balance equation tell you the ratio in which the substances react. C 3 H 8 + 5O 2  3CO 2 + 4H 2 O The coefficients also give the MOLE ratio in which the substances react.

14. C 3 H 8 + 5O 2  3CO 2 + 4H 2 O How many moles of water would be produced if 2.34 mole of propane reacted completely? How many moles of oxygen would be needed to produce 0.897 mole of carbon dioxide?

15. Mass – mass problems Always make sure your answer has the correct number of significant figures and units!

16. 8KClO 3 + C 12 H 22 O 11  8KCl + 11H 2 O + 12CO 2 Example: What mass of sugar would react with 25.00 g of potassium chlorate? What volume of carbon dioxide would be produced at STP if 13.76 g of sucrose reacted with excess potassium chlorate?

17. Percent yield = Actual yield x 100 Theoretical yield The actual yield is the amount made in the lab. The theoretical yield is how much should have been made according to a mass-mass problem.

18. Zn(s) + 2HCl(aq)  H 2 (g) + ZnCl 2 (aq) What is the percent yield if a student made 4.7g of hydrogen when she mixed 115.6 g of zinc with excess hydrochloric acid?

19. The limiting reactant is the reactant that is completely used up in the reaction. It limits the amount of product. Once a reactant is gone, no more product can be made. The other reactant is said to be “in excess”, or it is left over after the reaction is in excess.

20. (c) 2006, Mark Rosengarten Molarity What is the molarity of a 500.0 mL solution of NaOH (FM = 40.0) with 60.0 g of NaOH (aq)? – Convert g to moles and mL to L first! – M = moles / L = 1.50 moles / 0.5000 L = 3.00 M How many grams of NaOH does it take to make 2.0 L of a 0.100 M solution of NaOH (aq)? – Moles = M X L = 0.100 M X 2.0 L = 0.200 moles – Convert moles to grams: 0.200 moles X 40.0 g/mol = 8.00 g

21. (c) 2006, Mark Rosengarten Percent by Mass A 50.0 gram sample of a solution is evaporated and found to contain 0.100 grams of sodium chloride. What is the percent by mass of sodium chloride in the solution? % Comp = (0.100 g) / (50.0 g) X 100 = 0.200%

22. Gas Laws The volume of a sample of gas is affected by pressure, temperature and the amount of gas in the sample. There are laws that define each of these relationships.

23. Boyles Law: The volume of a sample of gas is inversely proportional to the pressure at constant temperature. Relates pressure and volume at constant temperature. That is as pressure increases the volume decreases. EQUATION:Graph: P 1 V 1 =P 2 V 2

24. Charles Law: The volume of a sample of gas at constant pressure is directly proportional to the thermodynamic temperature. Relates volume and the KELVIN temperature at constant pressure. As temperature increases, the volume increases. ALL gas laws must use the KELVIN temperature scale. K = C + 273 EQUATIONGraph: V 1 = V 2 T 1 T 2

25. Combined Gas Law: Combines Boyles and Charles Laws. P 1 V 1 = P 2 V 2 Don’t forget to use Kelvin! T 1 T 2

26. Avogadro’s Law: For a sample of gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (n). As the amount of gas increases the volume of the gas increases. Coefficients of the balanced equation give the volume ratio in which the substances react.

27. Ideal Gas Law: An ideal gas is a gas that follows the assumptions of KMT exactly. Ideal gases do not exist. The gas molecules do have some volume and some intermolecular attractions. However, real gases behave IDEALLY at low pressures and high temperatures, and we can use the ideal gas law. The ideal gas law relates T, P, V, and n. PV = nRT R is the universal gas constant. R = 8.31 L kPa/mol K or R = 0.08206 L atm/mol K The units for T, P, V and n must match the unit of R.

28. Dalton’s Law of Partial Pressures: For a mixture of gasses in a container, the total pressure exerted is the sum of the partial pressures of the gases present. The partial pressure is the pressure that each gas would exert if it were alone in the container. Basically each gas behaves as if it were alone in the mixture, but the total pressure of the mixture depends on all the collisions of all the gas molecules. P total = P 1 + P 2 + P 3 … The partial pressure of each gas depends on the moles of that gas. P 1 = n 1 Ptotal n total

30. (c) 2006, Mark Rosengarten Gas Laws Make a data table to put the numbers so you can eliminate the words. Make sure that any Celsius temperatures are converted to Kelvin (add 273). Rearrange the equation before substituting in numbers. If you are trying to solve for T 2, get it out of the denominator first by cross-multiplying. If one of the variables is constant, then eliminate it. Try these problems!

31. (c) 2006, Mark Rosengarten Gas Law Problem 1 A 2.00 L sample of N 2 gas at STP is compressed to 4.00 atm at constant temp- erature. What is the new volume of the gas? V 2 = P 1 V 1 / P 2 = (1.00 atm)(2.00 L) / (4.00 atm) = 0.500 L

32. (c) 2006, Mark Rosengarten Gas Law Problem 2 To what temperature must a 3.000 L sample of O 2 gas at 300.0 K be heated to raise the volume to 10.00 L? T 2 = V 2 T 1 /V 1 = (10.00 L)(300.0 K) / (3.000 L) = 1000. K

33. (c) 2006, Mark Rosengarten Gas Law Problem 3 A 3.00 L sample of NH 3 gas at 100.0 kPa is cooled from 500.0 K to 300.0 K and its pressure is reduced to 80.0 kPa. What is the new volume of the gas? V2 = P 1 V 1 T 2 / P 2 T 1 = (100.0 kPa)(3.00 L)(300. K) / (80.0 kPa)(500. K) = 2.25 L