1. FORMULAS, EQUATIONS AND MOLES Mole Calculation Chapter 3

2. Names associated with an amount

3. The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C 1 mol = N A = 6.0221367 x 10 23 Avogadro’s number (N A ) The Mole…..

4. Molar mass is the mass of 1 mole of in grams eggs shoes marbles atoms 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams)

5. What is the mass of one mole of: C S Cu Fe Hg

6. How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 10 23 atoms K 0.551 g K 1 mol K 39.10 g K x x 6.022 x 10 23 atoms K 1 mol K = 8.49 x 10 21 atoms K

7. Molar Mass The molar mass of a compound is found by adding Together the molar masses of all of its elements, taking Into account the number of moles of each element present.

8. Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2 1S32.07 amu 2O+ 2 x 16.00 amu SO 2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2

9. How many H atoms are in 72.5 g of C 3 H 8 O ? 1 mol C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 g C 3 H 8 O 1 mol H = 6.022 x 10 23 atoms H 5.82 x 10 24 atoms H 1 mol C 3 H 8 O molecules = 8 mol H atoms 72.5 g C 3 H 8 O 1 mol C 3 H 8 O 60 g C 3 H 8 O x 8 mol H atoms 1 mol C 3 H 8 O x 6.022 x 10 23 H atoms 1 mol H atoms x =

10. STOICHIOMETRY Quantitative relationship among reactants and products are called stoichiometry. Stoichiometry problems are easily solved when amounts of substances are converted from mass (in units of g, kg etc), volume (L), into moles. Amounts in moles depend on how the chemical formula or the chemical reaction equations are written. Chemical reaction equations are the basis for reaction stoichiometry, but when the reactants are not stoichiometric mixtures, some reactants will be in excess whereas others will be in limited supply.

11. 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the REQUIRED quantity 4.Convert moles of REQUIRED quantity into desired units Mass Changes in Chemical Reactions 3.8

12. Stoichiometric Ratio Coefficient B Coefficient A Moles B = X Moles A The stoichiometric ratio is used to determine amounts of compounds consumed or produced in a balanced chemical reaction

13. Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x = 235 g H 2 O

14. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100

15. Limiting Reagent The number of bicycles that can be assembled is limited by whichever part runs out first. In the inventory shown in this figure, wheels are that part. The reactant that is completely consumed by the reaction

16. A molecular view of a Limiting reactant situation for the ammonia Synthesis. To make 4 molecules of NH 3 requires 2 molecules of N 2 and 6 molecules of H 2. If we start with 4 molecules of N 2 and 6 molecules of H 2, H 2 is the limiting reactant.

17. Limiting Reagent: It is the reactant whose number of moles divided By it’s stoichiometric coefficient is the smallest Example: Given: If 1.00 Kg of each of the reactant is allowed to react. How many Kg of POCl 3 will be formed? 1.Find the limiting reagent a.Find moles of each b.Divide by stoichiometric coefficient c.The compound with the smallest number of moles is the LR PCl 3 : Cl 2 : P 4 O 10 : 2. Determine the amount of POCl 3 based on the limiting reagent PCl 3. Note that the conversions of grams to Kg is not shown.

18. Solution Concentrations Molarity: The most useful way of expressing the amount of a substance dissolved in a solution is with molarity Molarity= Moles of solute Volume of solution in L

19. Making Molar Solutions From Solids

20. What are molar solutions? A molar solution is one that expresses “concentration” in moles per volume Usually the units are in mol/L mol/L can be abbreviated as M or [ ] Molar solutions are prepared using: a balance to weigh moles (as grams) a volumetric flask to measure litres L refers to entire volume, not water! Because the units are mol/L, we can use the equation M = n/L Alternatively, we can use the factor label method Solution Concentrations

21. Practice making molar solutions 1.Calculate # of grams required to make 100 mL of a 0.10 M solution of NaOH (see above). 2.Get volumetric flask, plastic bottle, 100 mL beaker, eyedropper. Rinse all with tap water. 3.Fill a beaker with distilled water. 4.Pour 20 - 30 mL of H 2 O from beaker into flask. 5.Weigh NaOH. Add it to flask. Do step 5 quickly. 6.Mix (by swirling) until the NaOH is dissolved. 7.Add distilled H 2 O to just below the colored line. 8.Add distilled H 2 O to the line using eyedropper.

22. Making Molar Solutions From Liquids

23. Dilution Formula 1.Calculate volume of solution of known molarity required to make 100 mL of a 0.10 M solution of HCl. M i V i =M f V f 2.Get volumetric flask, plastic bottle, 100 mL beaker, eyedropper. Rinse all with tap water. 3.Fill a beaker with distilled water. 4.Pour 20 - 30 mL of H 2 O from beaker into flask. 5.Measure HCl (V i ). Add it to flask and mix 6.Add distilled H 2 O to just below the colored line (100mL). 7.Add distilled H 2 O to the line using eyedropper.