1. Mass Relationships in Chemical Reactions Chapter 3

2. By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00 amu Atomic mass is the mass of an atom in atomic mass units (amu) Micro World atoms & molecules Macro World grams 3.1

3. Natural lithium is: 7.42% 6 Li (6.015 amu) 92.58% 7 Li (7.016 amu) (7.42% x 6.015) + (92.58% x 7.016) 100 = 6.941 amu 3.1 Average atomic mass of lithium:

4. 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 3.2 1 mol = N A = 6.0221367 x 10 23 Avogadro’s number (N A )

5. 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) 3.2

6. One Mole of: C S Cu Fe Hg 3.2

7. 1 g = 6.022 x 10 23 amu 3.2 M = molar mass in g/mol N A = Avogadro’s number 1 amu = 1.66 x 10 -24 g

8. NMSI Renee Mc Cormick

9. Do You Understand Molar Mass? 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 3.2

10. 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 3.3

11. Do You Understand Molecular Mass? 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 3.3 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 =

12. Mole fractions: X a = n a _____ Where n is the number of moles n a +n b.... Use 1 mole to determine the ratio amounts: What is the mole fraction of Na in Na 2 O. Given the answer can you work backwards and find the moles of O?

13. mass spectrometer—a device for measuring the mass of atoms or molecules  atoms or molecules are passed into a beam of high-speed electrons  this knocks electrons OFF the atoms or molecules transforming them into cations o apply an electric field  this accelerates the cations since they are repelled from the (+) pole and attracted toward the (−) polesend the accelerated cations into a magnetic field  an accelerated cation creates it’s OWN magnetic field which perturbs the original magnetic fieldthis perturbation changes the path of the cation  the amount of deflection is proportional to the mass; heavy cations deflect little ions hit a detector plate where measurements can be obtained.

14. KE = 1/2 x m x v 2 v = (2 x KE/m) 1/2 F = q x v x B 3.4 Light Heavy

16. Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C2H6OC2H6O %C = 2 x (12.01 g) 46.07 g x 100% = 52.14%H = 6 x (1.008 g) 46.07 g x 100% = 13.13%O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0% 3.5

17. Types of Formulas Empirical FormulaEmpirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Ionic formula are always empirical formula Molecular FormulaMolecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

18. To obtain an Empirical Formula 1.Determine the mass in grams of each element present, if necessary. 2.Calculate the number of moles of each element. 3.Divide each by the smallest number of moles to obtain the simplest whole number ratio. 4.If whole numbers are not obtained * in step 3), multiply through by the smallest number that will give all whole numbers * Be careful! Do not round off numbers prematurely

19. A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. require mole ratios so convert grams to moles moles of N = 2.34g of N = 0.167 moles of N 14.01 g/mole 14.01 g/mole moles of O = 5.34 g = 0.334 moles of O 16.00 g/mole 16.00 g/mole Formula: Formula:

20. Calculation of the Molecular Formula A compound has an empirical formula of NO 2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance? empirical formula mass: 14.01+2 (16.00) = 46.01 g/mol n = molar mass = 92.0 g/mol emp. f. mass 46.01 g/mol n = 2 N2O4N2O4N2O4N2O4

21. Empirical Formula from % Composition A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance? Consider a sample size of 100 grams This will contain: 60.80 grams of Na, 28.60 grams of B, and 10.60 grams H Determine the number of moles of each Determine the simplest whole number ratio

22. Find the empirical formula using a equation.

23. 3.6 g CO 2 mol CO 2 mol Cg C g H 2 O mol H 2 Omol Hg H g of O = g of sample – (g of C + g of H) Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O

24. 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 sought quantity 4.Convert moles of sought quantity into desired units Mass Changes in Chemical Reactions 3.8

25. Other units Molarity –Moles solute / L solution Gases –22.4 L = 1 mole of ANY GAS at STP

26. 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 3.8

27. 6 green used up 6 red left over Limiting Reagents 3.9

28. Method 1 Pick A Product Try ALL the reactants The lowest answer will be the correct answer The reactant that gives the lowest answer will be the limiting reactant

29. This is the method I use. Pick your favorite and use it. Limiting reagent To determine the limiting reagent requires that you do two stoichiometry problems. Figure out how much product each reactant makes. The one that makes the least is the limiting reagent.

30. Limiting Reactant: Method 1 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2  2 AlCl 3 Start with Al: Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.5 g AlCl 3 27.0 g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl 3 35.0g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 71.0 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3 Limiting Reactant

31. Method 2 Convert one of the reactants to the other REACTANT See if there is enough reactant “A” to use up the other reactants If there is less than the GIVEN amount, it is the limiting reactant Then, you can find the desired species

32. Limiting Reagents: shortcut Limiting reagent problems can be solved another way (without using a chart)... Do two separate calculations using both given quantities. The smaller answer is correct. Q-How many g NO are produced if 20gNH3 is burned in 30g O2 ?

33. Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. g Almol Almol Fe 2 O 3 neededg Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al neededg Al needed 124 g Al 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al x 160. g Fe 2 O 3 1 mol Fe 2 O 3 x = 367 g Fe 2 O 3 Start with 124 g Alneed 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent 3.9

34. Use limiting reagent (Al) to calculate amount of product that can be formed. g Almol Almol Al 2 O 3 g Al 2 O 3 124 g Al 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al x 102. g Al 2 O 3 1 mol Al 2 O 3 x = 234 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 3.9

35. Finding Excess Practice 10.0g of aluminum reacts with 35.0 grams of chlorine gas 2 Al + 3 Cl 2  2 AlCl 3 We found that chlorine is the limiting reactant, and 43.8 g of aluminum chloride are produced. 35.0 g Cl 2 1 mol Cl 2 2 mol Al 27.0 g Al 71 g Cl 2 3 mol Cl 2 1 mol Al = 8.8 g Al USED! 10.0 g Al – 8.8 g Al = 1.2 g Al EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

36. 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 3.10 % Yield = what you got what you could have got x 100

37. Chemistry In Action: Chemical Fertilizers Plants need: N, P, K, Ca, S, & Mg 3H 2 (g) + N 2 (g) 2NH 3 (g) NH 3 (aq) + HNO 3 (aq) NH 4 NO 3 (aq) 2Ca 5 (PO 4 ) 3 F (s) + 7H 2 SO 4 (aq) 3Ca(H 2 PO 4 ) 2 (aq) + 7CaSO 4 (aq) + 2HF (g) fluorapatite