1. Vanessa Prasad-Permaul Valencia College CHM 1045

2. Atomic and Molecular Mass  The molecular mass is the sum of the masses of the atoms making up the molecule (units amu) (molecular compounds only!) calculate the molecular mass: H 2 O C 2 H 4 O 2

3. Formula Mass  Formula Mass is the sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic (units amu) calculate the formula mass: NaCl MgCO 3

4. Exercise 3.1  Calculate the formula masses of the following compounds, using the table of atomic masses. Give the answers to 3 significant figures:  A) NO 2  B) glucose, C 6 H 12 O 6  C) sodium hydroxide  D) magnesium hydroxide

5. A. NO 2 1 x AM of N= 14.0067 2 x AM of O=2 x 15.9994= 31.9988 MM of NO 2 = 46.0055 46.0 amu B. C 6 H 12 O 6 6 x AM of C=6 x 12.011= 72.066 12 x AM of H=12 x 1.0079= 12.0948 6 x AM of O =6 x 15.9994= 95.9964 MM of C 6 H 12 O 6 = 180.1572 180. amu C. NaOH 1 x AM of Na=22.98977 1 x AM of O=15.9994 1 x AM of H= 1.0079 MM of NaOH = 39.9971 40.0 amu D. Mg(OH) 2 1 x AM of Mg=24.305 2 x AM of O=2 x 15.9994=31.9988 2 x AM of H=2 x 1.0079 = 2.0158 MM of Mg(OH) 2 = 58.3196 58.3 amu

6. Avogadro  One mole of a substance is the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12g of C-12.  One mole of carbon = 6.02 x 10 23 molecules of carbon  Avogadro’s Number (N A = 6.022 x 10 23 molecules) is the numerical value assigned to the unit, 1 mole

7. Avogadro’s Number Li 2 SO 4 1. How many Li 2 SO 4 units are in 3 moles of Li 2 SO 4 ? 2. How many Li atoms are in 3 moles of Li 2 SO 4 ? 3. How many S atoms are in 3 moles of Li 2 SO 4 ? 4. How many O atoms are in 3 moles of Li 2 SO 4 ?

8. Molar Mass of a substance is the mass of one mole of that substance C-12 has a molar mass of exactly 12g/mol For all substances, the molar mass (g/mol) is numerically equal to the formula mass (amu). For example: 1 mole O contains 6.02 x 10 23 oxygen atoms 1 mole of O 2 contains 2x (6.02 x 10 23 ) oxygen atoms

9. Molar Mass Exercise 3.3 What is the mass in grams of a calcium atom? What is the mass in grams of an ethanol molecule?

10. The atomic mass of Ca = 40.08 amu; thus, the molar mass = 40.08 g/mol, and 1 mol Ca = 6.022 x 10 23 Ca atoms. Mass of one Ca = = 6.6556 x 10 −23 = 6.656 x 10 −23 g/atom x The molecular mass of C 2 H 5 OH, or C 2 H 6 O, = (2 x 12.01) + (6 x 1.008) + 16.00 = 46.068. Its molar mass = 46.07 g/mol, and 1 mol = 6.022 x 10 23 molecules of C 2 H 6 O. Mass of one C 2 H 6 O = = = 7.6503 x 10 −23 = 7.650 x 10 −23 g/molecule x Molar Mass 1 mol 6.022 x 10 23 molecules

11. Mole Calculations Consider ethanol with a molar mass of 46.1g/mol. Conversion to moles of ethanol: 1 mol C 2 H 5 OH = 46.1g/mol Prepare acetic acid from 10.0g of ethanol. How many moles of ethanol is this? 10.0g C 2 H 5 OH x 1 mol C 2 H 5 OH = 0.217 mol C 2 H 5 OH 46.1 g C 2 H 5 OH

12. Mole Calculations Exercise 3.4 Dilute aqueous hydrogen peroxide is a colorless liquid used as a bleach. Analysis of a solution shows it contains 0.909 mol of H 2 O 2 in 1.00L of solution. What is the mass of hydrogen peroxide in this volume of solution? The molar mass of H 2 O 2 is 34.02 g/mol. Therefore,

13. Mole Calculations Exercise 3.5 HNO 3 is a corrosive liquid used for the manufacture of nitrogen fertilizers and explosives. A sample containing 28.5g of nitric acid was poured into a beaker. How many moles of HNO 3 are there? The molar mass of HNO 3 is 63.01 g/mol. Therefore,

14. Mole Calculations Exercise 3.6 HCN is a volatile colorless liquid. The compound is highly poisonous. How many molecules are there in 56mg HCN, the average toxic does? Convert the mass of HCN from milligrams to grams. Then convert grams of HCN to moles of HCN. Finally, convert moles of HCN to the number of HCN molecules. 56 mg HCN x = 1.248 x 10 21 = 1.2 x 10 21 HCN molecules xx

15. Chemical Formulas Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass. Mass % A = mass A in the whole X 100% whole mass

16. Exercise 3.7 Ammonium nitrate, which is prepared from nitric acid is used as a nitrogen fertilizer. Calculate the mass percentages of the elements in ammonium nitrate to 3 sig. figs. The molecular mass of NH 4 NO 3 = 80.05; its molar mass = 80.05g/mol. Hence Percent N = 28.02g N x 100% = 35.00 = 35.0% 80.05g NH 4 NO 3 Percent H = 4.032g H x 100% = 5.036 = 5.04% 80.05g NH 4 NO 3 Percent O = 48.00g O x 80.05g NH 4 NO 3 100% = 59.96 = 60.0%

17. Exercise 3.8 How many grams of nitrogen are there in a fertilizer containing 48.5g of NH 4 NO 3 ? From the previous exercise, NH 4 NO 3 is 35.0% N (Fraction N = 0.350), so the mass of N in 48.5 g of NH 4 NO 3 is: 48.5 g NH 4 NO 3 x 0.350 g N = 16.975 = 17.0 g N 1 g NH 4 NO 3

18. Exercise 3.9 A 3.87 mg of ascorbic acid (vitamin C) gives 5.80mg of CO 2 and 1.58mg of H 2 O on combustion. What is the % composition of this compound? First convert the mass of CO 2 to moles of CO 2. Next, convert this to moles of C (1 mol CO 2 is equivalent to 1 mol C). Finally, convert to mass of C, changing milligrams to grams first: 5.80 x 10 −3 g CO 2 x 1 mol CO 2 x 1 mol C x 12.01g C = 1.583 x 10 −3 g C 44.01g CO 2 1 mol CO 2 1 mol C 1.583mg C x 100% = 40.9% C 3.87mg 1.58 x 10 -3 g H 2 O x 1 mol H 2 O x 2 mol H x 1.0079g H = 1.767 x 10 -4 g H 18.02g H 2 O 1 mol H 2 O 1 mol H 1.767 x 10 -4 g H x 100% = 4.57% H 3.87 x 10 -3 g

19. Now that we have the % composition of C & H, we can calculate % O: 1.583mg C x 100% = 40.9% C 3.87mg 1.767 x 10 -4 g H x 100% = 4.57% H 3.87 x 10 -3 g The mass percentage of O can be determined by subtracting the sum of the above percentages from 100%: Percent O = 100.000% − (40.90 + 4.5658) = 54.5342 = 54.5% O

20. Chemical Formulas Empirical Formula: Determined from data about percent composition, tells only the smallest whole number ratio of atoms in a compound. Tells the ratio of numbers of atoms in the compound hydrogen peroxide Molecular formula is H 2 O 2 Empirical formula is HO

21. Chemical Formulas Exercise 3.10 A sample of compound weighing 83.5g contains 33.4g of sulfur. The rest is oxygen. What is the empirical formula? Convert the masses to moles that are proportional to the subscripts in the empirical formula: 83.5g compound – 33.4g S = 50.1g O 33.4g S x 1 mol S = 1.0414 mol S 50.1g O x 1 mol O = 3.1312 mol O 32.07g S 15.9994g O Next, obtain the smallest integers from the moles by dividing each by the smallest number of moles: For O: 3.1312 = 3.01 (3 oxygen) For S: 1.0414 = 1.00 (1 sulfur) 1.0414 1.0414 The empirical formula is SO 3

22. Chemical Formulas Exercise 3.11 Benzoic acid is a white crystalline powder use a food preservative. The compound contains 68.8% C, 5.0% H and 26.2% O by mass. What is the empirical formula? For a 100.0-g sample of benzoic acid, 68.8 g are C, 5.0 g are H, and 26.2 g are O. Using the molar masses, convert these masses to moles: 68.8g C x 1 mol C = 5.729 mol C 12.01g C 5.0g H x 1 mol H = 4.961 mol H 1.0079g H 26.2g O x 1 mol O = 1.638 mol O 15.9994 g O 5.759 = 3.49 C 4.961 = 3.03 H 1.638 = 1.00 O 1.638 1.638 1.638 Rounding off, we obtain C 3.5 H 3.0 O 1.0. Multiplying the numbers by 2 gives whole numbers, for an empirical formula of C 7 H 6 O 2.

23. Molecular Formula: a multiple of the empirical formula which is obtained by summing the atomic masses of the atoms in the empirical formula. Molecular mass = n x empirical formula mass n = the number of empirical formula units in the molecule. The molecular formula is obtained by multiplying the subscripts of the empirical formula by n.

24. Exercise 3.12 The % composition of acetaldehyde is 54.5% C, 9.2% H and 36.3% O. The molecular mass is 44 amu. What is the molecular formula of acetaldehyde? For a 100.0-g sample of acetaldehyde, 54.5 g are C, 9.2 g are H, and 36.3 g are O. Using the molar masses, convert these masses to moles: 54.5g C x 1 mol C = 4.537 mol C 4.537 = 2.0 C 12.01g C 2.270 9.2g H x 1 mol H = 9.127 mol H 9.127 = 4.0 H 1.0079g H 2.270 36.3g O x 1 mol O = 2.270 mol O 2.270 = 1.0 O 15.994 g O 2.270 The empirical formula of acetaldehyde is C 2 H 4 O

25. The empirical formula of acetaldehyde is C 2 H 4 O. The empirical formula mass is 2C x 12 amu = 24 amu 4H x 1 amu = 4 amu 1O x 16 amu = 16 amu 44 amu To find n = molecular mass = 44 amu = 1 empirical formula mass 44 amu Therefore the molecular formula is the same as the empirical formula

26. Stoichiometry Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities involved in a chemical reaction. Grams of A Convert grams of A to moles of A Convert moles of A to moles of B Convert moles of B to grams of B Grams of B

27. Stoichiometry Exercise 3.14 Sodium is a soft reactive metal that instantly reacts with water to give hydrogen gas and a solution of sodium hydroxide. How many grams of sodium metal is needed to give 7.81g of hydrogen by this reaction: Equation: 2Na + 2H 2 O  H 2 + 2NaOH From this equation, 2 moles of Na corresponds to one 1 mole of H 2. Therefore, 7.81 g H 2 x 1 mol H 2 x 2 mols Na x 22.9898g Na = 178.1 = 178 g Na 2.0158g H 2 1 mol H 2 1 mol Na

28. Stoichiometry Exercise 3.15 Zinc sulfide reacts with oxygen to give zinc oxide and sulfur Dioxide. How many kg of oxygen gas combines with 5.00 x10 3 g of zinc sulfide in this reaction? 2 ZnS + 3 O 2 2 ZnO + 2 SO 2 5.00 x 10 3 g ZnS x 1 mol ZnS x 3 mols O 2 x 31.999g O 2 97.474g ZnS 2 mols ZnS 1 mol O 2 = 2462g O 2 x 1kg 1000g = 2.46kg O 2

29. Stoichiometry Limiting Reagent: The reactant or the reagent that is completely consumed in a chemical reaction –I need to make a fruit salad that is 1/2 apples and 1/2 oranges. I have 10 apples but only 7 oranges. What is the limiting fruit? How many apples and oranges can I use?

30. Exercise 3.17 Aluminum chloride is used as a catalyst in various industrial reactions. It is prepared from hydrogen chloride gas and aluminum metal shavings. A reaction vessel contains 0.15 mol Al and 0.35 mol HCl. How many moles of AlCl 3 can be prepared from this mixture? 2 Al (s) + 6 HCl (g) 2 AlCl 3 (s) + 3 H 2 (g) First, determine the limiting reactant by calculating the moles of AlCl 3 that would be obtained if Al and HCl were totally consumed: 0.15 mol Al x 2 mol AlCl 3 = 0.15 mol AlCl 3 2 mol Al 0.35 mol HCl x 2 mol AlCl 3 = 0.12 mol AlCl 3 (HCl is the limiting reactant) 6 mol HCl Because the HCl produces the smaller amount of AlCl 3, the reaction will stop when HCl is totally consumed but before all the Al is consumed. The limiting reactant is therefore HCl. The amount of AlCl 3 produced must be 0.1166, or 0.12 mol.

31. Stoichiometry Limiting Reagent Calculation: Lithium oxide is a drying agent used on the space shuttle. If 80.0 kg of water is to be removed and 65 kg of lithium oxide is available, which reactant is limiting? Li 2 O (s) + H 2 O (l)  2 LiOH (s) Li 2 O MM = 29.88 g/mol H 2 O MM = 18.02 g/mol

32. Stoichiometry Yields of Chemical Reactions: If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate a percentage yield. Theoretical Yield: maximum amount of product that can be obtained by a reaction from the given amounts of reactants. Percent Yield: the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). % Yield = Actual yield X 100% Theoretical yield

33. Stoichiometry Exercise 3.19 New industrial plants for acetic acid react liquid methanol with carbon monoxide in the presence of a catalyst. CH 3 OH(l) + CO (g) HC 2 H 3 O 2 (l) 15.0g of methanol and 10.0g of carbon monoxide were placed in a reaction vessel. What is the theoretical yield of acetic acid. If the actual yield is 19.1g, what is the percent yield?

34. Stoichiometry Exercise 3.19 Convert grams of each reactant to moles of acetic acid: 15.0g CH 3 OH x 1 mol CH 3 OH x 1 mol HC 2 H 3 O 2 = 0.468mol HC 2 H 3 O 2 32.04g CH 3 OH 1 mol CH3OH 10.0g CO x 1 mol CO x 1 mol HC 2 H 3 O 2 = 0.357mol HC 2 H 3 O 2 (limiting) 28.01g CO 1 mol CO Thus, CO is the limiting reactant, and 0.0357 mol HC 2 H 3 O 2 is obtained. The mass of product is 0.357mol HC 2 H 3 O 2 x 60.05g HC 2 H 3 O 2 = 21.44g HC 2 H 3 O 2 (theoretical) 1 mol HC 2 H 3 O 2 The percentage yield is 19.1g x 100% = 89.08 = 89.1% 21.44g

35. Stoichiometry CH 2 Cl 2 is prepared by reaction of CH 4 with Cl 2 giving HCl. How many grams of CH 2 Cl 2 result from the reaction of 1.85 kg of CH 4 if the yield is 43.1%? CH 4(g) + 2 Cl 2(g)  CH 2 Cl 2(l) + 2 HCl(g) 1.85kg CH 4 x 1000 g x 1 mol CH 4 x 1 mol CH 2 Cl 2 = 115.62 mol CH 2 Cl 2 1 kg 16 g 1 mol CH 4 115.62 mol CH2Cl2 x 84 g = 9712.5 g or 9.71kg CH2Cl2 produced 1 mol 43.1% =

36. Stoichiometry K 2 PtCl 4 + 2 NH 3  Pt(NH 3 ) 2 Cl 2 + 2 KCl If 10.0 g of K 2 PtCl 4 and 10.0 g of NH 3 are allowed to react: a) which is limiting reagent? b) how many grams of the excess reagent are consumed? c) how many grams of cisplatin are formed? K 2 PtCl 4 = 415.08 g/mol NH 3 = 17.04 g/mol Pt(NH 3 ) 2 Cl 2 = 300.06 g/mol

37. Percent Composition What is glucoses, C 6 H 12 O 6, empirical formula, and what is the percentage composition of glucose?

38. Percent Composition Saccharin has the molecular formula C 7 H 5 NO 3 S, what is its empirical formula and the percent composition?

39. Empirical Formula A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. The molar mass is 58.11 g/mol. –What is the empirical formula and molecular formula?