# The Mole IT IS JUST A STANDARD NUMBER OF ATOMS/MOLECULES

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1 The Mole S 6.02 X IT IS JUST A STANDARD NUMBER OF ATOMS/MOLECULES Relative Masses To understand relative scales, let s s ignore electrons and compare atoms by total number of nuclear particles. Hydrogen, H 1 particle Helium, He Lithium, Li Carbon, C Oxygen, O 4 particles 7 particles 12 particles 16 particles 1

2 Atomic Mass The atomic mass of an element is the mass of an atom of the element compared with the mass of an atom of carbon taken as 12 atomic mass units (u). Hydrogen, H = 1.0 u Atomic masses of a few elements: Carbon, C = 12.0 u Helium, He = 4.0 u Nitrogen, N = 14.0 u Lithium, Li = 7.0 u Neon, Ne = 20.0 u Beryllium, Be = 9.0 u Copper, Cu = 64.0 u Molecular Mass The molecular mass M is the sum of the atomic masses of all the atoms making up the molecule. Consider Carbon Dioxide (CO 2 ) The molecule has one carbon atom and two oxygen atoms 1 C = 1 x 12 u = 12 u 2 O = 2 x 16 u = 32 u CO 2 = 44 u 2

3 Definition of a Mole One mole is that quantity of any substance that contains the same number of particles as there are in 12 g of carbon-12. (6.022 x particles) 1 mole of Carbon has a mass of 12 g 1 mole of Helium has a mass of 4 g 1 mole of Neon has a mass of 20 g 1 mole of Hydrogen (H 2 ) = = 2 g 1 mole of Oxygen (O 2 ) is = 32 g A Mole of Particles Contains 6.02 x particles 1 mole C 1 mole H 2 O 1 mole NaCl = 6.02 x C atoms = 6.02 x H 2 O molecules = 6.02 x NaCl molecules (technically, ionics are compounds not molecules so they are called formula units) 6.02 x Na + ions and 6.02 x Cl ions 3

4 Just How Big is a Mole? Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles. If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. The Mole 1 dozen cookies = 12 cookies 1 mole of cookies = 6.02 X cookies 1 dozen cars = 12 cars 1 mole of cars = 6.02 X cars 1 dozen Al atoms = 12 Al atoms 1 mole of Al atoms = 6.02 X atoms Note that the NUMBER is always the same, but the MASS is very different! Mole is abbreviated mol (gee, that s a lot quicker to write, huh?) 4

5 Molecular Mass in grams/mole The unit of molecular mass M is grams per mole. Hydrogen, H = 1.0 g/mol Helium, He = 4.0 g/mol H 2 O 2 = 2.0 g/mol = 32.0 g/mol Carbon, C = 12.0 g/mol Oxygen, O = 16.0 g/mol H 2 O = 18.0 g/mol CO 2 = 44.0 g/mol Each mole has x molecules A Mole of Particles Suppose we invented a new collection unit called a crideroo. One crideroo contains 8 objects. 1. How many paper clips in 1 crideroo? a) 1 b) 4 c) 8 2. How many oranges in 2.0 crideroos? a) 4 b) 8 c) How many crideroos in 40 gummy bears? a) 5 b) 10 c) 20 5

6 Moles and # of Molecules Finding the number of moles n in a given number of N molecules: n = N N A Avogadro s number: N A = x particles/mol Example 2: How many moles of any gas will contain 20 x molecules? 23 N 20 x 10 molecules n = = n = N A x 10 molecules mol mol Moles and Molecular Mass M Finding the number of moles n in a given mass m of a substance: n = m M Molecular mass M is expressed in grams per mole. Example 3: How many moles are there in 200 g of oxygen gas O 2? (M = 32 g/mol) m 200 g n = = n = 6.25 mol M 32 g/mol 6

7 A Mole of Particles Contains 6.02 x particles 1 mole C 1 mole H 2 O 1 mole NaCl = 6.02 x C atoms = 6.02 x H 2 O molecules = 6.02 x NaCl molecules (technically, ionics are compounds not molecules so they are called formula units) 6.02 x Na + ions and 6.02 x Cl ions Avogadro s s Number as Conversion Factor 6.02 x particles 1 mole or 1 mole 6.02 x particles Note that a particle could be an atom OR a molecule! 7

8 1. Number of atoms in mole of Al a) 500 Al atoms b) 6.02 x Al atoms c) 3.01 x Al atoms 2.Number of moles of S in 1.8 x S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1 x mole S atoms Molar Mass The Mass of 1 mole (in grams) Equal to the numerical value of the average atomic mass (get from periodic table) 1 mole of C atoms = 12.0 g 1 mole of Mg atoms = 24.3 g 1 mole of Cu atoms = 63.5 g 8

9 Other Names Related to Molar Mass Molecular Mass/Molecular Weight: If you have a single molecule, mass is measured in amu s instead of grams. But, the molecular mass/weight is the same numerical value as 1 mole of molecules. Only the units are different. (This is the beauty of Avogadro s Number!) Formula Mass/Formula Weight: mass is measured in grams, just like with atoms but for compounds. But again, the numerical value is the same. Only the units are different. THE POINT: You may hear all of these terms which mean the SAME NUMBER just different units Find the molar mass (usually we round to the tenths place) A. 1 mole of Br atoms B. 1 mole of Sn atoms = 79.9 g/mole = g/mole 9

10 Molar Mass of Molecules and Compounds Mass in grams of 1 mole equal numerically to the sum of the atomic masses How much does 1 mole of CaCl 2 weigh 1 mole Ca x 40.1 g/mol = + 2 moles Cl x 35.5 g/mol = 40.1g 71.0 g g/mol CaCl 2 How much does 1 mole of N 2 O 4 = 92.0 g/mol A. Molar Mass of K 2 O = g/mole K 2 O B. Molar Mass of antacid Al(OH) 3 = g/mole Al(OH) 3 C. Molar Mass of water H 2 O = g/mole H 2 O 10

11 Prozac, C 17 H 18 F 3 NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. Find its molar mass g / mol Calculations with Molar Mass Grams molar mass Moles 11

12 The artificial sweetener aspartame (Nutra-Sweet) formula C 14 H 18 N 2 O 5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame? g/mol.764 moles Converting Moles and Grams Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al? g Al 12

13 Molar mass of Al 1 mole Al = g Al Setup 3.00 moles Al x g Al 1 mole Al Answer = g Al Calculations molar mass Avogadro s number Grams Moles particles Everything must go through Moles!!! 13

14 Atoms/Molecules and Grams How many atoms of Cu are present in 35.4 g of Cu? 35.4 g Cu 1 mol Cu 6.02 X atoms Cu g Cu 1 mol Cu = 3.35 X atoms Cu How many atoms of K are present in 78.4 g of K? 2.01 moles 1.21 X atoms Cu 14

15 Atoms/Molecules and Grams How much does one molecule of water weigh? 18.0 g H mol H20 1 mol H X 1023 atoms H20 = 2.99 X g for each molecule 15

16 What is the mass (in grams) of 1.20 X molecules of glucose (C 6 H 12 O 6 )? How many atoms of O are present in 78.1 g of oxygen? 78.1 g O 1 mol O X molecules O g O 1 mol O = 2.94 X molecules 16

17 RELATIVE MASS (ISOTOPE) HOW DO WE COME UP WITH THE MASS OF EACH ELEMENT ON THE PERIODIC TABLE? Average Atomic Masses (as shown on the periodic table: a Weighted average of all the Isotopes of an element. Grades are an example of weighted averages If you had a "C" on a 10 point paper And an "A" on a 100 point paper Would you want the teacher to do a simple average of "A" and "C" to make a "B"? The points for work act as a like weight on the periodic table. So the 100 point assignment counts more than the 10 point assignment in the averaging of the grade. A 99% isotope has more impact on the recognized atomic weight than the 1% isotope 17

18 With element masses, the % Abundance is the weight, in the weighted average. Example Carbon: 12 C, 13 C, 14 C Carbon Isotopes: Carbon-12 = Carbon-13 = Carbon-14 = 12 C 13 C 14 C % Abundance in Nature = 99% = 0.99% = 0.01% Carbon: 12 C, C, 13 C, 14 C C, 12 C = 12 x 0.99 = X + 13 C = 13 x = Y + 14 C = 14 x = Z Carbon Average Mass = X+Y+Z 18

19 RELATIVE MASS (ISOTOPE) The General Formula for Calculating Atomic Masses Is (mass of isotope 1)(% abundance/100) +(mass of isotope 2)(% abundance/100) +(mass of isotope 3)(% abundance/100) + keep going if there are more isotopes RELATIVE MASS (ISOTOPE) Calculate the Atomic Mass of Carbon Isotope Mass % abundance # #

20 RELATIVE MASS (ISOTOPE) (98.90 / 100) (1.10 / 100) = RELATIVE MASS (ISOTOPE) Isotope Mass % abundance # #

21 RELATIVE MASS (ISOTOPE) 10.0 (20.0/100) (80.0 / 100) = AMU BORON RELATIVE MASS (ISOTOPE) Calculate the Atomic Mass Isotope Mass % abundance # # # #

22 RELATIVE MASS (ISOTOPE) For Chromium (4.35/100) (83.79/100) (9.50/100) (2.36/100) AMU CHROMIUM RELATIVE MASS (ISOTOPE) Calculate the Atomic Mass Isotope Mass % abundance # % # % 22

23 RELATIVE MASS (ISOTOPE) For Gallium (60.11 /100) (39.89 /100) AMU GALLIUM PERCENT COMPOSITION Always Means By Mass A sample contains 26.0 grams of carbon, 6.5g of Hydrogen, and 17.5g of oxygen. What is the mass percentage (AKA PERCENT COMPOSITION) of each 23

24 PERCENT COMPOSITION % C: 26.0 g = 50.0g % H: 6.5 g = 50.0g % O: 17.5 g = 50.0g.52 X 100 = 52% C.13 X 100 = 13% H.35 X 100 = 35% O 100% PERCENT COMPOSITION A 11.6 gram sample of Iron reacts with 4.86g of oxygen to make FeO. What is the Percent composition of the compound? 24

25 PERCENT COMPOSITION = 16.5g % Fe: 11.6 g =.703 X 100 = 70.3% Fe 16.5g % O: 4.86 g =.295 X 100 = 29.5% O 16.5g 99.8% If you know two of the three you can find the other one. But if you Know all three, do all three to limit computational error. PERCENT COMPOSITION An 8.44g sample of Silver (Ag) reacts with Sulfur that weights 9.69g. What is the Percent composition of the compound? 25

26 PERCENT COMPOSITION % Ag: 8.44g =.466 X 100 = 46.6% Ag 18.13g % S: 9.69g S = 18.13g.534 X 100 = 53.4 % S 100% PERCENT COMPOSITION A compound has a formula C 3 H 6 O 2. What is the Percent Composition of the compound? use one mole to find molar mass 26

27 Percent Composition 3C = 3(12.01) = 36.03g 6H = 6(1.01) = 6.06g 2O = 2(16) = 32.00g Total 74.09g/mol MOLAR MASS % C: 36.0 g = 74.0g % H: 6.5 g = 74.0g % O: 32.0 g = 74.0g.486 X 100 = 48.6% C.081 X 100 = 8.1% H.432 X 100 = 43.2% O 99.9% Percent Composition Always Means By Mass A compound has a formula K 2 CrO 4. What is the Percent composition of the potassium chromate? Don t forget: use one mole to find molar mass 27

28 Percent Composition 2K = 2(39.1) = 78.2g Cr = 1(52.0) = 52.0g 4O = 4(16) = 64.0g % K: 78.2 g =.402 X 100 = 40.2% K 194.2g % Cr: 52.0 g =.267 X 100 = 26.7% Cr 194.2g % O: 64.0 g =.329 X 100 = 32.9% O 194.2g Total 194.2g/mol MOLAR MASS 99.8% EMPIRICAL VS. MOLECULAR Oreo Cookies can be combined in many ways but as long as we add in whole numbers (CAN T SPLIT AN ATOM) the ratio of cream to cookie wont change. THE SAME HAPPENS IN CHEMISTRY MASSES: WHITE = 1 g BLACK = 2 g We can weigh a substance, get the percentage of weights for each substance and Then determine the mole relationships and then the most basic formula. 28

29 MOLECULAR FORMULA B 16 W 16 MM 48 EMPIRICAL FORMULA B 1 W 1 MM 3 MOLECULAR FORMULA B 11 W 11 MM 33 EMPIRICAL FORMULA B 1 W 1 MM 3 29

30 MOLECULAR FORMULA B 8 W 8 MM 24 EMPIRICAL FORMULA B 1 W 1 MM 3 MOLECULAR FORMULA B 9 W 9 MM 27 EMPIRICAL FORMULA B 1 W 1 MM 3 30

31 MOLECULAR FORMULA B 18 W 9 MM 45 EMPIRICAL FORMULA B 2 W 1 MM 5 MOLECULAR FORMULA B 12 W 6 MM 30 EMPIRICAL FORMULA B 2 W 1 MM 5 31

32 MOLECULAR FORMULA B 10 W 5 MM 25 EMPIRICAL FORMULA B 2 W 1 MM 5 MOLECULAR FORMULA B 3 W 3 MM 9 EMPIRICAL FORMULA B 1 W 1 MM 3 32

33 MOLECULAR FORMULA B 9 W 3 MM 21 EMPIRICAL FORMULA B 3 W 1 MM 7 MOLECULAR FORMULA B 12 W 3 MM 27 EMPIRICAL FORMULA B 4 W 1 MM 9 33

34 MOLECULAR FORMULA B 8 W 2 MM 18 EMPIRICAL FORMULA B 4 W 1 MM 9 MOLECULAR FORMULA B 6 W 2 MM 14 EMPIRICAL FORMULA B 3 W 1 MM 7 34

35 MOLECULAR FORMULA B 4 W 2 MM 10 EMPIRICAL FORMULA B 2 W 1 MM 5 MOLECULAR FORMULA B 2 W 2 MM 6 EMPIRICAL FORMULA B 1 W 1 MM 3 35

36 MOLECULAR FORMULA B 1 W 1 MM 3 EMPIRICAL FORMULA B 1 W 1 MM 3 MOLECULAR FORMULA B 2 W 1 MM 5 EMPIRICAL FORMULA B 2 W 1 MM 5 36

37 MOLECULAR FORMULA B 5 W 1 MM 11 EMPIRICAL FORMULA B 5 W 1 MM 11 MOLECULAR FORMULA B 12 W 3 MM 27 EMPIRICAL FORMULA B 4 W 1 MM 9 37

38 MOLECULAR FORMULA B 10 W 10 MM 30 EMPIRICAL FORMULA B 1 W 1 MM 3 MOLECULAR FORMULA B 16 W 8 MM 40 EMPIRICAL FORMULA B 2 W 1 MM 5 38

39 MOLECULAR FORMULA B 24 W 8 MM 56 EMPIRICAL FORMULA B 3 W 1 MM 7 SO WHAT IF MOLECULES HAVE THE Molecular Formula BIG OREO Empirical Formula IS AN OREO Molecular Formula LITTLE OREO SAME REL WT, WE JUST HAVE A BIGGER STRUCTURE THAT APPEARS THE SAME 39

40 Common Decimals and how to turn them into whole numbers FROM DIANE 40

41 In chemistry, the empirical formula of a chemical compound is a simple expression of the relative numbers of each type of atom in it. An empirical formula makes no reference to isomerism, structure, or absolute number of atoms. The empirical formula is used as standard for most ionic compounds, such as CaCl2, and for macromolecules, such as SiO2. The term empirical refers to the process of elemental analysis, a technique of analytical chemistry used to determine the relative amounts of each element in a chemical compound. Empirical Formula Smallest ratio of atoms of the compound Actual (Molecular) H 2 O 2 (hydrogen peroxide) C 6 H 12 O 6 (glucose) C 4 H 8 O 2 (butanoic acid) C 8 H 10 N 4 O 2 H 2 O C 12 H 22 O 11 (sucrose=sugar) C 12 H 22 O 11 (NCM) Empirical HO CH 2 O C 2 H 4 O C 4 H 5 N 2 O H2O (NCM) 41

42 PERCENT COMPOSITION Law of Definite Proportions: You can have five thousand pounds or five pounds the percent composition will not change. You can assume any sample size so assume you have a mole of the substance. Hexane's molecular formula is C6H14, and its empirical formula would be C3H7 showing a C:H ratio of 3:7. C6H14, and its empirical formula would be C3H7 if the empirical formula of a compound is C3H8, its molecular formula may be C3H8, C6H16, etc. When teaching the method for converting percentage composition to an empirical formula, I have devised the follo Percent to mass Mass to mole Divide by small Multiply 'til whole 42

43 If I told you that a compounds percent composition is C 40.0%, 6.7 %, 53.3 O% what is the compound s empirical formula? The percent composition is only for mass. Chemical formulas are in moles of atoms. If I told you that a compounds percent composition is C 40.0%, 6.7 H %, 53.3 O% what is the compound s empirical formula? The percent composition is only for mass. Chemical formulas are in moles of atoms. 43

44 Empirical Formula 40.0% x 100g = 40.0g C 6.7% x 100g = 6.7 g H 53.3% x 100g = 53.3g O mol C: 40.0g C = 3.33 mol C 12.0g/mol C mol H: 6.7g H = 6.66 mol H 1.007g/mol H mol O: 53.3g O = 3.33 mol O 16.0g/mol O Greatest common multiple 3.33 C 1 H 2 O 1 A substance has the following composition by mass: % Na ; % B ; % H Find the empirical formula. 44

45 Empirical Formula 60.80% x 100g = g Na 28.60% x 100g = g B % x 100g = g H mol Na: 60.80g Na = mol Na g/mol Na mol B: g B = mol B g/mol B mol H: H = mol H 1.007g/mol H Na 1 B 1 H 4 Greatest common multiple A compounds mass is 54.1% Ca, 2.70% H, and 40.8% O. What is the compound s empirical formula? 45

46 Empirical Formula 54.1 g Ca 2.70 g H 40.8 g O mol Ca: 54.1 g Ca = mol Ca g/mol Ca mol H: 2.7 g H = 2.68 mol H 1.007g/mol H mol O: 40.8 g O = 2.55 mol O 16.00g/mol O Greatest common multiple = = 2 If we know Empirical Formula (relative amounts of each by weight and Experimental molar mass (how much that formula weighs) we can find the molecular formula The empirical formula of a substance simply gives the most basic ratio of the combination of the constituent elements. Experimental MM = Ratio to Increase EF Empirical MM Obviously, an empirical formula does not contain as much information as does a molecular formula. 46

47 a. A compound with a molecular mass of 70µ and an empirical formula of CH2. b. A compound with a molecular mass of 46.0µ and an empirical formula of NO2. A sample has empirical formula of CH 2 O (Remember from previous problem) and a Molecular Mass (Experimental) = g. What is the molecular formula? Molecular Weight = g = Empirical Weight 30.03g C 4 H 8 O4 47

48 We have determined the percentage composition of benzene: 92.3% C and 7.7% H. What is the empirical formula of benzene? Empirical Formula 92.3% x 100 = 92.3 g C 7.7% x 100 = 7.7 g H mol C: 92.3g C = 7.69 mol C g/mol C mol H: 7.7g H = 7.65 mol H 1.007g/mol H Least common multiple 7.65 C H 48

49 Benzene has empirical formula of CH and a Experimental Weight = 78 g. What is the molecular formula? Molecular Weight = 78 = 6.0 Empirical Weight C 6 H 6 A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and g H. What is the empirical formula of the compound? 49

50 50

51 NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O. What is the compound s empirical formula? Empirical Formula % x 100g = g C 6.16 % x 100g = 6.16 g H 9.52 % x 100g = 9.52 g N % x 100g = g O mol C: g C = mol C 12.01g/mol C mol H: 6.16 g H = 6.11 mol H 1.007g/mol H mol N: 9.52 g N = 0.68 mol N 14.01g/mol N mol O: g O = mol O 16.00g/mol C Greatest common multiple =

52 A sample has empirical formula of C 14 H 18 N 2 O 5 and a Molecular Weight = g. What is the molecular formula? Experimental MM = g = Empirical MM C 14 H 18 N 2 O5 Write the Empirical Formula for Each of the Following: a. P4O6 a. P2O3 b. C6H9 b. C2H3 c. CH2OHCH2OH c. CH3O d. BrCl2 d. BrCl2 e. C6H8O6 e. C3H4O3 f. C10H22 f. C5H11 g. Cu2C2O4 g. CuCO2 h. Hg2F2 h. HgF 52

53 The formulas for Ionic compounds are empirical formulas. A molecular formula is "a whole number multiple of an empirical formula."3 Converting to Empirical Formula from Molecular Formulas is slightly more difficult than vice versa. Hydrogen Peroxide Step 1: Molecular Formula = H 2 O 2 Step 2: All subscripts are divisible by 2. Step 3: Empirical Formula = HO Glucose Step 1: Molecular Formula = C 6 H 12 O 6 Step 2: All subscripts are divisible by 6. Step 3: Empirical Formula = CH 2 O Acetic Acid Step 1: Molecular Formula = C 2 H 3 COOH -- Step 1b: Standard Molecular Formula = C2H4O2 Step 2: All subscripts are divisible by 2. Step 3: Empirical Formula = CH2O 53

54 COMBINATION % COMPOSITION, EMPIRCAL FORMULA AND MOLECULAR FORMULA A 9.2 gram sample of a compound is 2.8 grams nitrogen and 6.4 grams oxygen. 1. Find the percent composition of the compound 2. Find the Empirical Formula of the compound 3. Find Molecular Formula if molecular (experimental) weight is 92g. 54

55 PERCENT COMPOSITION % N: 2.8 g =.304 X 100 = 30.4%=30.% N 9.2 g % O: 6.4 g =.696 X 100 = 69.6%= 70.% O 9.2 g 100% Empirical Formula _30.4_ % x 100g = _30.4_ g N _69.6_% x 100g = 69.6_ g O mol N: 30.4 g N = 2.17 mol N 14.01g/mol N mol O: 69.6 g O = mol O g/mol O Greatest common multiple 2.17 Molar Mass _46g _ 1 2 NO 2 92 / 46 = 2.0 = N 2 O 4 55

56 A compound containing 74.0% carbon (C), 8.65% hydrogen (H), and 17.3% nitrogen (N) by mass. 4. Determine the Empirical Formula (4) 5. Find the Empirical Molar Mass (3) 6. Find Molecular Formula if molecular (2) (experimental) mass is g mol C: 74.0 g C Empirical Formula 12.01g/mol C mol H: 8.65 g H 1.01g/mol H mol N: 17.3 g O 14.01g/mol O Greatest common multiple 1.24 Molar Mass % x 100g = 74.0 g C 8.65% x 100g = 8.65 g H 17.3% x 100g = 17.3 g N = 6.16 mol C = 8.56 mol H = 1.24 mol N C 5 H 7 N /81.13 = 6.0 C 30 H 42 N

57 A compound has an empirical formula of CH 2 and a molecular mass of 42g = / = 3.0 C 3 H 6 57

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