# Ch. 10 The Mole I. Molar Conversions

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1 Ch. 10 The Mole I. Molar Conversions I II III IV

2 A. What is the Mole? A counting number (like a dozen) Avogadro s number (N A ) 1 mole = representative particles

3 B. Mole/Particle Conversions MOLES N A (particles/mol) N A atoms/mol N A molecules/mol NUMBER OF PARTICLES Particles = atoms, molecules, formula units, ions, etc

4 B. Mole/Particle Conversion Examples How many molecules are in 2.50 moles of C 12 H 22 O 11? 2.50 mol molecules C 12 H 22 O 11 C 12 H 22 O 11 1 mol C 12 H 22 O 11 = molecules C 12 H 22 O 11

5 B. Mole/Particle Conversion Examples If you have 2.23 x atoms of sodium, how many moles is that? atoms Na 1 mole Na atoms Na = 3.70 x 10-6 moles Na

6 B. Mole/Particle Conversion Examples How many formula units is 3.75 moles of calcium hydroxide? 3.75 mol Ca(OH) formula units Ca(OH) 2 1 mol Ca(OH) 2 = formula units Ca(OH) 2

7 C. Molar Mass Avogadro discovered the relationship between number of particles and volume of a gas This was used to find the relationship for particles in a mole

8 Representative Particles & Moles Substance Chemical Formula Representative Particle Rep Particles in 1.00 mole Carbon C Atom 6.02 x Nitrogen gas N 2 Molecule 6.02 x Calcium ion Ca 2+ Ion 6.02 x Magnesium fluoride MgF 2 Formula unit 6.02 x 10 23

9 C. Molar Mass Mass of 1 mole of an element or compound Atomic mass (on the PT) tells the... mass of each atom (amu) grams per mole (g/mol)

10 C. Molar Mass Examples carbon g/mol aluminum g/mol zinc g/mol

11 C. Molar Mass Examples water H 2 O 2(1.01) sodium chloride NaCl = g/mol = g/mol

12 C. Molar Mass Examples sodium bicarbonate NaHCO (15.99) sucrose = g/mol C 12 H 22 O 11 12(12.01) + 22(1.01) + 11(15.99) = g/mol

13 D. Molar Conversions molar mass N A MASS IN MOLES NUMBER OF GRAMS (g/mol) (particles/mol) PARTICLES N A atoms/mol Particles = atoms, molecules, formula units, ions, etc N A molecules/mol

14 D. Molar Conversion Examples How many moles of carbon atoms are in 26 g of carbon? 26 g C 1 mol C g C = 2.2 mol C

15 D. Molar Conversion Examples Find the mass of molecules of NaHCO Molecules NaHCO 3 1 mol NaHCO Molecules g NaHCO 3 1 mol NaHCO 3 NaHCO 3 = 290 g NaHCO 3

16 D. Molar Conversion Examples How many atoms are in 22.5 grams of potassium? 22.5 g K 1 mol K g K atoms K 1 mol K = 3.46 x atoms K

17 Mole Ratios The ratio of the moles of one molecule to the atoms within that molecule. molecular formulas give atom-to-atom and moleto-mole ratios example: molecular formula C 6 H 12 O 6 atom-to-atom ratios atom-to-molecule ratios 6 atoms C 12 atoms H 6 atoms C 6 atoms O 12 atoms H 6 atoms O 6 atoms C 1 molecule 12 atoms H 1 molecule 6 atoms O 1 molecule

18 Mole Ratios Example: molecular formula C 6 H 12 O 6 mole-to-mole ratios (elements) 6 mol C 12 mol H 6 mol C 6 mol O 12 mol H 6 mol O mole-to-mole ratios (compound) 6 mol C 1 mol C 6 H 12 O 6 12 mol H 1 mol C 6 H 12 O 6 6 mol O 1 mol C 6 H 12 O 6 problems that ask you to relate one substance to another require mole-to-mole ratios

19 Ch. 10 The Mole II. Formula Calculations I II III IV

20 A. Percent Composition the percentage by mass of each element in a compound % mass of element total mass of element total mass of compound 100 % mass of element mass of element in 1 mol compound molar mass of compound 100

21 A. Percent Composition Find the percent composition of a sample that is 28 g Fe and 8.0 g O. Known: Unknown: Mass of Fe = 28 g % Fe =? Mass of O = 8.0 g % O =? Total Mass = g = 36 g %Fe = %O = 28 g 36 g 8.0 g 36 g 100 = 78% Fe 100 = 22% O Check: 78% + 22% = 100%

22 A. Percent Composition Find the % composition of Cu 2 S. Use this formula when finding % composition from a chemical formula: mass of element in 1 mol %Cu = % mass of element g Cu g Cu 2 S 100 = compound molar mass of compound 100 Known: Unknown: Mass of Cu in 1 mol Cu 2 S = % Cu =? 2(63.55g) = g Cu % S =? Mass of S in 1 mol Cu 2 S = g S Molar Mass = g g = g/mol 79.85% Cu %S = g S g Cu 2 S 100 = 20.15% S

23 A. Percent Composition How many grams of copper are in a 38.0-gram sample of Cu 2 S? Use answer from last question as a conversion factor. Cu 2 S is 79.85% Cu = g Cu 100 g Cu 2 S 38.0 g Cu 2 S g Cu 100 g Cu 2 S = 30.3 g Cu

24 A. Percent Composition Find the percent composition of Cu 2 SO 4. Known: Mass of Cu in 1 mol Cu 2 SO 4 = 2(63.55 g) = g Cu Mass of S in 1 mol Cu 2 SO 4 = g S Mass of 0 in 1 mol Cu 2 SO 4 = 4(16.00 g) = g O Molar Mass = g g g = g/mol Unknown: % Cu =? % S =? % O =?

25 A. Percent Composition Find the percent composition of Cu 2 SO 4. %Cu = g g %S = %O = 100 = 56.95% Cu g g 100 = 14.37% S g g 100 = 28.68% O Check: 56.95% % % = 100%

26 B. Empirical Formula Smallest whole number ratio of atoms in a compound C 2 H 6 reduce subscripts CH 3

27 B. Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole # s.

28 EX 1 Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O g N 1 mol N g N = 1.85 mol N 1.85 mol = 1 N 74.1 g O 1 mol O g O N 1.85 O 4.63 = 4.63 mol O 1.85 mol = 2.5 O

29 EX 1 Empirical Formula N 1 O 2.5 Need to make the subscripts whole numbers multiply by 2 N 2 O 5

30 Mole Ratios that do not end up in whole numbers Some mole ratios that are close enough to a whole number to round off. (1.04 1; ) Other ratios are too far from a whole number to round. ( ) Example: It was found that iron and oxygen combine in the following molar ratios: Fe O Fe 1 O (Fe 1 O 1.49 ) » = Fe 2 O 3

31 Mole Ratios that do not end up in whole numbers Since we cannot have "fractional" atoms in a compound, we need to normalize the relative amount of oxygen to be equal to a whole number would appear to be 1 and 1/2, so if we multiply the relative amounts of each atom by '2', we should be able to get whole number values for each atom. A multiple can be used to convert to a whole number. (1.25 x 4 = 5 ; 1.33 x 3 = 4 ; 1.50 x 2 = 3)

32 EX 2 Empirical Formula Find the empirical formula for a sample of 94.1% O and 5.9% H g O 1 mol O g O = 5.88 mol O 5.84 mol = 1 O 5.9 g H 1 mol H 1.01 g H = 5.84 mol H 5.84 mol = 1 H EF = OH

33 Molecular Formulas True Formula - the actual number of atoms in a compound, either the same as or a whole-number multiple of the empirical formula empirical formula CH 3? molecular formula C 2 H 6

34 Steps to a Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molar mass by the empirical formula mass. 4. Multiply each subscript in your EF by the answer from step 3. molar mass EF mass EF n n

35 EX 1 Molecular Formula The empirical formula for ethylene is CH 2. Find the molecular formula if the molar mass is 28.1 g/mol? empirical formula mass = g/mol n = 28.1 g/mol g/mol = 2.00 (EF) n = (CH 2 ) 2 C 2 H 4

36 Put it all together!! 1,6 - Diaminohexane is 62.1% C, 13.8% H, and 24.1% N. What is the empirical formula? If the molar mass is g/mol, what is the molecular formula? 62.1 g C 1 mol C g C 13.8 g H 1 mol H 1.01 g H = 5.17 mol C 1.72 mol = 13.7 mol H 1.72 mol = 3 C = 8 H 24.1 g N 1 mol N g N = 1.72 mol N = 1 N 1.72 mol EF = C 3 H 8 N

37 Put it all together!! C 3 H 8 N empirical formula mass = g/mol n = g/mol g/mol = 2.00 (C 3 H 8 N) 2 C 6 H 16 N 2

38 Combustion Analysis When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO 2 and all the hydrogen to H 2 O. The amount of carbon produced can be determined by measuring the amount of CO 2 produced. The amount of H produced by the amount of H 2 O produced.

39 Example 1 Consider the combustion of isopropyl alcohol. The sample is known to contain only C, H and O. The mass of C x H y O z is the total mass of the C + H + O. Combustion of grams of isopropyl alcohol produces grams of CO 2 and grams of H 2 O. What is the empirical formula of isopropyl alcohol? The reaction looks like: C x H y O z + O 2 CO 2 + H 2 O

40 Determine the amount of C.561grams CO 2 1 mole CO g CO 2 1 mole C 1 mole CO 2 =.0128 mole C g C =.154 g C 1 mole C

41 Determine the amount of H.306 grams H 2 O 1 mole H 2 O g H 2 O 2 mole H 1 mole H 2 O =.034 mole H g H =.034 g H 1 mole H

42 If we have.0128 moles of CO 2 in our sample, then we know we have moles of C in the sample. How many grams of C is this? grams C (as done above). If we have moles of H 2 O, then we have 2 x (0.017) = moles of hydrogen. Since hydrogen is about gram/mole, we must have grams of hydrogen in our original sample. When we add our carbon and hydrogen together we get: grams (C) grams (H) = g

43 What about oxygen? We know we combusted grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol: g g = grams oxygen =.067 grams O 1 mole O =.0042 mol O g O

44 Find the Empirical Formula Overall therefore, we have: C H O C 3.05 H 8.1 O 1.0 C 3 H 8 O Within experimental error, the most likely empirical formula for propanol would be:»c 3 H 8 O

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