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1 The Mole 6.02 X 10 23 Chemistry I Mr. Trotts Summer 2008.

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Presentation on theme: "1 The Mole 6.02 X 10 23 Chemistry I Mr. Trotts Summer 2008."— Presentation transcript:

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2 1 The Mole 6.02 X 10 23 Chemistry I Mr. Trotts Summer 2008

3 2 What we’re going to Learn 1.The concept of the mole 2.Given moles calculate number of particles 3.Be able to calculate molecular mass and gram formula mass and weight 4.Be able to convert moles to mass, particles, and volume and convert mass to moles, volume, and particles

4 3 5.Calculate the percent composition of a compound by mass of each element. and 6.Calculate both the empirical (simplest) formula and molecular formula of a compound. NOTE: As we encounter Learning Checks write the answers down for a daily grade

5 4 STOICHIOMETRYSTOICHIOMETRY -the study of the quantitative aspects of chemical reactions. -Combines mole problems with balanced equations

6 5 The Mole A counting unit Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000 6.02 X 10 23 (in scientific notation) Amedeo Avagadro (1776 – 1856)This number is named in honor of Amedeo Avagadro (1776 – 1856), who studied quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present

7 6 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.

8 7 Everybody Has Avogadro’s Number! But Where Did it Come From? It was NOT just picked! It was MEASURED. One of the better methods of measuring this number was the Millikan Oil Drop Experiment Since then we have found even better ways of measuring using x- ray technology

9 8 Suppose we invented a new collection unit called a rapp. One rapp contains 8 objects. 1. How many paper clips in 1 rapp? a) 1b) 4c) 8 2. How many oranges in 2.0 rapp? a) 4b) 8c) 16 3. How many rapps contain 40 gummy bears? a) 5b) 10c) 20 Learning Check

10 9 Suppose we invented a new collection unit called a mep. One mep contains 8 objects. A. How many paper clips in 1 mep? 3) 8 B. How many oranges in 2.0 meps? 3) 16 C. How many meps contain 40 gummy bears? 1) 5 Solution

11 10 The Mole 1 dozen cookies = 12 cookies 1 mole of cookies = 6.02 X 10 23 cookies 1 dozen cars = 12 cars 1 mole of cars = 6.02 X 10 23 cars 1 dozen Al atoms = 12 Al atoms 1 mole of Al atoms = 6.02 X 10 23 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?)

12 11 = 6.02 x 10 23 C atoms = 6.02 x 10 23 H 2 O molecules = 6.02 x 10 23 NaCl “molecules” (technically, ionics are compounds not molecules so they are called formula units) 6.02 x 10 23 Na + ions and 6.02 x 10 23 Cl – ions A Mole of Particles A Mole of Particles Contains 6.02 x 10 23 particles 1 mole C 1 mole H 2 O 1 mole NaCl

13 12 6.02 x 10 23 particles 1 mole or 1 mole 6.02 x 10 23 particles Note that a particle could be an atom OR a molecule! Avogadro’s Number as Conversion Factor

14 13 Example: How many moles of Calcium are in 2.408 x 10 24 calcium atoms? 2.408 x 10 24 Ca atoms x 1 mole Ca/6.02 x 10 23 atoms Ca = 4.0 x 10 1 moles Ca Example: How many atoms of Calcium are in 40 moles of Calcium 40 moles Ca x 6.02 x 10 23 atoms Ca/1 mole Ca = 2.408 x 10 24 atoms Ca

15 14 4. Number of atoms in 0.500 mole of Al a) 500 Al atoms b) 6.02 x 10 23 Al atoms c) 3.01 x 10 23 Al atoms 5.Number of moles of S in 1.8 x 10 24 S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1 x 10 48 mole S atoms Learning Check

16 15 1. Number of atoms in 0.500 mol of Al 3) 3.01 x 10 23 Al atoms 0.500 mol Al x 6.02 x 10 23 Al atoms 1 mol Al 2. Number of moles of S if a sample of S contains 4.50 x 10 24 S atoms 2) 3.0 mole S atoms 4.50 x 10 24 S atoms x 1 mol S 6.02 x 10 23 S atoms Solution

17 16 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 Molar Mass

18 17 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: Same goes 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

19 18 Find the molar mass (usually we round to the tenths place) Learning Check! 6. 1 mole of Br atoms 7. 1 mole of Sn atoms =79.9 g/mole = 118.7 g/mole

20 19 Mass in grams of 1 mole equal numerically to the sum of the atomic masses 1 mole of CaCl 2 = 111.1 g/mol 1 mole Ca x 40.1 g/mol + 2 moles Cl x 35.5 g/mol = 111.1 g/mol CaCl 2 1 mole of N 2 O 4 = 92.0 g/mol Molar Mass of Molecules and Compounds

21 20 8. Molar Mass of K 2 O = ? Grams/mole 9. Molar Mass of antacid Al(OH) 3 = ? Grams/mole Learning Check!

22 21 A. 1 mole of K 2 O= 94.2 g 2 K x 39.1 g/mole + 1 O x 16.0 g/mole B. 1 mole of antacid Al(OH) 3 = 78.0 g 1 Al x 27.0 g/mole + 3 O x 16.0 g/mole + 3 x 1.0 Solution

23 22 10. 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. Complete the handout GRAM FORMULA MASS Learning Check

24 23 Prozac, C 17 H 18 F 3 NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. It has a molar mass of 309 g/mole 17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) Solution

25 24 molar mass Grams Moles Calculations with Molar Mass

26 25 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 Converting Moles and Grams

27 26 1. Molar mass of Al1 mole Al = 27.0 g Al 2. Conversion factors for Al 27.0g Al or 1 mol Al 1 mol Al 27.0 g Al 3. Setup3.00 moles Al x 27.0 g Al 1 mole Al Answer = 81.0 g Al

28 27 11. 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? Learning Check!

29 28 Atoms/Molecules and Grams Since 6.02 X 10 23 particles = 1 mole AND 1 mole = molar mass (grams) You can convert atoms/molecules to moles and then moles to grams! (Two step process) You can’t go directly from atoms to grams!!!! You MUST go thru MOLES. That’s like asking 2 dozen cookies weigh how many ounces if 1 cookie weighs 4 oz? You have to convert to dozen first!

30 29 molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!! Calculations

31 30 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 10 23 atoms Cu 63.5 g Cu 1 mol Cu = 3.4 X 10 23 atoms Cu

32 31 Methane CH 4 known as natural gas is used in gas cook tops and gas heaters. Express the molar mass of methane in the form of conversion factors. Molar mass of CH 4 = 16.0 g 16.0 g CH 4 and 1 mole CH 4 1 mole CH 4 16.0 g CH 4 Molar Mass Factors

33 32 Methane CH 4 known as natural gas is used in gas cook tops and gas heaters. Express the molar mass of methane in the form of conversion factors. Molar mass of CH 4 = 16.0 g 16.0 g CH 4 and 1 mole CH 4 1 mole CH 4 16.0 g CH 4 Molar Mass Factors

34 33 12. Acetic acid CH 3 COOH is the acid in vinegar. It has a molar mass of 60.0 g/mole. 1 mole of acetic acid = ____________ 1 mole acetic acid or g acetic acid g acetic acid 1 mole acetic acid Learning Check

35 34 Acetic acid CH 3 COOH is the acid in vinegar. It has a molar mass of 60.0 g/mole. 1 mole of acetic acid = 60.0 g acetic acid 1 mole acetic acid or 60.0 g acetic acid 60.0 g acetic acid1 mole acetic acid Solution

36 35 Learning Check! 13. How many atoms of K are present in 78.4 g of K?

37 36 Learning Check! 14. What is the mass (in grams) of 1.20 X 10 24 molecules of glucose (C 6 H 12 O 6 )?

38 37 More mole equivalents 1 mole of a gas at standard temperature (0 Celsius) and pressure (1 atmosphere) occupies 22.4 liters by volume Example: 2 moles of oxygen at STP would occupy 2 moles x 22.4L/1 mole or 44.8L

39 38 Learning Check! 15. How many atoms of O are present in 78.1 g of oxygen? 78.1 g O 2 1 mol O 2 6.02 X 10 23 molecules O 2 2 atoms O 32.0 g O 2 1 mol O 2 1 molecule O 2 Complete the handout Mixed Mole Problems Given 1 mole = 6.02 x 10 23 particles = gfm = 22.4L @ STP Complete Atoms Worksheet #2 (The Mole)

40 39 16. What is the percent carbon in C 5 H 8 NO 4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? a) 8.22 %C b) 24.3 %C c) 41.1 %C Percent Composition Complete Percentage Composition Worksheet

41 40 Chemical Formulas of Compounds Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions).Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions). NO 2 2 atoms of O for every 1 atom of N NO 2 2 atoms of O for every 1 atom of N 1 mole of NO 2 : 2 moles of O atoms to every 1 mole of N atoms If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

42 41 Types of Formulas Empirical FormulaEmpirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. The Ionic formula is always the empirical formula Molecular FormulaMolecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

43 42 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

44 43 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:

45 44 Empirical Formula from % Composition 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 28.60 grams of B and 10.60 grams H Determine the number of moles of each Determine the simplest whole number ratio

46 45 Empirical Formulas from Percent Composition Given a compound that is 32.4% Na, 22.5% S, and 45.1% O 32.4gNa x 1 mole/23.0g = 1.41 mole Na 22.5gS x 1 mole/32g =.70 mole S 45.1gO x 1 mole/16g = 2.82 mole O Na = 1.41/.70 = 2 S =.70/.70 = 1 O = 2.82/.70 = 4 therefore = Na 2 SO 4 Complete the Determining Empirical Formula worksheet

47 46 Calculation of the Molecular Formula 17. 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? N = 14.0g/mole O 2 = 2 x 16 g/mole = 32g NO 2 = 46g 92.0g/46g = 2 2 x (NO 2 ) = N 2 O 4 Complete Determining Molecular Formulas Worksheet


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