Chapter 10 The Molar Mass Mole Relationships in

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 Chapter 10 The Molar Mass Mole Relationships in Chemical Equations Mass Calculations for

Chapter 10 The Molar Mass Mole Relationships in Chemical Equations Mass Calculations for Reactions Molar Volume Percentage Composition Empirical and Molecular Formulas 1

THE MOLE CONCEPT n n n A sensitive balance can weight to the nearest

THE MOLE CONCEPT n n n A sensitive balance can weight to the nearest 0. 0001 g, but a typical atom has a mass of only 0. 000 000 01 g Is it possible to keep track of atoms by counting them? Yes, but not directly. We count them in groups. That is we count atoms in the same way we count eggs by the dozen (12), pencils by the gross (144), pop by the case (24), and sheets of paper by the ream (500). The relative mass of an atom is compared to a reference atom using an instrument called a mass spectrometer. Carbon-12 has been chosen as a reference and it is assigned a mass of exactly 12 amu. The masses of other atoms are compared relative to carbon-12.

Experiments have shown that 12. 01 g of carbon contains 6. 02 x 1023

Experiments have shown that 12. 01 g of carbon contains 6. 02 x 1023 atoms of carbon. n In fact, the gram atomic mass of each element contains 6. 02 x 1023 atoms of that element. AVOGADRO’S Number: n Number of atoms in 12. 01 grams of carbon n 6. 02 x 1023 atoms n

MOLE: n n n Amount of a substance containing 6. 02 x 1023 representative

MOLE: n n n Amount of a substance containing 6. 02 x 1023 representative particles Why do we need this tool (the mole)? Chemists and many others need to know relative amounts of substances.

A Moles of Particles Contains 6. 02 x 1023 particles 1 mole C 1

A Moles of Particles Contains 6. 02 x 1023 particles 1 mole C 1 mole H 2 O = 6. 02 x 1023 C atoms = 6. 02 x 1023 H 2 O molecules 1 mole Na. Cl = 6. 02 x 1023 Na+ ions and 6. 02 x 1023 Cl– ions 8

Examples of Moles of elements 1 mole Mg = 6. 02 x 1023 Mg

Examples of Moles of elements 1 mole Mg = 6. 02 x 1023 Mg atoms 1 mole Au = 6. 02 x 1023 Au atoms Moles of compounds 1 mole NH 3 = 6. 02 x 1023 NH 3 molecules 1 mole C 9 H 8 O 4 = 6. 02 x 1023 aspirin molecules 9

 Avogadro’s Number 6. 02 x 1023 particles 1 mole or 1 mole 6.

Avogadro’s Number 6. 02 x 1023 particles 1 mole or 1 mole 6. 02 x 1023 particles 10

MOLAR MASS: n n n n Mass in grams of one mole of any

MOLAR MASS: n n n n Mass in grams of one mole of any substance, element, or compound Has the same numerical value as formula mass. Molar mass of any element = the gram atomic mass (gam) of that element Example: 1 mole H = 1. 01 g H 1 mole Ca = 40. 1 g Ca 1 mole O = 16. 0 g O 1 mole C = 12. 0 g C

Molar Mass n Number of grams in 1 mole n Equal to the numerical

Molar Mass n Number of grams in 1 mole n Equal to the numerical value of the atomic mass 1 mole of C atoms = 12. 0 g 1 mole of Mg atoms = 24. 3 g 1 mole of Cu atoms = 63. 5 g 12

Learning Check Give the molar mass to 0. 1 g A. 1 mole of

Learning Check Give the molar mass to 0. 1 g A. 1 mole of Br atoms = ____ B. 1 mole of Sn atoms = ____ 13

Solution Give the molar mass to 0. 1 g A. 1 mole of Br

Solution Give the molar mass to 0. 1 g A. 1 mole of Br atoms = 79. 9 g/mole B. 1 mole of Sn atoms = 118. 7 g/mole 14

 Problems What is the molar mass for the following elements? 1. Na 2.

Problems What is the molar mass for the following elements? 1. Na 2. Cl 3. Fe 4. Hg

FORMULA MASS: (Molar Mass) n n n The sum of the atomic masses, for

FORMULA MASS: (Molar Mass) n n n The sum of the atomic masses, for each element, found in the compound Exp: Calculate the formula mass of H 2 O 2 hydrogen atoms @ 1. 008 amu 1 oxygen atom @ 15. 999 amu 2. 016+15. 9999= 18. 015 amu One molecule of water has a mass of 18. 015 amu

Example: H 2 S n 2 atoms H x atomic mass 1. 0 g

Example: H 2 S n 2 atoms H x atomic mass 1. 0 g = 2. 0 g/mole n 1 atom S x atomic mass 32. 1 g = 32. 1 g/mole (Molar mass) = 34. 1 g/mole 2(1. 0) + 32. 1 = 34. 1 g/mole 1 mole = 6. 02 x 1023 (how many) = molar mass (grams)

Molar Mass of Compounds Mass in grams of 1 mole equal numerically to the

Molar Mass of Compounds Mass in grams of 1 mole equal numerically to the sum of the atomic masses 1 mole of Ca. Cl 2 = 111. 1 g/mole 1 mole Ca x 40. 1 g/mole = 40. 1 g/mole + 2 moles Cl x 35. 5 g/mole = 71. 0 g/mole 1 mole of N 2 O 4 = 74. 0 g/mole 2 moles N x 14. 0 g/mole = 28. 0 g/mole + 4 moles O x 16. 0 g/mole = 64. 0 g/mole 18

Learning Check A. 1 mole of K 2 O = ______g B. 1 mole

Learning Check A. 1 mole of K 2 O = ______g B. 1 mole of antacid Al(OH)3 = ______g 19

Solution A. 1 mole of K 2 O = 94. 2 g/mole 2 atoms

Solution A. 1 mole of K 2 O = 94. 2 g/mole 2 atoms K x 39. 1 g/mole + 1 atom O x 16. 0 g/mole B. 1 mole of antacid Al(OH)3 = 78. 0 g/mole 1 atom Al x 27. 0 g/mole + 3 atoms O x 16. 0 g/mole + 3 atoms x 1. 0 g/mole 20

Learning Check Prozac, C 17 H 18 F 3 NO, is a widely used

Learning Check 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 1) 40. 0 g/mole 2) 262 g/mole 3) 309 g/mole 21

Solution Prozac, C 17 H 18 F 3 NO, is a widely used antidepressant

Solution 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 3) 309 g/mole 17 atoms C (12. 0) + 18 atoms H (1. 0) + 3 atoms F (19. 0) + 1 atom N (14. 0) + 1 atom O (16. 0) 22

Problems State how many moles and calculate the molar mass of each substance. 1.

Problems State how many moles and calculate the molar mass of each substance. 1. 2 Cu 2. 5. 00 Na. Cl 3. 2. 5 H 2 SO 4 23

Molar Mass Factors Methane CH 4 known as natural gas is used in gas

Molar Mass Factors 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 CH 4 and 1 mole CH 4 16. 0 g CH 4 24

Learning Check Acetic acid CH 3 COOH is the acid in vinegar. It has

Learning Check 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 1 mole acetic acid 25

Solution Acetic acid CH 3 COOH is the acid in vinegar. It has a

Solution 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 1 mole acetic acid 26

Problems 1. 2. n n 3. The formula of sodium hydrogen carbonate (baking soda)

Problems 1. 2. n n 3. The formula of sodium hydrogen carbonate (baking soda) is Na. HCO 3. What is its formula mass? Answer the following: Write the formula for Magnesium Acetate. Determine its formula mass. Determine the molar mass for the following: a) Na. Cl b) Ca. CO 3 c) H 2 SO 4

Calculations with Molar Mass molar mass Grams 1 mole = molar mass Moles 28

Calculations with Molar Mass molar mass Grams 1 mole = molar mass Moles 28

Moles and Grams Aluminum is often used for the structure of light-weight bicycle frames.

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 29

Known: n Given: 3. 00 Al n Molar mass of Al 1 mole Al

Known: n Given: 3. 00 Al n Molar mass of Al 1 mole Al = 27. 0 g Al n Conversion factors for Al 27. 0 g Al or 1 mol Al 27. 0 g Al 30

Unknown: n ? g Al M(mass) = n(moles) x molar mass or n(moles) =

Unknown: n ? g Al M(mass) = n(moles) x molar mass or n(moles) = M(mass) / molar mass n Setup: M(mass) = n(moles) x molar mass 31

Solution 1. Molar mass of Al 1 mole Al = 27. 0 g Al

Solution 1. Molar mass of Al 1 mole Al = 27. 0 g Al 2. Conversion factors for Al 27. 0 g Al or 1 mol Al 27. 0 g Al 3. Setup 3. 00 moles Al x Answer 27. 0 g Al 1 mole Al = 81. 0 g Al 32

Learning Check The artificial sweetener aspartame (Nutri- Sweet) formula C 14 H 18 N

Learning Check The artificial sweetener aspartame (Nutri- 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? 33

Solution 1. Molar mass of Aspartame C 14 H 18 N 2 O 5

Solution 1. Molar mass of Aspartame C 14 H 18 N 2 O 5 (14 x 12. 0) + (18 x 1. 01) + (2 x 14. 0) + (5 x 16. 0) = 294 g/mole 2. Setup 225 g aspartame x 1 mole aspartame 294 g aspartame = 0. 765 mole aspartame 34

Calculations with Molar Volume Molar volume Liter Moles 1 mole = 22. 4 L

Calculations with Molar Volume Molar volume Liter Moles 1 mole = 22. 4 L 35

RELATE MOLES TO PARTICLES AND PARTICLES TO MOLES Factor Label Method: unit needed =

RELATE MOLES TO PARTICLES AND PARTICLES TO MOLES Factor Label Method: unit needed = given x conversion factor 1. Determine the given and unit needed. 2. Find relationship between given unit and needed unit that is your conversion factor. 3. Substitute into factor label equation. 4. Unit that are identical cancel out. 36

How many H 2 molecules are there in 0. 025 moles of H 2?

How many H 2 molecules are there in 0. 025 moles of H 2? Known: n Given = 0. 025 moles of H 2 n 1 mole H 2 = 6. 022 x 1023 molecules H 2 Unknown: n unit needed = H 2 molecules n Setup: molecules of H 2 = 37

Learning Check 1. Number of atoms in 0. 500 mole of Al 1) 500

Learning Check 1. Number of atoms in 0. 500 mole of Al 1) 500 Al atoms 2) 6. 02 x 1023 Al atoms 3) 3. 01 x 1023 Al atoms 2. Number of moles of S in 1. 8 x 1024 S atoms 1) 1. 0 mole S atoms 2) 3. 0 mole S atoms 3) 1. 1 x 1048 mole S atoms 38

Solution 1. Number of atoms in 0. 500 mol of Al 3) 3. 01

Solution 1. Number of atoms in 0. 500 mol of Al 3) 3. 01 x 1023 Al atoms 0. 500 mol Al x 6. 02 x 1023 Al atoms 1 mol Al 2. Number of moles of S if a sample of S contains 4. 50 x 1024 S atoms 2) 3. 0 mole S atoms 4. 50 x 1024 S atoms x 1 mol S 6. 02 x 1023 S atoms 39

Percent Composition The percent composition is a list of the mass percent of each

Percent Composition The percent composition is a list of the mass percent of each element in a compound. Percent Composition of a Compound = n Mass of the element Mass of the compound x 100 n 40

Calculate the percent composition of proply chloride (C 3 H 7 Cl) First, calculate

Calculate the percent composition of proply chloride (C 3 H 7 Cl) First, calculate the molar mass for the compound by adding up the masses of all the elements: a) 3 carbons at 12. 01 amu each = 36. 03 amu b) 7 hydrogen at 1. 01 amu each= 7. 07 amu c) 1 chlorine atom at 35. 45 amu Summing these values yield: n 36. 03+7. 07+35. 45= 78. 55 amu = formula mass of C 3 H 7 Cl 1. 41

2. a) b) c) n Second, for each element found in the compound set

2. a) b) c) n Second, for each element found in the compound set up ratio of its mass over the total mass times 100 to convert to percent. Carbon (36. 03 / 78. 55) x 100 = 45. 87% Carbon Hydrogen (7. 07 / 78. 55) x 100 = 9. 00% Hydrogen Chlorine (35. 45 / 78. 55) x 100 = 45. 13% Chlorine These values add up to a hundred percent (100%). 42

Learning Check What is the percent carbon in C 5 H 8 NO 4

Learning Check What is the percent carbon in C 5 H 8 NO 4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? 1) 8. 22 %C 2) 24. 3 %C 3) 41. 1 %C 43

Solution Molar mass = 146. 0 g/mole % = total g C x 100

Solution Molar mass = 146. 0 g/mole % = total g C x 100 total g compound = 60. 0 g C x 100 = 41. 1% C 146. 0 g MSG 44

n n Empirical formula Simplest formula for a compound; it is the smallest whole-number

n n Empirical formula Simplest formula for a compound; it is the smallest whole-number ratio of the atoms present. Molecular formula Expression of the formula for a compound; it shows the actual number of atoms of each element present in one molecule of the compound. 45

Calculating Empirical Formulas Steps: 1. Change % to g (change signs). 2. Convert to

Calculating Empirical Formulas Steps: 1. Change % to g (change signs). 2. Convert to Moles. 3. Divide the mass by 1 mol/atomic mass of each element. 4. Find mole ratio. 5. Divide each by the lowest moles. n If ratio is not a whole number multiple by a number that converts the fraction to a whole 46 number.

Calculate the empirical formula for a compound composed of 26. 6% potassium, 35. 4%

Calculate the empirical formula for a compound composed of 26. 6% potassium, 35. 4% chromium, and 38. 1% oxygen. n Percent means a fraction of 100. Thus, if we assume 100 grams of sample we can then use the percent values as grams of each element. From grams we find moles of each element in the formula. 47

n Find moles potassium. Next find moles of chromium. n Then find moles oxygen.

n Find moles potassium. Next find moles of chromium. n Then find moles oxygen. 48

n n Next, set up a mole ratio that relates the moles of each

n n Next, set up a mole ratio that relates the moles of each element to the moles of the element that is least present. Because a chemical formula must have only whole numbers we multiply by 2 to yield: K 2 Cr 2 O 7 as the empirical formula. 49

Calculating Molecular Formulas n n n Molecular mass must be known before a molecular

Calculating Molecular Formulas n n n Molecular mass must be known before a molecular formula can be determined. Molecular mass is a whole-number multiple of the empirical formula mass. (empirical formula mass)x = molecular mass 50

Molecular formulas Solve for x. 2. Multiply empirical formula by whatever is x. Examples:

Molecular formulas Solve for x. 2. Multiply empirical formula by whatever is x. Examples: (CH 4)x (CH 4 mass)x = 16 {12. 0 + 4(1. 0)}x = 16 (16. 0)x = 16 x = 16. 0 x = 1 (CH 4)x (CH 4)1 = CH 4 1. 51

(HO)x (HO mass)x = 34. 0 (1. 0 + 16. 0)x = 34. 0

(HO)x (HO mass)x = 34. 0 (1. 0 + 16. 0)x = 34. 0 (17. 0)x = 34. 0 17. 0 x = 2 (HO)x (HO)2 = H 2 O 2 52

Problems 1. 2. a) b) c) A certain hydrocarbon compound is found by analysis

Problems 1. 2. a) b) c) A certain hydrocarbon compound is found by analysis to have the empirical formula CH 3. Its molecular weight is 30. 0. What is the molecular formula? A compound is found by analysis to consist of 40. 1% S and 59. 9% O. Its molecular weight is 80. 1. (Change % to g, just change the signs) What is its empirical formula? What is its molecular formula? What is the name of the compound?

3. a) b) A compound of nitrogen and sulfur is found by analysis to

3. a) b) A compound of nitrogen and sulfur is found by analysis to have the empirical formula NS 2. Its molecular weight is 156. 4. What is its molecular formula? What is the name of the compound?