nwtc general chemistry ch 07

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Chapter 7 Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena Quantitative Composition of Compounds These black pearls are made of layers of calcium carbonate. They can be measured by counting or weighing.

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Page 1: NWTC General Chemistry Ch 07

Chapter 7

Introduction to General, Organic, and Biochemistry 10e

John Wiley & Sons, Inc

Morris Hein, Scott Pattison, and Susan Arena

Quantitative Composition of Compounds

These black pearls are made of layers of calcium carbonate. They can be measured by counting or weighing.

Page 2: NWTC General Chemistry Ch 07

Chapter Outline

Copyright 2012 John Wiley & Sons, Inc

7.1 The Mole

7.2 Molar Mass of Compounds

7.3 Percent Composition of Compounds

7.4 Empirical Formula versus Molecular Formula

7.5 Calculating Empirical Formulas

7.6 Calculating the Molecular Formula from the Empirical Formula

Page 3: NWTC General Chemistry Ch 07

Convenient Ways of Counting Items

• 1 dozen eggs = 12 eggs• 1 ream paper = 500 sheets• 1 gross pencils = 144 pencils

How do chemists count really small things, like atoms?

1 mole atoms =

6.022×1023 atoms

This number is known as Avogadro’s number.

:

Copyright 2012 John Wiley & Sons, Inc

1×10 -10 m

Review Question 1

Page 4: NWTC General Chemistry Ch 07

The Mole

1 mole of anything contains Avogadro’s number (6.022×1023 ) of particles.

• 1 mol atoms = 6.022×1023 atoms• 1 mol molecules = 6.022×1023 molecules• 1 mol ions = 6.022×1023 ions

Avogadro’s number can be used as a conversion factor:

Copyright 2012 John Wiley & Sons, Inc

236.022 10 particles

1 mol

23

1 mol

6.022 10 particles

Review Question 7 & 8

Page 5: NWTC General Chemistry Ch 07

The Mole

Calculate the number of atoms in 2.4 mol Na.

Copyright 2012 John Wiley & Sons, Inc

2.4 mol Na236.022 x 10 Na atoms

1 mol Na

24= 1.4 10 Na atoms

Plan

Calculate

2.5 mol Na Na atoms

236.022 x 10 Na atoms

1 mol Na23

1 mol Na

6.022 x 10 Na atoms

Page 6: NWTC General Chemistry Ch 07

Your Turn!

How many moles of HCl are present in 4.3×1023 molecules?

a. 2.6×1047 mol

b. 0.71 mol

c. 2.6 mol

d. 7.1×1045 mol

Copyright 2012 John Wiley & Sons, Inc

4.3 x10 23 molecules 1 mol = 0.71 mol

6.022 x10 23 molecules

Page 7: NWTC General Chemistry Ch 07

Convenient Ways of Counting

It is often convenient to count using mass.

For example, a canning recipe calls for 150 apples to be peeled and cored.

The average mass of an individual apple is 235 g. How many kg are needed to complete this recipe?

The canner should buy 35 kg of apple.

Copyright 2012 John Wiley & Sons, Inc

235 g 1 kg150 apples × = 35 kg

1 apple 1000 g

Page 8: NWTC General Chemistry Ch 07

Counting Atoms with Mass

• Chemists count atoms by using mass since individual atoms are too small to count.

• Average mass of an atom = atomic mass in amu• Average mass of 1 mole of atoms = atomic mass

expressed in grams (molar mass).

1 mol atoms = atomic mass in grams

1 mol atoms = 6.022×1023 atoms

atomic mass in grams = 6.022×1023 atoms

Copyright 2012 John Wiley & Sons, Inc

Review Question 5

Page 9: NWTC General Chemistry Ch 07

The Mole

Copyright 2012 John Wiley & Sons, Inc

Page 10: NWTC General Chemistry Ch 07

Your Turn!

Which of these is not correct?

a. The mass of 1 atom of C is 12.01 amu.

b. The mass of 1 mole of C atoms is 12.01 g.

c. Avogadro’s number of atoms has a mass of 12.01 amu.

d. Avogadro’s number of atoms has a mass of 12.01 g.

Copyright 2012 John Wiley & Sons, Inc

C6

12.01

Page 11: NWTC General Chemistry Ch 07

The Mole

Calculate the mass of 2.4 mol C.

Copyright 2012 John Wiley & Sons, Inc

Plan

Calculate

12.01g C

1 mol C

1 mol C

12.01g C

2.4 mol C g

2.4 mol C12.01g C

1 mol C

= 29 g C

atomic mass

C 12.01

Molar mass is a conversion factor:

Page 12: NWTC General Chemistry Ch 07

Your Turn!

What is the correct set up to calculate number of moles of atoms contained in 3.52 g Al?

a.

b.

c.

Copyright 2012 John Wiley & Sons, Inc

1 mol Al3.52g Al

26.98 g Al

26.98g Al3.52g Al

1 mol Al

236.022 x 10 Al atoms3.52g Al

1 g Al

atomic mass

Al 26.98

Page 13: NWTC General Chemistry Ch 07

The Mole

Copyright 2012 John Wiley & Sons, Inc

Calculate the number of atoms in 36.0 g C.

Plan

Calculate

236.022 10 atoms C

1 mol C

36.0 g C moles C atoms C

36.0 g C236.022 10 atoms C

1 mol C

24= 1.8 10 atoms C

atomic mass

C 12.01

1 mol C

12.01 g C

1 mol C

12.01 g C

Page 14: NWTC General Chemistry Ch 07

Your Turn!

What is the mass of 3.01 ×1023 atoms of lead?

a. 104 g

b. 414 g

c. 0.500 g

d. 1.04×1048 g

Copyright 2012 John Wiley & Sons, Inc

atomic mass

Pb 207.2

3.01 x10 23 atoms 1 mol 207.2 gram = 104 gram

6.022 x10 23 atoms 1 mol

Page 15: NWTC General Chemistry Ch 07

Your Turn!

1 gram of which of the following elements would contain the largest number of atoms?

a. nitrogen

b. hydrogen

c. phosphorus

d. oxygen

Copyright 2012 John Wiley & Sons, Inc

atomic mass

N 14.01

H 1.01

P 30.97

O 16.00

1 gram H 1 mol 6.022 x10 23 atom = 5.96 x10 23 atoms

1.01 g 1 mol

1 gram P 1 mol 6.022 x10 23 atom = 1.94 x10 22 atoms

30.97 g 1 mol

Review Question 3

Page 16: NWTC General Chemistry Ch 07

Your Turn!

Which has a higher mass, mol of K or mol of Au?

a. potassium

b. gold

Copyright 2012 John Wiley & Sons, Inc

atomic mass

K 39.0983

Au 196.96654

1 Mol of K 39.0983 gram = 39.0983 gram

1 mol

1 Mol of Au 196.9665 gram = 196.9665 gram

1 mol

Review Question 2

Page 17: NWTC General Chemistry Ch 07

Your Turn!

Which has more electrons, mol of K or mol of Au?

a. potassium

b. gold

Copyright 2012 John Wiley & Sons, Inc

1 Mol K 6.022 x10 23 atom 19 e- = 1.144 x10 25 E-

1 mol 1 atom

1 Mole Au 6.022 x10 23 atom 79 e- = 4.757 x10 25 E-

1 mol atom

atomic number

K 19

Au 79

Review Question 4

Page 18: NWTC General Chemistry Ch 07

Review Question 9

6.022 x10 23

6.022 x10 23

1.204 x10 24

15.99 g

31.98 g

Copyright 2012 John Wiley & Sons, Inc

1 Mole of O 15.99 gram = 15.99 gram

1 mole

1 Mole O 2 2 Mole of O 15.99 gram = 31.98 gram

1 Mole O 2 1 mole

Review Question 9

a. Mole of O atoms = ? Atoms

b. Mole of O2 molecules = ? Molecules

c. Mole of O2 molecules = ? atoms

d. Mole of O atoms= ? grams

e. Mole of O2 molecules = ? grams

Page 19: NWTC General Chemistry Ch 07

Molar Mass of Compounds

1 mol compound = 6.022×1023 formula units compound

Molar mass of compound = mass of 1 mol compound

The molar mass of a compound is the sum of the atomic masses of each atom in the compound.

What is the molar mass of CO2?

1C 1(12.01g)

2O 2(16.00g)

CO2 44.01g/mol

Copyright 2012 John Wiley & Sons, Inc

atomic mass

C 12.01

O 16.00

Page 20: NWTC General Chemistry Ch 07

Molar Mass of Compounds

What is the molar mass of Al2(CO3)3?

2Al 2(26.98g)

3C 3(12.01g)

9O 9(16.00g)

Al2(CO3)3 233.99g/mol

Copyright 2012 John Wiley & Sons, Inc

atomic mass

Al 26.98

C 12.01

O 16.00

Page 21: NWTC General Chemistry Ch 07

Your Turn!

What is the molar mass of (NH4)3PO4?

a. 141.04g/mol

b. 144.07g/mol

c. 146.09g/mol

d. 149.12g/mol

Copyright 2012 John Wiley & Sons, Inc

atomic mass

N 14.01

H 1.01

P 30.97

O 16.00

3*1 N 3(14.01g)3*4 H 12(1.01g)1 P 1(30.97g)4 O 4(16.00g)

Page 22: NWTC General Chemistry Ch 07

Your Turn!

What is the molar mass of Mg(ClO4)2?

a. 301.01g/mol

b. 191.21g/mol

c. 223.21g/mol

d. 123.76g/mol

Copyright 2012 John Wiley & Sons, Inc

atomic mass

Mg 24.31

Cl 35.45

O 16.00

__ Mg __(____g)__ Cl __(____g)__ O __(____g)

Page 23: NWTC General Chemistry Ch 07

Using Molar Masses of Compounds

Molar mass = 1 mol = 6.022×1023 formula units

Calculate the mass of 0.150 mol Mg(ClO4)2

Copyright 2012 John Wiley & Sons, Inc

Plan

Calculate

4 2

4 2

1 mol Mg(ClO )

223.21 g Mg(ClO )4 2

4 2

223.21 g Mg(ClO )

1 mol Mg(ClO )

0.150 mol Mg(ClO4)2 g Mg(ClO4)2

4 20.150 mol Mg(ClO ) 4 2= 33.5 g Mg(ClO )4 2

4 2

223.21 g Mg(ClO )

1 mol Mg(ClO )

Page 24: NWTC General Chemistry Ch 07

Using Molar Masses of Compounds

Molar mass = 1 mol = 6.022×1023 formula units

Calculate the number of moles in 35 g H2O.

Copyright 2012 John Wiley & Sons, Inc

Plan

Calculate

2

2

1 mol H O

18.02 g H O2

2

18.02 g H O

1 mol H O

35 g H2O moles H2O

235 g H O 2= 1.9 g H O 2

2

1 mol H O

18.02 g H O

atomic mass

H 1.01

O 16.00

Page 25: NWTC General Chemistry Ch 07

Using Molar Masses of Compounds

Molar mass = 1 mol = 6.022×1023 formula units

Calculate the number of molecules in 35 g H2O.

Copyright 2012 John Wiley & Sons, Inc

Plan

Calculate

232

2

6.022 10 molecules H O

1 mol H O

2

2

18.02 g H O

1 mol H O

35 g H2O mol H2O molecules H2O

235.0 g H O

242= 1.2 10 molecules H O

2

2

1 mol H O

18.02 g H O

232

2

6.022 10 molecules H O

1 mol H O

Page 26: NWTC General Chemistry Ch 07

Preview Question 10

How many molecules are present in 1 molar mass of sulfuric acid (H2SO4)?

1. What is the molar mass of H2SO4?

2 H 2(1.01g)

1 S 1(32.06g)

4 O 4(16.00g) = 98.08g/mol

2. How many atoms in one molecule? 2+1+4=7

How many atoms in 1 molar mass? ___

Copyright 2012 John Wiley & Sons, Inc

atomic mass

S 32.06

H 1.01

O 16.00

98.08 g 1 mol 6.022 x10 23 molecules = 6.022 x10 23 molecules

98.08 g 1 mol

Page 27: NWTC General Chemistry Ch 07

Your Turn!

How many moles of molecules are present in 146 g of glucose (C6H12O6)?

a. 180. mol

b. 0.810 mol

c. 26300 mol

d. 4.88×1023 mol

Copyright 2012 John Wiley & Sons, Inc

atomic mass

C 12.01

H 1.01

O 16.00

g = molecules

__ C __(____g)__ H __(____g)__ O __(____g) = _____ g/mol

Page 28: NWTC General Chemistry Ch 07

Your Turn!

What is the mass of 1.20 ×1023 molecules of CH3OH?

a. 161g

b. 38.5g

c. 32.1g

d. 6.39g

Copyright 2012 John Wiley & Sons, Inc

atomic mass

C 12.01

H 1.01

O 16.00

molecules = g

__ C __(____g)__ H __(____g)__ O __(____g) = _____ g/mol

Page 29: NWTC General Chemistry Ch 07

Your Turn!

How many molecules are present in 4.21 moles of HBr?

a. 2.53×1023 molecules

b. 2.53×1024 molecules

c. 6.99×10-24 molecules

d. 3.97×102 molecules

e. 6.99×1024 molecules

 

Copyright 2012 John Wiley & Sons, Inc

atomic mass

H 1.01

Br 79.90

Page 30: NWTC General Chemistry Ch 07

Percent Composition of Compounds

Copyright 2012 John Wiley & Sons, Inc

Percent composition is a list of the mass percent of each element in a compound.

Na2CO3 is

43.38% Na 11.33% C 45.29% O

How do we calculate the mass percent of Na2CO3?

Page 31: NWTC General Chemistry Ch 07

First determine the molar mass of Na2CO3

2(22.99g Na) + 1(12.01g C) + 3(16.00 g O) = 105.99g/mol Na2CO3

Then find ratio of the mass of each element to the mass of the compound.

2 3

2 22.99 g Nax100 43.38% Na

105.99 g Na CO

2 3

1 12.01 g Cx100 11.33% C

105.99 g Na CO

2 3

3 16.00 g Ox100 45.29% O

105.99 g N a CO

Calculating Percent Composition

Page 32: NWTC General Chemistry Ch 07

Calculating Percent Composition

A compound is found to consist of 2.74g of iron and 5.24g of chlorine. What is the percent composition of the compound?

1. Calculate the mass of the product formed:

2.74g Fe + 5.24g Cl = 7.98g product

2. Calculate the percent for each element.

Copyright 2012 John Wiley & Sons, Inc

2.74 g Fex100 = 34.3% Fe

7.98 g

5.24 g Clx100 = 65.7% Cl

7.98 g

Page 33: NWTC General Chemistry Ch 07

Your Turn!

What is the percent carbon in acetic acid, HC2H3O2?

a. 41.01% C

b. 20.00% C

c. 6.73% C

d. 39.99% C

Copyright 2012 John Wiley & Sons, Inc

atomic mass

C 12.01

H 1.01

O 16.00

__ C __(____g)__ H __(____g)__ O __(____g) = _____ g/mol

First determine the molar mass of Na2CO3

Page 34: NWTC General Chemistry Ch 07

Your Turn!

A 6.00g sample of calcium sulfide is found to contain 3.33g of calcium. What is the percent by mass of sulfur in the compound?

a. 80.2% S

b. 55.5% S

c. 44.5% S

d. 28.6% S

Copyright 2012 John Wiley & Sons, Inc

Page 35: NWTC General Chemistry Ch 07

Empirical Formula versus Molecular Formula

Copyright 2012 John Wiley & Sons, Inc

The empirical formula is the lowest whole number ratio of atoms in a compound.

C6H12O6

CH2O

(CH2O)6

The molecular formula for a substance is the actual number of atoms of each element.

Note that the molecular formula is a whole number multiple of the empirical formula.

Review Question 11

Page 36: NWTC General Chemistry Ch 07

Empirical Formula versus Molecular Formula

It is possible for several different molecules to have the same empirical formula.

Copyright 2012 John Wiley & Sons, Inc

Page 37: NWTC General Chemistry Ch 07

Your Turn!

What is the empirical formula for the compound P4O10?

a. P4O10

b. P2O5

c. PO2.5

d. PO3

Copyright 2012 John Wiley & Sons, Inc

Page 38: NWTC General Chemistry Ch 07

Calculating Empirical Formulas

Steps for calculating an empirical formula:

1. Assume 100g of compound and express the mass of each element in grams.

2. Convert the grams of each element to moles.

3. Find the mole ratio of each element. Round to nearest whole number if it is close to the whole number.

4. If necessary, multiply the ratios by the smallest whole number that will convert them to a whole number.

Copyright 2012 John Wiley & Sons, Inc

Review Question 12

Page 39: NWTC General Chemistry Ch 07

Calculating Empirical Formulas

Calculate the empirical formula of a compound that is 63.19% Mn and 36.81% O.

1. Assume 100 g of material.

63.19 g Mn

36.81 g O

2. Convert grams of each element to moles:

Copyright 2012 John Wiley & Sons, Inc

atomic mass

Mn 54.94

0 16.00

1 mol Mn63.19 g Mn× =1.1502 mol Mn

54.94 g Mn1 mol O

36.81 g O× = 2.3006 mol O16.00 g O

Page 40: NWTC General Chemistry Ch 07

Calculating Empirical Formulas

3. Change the numbers of atoms to whole numbers by dividing by the smallest number.

The simplest ratio of Mn:O is 1:2.

Empirical formula = MnO2

Copyright 2012 John Wiley & Sons, Inc

2.3006 mol OO = = 2.000

1.1502 mol Mn1.1502 mol Mn

Mn = = 1.0001.1502 mol Mn

Page 41: NWTC General Chemistry Ch 07

Calculating Empirical Formulas

Calculate the empirical formula of a compound that is 72.2% Mg and 27.8% N.

1. Assume 100 g of material.

72.2 g Mg

27.8 g O

2. Convert grams of each element to moles:

Copyright 2012 John Wiley & Sons, Inc

atomic mass

Mg 24.31

N 14.01

1mol Mg72.2g Mg× =2.970 mol Mg

24.31g Mg1mol N

27.8g N× = 1.984 mol N14.01g N

Page 42: NWTC General Chemistry Ch 07

Calculating Empirical Formulas

3. Change the numbers of atoms to whole numbers by dividing by the smallest number.

4. Multiply by a number that will give whole numbers.

Mg: (1.500)2 = 3.00 N: (1.000)2 = 2.00

Empirical formula = Mg3N2

Copyright 2012 John Wiley & Sons, Inc

2.970 mol MgMg = = 1.500

1.984 mol N

1.984 mol NN = = 1.000

1.984 mol N

Page 43: NWTC General Chemistry Ch 07

Your Turn!

What is the empirical formula of an alcohol that is 52.13% C, 13.15% H and 34.72% O.

a. CH2O

b. C4HO3

c. C2H6O

d. C2H3O2

Copyright 2012 John Wiley & Sons, Inc

atomic mass

C 12.01

H 1.01

O 16.00

1. Assume 100 g so52.13g C, 13.15g H and 34.72g O

2. Moles52.13g C/12.01= 4.34 mol C13.15g H/1.01= 13.02 mol H34.72g O/16.00= 2.17mol O

3. Fractional part4.34 mol C /2.17= 213.02 mol H /2.17= 62.17mol O /2.17= 1

4. Whole numubersAlready done

5. Empirical formula: 2 C & 6 H & 1 O, so …

Give it a try

Page 44: NWTC General Chemistry Ch 07

Calculating the Molecular Formulafrom the Empirical Formula

Copyright 2012 John Wiley & Sons, Inc

The molecular formula will be either equal to the empirical formula or some integer multiple of it.

The ratio of the molecular mass to the mass predicted by the empirical formula tells us how many times larger the molecular formula is.

molecular formula mass = = number of empirical formula units

empirical formula massn

Page 45: NWTC General Chemistry Ch 07

Calculating the Molecular Formulafrom the Empirical Formula

Copyright 2012 John Wiley & Sons, Inc

Determine the molecular formula for glyceraldehyde which has a molar mass of 90.08 g/mol and an empirical formula of CH2O.

2

90.09 g n = = 3

30.03 g CH O(CH2O)3 = C3H6O3

molecular formula massn=

empirical formula mass

Page 46: NWTC General Chemistry Ch 07

Calculating the Molecular Formulafrom the Empirical Formula

Copyright 2012 John Wiley & Sons, Inc

Determine the molecular formula of a nitrogen oxide compound (NxOy) with a molar mass of 92.011 g/mol and a empirical formula of NO2.

(NO2)2 = N2O4

2

92.011 g N On 2

46.0055 g NOx y

molecular formula massn=

empirical formula mass

Page 47: NWTC General Chemistry Ch 07

Your Turn!

Copyright 2012 John Wiley & Sons, Inc

atomic mass

C 12.01

H 1.01

Cl 35.45

A. 1(12.01g C) + 2(1.01g H) + 1(35.45 g Cl) = 49.48g/mol CH2Cl

B. 197.92 g/mol / 49.48 g/mol = 4

C. (CH2Cl) *4=

PLAN THE STEPS

A. Empirical formula mass

B. Divide molar by empirical

C. Multiple result by formula

Plan – Set up - Calcualte

What is the molecular formula of a compound with the empirical formula CH2Cl and molar mass of 197.92 g/mol?

a. CH2Cl

b. C2H4Cl2

c. C3H6Cl3

d. C4H8Cl4

 

Page 48: NWTC General Chemistry Ch 07

Calculating the Molecular Formula

A disinfectant is known to be 76.57% C , 6.43% H, and 17.00% O. It has a molar mass of 188.24 g/mol Determine its molecular formula.

Determine the mass and moles of C, H and O.

C: 76.57%(188.24g) = 144.1 g C/(12.01g/mol) = 12

H: 6.43% (188.24g) = 12.1 g H/(1.01g/mol) = 12

O: 17.00%(188.24g) = 32.00 g O/(16.00 g/mol) = 2

Molecular formula: C12H12O2

Copyright 2012 John Wiley & Sons, Inc

Page 49: NWTC General Chemistry Ch 07

Your Turn!

What is the molecular formula of a substance that consists of 85.60% C and 14.40% H and has a molar mass of 28.08 g/mol?

a. CH2

b. C2H2

c. CH3

d. C2H4

e. C2H6

Copyright 2012 John Wiley & Sons, Inc

atomic mass

C 12.01

H 1.01

Page 50: NWTC General Chemistry Ch 07

Questions

Review Questions – Did in class

Paired Questions (pg 139)– Do 1, 3, 7, 9, 11, 15, 21, 27, 29, 31, 35 , 39, 43– Practice later 2, 6, 12, 16, 20, 24, 28, 32, 36, 40, 44

Copyright 2012 John Wiley & Sons, Inc 1-50