chapter 7 chemical quantities

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Chapter 7 Chemical Quantities

7.1 The Mole7.2 Molar Mass7.3 Calculations Using Molar Mass7.4 Percent Composition and Empirical

Formulas

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A mole contains 6.02 x 1023 particles (atoms, ions, molecules, formula unit)

The number 6.02 x 1023 is known as Avogadro’s number.

One mole of any element contains Avogadro’s number of atoms.

1 mole Na = 6.02 x 1023 Na atoms1 mole Au = 6.02 x 1023 Au atoms

7.1 A Mole

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One mole of a covalent compound contains Avogadro’s number of molecules.1 mole CO2 = 6.02 x 1023 CO2 molecules

1 mole H2O = 6.02 x 1023 H2O molecules

One mole of an ionic compound contains Avogadro’s number of formula units.1 mole NaCl = 6.02 x 1023 NaCl formula units

A Mole of Molecules

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Samples of One Mole Quantities

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Avogadro’s number is written as conversion factors.

6.02 x 1023 particles and 1 mole 1 mole 6.02 x 1023 particles The number of molecules in 0.50 mole of CO2

molecules is calculated as

0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules

1 mole CO2 molecules

= 3.0 x 1023 CO2 molecules

Avogadro’s Number

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A. Calculate the number of atoms in 2.0 moles of Al.

B. Calculate the number of moles of S in 1.8 x 1024 S.

Learning Check

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A. Calculate the number of atoms in 2.0 moles of Al. 2.0 moles Al x 6.02 x 1023 Al atoms

1 mole Al=1.2 x 1024 Al atoms

B. Calculate the number of moles of S in 1.8 x 1024 S. 1.8 x 1024 S atoms x 1 mole S

6.02 x 1023 S atoms

= 3.0 mole S atoms

Solution

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The mass of one mole is called molar mass (g/mole).

The molar mass of an element is the atomic mass expressed in grams.

7.2 Molar Mass

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Give the molar mass to the nearest 0.1 g.

A. 1 mole of K atoms = ________

B. 1 mole of Sn atoms = ________

Learning Check

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Give the molar mass to the nearest 0.1 g.

A. 1 mole of K atoms =39.1 g

B. 1 mole of Sn atoms = 118.7 g

Solution

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Molar Mass of CaCl2

For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl2 to the nearest 0.1 g as follows.

Formula mass of CaCl2 = [40.1 + 2(35.45)] = 111.1amu Formula mass = molar mass, so

111.1amu = 111.1g/mol

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Molar Mass of K3PO4

Determine the molar mass of K3PO4 to 0.1 g.

Molar Mass = [3(39.1) + 31.0 + 4(16)] = 212.3g/mol

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One-Mole Quantities

32.1 g 55.9 g 58.5 g 294.2 g 342.3 g

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A. 1 mole of K2O = ______g

B. 1 mole of antacid Al(OH)3 = ______g

Learning Check

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A. 1 mole of K2O

2 (39.1) + 1 (16.0) = 94.2 g/mol

1mole K2O x 94.2g/mol = 94.2g

B. 1 mole of antacid Al(OH)3

1 (27.0) + 3 (16.0) + 3 (1.0) = 78.0 g/mol1mole Al(OH)3 x 78.0 g/mol = 78.0g

Solution

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Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?

Learning Check

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Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?

17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) =

204 + 18 + 57.0 + 14.0 + 16.0

= 309 g/mole

Solution

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Methane CH4 known as natural gas is used in gas cook tops and gas heaters.

1 mole CH4 = 16.0 g

The molar mass of methane can be written as conversion factors.

16.0 g CH4 and 1 mole CH4

1 mole CH4 16.0 g CH4

Molar Mass Factors

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Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.

Learning Check

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Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid.

2(12.0) + 4(1.0) + 2(16) = 60.0g/mol

1 mole of acetic acid = 60.0 g acetic acid

1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid

Solution

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Mole factors are used to convert between the grams of a substance and the number of moles.

7.3 Calculations with Molar Mass

Grams Mole factor Moles

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Aluminum is often used for the structure oflightweight bicycle frames. How many gramsof Al are in 3.00 moles of Al?

3.00 moles Al x 27.0 g Al = 81.0 g Al1 mole Al

mole factor for Al

Calculating Grams from Moles

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The artificial sweetener aspartame (Nutri-Sweet) C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?

Learning Check

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Calculate the molar mass of C14H18N2O5.

14 (12.0) + 18 (1.0) + 2 (14.0) + 5(16.0) = 294 g/mole

Set up the calculation using a mole factor. 225 g aspartame x 1 mole aspartame

294 g aspartame

mole factor(inverted)

= 0.765 mole aspartame

Solution

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7.4 Percent Composition In a compound, the percent composition is the

percent by mass of each element in the formula. In one mole of CO2 there are 12.0 g of C and

32.0 g of O (molar mass 44.0 g/mol),

12.0 g C x 100 = 27.3 % C

44.0 g CO2 32.0 g O x 100 = 72.7 % O

44.0 g CO2 100.0 %

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What is the percent carbon in C5H8NNaO4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats?

1) 7.10 %C

2) 35.5 %C

3) 60.0 %C

Learning Check

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2) 35.5 %C

Molar mass = 169.1 g

% = total g C x 100 total g MSG

= 60.0 g C x 100 = 35.5 % C 169.1 g MSG

Solution

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The molecular formula is the true or actual number of the atoms in a molecule.

The empirical formula is the simplest whole number ratio of the atoms.

The empirical formula is calculated by dividing the subscripts in the molecular formula by a whole number to give the lowest ratio.

C5H10O5 5 = C1H2O1 = CH2Omolecular empirical formulaformula

Types of Formulas

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Some Molecular and Empirical Formulas

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A. What is the empirical formula for C4H8?

1) C2H4 2) CH2 3) CH

B. What is the empirical formula for C8H14?

1) C4H7 2) C6H12 3) C8H14

C. Which is a possible molecular formula for CH2O?

1) C4H4O4 2) C2H4O2 3) C3H6O3

Learning Check

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A. What is the empirical formula for C4H8?

2) CH2 C4H8 4

B. What is the empirical formula for C8H14?

1) C4H7 C8H14 2

C. Which is a possible molecular formula for CH2O?

2) C2H4O2 3) C3H6O3

Solution

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If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.

1) SN

2) SN4

3) S4N4

Learning Check

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If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.

3) S4N4

If the molecular formula has 4 atoms of N, and S and N are related 1:1, then there must also be 4 atoms of S.

Solution

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A molecular formula is equal to or a multiple of the empirical formula.

Thus, the molar mass is equal to or a multiple of the empirical mass. molar mass = a whole number empirical mass

Multiply the empirical formula by the whole number to determine the molecular formula.

Relating Empirical and Molecular Formulas

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Determine the molecular formula of a compoundthat has a molar mass of 78.0 and an empiricalformula of CH. 1. Empirical mass of CH = 13.0 g/mol2. Divide the molar mass by the empirical mass.3. 78.0 g/mol = 6.00 13.0 g/mol4. Multiply the subscripts in CH by 6.5. Molecular formula = (CH)6 = C6H6

Finding the Molecular Formula

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A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?

Learning Check

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A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?

C3H4O3 = 3(12.0) + 4(1.0) + 3(16.0) = 88.0 g/mol

176.0 g/mol (molar mass) = 2.00 88.0 g/mol (empirical mass)

Molecular formula = 2 (empirical formula) (C3H4O3 )2 = C6H8O6

Solution

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A compound is C 24.27%, H 4.07%, and Cl 71.65%. The molar mass is known to be 99.0 g. What are the empirical and molecular formulas? 1. Write the mass percents as the grams in a 100.00-g sample of the compound.

C 24.27 g H 4.07 g Cl 71.65 g

Finding the Molecular Formula

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Finding the Molecular Formula Continued

2. Calculate the number of moles of each element.

24.27 g C x 1 mole C = 2.02 moles C 12.0 g C

4.07 g H x 1 mole H = 4.03 moles H 1.01 g H

71.65 g Cl x 1 mole Cl = 2.02 moles Cl 35.5 g Cl

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Finding the Molecular Formula (continued)

3. Divide each by the smallest 2.02 moles C = 1 mole C 2.024.03 moles H = 2 moles H 2.022.02 moles Cl = 1 mole Cl 2.02

Empirical formula = C1H2Cl1 = CH2Cl

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Finding the Molecular Formula (continued)

4. Calculate empirical mass (EM)empirical mass CH2Cl = 49.5 g/mol

5. Divide molar mass by empirical mass Molar mass = 99.0 g/mol = 2Empirical mass 49.5 g/mol

6. Determine Molecular formula(CH2Cl)2 = C2H4Cl2

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Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula.

Learning Check

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In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O.

60.0 g C x 1 mole C = 5.00 moles C 12.0 g C

4.5 g H x 1 mole H = 4.5 moles H 1.01 g H

35.5 g O x 1mole O = 2.22 moles O 16.0 g O

Solution (continued)

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Solution (continued)

Divide by the smallest number of moles.5.00 moles C = 2.25 moles C

2.22 4.5 moles H = 2.0 moles H 2.222.22 moles O = 1.00 mole O 2.22Note that the results are not all whole numbers. To obtain whole numbers, multiply by a factor to give whole numbers.

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Solution (continued)

Multiply each number of moles by 4C: 2.25 moles C x 4 = 9 moles CH: 2.0 moles H x 4 = 8 moles HO: 1.00 mole O x 4 = 4 moles O

Use the whole numbers as subscripts toobtain the simplest formula

C9H8O4

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A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula?

Learning Check

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In 100.0 g, there are 27.4 g S, 12.0 g N and 60.6 g Cl. 27.4 g S x 1 mole S = 0.854 mole S

32.1 g S

12.0 g N x 1 mole N = 0.857 moles N 14.0 g N

60.6 g Cl x 1mole Cl = 1.71 moles Cl 35.5 g Cl

Solution

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Dividing by the smallest number of moles

0.854 mole S /0.854 = 1.00 mole S

0.857 mole N/0.854 = 1.00 mole N

1.71 moles Cl/0.854 = 2.00 moles Cl

Empirical formula = SNCl2 = 117.1 g/mol

Molar Mass/ Empirical mass

351 = 3117.1 Molecular formula = (SNCl2)3 = S3N3Cl6

Solution (continued)