chapter 7 chemical quantities

1Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings.

Chapter 7 Chemical Quantities

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

Formulas


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


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


Samples of One Mole Quantities


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


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


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


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


Give the molar mass to the nearest 0.1 g.

A. 1 mole of K atoms = ________

B. 1 mole of Sn atoms = ________

Learning Check


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


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


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


One-Mole Quantities

32.1 g 55.9 g 58.5 g 294.2 g 342.3 g


A. 1 mole of K2O = ______g

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

Learning Check


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


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

Learning Check


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


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


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

Learning Check


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


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


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


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


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


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 %


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


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


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


Some Molecular and Empirical Formulas


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


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


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


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


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


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


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

Learning Check


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


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


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


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


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


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

Learning Check


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)



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.



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


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


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


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


chapter 7 chemical quantities

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