03a Stoichiometry 6.pdf

Chapter 3

Stoichiometry

Average Atomic Mass

Atomic massis the mass of an atom in atomic mass units (amu). 1 amu = 1/12 the mass of a carbon-12 atom

The average atomic masson the periodic table represents the average mass of the

naturally occurring mixture of isotopes.

Average mass (C) = (0.9893)(12.00000 amu) + (0.0107)(13.003355 amu)

Isotope Isotopic mass (amu) _{abundance (%)} Natural

12_{C 12.00000 98.93}

13_{C 13.003355 1.07}

= 12.01 amu

Oxygen is the most abundant element in both Earth’s crust and the human body. The atomic masses of its three stable isotopes, O (99.757 percent), O (0.038 percent), O (0.205 percent), are 15.9949, 16.9991, and 17.9992 amu, respectively. Calculate the average atomic mass of oxygen using the relative abundances given in parentheses.

Calculating the Average Mass of an Element

Solution

(0.99757)(15.9949 amu) + (0.00038)(16.9991 amu) + (0.00205)(17.992 amu) = 15.9994 amu

16

8 17 8

18 8

Think About It The average atomic mass should be closest to the atomic mass of the most abundant isotope (in this case, oxygen-16) and, to four significant figures, should be the same number that appears in the periodic table on the inside front cover of your textbook (in this case, 16.00 amu).

Strategy Each isotope contributes to the average atomic mass based on its relative abundance. Multiplying the mass of each isotope by its fractional abundance (percent value divided by 100) will give its contribution to the average atomic mass.

What is a Mole?

2 H2(g) + O2(g)  2 H2O(g)

We can multiply a balanced equation by any # and still

have a balanced equation.

2(pair) H2(g) + 1 (pair) O2(g)  2(pair) H2O(g)

2(dozen) H2(g) + 1 (dozen) O2(g)  2(dozen) H2O(g)

2(mole) H2(g) + 1 (mole) O2(g)  2(mole) H2O(g)

A mole is 6.02 x 1023_{things. This # is called Avogadro’s #.}

Avogadro’s # is a collective noun used to count things (just like a dozen or a pair).

Mole: the amount of substance that contains as many particles as exactly 12 g of C; 6.02 x 1023_things.

Avogadro

’

s Number: The Mole

and Avogadro’s Number

 Avogadro’s number, named after Amadeo Avogadro, is given a specific name: the mole (mol). The mole is the metric unit for amount.

 Think of it like any other counting number.

 Pair……….. 2  Dozen………... 12  Grand………... 1000  Mole………… 602,214,199,000,000,000,000,000

 It is so large that it is almost always represented in scientific notation

A Mole of Particles

A moleis a collection that contains

 the same number of particles as there are carbon atoms

in 12.0 g of carbon 12_C

 6.02 x 1023 _{atoms of an element (Avogadro’s number)}

 A mole of an element has Avogadro’s number of atoms

1 mole of C = 6.02 x 1023C atoms

 A mole of a covalent compound has Avogadro’s number of

molecules

Avogadro’s Number May Be Used as a

Conversion Factor to Convert

Particles

Moles

Avogadro’s number (6.02 x 1023_{) can be written as an}

equality and two conversion factors.

Equality:

1 mole = 6.02 x 1023 _particles

Conversion Factors:

6.02 x 1023 _{particles and 1 mole}

1 mole 6.02 x 1023 _particles

Using Avogadro’s Number to Convert

Between Moles and # of Particles

 How many Cu atoms are in 0.50 mole of Cu?

0.50 mole Cu x 6.02 x 1023 _{Cu atoms}

1 mole Cu = 3.0 x 1023 _{Cu atoms}

 How many moles of CO2 are in 2.50 x 1024 molecules of CO2?

2.50 x 1024 _{molecules CO2 x 1 mole CO2}

6.02 x 1023 _{molecules CO}

2

= 4.15 moles of CO2

Subscripts in Formulas and Moles

The subscriptsin a formula give  the relationship of atoms in the formula

 the moles of each element in 1 mole of a compound

 Example: carbon dioxide, CO2

 1 molecule contains 1 C atom, and 2 O atoms

 1 mole of CO2contains 1 mole C atoms, 2 moles O atoms

 Example: glucose, C6H12O6

 1 molecule contains 6 atoms C, 12 atoms H, 6 atoms O  1 mole C6H12O6 contains 6 moles C, 12 moles H, 6 moles O

Factors from Subscripts

Subscripts used for conversion factors

 relate moles of each element in 1 mole compound

 for aspirin, C9H8O4 ,can be written as:

9 moles C 8 moles H 4 moles O

1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4

and

1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4

9 moles C 8 moles H 4 moles O

Using Avogadro’s Number to Convert

Between Moles and # of Particles

How many Ca2+ ions are in a mole of CaCl2? How many Cl– ions?

1 CaCl2 formula unit contains 1 Ca2+ ion, and 2 Cl– ions.

(1 mol Ca2+ _{ions)(6.02 x 10}23 _{ions/mol) = 6.02 x 10}23 _Ca2+ _ions

(2 mol Cl– _{ions)(6.02 x 10}23 _{ions/mol) = 1.20 x 10}24 _Cl– _ions

1 mole CaCl2 formula units contain 1 mole Ca2+ ions, 2 moles 2 Cl– ions.

Now convert from moles to numbers by multiplying by Avogadro’s number.

Practice Using Avogadro’s Number to

Convert Between Moles and # of Particles

How many O atoms are in 0.150 mole of aspirin, C9H8O4?

Write a plan: moles aspirin  moles of O  atoms of O

Molecular Mass is the Mass of

One Molecule of a Compound

To calculate the molecular mass of a compound, add the atomic masses of all the atoms in the formula in terms of the atomic mass

unit.

Remember that amu is also represented as just u.

mass of

hydrogen mass of oxygen

2´1.008

(

)

(

)

Fe2O3

2´55.85

(

)

(

)

(

)

CaCO3

1´40.08

(

)

(

)

(

)

The Mole is a Connection Between the

Atomic-Molecular Scale and the Real World

 The molar mass is the mass of one mole of a substance in grams; numerically equal to the atomic or molecular mass.

1 u = 1/12 the mass of a carbon-12 atom 1 u = 1 g/mol

Atomic weight of carbon = 12.01 u = 12.01 g/mol

Therefore, 1 mole of carbon atoms weighs 12.01 g.

Molecular weight of CaCO3 = 100.09 u =100.09 g/mol

1 mole of CaCO3 weighs 100.09 g.

For a Compound, the Molar Mass is the

Sum of the Molar Masses of the Elements

in the Formula

What is the molar mass of ammonium dichromate, (NH4)2Cr2O7?

In other words, what is the mass (in grams) of 1 mole of ammonium dichromate?

(2 x 14.0) + (8 x 1.0) + (2 x 52.0) + (7 x 16.0) = 252.0 g/mol

nitrogen hydrogen chromium oxygen

Therefore, 1 mol of ammonium dichromate

weighs

252.0 g

Molar Mass May Be Used as a

Conversion Factor to Convert

Moles

Mass

Molar Mass can be written as an equality and two conversion factors.

Equality:

1 mole F = 19.0 g F

Conversion Factors:

19.0 g F and 1 mole

1 mole 19.0 g F

Using Molar Mass to Convert Between

Moles and Mass

Juglone is a dye that is produced from the husks of black walnuts. It is also a natural herbicide (weed killer) that kills off competative plants around the black walnut tree but does not affect grass and other

noncompetative plants. The formula for Juglone is C10H6O3. A

sample of 1.56 x 10-2 _{g of juglone}

was extracted from black walnut husks. How many moles of juglone

Using Molar Mass to Convert Between

Moles and Mass

Calcium carbonate (CaCO3), also known as calcite, is the

principal mineral found in limestone, marble, chalk, pearls, and the shells of marine animals such as clams.



number

Mass/Moles/Particles Conversions

Grams of

Substance Substance Moles of or Molecules # of Atoms



molar mass



Avogadro’s number



molar mass

Divide by Molar Mass to Convert

Mass to Moles

Caffeine has the formula C8H10N4O2. If an average cup

of coffee contains approximately 125 mg of caffeine, how many moles of caffeine are in one cup?

125 mg 1000 mg/g æ è

ç ö

ø

÷

194.20 g/mol = 6.436663234´10 -4

(

)

Grams of

Substance Substance Moles of



molar mass

# of Atoms or Molecules



Avogadro’s number



Avogadro’s number



molar mass

Multiply by Molar Mass to Convert

Moles to Mass

What is the mass in grams of

0.0015 mol of aspirin, C

H

O

?

Grams of

Substance Substance Moles of



molar mass

# of Atoms or Molecules



Avogadro’s number



Avogadro’s number



molar mass

0.0015 mol

(

)

(

)

To Convert Mass to Particles you

Must Go Through Moles!

Isopentyl acetate (C7H14O2) is the compound responsible for the

scent of bananas. Bees release about 1 mg (1 x 10-3 _{g) of this}

compound when they sting. The resulting scent attracts other bees to join the attack. How many molecules of isopentyl acetate

are released in a typical bee sting?



number

To Convert Mass to Particles you

Must Go Through Moles!

A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg.

How many silicon (Si) atoms are present in the chip?

Grams of

Substance Substance Moles of or Molecules # of Atoms



mol. wt.



Avogadro’s number



mol. wt.

How Many Atoms in a Given Mass?

g



mol



atoms

Consider that a diamond in a jewelry item is made of only carbon atoms. If the diamond had a mass of 0.20 grams (about 1 carat),

how many atoms of carbon would be present?

1. 12

2. 2.4

3. 1.0  1022

Calculate Mass of Atoms

atoms



mol



g

Since atoms are so very small there will be a large number even in a small sample of an element. For example, if you just received 10 billion atoms of gold would this be better than winning a lottery prize?

Prove by reporting the mass of gold.

1. The mass of gold would be 1970 grams.

2. The mass of gold would be 3  10–22 _grams.

3. The mass of gold would be 1.7  10–17 _grams.

4. The mass of gold would be 3  10–12 _grams.

Calculate Mass of Element From

Mass of Compound

The compound Cr2O3 (chromium (III) oxide) is one of the key components responsible for the red color of ruby gems.

If you had 34.8 grams of Cr2O3, how many grams of chromium (atomic number = 24) metal would be present?

1. 11.9 grams of Cr

2. 5.85 grams of Cr

3. 23.8 grams of Cr

4. 69.8 grams of Cr

Percent Composition

 Mass percent of an element:

 For iron in iron (III) oxide, (Fe2O3)

mass mass of element in compound mass of compound

% 100%

mass%Fe .

. .

 11169 

159 69 100% 69 94%

Percentage Composition

compound

 Calculate the percentage composition of borax, Na2B2O7, used

in commercial laundry processes.

% Na =

( )

(

)

179.60 ´100% = 25.60%

% B =

( )

(

)

179.60 ´100% = 12.04%

% O =

( )

(

)

179.60 ´100% = 62.36%

Note: sum of the percentages equals 100%

% element =( number of atoms of that element )( atomic weight of element ) molecular weight of compound ´100%

Calculating Percent By Mass of

Elements in a Compound

Morphine, derived from opium plants, has the potential for use and abuse. It’s formula is C17H19NO3. What percent, by mass,

is the carbon in this compound?

1. 42.5%

2. 27.9%

3. 71.6%

4. This cannot be solved until the mass of

Determining the Formula of a

Compound

Molecular and Empirical Formulae

molecular formula = (CH)6 = C6H6 [n = 6] Molecular Formula

The chemical formula that gives the actual numbers and types of atoms in a molecule.

Empirical Formula

The chemical formula of a compound written with the smallest integer subscripts.

Example: benzene

C

H

CH

Substances Whose Empirical

and Molecular Formulas Differ

Empirical Formula Problem #1

66.07 g of hydrogen, and 524.2 g of oxygen.

There is a saying to help you remember the steps to follow to solve this problem:

Percent to mass Mass to moles Divide by small Multiply till whole

Empirical Formula Problem

#1

Calculate the empirical formula of glucose, a simple sugar, if a certain sample contains 393.4 g of carbon, 66.07 g of hydrogen,

and 524.2 g of oxygen.

We can skip percent to mass and convert each of the masses to moles.

393.4 12.01 æ è

ç ö_ø÷ = 32.76 mol C

66.07 1.008 æ è

ç ö_ø÷ = 65.55 mol H

524.2 16.00 æ è

ç ö

ø

÷ = 32.76 mol O

Empirical Formula Problem

#1

Calculate the empirical formula of glucose, a simple sugar, if a certain sample contains 393.4 g of carbon, 66.07 g of

hydrogen, and 524.2 g of oxygen.

Divide each by the smallest number of moles.

The empirical formula is CH2O.

32.76 mol C 32.76 æ è

ç ö_ø÷ = 1.000 mol C

65.55 mol H 32.76 æ è

ç ö_ø÷ = 2.001 mol H

32.76 mol O 32.76 æ è

ç ö

ø

Empirical Formula Problem

#2

Calculate the empirical formula of a compound composed of 52.9% carbon and 47.1% oxygen.

Step 1: “Percent to Mass” Convert the percentages to grams. Assume you have 100.0 g of the compound.

52.9% ´100.0 = 52.9 g C

47.1% ´100.0 = 47.1 g O

Empirical Formula Problem

#2

Calculate the empirical formula of a compound composed of 52.9% carbon and 47.1% oxygen.

Step 2: “Mass to moles” Convert each of the masses to moles.

52.9 12.01 æ è

ç ö_ø÷ = 4.40 mol C

47.1 16.00 æ è

ç ö

ø

÷ = 2.94 mol O

Empirical Formula Problem

#2

Calculate the empirical formula of a compound composed of 52.9% carbon and 47.1% oxygen.

Step 3: “Divide by small” Divide each by the smallest number of moles.

4.40 mol C 2.94 æ è

ç ö

ø

÷ = 1.50 mol C

2.94 mol O 2.94 æ è

ç ö_ø÷ = 1.00 mol O

Empirical Formula Problem

#2

Calculate the empirical formula of a compound composed of 52.9% carbon and 47.1% oxygen.

Step 4: “Multiply till whole” Find thesmallest whole-number ratio.

4.40 mol C 2.94 æ è

ç ö

ø

÷ = 1.50 mol C

2.94 mol O 2.94 æ è

ç ö_ø÷ = 1.00 mol O

The empirical formula is C3O2.

Empirical/Molecular Formula Problem

A compound with a molecular weight of 180 u consists of 40.0% carbon, 6.70% hydrogen, and 53.3% oxygen. What is its empirical

formula? What is its molecular formula?

Step 1: “Percent to mass” Convert the percentages to grams. Assume you have 100.0 g of the compound.

40.0% ´100.0 = 40.0 g C

6.70% ´100.0 = 6.70 g H

53.3% ´100.0 = 53.3 g O

Empirical/Molecular Formula Problem

A compound with a molecular weight of 180 u consists of 40.0% carbon, 6.70% hydrogen, and 53.3% oxygen. What is its

molecular formula?

Step 2: “Mass to moles” Convert each of the masses to moles.

40.0 12.01 æ è

ç ö_ø÷ = 3.33 mol C

6.70 1.008 æ è

Empirical/Molecular Formula Problem

A compound with a molecular weight of 180 u consists of 40.0% carbon, 6.70% hydrogen, and 53.3% oxygen. What is its

molecular formula?

Step 3:”Divide by small” Divide each by the smallest number of moles.

3.33 mol C 3.33 æ è

ç ö_ø÷ = 1.00 mol C

3.33 mol O 3.33 æ è

ç ö_ø÷ = 1.00 mol O

6.65 mol H 3.33 æ è

ç ö_ø÷ = 2.00 mol H formula is CHThe empirical 2O.

Empirical/Molecular Formula Problem

A compound with a molecular weight of 180 u consists of 40.0% carbon, 6.70% hydrogen, and 53.3% oxygen. What is its

molecular formula?

Step 4: Determine the mass of the empirical formula and compare it to the molecular formula.

Empirical formula CH2O 30.03 u

180

30.03 = 5.99 » 6