Dr. S. M. Condren Chapter 3 Calculations with Chemical Formulas and Equations.

Dr. S. M. Condren

Chapter 3

Calculations with

Chemical Formulas

and Equations

Dr. S. M. Condren

Molar Mass

Sum

atomic masses

represented by formula

atomic masses => gaw

molar mass => MM

Dr. S. M. Condren

One Mole of each Substance

Clockwise from top left:1-Octanol, C8H17OH;Mercury(II) iodide, HgI2;Methanol, CH3OH; andSulfur, S8.

Dr. S. M. Condren

ExampleWhat is the molar mass of ethanol, C2H5O1H1?

MM = 2(gaw)C + (5 + 1)(gaw)H + 1(gaw)O

= 2(12.011)C + 6(1.00794)H + 1(15.9994)O

= 24.022 + 6.04764 + 15.9994

= 46.069 g/molSignificant figures rule for multiplication

Significant figures rule for addition

Sequence – multiplication then addition,apply significant figure rules in proper sequence

Dr. S. M. Condren

The Mole

• a unit of measurement, quantity of matter present

• Avogadro’s Number6.022 x 1023 particles

• Latin for “pile”

Dr. S. M. Condren

Example How many moles of carbon dioxide molecules are there in 10.00g of carbon dioxide? CO2

MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

#mol CO2 = (10.00g)(1 mol/44.01g)

Dr. S. M. Condren


MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

#mol CO2 = (10.00g)(1 mol/44.01g)

Dr. S. M. Condren


MM = 1(gaw)C + 2(gaw)O = 44.01 g/mol

#mol CO2 = (10.00)(1 mol/44.01)

= 0.2272 mol

Dr. S. M. Condren

Combustion Analysis

Dr. S. M. Condren

Percentage Composition

description of a compound based on the relative amounts of each element in the compound

Dr. S. M. Condren

EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?

MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl

= (12.011 + 1.00797 + 3*35.453)amu

= 119.377amu

Dr. S. M. Condren



= (12.011 + 1.00797 + 3*35.453)amu

= 119.377amu

1(gaw)%C = ------------ X 100

MM 1(12.011)

%C = -------------- X 100 = 10.061% C 119.377

Dr. S. M. Condren



= (12.011 + 1.00797 + 3*35.453)amu

= 119.377amu

1(1.00797)%H = ---------------- X 100 = 0.844359% H

119.377

Dr. S. M. Condren



= (12.011 + 1.00797 + 3*35.453)amu

= 119.377amu

3(35.453)%Cl = -------------- X 100 = 89.095% Cl

119.377

Dr. S. M. Condren

EXAMPLE: What is the percent composition of chloroform, CHCl3, a substance once used as an anesthetic?MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl

= 119.377amu

%C = 10.061% C

%H = 0.844359% H

%Cl = 89.095% Cl

Dr. S. M. Condren

Simplest (Empirical) Formula

• formula describing a substance based on the smallest set of subscripts

Dr. S. M. Condren

Acetylene, C2H2, and benzene, C6H6, have the same empirical formula. Is the correct empirical formula:

C2H2

CH

C6H6

Dr. S. M. Condren

EXAMPLE: A white compound is formed when phosphorus burns in air. Analysis shows that the compound is composed of 43.7% P and 56.3% O by mass. What is the empirical formula of the compound?

Element

P

O

%

43.7 56.3

Relative Number of Atoms(%/gaw)

43.7/30.9756.3/15.9994

= 1.41= 3.52

Divide by Smaller1.41/1.41 = 1.003.52/1.41 = 2.50

Dr. S. M. Condren

EXAMPLE: A white compound is formed when phosphorus burns in air. Analysis shows that the compound is composed of 43.7% P and 56.3% O by mass. What is the empirical formula of the compound?

Relative Number of Atoms Multiply % (%/gaw) Divide by Smaller by Integer P 43.7 43.7/30.97 = 1.41 1.41/1.41 = 1.00 2*1.00 => 2

O 56.3 56.3/15.9994 = 3.52 3.52/1.41 = 2.50 2*2.50 => 5

Empirical Formula => P2O5

Dr. S. M. Condren

EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound?

2.34 %N = ----------------- X 100 = 30.5% N 2.34 + 5.34

5.34%O = ----------------- X 100 = 69.5% O 2.34 + 5.34

Dr. S. M. Condren

EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound?%N = 30.5% N %O = 69.5% O

Element N O

%30.5 69.5

Relative # Atoms (%/gaw)30.5/14.0067 = 2.1869.5/15.9994 = 4.34

Divide by Smaller 2.18/2.18 = 1.00 4.34/2.18 = 1.99

Dr. S. M. Condren

EXAMPLE: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the empirical formula of the compound?%N = 30.5% N %O = 69.5% O

Element N O

%30.5 69.5

Relative # Atoms (%/gaw)30.5/14.0067 = 2.1869.5/15.9994 = 4.34

Divide by Smaller 2.18/2.18 = 1.00 4.34/2.18 = 1.99

Multiplyby Integer1*1.00=>11*1.99=>2

Empirical Formula => NO2

Dr. S. M. Condren

Molecular Formula

• the exact proportions of the elements that are formed in a molecule

Dr. S. M. Condren

Molecular Formula from Simplest Formula

empirical formula => EF

molecular formula => MF

MF = X * EF

Dr. S. M. Condren


formula mass => FM

sum of the atomic weights represented by the formula

molar mass = MM = X * FM

Dr. S. M. Condren


first, knowing MM and FM

X = MM/FM

then

MF = X * EF

Dr. S. M. Condren

EXAMPLE: A colorless liquid used in rocket engines, whose empirical formula is NO2, has a molar mass of 92.0. What is the molecular formula?

FM = 1(gaw)N + 2(gaw)O = 46.0 MM 92.0X = ------- = -------- = 2 FM 46.0

thus MF = 2 * EF

Dr. S. M. Condren

What is the correct molecular formula for this colorless liquid rocket fuel?

2NO2

NO2

N2O4

Dr. S. M. Condren

Stoichiometry

stoi·chi·om·e·try noun1. Calculation of the quantities of

reactants and products in a chemical reaction.

2. The quantitative relationship between reactants and products in a chemical reaction.

Dr. S. M. Condren

The Mole and Chemical Reactions:The Macro-Nano Connection

2 H2 + O2 -----> 2 H2O 2 H2 molecules 1 O2 molecule 2 H2O molecules

2 H2 moles molecules 1 O2 mole molecules 2 H2O moles molecules

4 g H2 32 g O2 36 g H2O

Dr. S. M. Condren

EXAMPLE

How much H2O, in moles results from burning an excess of H2 in 3.3 moles of O2?

H2 + O2 -----> H2O

2 H2 + O2 -----> 2 H2O

Dr. S. M. Condren

EXAMPLE


H2 + O2 -----> H2O

2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2O)

#mol H2O = ------------------------------------ (1 mol O2)

Dr. S. M. Condren

EXAMPLE


H2 + O2 -----> H2O

2 H2 + O2 -----> 2 H2O (3.3 mol O2) (2 mol H2O)

#mol H2O = ------------------------------------ (1 mol O2)

Dr. S. M. Condren

EXAMPLE


H2 + O2 -----> H2O

2 H2 + O2 -----> 2 H2O (3.3) (2 mol H2O)

#mol H2O = ------------------------ = 6.6 mol H2O (1)

Dr. S. M. Condren

Combination Reaction

PbNO3(aq) + K2CrO4(aq) PbCrO4(s) + 2 KNO3(aq)

Colorless yellow yellow colorless

Dr. S. M. Condren

Stoichiometric Roadmap

Dr. S. M. Condren

EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together?

Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron.

http://www.cbu.edu/~mcondren/Demos/Thermite-Welding.ppt

Dr. S. M. Condren

EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together?

Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additional 10% mass of iron.

The mass of iron in 1 inch of this rail is:

#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)

= 1.67 X 103 g/in

Dr. S. M. Condren

EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that

one inch of the rails will be covered by an additional 10% mass of iron.

The mass of iron in 1 inch of this rail is:#g/in = (132) (1/36 in) (454 g)

= 1.67 X 103 g/inThe mass of iron in a weld adding 10% mass:

#g = (1.67 X 103 g) (0.10) = 167 g

Dr. S. M. Condren

EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron.


#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)

= 1.67 X 103 g/in

The mass of iron in a weld adding 10% mass:

#g = (1.67 X 103 g) (0.10) = 167 g

Balanced chemical equation:

Fe2O3 + 2 Al ---> 2 Fe + Al2O3

Dr. S. M. Condren



#g/in = (132 #/yard) (1 yard/36 in) (454 g/#)

= 1.67 X 103 g/in


#g = (1.67 X 103 g) (0.10) = 167 g


Fe2O3 + 2 Al ---> 2 Fe + Al2O3

What mass of Fe2O3 is required for the thermite process?

Dr. S. M. Condren



#g = (1.67 X 103 g) (0.10) = 167 g


Fe2O3 + 2 Al ---> 2 Fe + Al2O3

What mass of Fe2O3 is required for the thermite process?

#g Fe2O3 = (167 g Fe) (1 mol Fe) * -------------- (55.85 g Fe)

(1 mol Fe2O3)----------------- (2 mol Fe)

(159.7 g Fe2O3)------------------- (1 mol Fe2O3)= 238 g Fe2O3

Dr. S. M. Condren

EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb/yard and that one inch of the rails will be covered by an additonal 10% mass of iron.The mass of iron in a weld adding 10% mass: #g Fe = 167 g Fe

Balanced chemical equation:Fe2O3 + 2 Al ---> 2 Fe + Al2O3

What mass of Fe2O3 is required for the thermite process?#g Fe2O3 = 238 g Fe2O3

What mass of Al is required for the thermite process?

#g Al = (167 g Fe) (1 mol Fe) * ---------------- (55.85 g Fe)

(2 mol Al) -------------(2 mol Fe)

(26.9815 g Al)------------------- (1 mol Al)

= 80.6 g Al

Dr. S. M. Condren



#g Fe = 167 g Fe

#g Fe2O3 = 238 g Fe2O3

#g Al = 80.6 g Al

Dr. S. M. Condren

Limiting Reactant

reactant that limits the amount of product that can be produced

Dr. S. M. Condren

EXAMPLE

What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?

2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

Dr. S. M. Condren

EXAMPLE



balanced equation relates:

2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)

Dr. S. M. Condren

EXAMPLE



balanced equation relates:

2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)

have only:

1Fe2S3(S) <=> 2H2O(l) <=> 3O2(g)

Dr. S. M. Condren

EXAMPLE What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

balanced equation relates:2Fe2S3(S) <=> 6H2O(l) <=> 3O2(g)

have only:1Fe2S3(S) <=> 2H2O(l) <=> 3O2(g)

not enough H2O to use all Fe2S3

plenty of O2

Dr. S. M. Condren

EXAMPLE What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?


if use all Fe2S3:

(1.0 mol Fe2S3) (4 mol Fe(OH)3)

#mol Fe(OH)3 = ------------------------------------------ (2 mol Fe2S3)

= 2.0 mol Fe(OH)3

Dr. S. M. Condren

EXAMPLE: What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?


if use all H2O:

(2.0 mol H2O) (4 mol Fe(OH)3) #mol Fe(OH)3 = -----------------------------------------

(6 mol H2O)= 1.3 mol Fe(OH)3

Dr. S. M. Condren

EXAMPLE: What is the number of moles of Fe(OH)3 (S) that can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O, and 3.0 mol O2 to react?2Fe2S3(S) + 6H2O(l) + 3O2(g) -----> 4Fe(OH)3(S) + 6S(S)

if use all O2

(3.0 mol O2) (4 mol Fe(OH)3) #mol Fe(OH)3 = ---------------------------------------

(3 mol O2)

= 4.0 mol Fe(OH)3

Dr. S. M. Condren



1.0 mol Fe2S3 => 2.0 mol Fe(OH)3

2.0 mol H2O => 1.3 mol Fe(OH)3

3.0 mol O2 => 4.0 mol Fe(OH)3

Dr. S. M. Condren




2.0 mol H2O => 1.3 mol Fe(OH)3

3.0 mol O2 => 4.0 mol Fe(OH)3

Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3, then H2O is the limiting reactant. Thus the correct number of moles of Fe(OH)3 is 1.33 moles.

Dr. S. M. Condren




2.0 mol H2O => 1.3 mol Fe(OH)3 least amount

3.0 mol O2 => 4.0 mol Fe(OH)3

Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3, then H2O is the limiting reactant.

Dr. S. M. Condren




2.0 mol H2O => 1.3 mol Fe(OH)3 least amount

3.0 mol O2 => 4.0 mol Fe(OH)3

Since 2.0 mol H2O will produce only 1.3 mol Fe(OH)3, then H2O is the limiting reactant. Thus the maximum number of moles of Fe(OH)3 that can be produced by this reaction is 1.3 moles.

Dr. S. M. Condren

Theoretical Yield

the amount of product produced by a reaction based on the amount of the limiting reactant

Dr. S. M. Condren

Actual Yield

amount of product actually produced in a reaction

Dr. S. M. Condren

Percent Yield

actual yield% yield = --------------------- * 100 theoretical yield

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?

2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O

(a) to calculate the theoretical yield, use the net equation for the overall process

Dr. S. M. Condren


2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) to calculate the theoretical yield, use the net equation for

the overall process

(1.00 kg Cl2) #kg N2H4 = ---------------------

Dr. S. M. Condren



the overall process

(1.00 kg Cl2) (1000 g Cl2)#kg N2H4 = -----------------------------------

(1 kg Cl2) metric conversion

Dr. S. M. Condren



the overall process

(1.00) (1000 g Cl2) (1 mol Cl2) #kg N2H4 = -----------------------------------------

(1) (70.9 g Cl2) molar mass

Dr. S. M. Condren



the overall process

(1.00)(1000)(1 mol Cl2) #kg N2H4 = -----------------------------------------

(1)(70.9)

Dr. S. M. Condren



the overall process

(1.00)(1000)(1 mol Cl2)(1 mol N2H4) #kg N2H4 = -------------------------------------------------

(1) (70.9) (1 mol Cl2)

Dr. S. M. Condren



the overall process

(1.00)(1000)(1)(1 mol N2H4) #kg N2H4 = -------------------------------------------------

(1) (70.9)(1)

Dr. S. M. Condren



the overall process

(1.00)(1000)(1)(1 mol N2H4) (32.0 g N2H4) #kg N2H4 = --------------------------------------------------------

(1)(70.9) (1) (1 mol N2H4)

molar mass

Dr. S. M. Condren



the overall process

(1.00)(1000)(1)(1) (32.0 g N2H4)(1 kg N2H4)#kg N2H4 = ----------------------------------------------------------

(1)(70.9)(1)(1) (1000 g N2H4)

metric conversion

Dr. S. M. Condren



the overall process(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)

#kg N2H4 = ----------------------------------------------------------(1)(70.9)(1)(1)(1000)

Dr. S. M. Condren



the overall process(1.00)(1000)(1)(1) (32.0)(1 kg N2H4)

#kg N2H4 = ----------------------------------------------------------(1)(70.9)(1)(1)(1000)

= 0.451 kg N2H4

Dr. S. M. Condren


2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield

(0.299 kg product) # kg N2H4 = --------------------------

Dr. S. M. Condren



(b) actual yield (0.299 kg product) (98.0 kg N2H4)

# kg N2H4 = -------------------------------------------- (100 kg product)

purity factor

Dr. S. M. Condren

EXAMPLE A chemical plant obtained 0.299 kg of 98.0% N2H4 for every 1.00 kg of Cl2 that is reacted with excess NaOH and NH3. What are the: (a) theoretical, (b) actual, and (c) percent yield of pure N2H4?2NaOH + Cl2 + 2NH3 -----> N2H4 + 2NaCl + 2H2O(a) theoretical yield #kg N2H4 = 0.451 kg N2H4

(b) actual yield (0.299 kg product) (98.0 kg N2H4)

# kg N2H4 = -------------------------------------------- (100 kg product)

= 0.293 kg N2H4

Dr. S. M. Condren



(b) actual yield # kg N2H4 = 0.293 kg N2H4

Dr. S. M. Condren



(b) actual yield # kg N2H4 = 0.293 kg N2H4

(c) percent yield 0.293 kg

% yield = -------------- X 100 = 65.0 % yield 0.451kg

Dr. S. M. Condren Chapter 3 Calculations with Chemical Formulas and Equations.

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