chapter 3 calculations with chemical formulas and...

Chapter 3

Calculations with

Chemical

Formulas and

Equations

Copyright © Cengage Learning. All rights reserved. 3 | 2

Contents and Concepts

Mass and Moles of Substances

Here we will establish a critical relationship

between the mass of a chemical substance and

the quantity of that substance (in moles).

1. Molecular Mass and Formula Mass

2. The Mole Concept


Determining Chemical Formulas

Explore how the percentage composition and

mass percentage of the elements in a chemical

substance can be used to determine the

chemical formula.

3. Mass Percentages from the Formula

4. Elemental Analysis: Percentages of C, H, and O

5. Determining Formulas


Stoichiometry: Quantitative Relations in

Chemical Reactions

Develop a molar interpretation of chemical

equations, which then allows for calculation

of the quantities of reactants and products.

6. Molar Interpretation of a Chemical Equation

7. Amounts of Substances in a Chemical Equation

8. Limiting Reactant: Theoretical and Percentage

Yield


Learning Objectives

Mass and Moles of Substances

1. Molecular Mass and Formula Mass

a. Define the terms molecular and formula mass of a substance.

b. Calculate the formula mass from a formula.

c. Calculate the formula mass from molecular models.


2. The Mole Concept

a. Define the quantity called the mole.

b. Learn Avogadro’s number.

c. Understand how the molar mass is related

to the formula mass of a substance.

d. Calculate the mass of atoms and molecules.

e. Perform calculations using the mole.

f. Convert from moles of substance to grams

of substance.

g. Convert from grams of substance to moles

of substance.

h. Calculate the number of molecules in a

given mass of a substance.


Determining Chemical Formulas

3. Mass Percentages from the Formula

a. Define mass percentage.

b. Calculate the percentage composition of

the elements in a compound.

c. Calculate the mass of an element in a

given mass of compound.


4. Elemental Analysis: Percentages of C, H, and O

a. Describe how C, H, and O combustion

analysis is performed.

b. Calculate the percentage of C, H, and O

from combustion data.


5. Determining Formulas

a. Define empirical formula.

b. Determine the empirical formula of a binary

compound from the masses of its elements.

c. Determine the empirical formula from the

percentage composition.

d. Understand the relationship between the

molecular mass of a substance and its

empirical formula mass.

e. Determine the molecular formula from the

percentage composition and molecular

mass.


Stoichiometry: Quantitative Relations in

Chemical Reactions

6. Molar Interpretation of a Chemical Equation

a. Relate the coefficients in a balanced

chemical equation to the number of

molecules or moles (molar interpretation).


7. Amounts of Substances in a Chemical Equation

a. Use the coefficients in a balanced chemical

equation to perform calculations.

b. Relate the quantities of reactant to the

quantity of product.

c. Relate the quantities of two reactants or two

products.


8. Limiting Reactant: Theoretical and Percentage Yield

a. Understand how a limiting reactant determines how many moles of product are formed during a chemical reaction and how much excess reactant remains.

b. Calculate with a limiting reactant involving moles.

c. Calculate with a limiting reactant involving masses.

d. Define and calculate the theoretical yield of chemical reactions.

e. Determine the percentage yield of a chemical reaction.


Molecular Mass

The sum of the atomic masses of all the atoms in a

molecule of the substance.

Formula Mass

The sum of the atomic masses of all atoms in a

formula unit of the compound, whether molecular

or not.


Calculate the formula weight of the

following compounds from their

formulas. Report your answers to three

significant figures.

calcium hydroxide, Ca(OH)2

methylamine, CH3NH2


3 significant figures

74.1 amu

3 significant figures

31.1 amu

Total 74.095

2 O 2(16.00) = 32.00 amu

Ca(OH)21 Ca 1(40.08) = 40.08 amu

2 H 2(1.008) = 2.016 amu

CH3NH21 C 1(12.01) = 12.01 amu

5 H 5(1.008) = 5.040 amu

1 N 1(14.01) = 14.01 amu

Total 31.060


What is the mass in grams of the nitric

acid molecule, HNO3?

First, find the molar mass of HNO3:

1 H 1(1.008) = 1.008

1 N 1(14.01) = 14.01

3 O 3(16.00) = 48.00

63.018 (2 decimal places)

63.02 g/mol


Mole, mol

The quantity of a given amount of substance that

contains as many molecules or formula units as

the number of atoms in exactly 12 g of carbon-12.

Avogadro’s Number, NA

The number of atoms in exactly 12 g of carbon-12

NA = 6.02 × 1023 (to three significant figures).


molecules10x6.02

mol1 x

mol

g63.0223

g10x41.04684385 22��

figures)tsignifican(3

g10x1.05ismoleculeHNOoneofmassThe 223 .��

Next, convert this mass of one mole to one

molecule using Avogadro’s number:


Molar Mass

The mass of one mole of substance.

For example:

Carbon-12 has a molar mass of 12 g or 12 g/mol


A sample of nitric acid, HNO3, contains

0.253 mol HNO3. How many grams is

this?

First, find the molar mass of HNO3:

1 H 1(1.008) = 1.008

1 N 1(14.01) = 14.01

3 O 3(16.00) = 48.00


63.02 g/mol


mole1

g63.02xmole0.253


g15.9��

Next, using the molar mass, find the mass of 0.253

mole:

= 15.94406 g


Calcite is a mineral composed of

calcium carbonate, CaCO3. A sample

of calcite composed of pure calcium

carbonate weighs 23.6 g. How many

moles of calcium carbonate is this?

First, find the molar mass of CaCO3:

1 Ca 1(40.08) = 40.08

1 C 1(12.01) = 12.01

3 O 3(16.00) = 48.00

100.09 2 decimal places

100.09 g/mol


g100.09

mole1xg23.6

g10x2.35787791 1��


g0.236org10x2.36 1��

Next, find the number of moles in 23.6 g:


The average daily requirement of the

essential amino acid leucine, C6H14O2N,

is 2.2 g for an adult. What is the average

daily requirement of leucine in moles?

First, find the molar mass of leucine:

6 C 6(12.01) = 72.06

2 O 2(16.00) = 32.00

1 N 1(14.01) = 14.01

14 H 14(1.008) = 14.112

132.182

2 decimal places

132.18 g/mol


g132.18

mole1xg2.2

mol10x1.6643 2��


mol0.017ormol10x1.7 2��

Next, find the number of moles in 2.2 g:


The daily requirement of chromium in

the human diet is 1.0 × 10-6 g. How

many atoms of chromium does this

represent?


mol1

atoms10x6.02x

g51.996

mol1xg10x1.0

236��

First, find the molar mass of Cr:

1 Cr 1(51.996) = 51.996

Now, convert 1.0 x 10-6 grams to moles:

=1.157781368 x 1016 atoms

1.2 x 1016 atoms

(2 significant figures)


Lead(II) chromate, PbCrO4, is used as

a paint pigment (chrome yellow). What

is the percentage composition of

lead(II) chromate?

First, find the molar mass of PbCrO4:

1 Pb 1(207.2) = 207.2

1 Cr 1(51.996) = 51.996

4 O 4(16.00) = 64.00

323.196 (1 decimal place)

323.2 g/mol


64.11%100%xg323.20

g207.2:Pb ��

16.09%100%xg323.20

g51.996:Cr ��

19.80%100%xg323.20

g64.00:O ��

Now, convert each to percent composition:

Check:

64.11 + 16.09 + 19.80 = 100.00


The chemical name of table sugar is

sucrose, C12H22O11. How many grams

of carbon are in 68.1 g of sucrose.

First, find the molar mass of C12H22O11:

12 C 12(12.01) = 144.12

11 O 11(16.00) = 176.00

22 H 22(1.008) = 22.176


342.30 g/mol



carbong26.0��

sucroseg342.30

carbong144.12xsucroseg61.8

Now, find the mass of carbon in 61.8 g sucrose:


Percentage Composition

The mass percentage of each element in the

compound.

The composition is determined by experiment,

often by combustion. When a compound is burned,

its component elements form oxides—for example,

CO2 and H2O. The CO2 and H2O are captured and

weighed to determine the amount of C and H in

the original compound.


Benzene is a liquid compound

composed of carbon and hydrogen; it is

used in the preparation of polystyrene

plastic. A sample of benzene weighing

342 mg is burned in oxygen and forms

1156 mg of carbon dioxide. What is the

percentage composition of benzene?


Strategy

1. Use the mass of CO2 to find the mass of carbon

from the benzene.

2. Use the mass of benzene and the mass of

carbon to find the mass of hydrogen.

3. Use these two masses to find the percent

composition.


g10

mg1Cg10x23.15463758

3

1

��

��

222

22

3

COmol1

Cg12.01x

COmol1

Cmol1x

COg44.01

COmol1COg10x1156 ��

= 315.5 mg C

First, find the mass of C in 1156 mg of CO2:


Now, we can find the percentage composition:

��x100%mg342

mg26.5

��x100%mg342

mg315.5C92.3%

H7.7%

342 mg benzene

-315.5 mg C

26.5 mg H

(the decimal is not significant)

Next, find the mass of H in the benzene sample:


Empirical Formula (Simplest Formula)

The formula of a substance written with the

smallest integer subscripts.

For example:

The empirical formula for N2O4 is NO2.

The empirical formula for H2O2 is HO


Determining the Empirical Formula

Beginning with percent composition:

1. Assume exactly 100 g so percentages convert

directly to grams.

2. Convert grams to moles for each element.

3. Manipulate the resulting mole ratios to obtain

whole numbers.


Manipulating the ratios:

Divide each mole amount by the smallest mole

amount.

If the result is not a whole number:

Multiply each mole amount by a factor.

For example:

If the decimal portion is 0.5, multiply by 2.

If the decimal portion is 0.33 or 0.67, multiply by 3.

If the decimal portion is 0.25 or 0.75, multiply by 4.


Benzene is composed of 92.3% carbon

and 7.7% hydrogen. What is the

empirical formula of benzene?

Hmol7.64Hg1.008

Hmol1Hg7.7

Cmol7.685Cg12.01

Cmol1Cg92.3

��

��

17.64

7.64

17.64

7.685

��

��

Empirical formula: CH


Molecular Formula

A formula for a molecule in which the subscripts

are whole-number multiples of the subscripts in the

empirical formula.


To determine the molecular formula:

1. Compute the empirical formula weight.

2. Find the ratio of the molecular weight to the

empirical formula weight.

3. Multiply each subscript of the empirical formula

by n.

weightformulaempirical

weightmolecular��n


Benzene has the empirical formula CH.

Its molecular weight is 78.1 amu. What

is its molecular formula?

613.02

78.1

amu13.02weightformulaEmpirical

��

��

Molecular formula

C6H6


Sodium pyrophosphate is used in

detergent preparations. It is composed

of 34.5% Na, 23.3% P, and 42.1% O.

What is its empirical formula?

Omol2.631Og16.00

Omol1Og42.1

Pmol0.7523Pg30.97

Pmol1Pg23.3

Namol1.501Nag22.99

Namol1Nag34.5

��

��

Empirical formula

Na4P2O7

72x

22x

42x

��

��

��

3.500.7523

2.631

1.000.7523

0.7523

2.000.7523

1.501

��

��

��


Hexamethylene is one of the materials

used to produce a type of nylon. It is

composed of 62.1% C, 13.8% H, and

24.1% N. Its molecular weight is 116

amu. What is its molecular formula?

Hmol1.720Ng14.01

Nmol1Ng24.1

Hmol13.69Hg1.008

Hmol1Hg13.8

Cmol5.171Cg12.01

Cmol1Cg62.1

��

��

��

11.720

1.720

81.720

13.69

31.720

5.171

��

��

��

Empirical formula

C3H8N


The empirical formula is C3H8N.

Find the empirical formula weight:

3(12.01) + 8(1.008) + 1(14.01) = 58.104 amu

Molecular formula: C6H16N2

258.10

116��n


Stoichiometry

The calculation of the quantities of reactants and

products involved in a chemical reaction.

Interpreting a Chemical Equation

The coefficients of the balanced chemical equation

may be interpreted in terms of either (1) numbers

of molecules (or ions or formula units) or (2)

numbers of moles, depending on your needs.


1. Convert grams of A to moles of A

� Using the molar mass of A

2. Convert moles of A to moles of B

� Using the coefficients of the

balanced chemical equation

3. Convert moles of B to grams of B

� Using the molar mass of B

To find the amount of B (one reactant or product)

given the amount of A (another reactant or

product):


Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:

C3H8(g) + 5O2(g) � 3CO2(g) + 4H2O(g)

How many grams of CO2 are produced when 20.0 g of propane is burned?


Molar masses

C3H8: 3(12.01) + 8(1.008) = 44.094 g

CO2: 1(12.01) + 2(16.00) = 44.01 g

59.9 g CO2


2

2

83

2

83

8383

COmol1

COg44.01

HCmol1

COmol3

HCg44.094

HCmol1HCg20.0

2COg359.8856987��


Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:

C3H8(g) + 5O2(g) � 3CO2(g) + 4H2O(g)

How many grams of O2 are required to burn 20.0 g of propane?


Molar masses:

O2 2(16.00) = 32.00 g

C3H8 3(12.01) + 8(1.008) = 44.094 g

2

2

83

2

83

8383

Omol1

Og32.00

HCmol1

Omol5

HCg44.094

HCmol1HCg20.0

2Og572.5722320��

72.6 g O2



Limiting Reactant

The reactant that is entirely consumed when a

reaction goes to completion.

Once one reactant has been completely

consumed, the reaction stops.

Any problem giving the starting amount for more

than one reactant is a limiting reactant problem.


All amounts produced and reacted are

determined by the limiting reactant.

How can we determine the limiting reactant?

1. Use each given amount to calculate the

amount of product produced.

2. The limiting reactant will produce the

lesser or least amount of product.


Magnesium metal is used to prepare

zirconium metal, which is used to make

the container for nuclear fuel (the nuclear

fuel rods):

ZrCl4(g) + 2Mg(s) � 2MgCl2(s) + Zr(s)

How many moles of zirconium metal can

be produced from a reaction mixture

containing 0.20 mol ZrCl4 and 0.50 mol

Mg?


ZrCl4 is the limiting reactant.

0.20 mol Zr will be produced.

Zrmol0.25Mgmol2

Zrmol1Mgmol0.50 ��

Zrmol0.20ZrClmol1

Zrmol1ZrClmol0.20

4

4 ��


Urea, CH4N2O, is used as a nitrogen

fertilizer. It is manufactured from

ammonia and carbon dioxide at high

pressure and high temperature:

2NH3 + CO2(g) � CH4N2O + H2O

In a laboratory experiment, 10.0 g NH3

and 10.0 g CO2 were added to a reaction

vessel. What is the maximum quantity (in

grams) of urea that can be obtained?

How many grams of the excess reactant

are left at the end of the reactions?


Molar masses

NH3 1(14.01) + 3(1.008) = 17.02 g

CO2 1(12.01) + 2(16.00) = 44.01 g

CH4N2O 1(12.01) + 4(1.008) + 2(14.01) +

1(16.00) = 60.06 g

CO2 is the limiting reactant.

13.6 g CH4N2O will be produced.

ONCHg13.6

ONCHmol1

ONCHg60.06

COmol1

ONCHmol1

COg44.01

COmol1COg10.0

24

24

24

2

24

2

22

��

ONCHg17.6

ONCHmol1

ONCHg60.06

NHmol2

ONCHmol1

NHg17.024

NHmol1NHg10.0

24

24

24

3

24

3

33

��


10.0 at start

-7.73 reacted

2.27 g remains

2.3 g NH3 is left unreacted.

(1 decimal place)

reactedNHg7.73 3��

3

3

2

3

2

22

NHmol1

NHg17.02

COmol1

NHmol2

COg44.01

COmol1COg10.0

3NHg17.73460577��

To find the excess NH3, we find how much NH3

reacted:

Now subtract the amount reacted from the

starting amount:


Theoretical Yield

The maximum amount of product that can be

obtained by a reaction from given amounts of

reactants. This is a calculated amount.


Actual Yield

The amount of product that is actually obtained.

This is a measured amount.

Percentage Yield

x100%yieldltheoretica

yieldactualyieldpercentage ��


2NH3 + CO2(g) � CH4N2O + H2O

When 10.0 g NH3 and 10.0 g CO2 are

added to a reaction vessel, the limiting

reactant is CO2. The theoretical yield is

13.6 of urea. When this reaction was

carried out, 9.3 g of urea was obtained.

What is the percent yield?

Theoretical yield = 13.6 g

Actual yield = 9.3 g

= 68% yield


100%x g13.6

g9.3

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