1

Binary compounds: only two different elements present.

Examples:

NaCl MgCl2 Al2O3Ionic

Covalent H2O CO SO2 CBr4

Ionic: metal ion (positive or cation) combined with a non-metal ion (negative or anion)

Covalent: non-metal atom combined with another non-metal atom No ions!

2

Ionic Naming

2 parts to the name of an ionic compound

1st part is the positive ion 2nd part is the negative ion

Positive ions will be named the same as the metallic element they are made from

Negative ions will have endings that depend on how many oxygen atoms are present in the ion

there are three possibilities:

no oxygen: name will be element name with the ending changed to –ide (three exceptions are oxide, hydroxide and peroxide)

fewer oxygen atoms: name will be poly atomic ion name ending with –ite

more oxygen atoms: name will be poly atomic ion name ending with –ate

3

Positive ions

Examples

sodium atom (Na) becomes sodium ion (Na+1)

calcium atom (Ca) becomes calcium ion (Ca+2)

aluminum atom (Al) becomes aluminum ion (Al+3)

Negative ions (non-metal atoms becoming negative ions) drop the ending of the element name and replace it with -ide

Hydrogen (H)Carbon (C)Nitrogen (N)Oxygen (O)Fluorine (F)Phosphorpus (P)Sulfur (S)Chlorine (Cl)Selenium (Se)Bromine (Br)Iodine (I)

becomesbecomesbecomesbecomesbecomesbecomesbecomesbecomesbecomesbecomesbecomes

Hydride (H-1)Carbide (C-4)Nitride (N-3)Oxide (O-2)Fluoride (F-1)Phosphide (P-3)Sulfide (S-2)Chloride (Cl-1)Selenide (Se-2)Bromide (Br-1)Iodide (I-1)

4

NaCl

CaS

AlN

sodium chloride

calcium sulfide

aluminum nitride

Example names

Na+1 and Cl-1

Ca+2 and S-2

Al+3 and N-3

5

If a metal comes from the “d” block (transition metal) or the “p” block in periods 4-6,the name of the compound must include the charge of the metal ion in that compound.

The charge is indicated by using (Roman Numerals).

FeCl3 iron (III) chlorideFe+3 and Cl-1

FeCl2 iron (II) chlorideFe+2 and Cl-1

Examples: you must figure out what the charge of the metal is based on the number of negative ions present and their charge

6

SnCl4

V2S5

CrN2

Example names

(Cl-1)* 4 = -4 so Sn must be +4

(S-2)* 5 = -10 so V must be +5

(N-3)* 2 = -6 so Cr must be +6

tin (IV) chloride

vanadium (V) sulfide

chromium (VI) nitride

7

Al2O3

Sodium Sulfide

Magnesium Fluoride

Aluminum Oxide

Fe2S3

Nickel (I) Bromide

Nickel (II) Bromide

Iron (III) Sulfide

Practice writing formulas from names

MgF2

NiBr2

Na2S

NiBr

Al+3 and O-2

Fe+3 and S-2

Mg+2 and F-1

Ni+2 and Br-1

Na+1 and S-2

Ni+1 and Br-1

8

Ionic Compound Naming Practice

Li +1

Ca +2

Al +3

Sn +4

V +5

Cr +6

Name the following Compounds

Cl -1 S -2 N -3

LiCl Li2S Li3N

CaCl2 CaS Ca3N2

AlCl3 Al2S3 AlN

SnCl4 SnS2 Sn3N4

VCl5 V2S5 V3N5

CrCl6 CrS3 CrN2

Lithium chloride Lithium sulfide Lithium nitride

Calcium chloride

Aluminum chloride

Tin (IV) chloride

Vanadium (V) chloride

Chromium (VI) chloride

Calcium sulfide Calcium nitride

Aluminum sulfide Aluminum nitride

Tin (IV) sulfide Tin (IV) nitride

Vanadium (V) sulfide Vanadium (V) nitride

Chromium (VI) sulfide Chromium (VI) nitride

9

Ionic Compound Naming Practice

Na +1

Ba +2

Fe +3

Pt +4

Mn +5

Co +6

Write formulas for ionic compounds and name the compounds.

Br -1 O -2 P -3

10

Polyatomic ions: ions with 2 or more different atoms

Polyatomic ions that contain oxygen are also called oxyanions

The ion with the fewer O atoms will end with “-ite”

The ion with the more O atoms will end with “-ate”

The rest of the polyatomic ion name comes from the element that is not oxygen.

NO2 -1

NO3 -1

Examples:

nitrite

nitrate

SO3 -2

SO4 -2

sulfite

sulfate

PO3 -3

PO4 -3

phosphite

phosphate

When there is more than one oxyanion in a family (same element):

Notice that for S and P oxyanions, the charge of the ion is the same as the one that the element forms (-2) for S and (-3) for P

If a polyatomic ion contains zero oxygen atoms, it will end in –ide!

11

If more than two oxyanions exist in a family, prefixes are used to separate the ions

the most O atoms will be “per- -ate”

the fewest O atoms will be “hypo- -ite”

ClO2 -1

ClO3 -1

ClO -1

ClO4 -1

Examples of the chlorine oxyanions

Least oxygen atoms

Most oxygen atoms

hypochlorite

chlorite

chlorate

perchlorate

Bromine (Br) and Iodine (I) both form similar families.

12

Examples of all of the extended families (four members):

ClO2 -1

ClO3 -1

ClO -1

ClO4 -1

BrO2 -1

BrO3 -1

BrO -1

BrO4 -1

IO2 -1

IO3 -1

IO -1

IO4 -1

Hypochlorite

Chlorite

Chlorate

Perchlorate

Hypobromite

Bromite

Bromate

Perbromate

Hypoiodite

Iodite

Iodate

Periodate

Notice that the oxyanions that come from the halogens all have the same ion charge as the halogens do (-1)

13

Polyatomic ions not in special families or groups

O2 -2

OH -1

CN -1

MnO4 -1

C2H3O2 -1

S2O3 -2

CO3 -2

Cr2O7 -2

CrO4 -2

Acetate

Permanganate

Carbonate

Chromate

Dichromate

Thiosulfate

Cyanide

Hydroxide

Peroxide

Notice that when there is only one oxyanion in a family, it always ends in –ate.For example: there is no such thing a carbonite.

14

Examples of polyatomic ions made by adding H+ to other polyatomic ions.

HCO3 -1

HSO4 -1

H2PO4 -1

HPO3 -2

SO3 -2 SO4

-2

CO3 -2

PO3 -3 PO4

-3

AsO3 -3

HSO3 -1

Sulfite Sulfate

Hydrogen sulfite Hydrogen sulfate

Hydrogen carbonate

Carbonate

Phosphite Phosphate

HPO4 -2

H2PO3 -1

Hydrogen phosphite Hydrogen phosphate

Dihydrogen phosphite Dihydrogen phosphate

A similar series exists for and AsO4 -3

15

We must know: 1) the formula 2) the charge 3) the nameof every ion in table 8-6 and of the additional ones included in these notes!!!!

-1 ions -2 ions -3 ions

NO2 -1

NO3 -1

O2 -2

OH -1

CN -1

MnO4 -1

ClO2 -1

ClO3 -1

ClO -1

ClO4 -1

BrO2 -1

BrO3 -1

BrO -1

BrO4 -1

IO2 -1

IO3 -1

IO -1

IO4 -1

C2H3O2 -1

HCO3 -1

HSO4 -1

H2PO4 -1

SO3 -2

SO4 -2

S2O3 -2

CO3 -2

Cr2O7 -2

HPO4 -2

PO3 -3

PO4 -3

AsO4 -3

CrO4 -2

NH4+1

The only positivepolyatomic ion:

16

K +1

Sr +2

Fe +3

Pb +4

Mn +5

Mo +6

ClO3 -1 SO3

-2 PO4 -3

Write formulas for ionic compounds and name the compounds.

17

Write formulas for the following named Compounds

Sodium iodide

Iron (III) bromide

Tin (IV) oxide

Copper (I) sulfate

Calcium oxide

Barium nitrite

Iron (II) phosphite

Potassium hydroxide

Zinc (II) cyanide

Strontium chlorate

Copper (II) acetate

Chromium (VI) fluoride

Lithium hydrogen sulfite

Lead (IV) carbonate

18

Naming Binary Molecules (covalent compounds)

Molecules are compounds held together with covalent bonds.

Covalent bonds occur when two non-metal atoms bond to each other.

Prefixes are used to indicate how many atoms of each type are present in the molecule.

mono is onedi is twotri is threetetra is fourpenta is five

hexa is sixhepta is sevenocta is eightnona is ninedeca is ten

The less electronegative element is named first as an element, the more electronegative element is named second as a nonoatomic anion.

Mono is never used for the first element named.

19

Examples: Write names for the covalent compounds.

CO

CO2

N2O4

PCl5

SF6

SO3

H2O

P4O10

SiCl4

SO2

Carbon monoxide

Carbon dioxide

Water

Dinitrogen tetroxide

Phosphorous pentachloride

Sulfur hexafluoride

Sulfur trioxide

Sulfur dioxide

Tetraphosphorous Decoxide

Silicon tetrachloride

20

Examples: Write formulas for the covalent compounds.

Carbon tetrachloride

Nitrogen trifluoride

Dinitrogen monoxide

Xenon tetrafluoride

Hydrogen moniodide

Sulfur hexafluoride

Carbon disulfide

Boron trichloride

Dihydrogen monosulfide

Ammonia

CCl4

NF3

N2O

XeF4

HI

SF6

CS2

BCl3

H2S

NH3

21

Naming Acids

Acids are aqueous solutions of molecules that produce H+ ions when they dissolve in water.

If H+ ions are added to the polyatomic ions, the compounds produced are acids when they are dissolved in water.

If H+ ions are added to the monatomic halogen ions (F -1, Cl -1, Br-1, or I -1) or to the sulfide ion, the compounds produced are acids when they are dissolved in water.

The name of an acid depends upon the name of the anion it was made from.

anions ending in –ate are named as –ic acids

anions ending in –ite are named as –ous acids

anions ending in –ide are named as hydro- -ic acids

Note: anions ending in –ide do not contain oxygen atoms in them!

22

Examples:

Sulfuric acid

Sulfurous acid

Hydrosulfuric acid

What acids would the following ions make?

PO4 -3

NO3 -1

ClO-1

ClO4 -1

Formula Name

H3PO4 phosphoric acid

HNO3 nitric acid

comes from the sulfate ion (SO4-2) H2SO4

comes from the sulfite ion (SO3-2) H2SO3

comes from the sulfide ion (S -2) H2S

HClO hypochloric

HClO4 perchloric acid

23

Examples: practice writing names of acids based on the formulas of the anions that made them

HCN

H2CO3

HNO2

HBr

HF

HBrO3

H2Se

Formula of anion Name of acidName of anion

CN-1 Cyanide hydrocyanic acid

CO3-2 Carbonate carbonic acid

Se-2 Sellenide hydrosellenic acid

NO2-1 Nitrite nitrous acid

Br-1 Bromide hydrobromic acid

F -1 Fluoride hydrofluoric acid

BrO3-1 bromate bromic acid

24

General Rules for Oxidation Numbers:

An element in its normal state has oxidation number = 0

An element that becomes an ion has an oxidation number equal to the charge of the ion.

The sum of the oxidation numbers of all atoms in a compound or ion must equal the charge of the compound or ion.

In most compounds, oxygen has an oxidation number of -2. Oxygen will have an oxidation number of -1 when oxygen is a peroxide. Peroxides must have an oxygen-oxygen bond in the molecule.

Fluorine always has an oxidation number of -1 when bonded to other elements.

In compounds where hydrogen is bonded to a nonmetal, hydrogen has an oxidation number of +1, when bonded to a metal it will have an oxidation number of -1.

If none of the above rules apply to the atoms in a problem, then the more electronegative atom is assumed to have an oxidation number equal to its charge if it became a negative ion.

25

Assign Oxidation Numbers to each atom or ion:

Ca S8 Mg+2

H2O NO3- Fe(NO3)3

Cr2(SO3)3 Na2O2 CH3F

CO2 NH4+ NH4NO2

LiH

Br2

NCl3

26

One mole is 6.0221367X1023 particles of any substance.

1 mole of Au atoms = 6.022X1023 atoms of Au

1 mole of Na+1 ions = 6.022X1023 ions of Na

1 mole of CO2 molecules = 6.022A. X1023 molecules of CO2

1 mole of NaCl formula units = 6.022X1023 formula units of NaCl

Only atomic scale particles should use the mole concept!

These particles include: atoms, ions, molecules and formula units.

Remember that the term “molecules” is used for covalent compounds and the term “formula units” is used for ionic compounds.

27

Using the mole concept is very much like using “dozen”.

Equivalence statement: 1 dozen eggs = 12 eggs

3.5 dozen eggs1 dozen

12 eggs( )( ) = 42 eggs

How many eggs are in 3.5 dozen eggs?

3.5 mole Mg ( )( ) = 2.1X1024 Mg atoms

How many Mg atoms are in 3.5 moles of Mg?

Equivalence statement: 1 mole Mg = 6.022X1023 Mg atoms

6.022X1023 Mg atoms

1 mole Mg

28

1 mole of Cl2 = 6.022X1023 molecules of Cl2

How many moles of Cl2 do you have if you have 3.423X1022 molecules of Cl2?

6.022X1023 molecules Cl2

1 mole Cl2( )3.423X1022 molecules Cl2 ( ) = 5.684X10-2 mol Cl2

6.022X1023 is known as Avogadro’s Number

= 0.05684 mol Cl2

or

29

Since a mole is such a large number of things, it really should only be used with very small particles. These particles are called representative particles. Representative Particles include: atoms, molecules, ions, and formula units.

1 mole of atoms = 6.022X1023 atoms

1 mole of molecules = 6.022X1023 molecules

1 mole of ions = 6.022X1023 ions

1 mole of formula units = 6.022X1023 formula units

Since we usually know what type of representative particles are being used, we often do not write the words: atoms, molecules, ions, and formula units, but we should remember that they are implied!.

1 mole of Ne = 6.022X1023 Ne

1 mole of H2O = 6.022X1023 H2O

1 mole of Ba+2 = 6.022X1023 Ba+2

1 mole of NaCl = 6.022X1023 NaCl

Element, so “atoms”

Covalent, so “molecules”

Ion, so “ions”

Ionic, so “formula units”

30

A large number of small things is very difficult to count by hand.

Imagine a dump truck full of jelly beans. How long might it take you to count them all by hand and how many mistakes would you make?

A faster way to “count” them would be to measure the mass of 100 jelly beans and then measure the mass of all the jelly beans in the truck.

Why didn’t we just measure the mass of 1 jelly bean?

Let’s say that 100 jelly beans has a mass of 235.32 g

31

Remember that 100 jelly beans = 235.32 g

How many jelly beans are in a dump truck if the mass of jelly beans in the truck is 4.532 X104 kg?

( )4.532X104 kg jelly beans1 kg jelly beans

1000 g jelly beans)( 235.32 g jelly beans

100 jelly beans( )= 1.926X107 jelly beans

= 19,260,000 jelly beans

or

This method is called “counting by weighing” and it is how we count large numbers of atoms, molecules, ions, and formula units.

32

When we count atoms by “weighing”, we need to know how much one mole of atoms “weighs”.

molar mass is the mass in grams of 1 mole of a pure substance

For elements, the molar mass is the same as the atomic mass expressed in grams. Remember that atomic mass is given on the periodic table.

Example: the atomic mass of Carbon is 12.011 amu

So 1 mole C = 12.011 g C

If we have 12.011 g of carbon we have 1 mole of C which means we have 6.022X1023 atoms of C.

33

1 mole C = 12.011 g C

0.382 mol C1 mol C

12.011 g C( )( ) = 4.58 g C

How many grams C are in 0.382 mol C?

42.573 g C ( )( ) = 3.5445 mol C

How many moles C are in 42.573 g C?

12.011 g C atoms

1 mole C

1 mole C = 12.011 g C

34

Avogadro’s number is the link between number of particles and moles.

Molar mass is the link between mass in grams and moles.

If we are given “mass” in grams, we can calculate the number of particles of a substance that are present.

Mass in g( ) 1 mol

atomic mass in g ( )( ) 1 mol

6.022X1023 atoms}Molar mass

}

Avogadro’s Number

= number of atoms

35

The same type of problem can be worked in reverse as well.

If we are given number of particles of a substance that are present, we can calculate the “mass” in grams.

= Mass in g( ) 1 mol

atomic mass in g( )( )1 mol

6.022X1023 atoms}

Molar mass

}

Avogadro’s Number

number of atoms

36

= Mass in g( ) 1 mol

atomic mass in g( )( )1 mol

6.022X1023 atoms}

Molar mass

}Avogadro’s Number

number of atoms

Mass in g( ) 1 mol

atomic mass in g ( )( ) 1 mol

6.022X1023 atoms}

Molar mass

}

Avogadro’s Number

= number of atoms

Notice the similarities and differences between these two related problems.

37

1 mole Xe = 131.3 g Xe

1 mol Xe

131.3 g Xe( )( )= 5,490 g Xe

How many grams Xe are 2.52X1025 atoms of Xe?

42.5 g Xe( )

How many atoms Xe are contained in 42.5 g Xe?

1 mole of Xe = 6.022X1023 atoms Xe

( )2.52X1025 atoms Xe1 mol Xe

6.022X1023 atoms Xe

1 mol Xe

131.3 g Xe ( )( ) 1 mol Xe

6.022X1023 atoms Xe

1 mole Xe = 131.3 g Xe 1 mole of Xe = 6.022X1023 atoms Xe

= 1.95X1023 atoms Xe

38

Moles and chemical formulas

If we have 10 molecules of P2O5

How many P atoms are present? How many O atoms are present?

We can create “molecule ratios” from the chemical formula to help us solve these problems.

2 atoms P( )1 molecule P2O5

This is what we use to convert from molecules of P2O5 to atoms of P

5 atoms O( )1 molecule P2O5

In a similar fashion, this is what we use to convert from molecules of P2O5 to atoms of O

What if we have 0.2 moles of P2O5?

39

2 moles P( )1 mole P2O5

This mole ratio converts from moles of P2O5 to

moles of P atoms

5 moles O( )1 mole P2O5

In a similar fashion, this mole ratio converts from moles of P2O5 to moles of O atoms

What if we have 0.2 moles of P2O5?

We could change the language.

Now we can write “mole ratios” from the formula.

Mole ratios of this type can be written for any formula!

40

Assume we have 1.0 mole of K2SO4

How many moles of K+1 ions are present?

How many moles of SO4-2 ions are present?

Mole ratios from formulas can be used as conversion factors!

2 mol K+1( )1 mole K2SO4

1 mol SO4-2( )1 mole K2SO4

If we have 0.381 mole of K2SO4, how many moles of K+1 ions do we have?

= 0.762 mol K+1( )0.381 mole K2SO4

2 mol K+1( )1 mole K2SO4

41

If we have 0.50 mole of C5H7Cl3

How many moles of C atoms are present?

How many moles of Cl atoms are present?

How many moles of H atoms are present?

5 mol C( )1 mole C5H7Cl3

7 mol H( )1 mole C5H7Cl3

3 mol Cl( )1 mole C5H7Cl3

= 2.5 mol C atoms( )0.50 mole C5H7Cl3

5 mol C( )1 mole C5H7Cl3

= 3.5 mol H atoms( )0.50 mole C5H7Cl3

7 mol H( )1 mole C5H7Cl3

= 1.5 mol Cl atoms( )0.50 mole C5H7Cl3

3 mol Cl( )1 mole C5H7Cl3

42

The atomic mass from the periodic table is the mass in grams of one mole of the element.

When we have a compound, we can find the molar mass of the compound by adding up the atomic masses for each atom present in the compound.

For example, H2O one mole of H2O has 2 moles of H and one mole of O in it.

Therefore the molar mass of H2O is: (2)*(1.008 g/mol)

+ (1)*(16.00 g/mol)

= 18.02 g/mol

18.02 g/mol is the molar mass of water

43

What would be the molar mass of C5H7Cl3?

1 mole of C5H7Cl3 contains 5 mol C, 7 mol H and 3 mol Cl

1 mol C = 12.01 g C

1 mol H = 1.008 g H

1 mol Cl = 35.45 g Cl

12.01 g X 5 = 60.05 g

1.008 g X 7 = 7.056 g

35.45 g X 3 = 106.4 g

60.05 g+ 7.056 g+ 106.4 g

173.5 g

So, 1 mole of C5H7Cl3 = 173.5 g C5H7Cl3

Molar Mass is used as a conversion factor in many problems!

This is also written as 173.5 g/mol

44

( )37.32 g C5H7Cl3 ( )1 mole C5H7Cl3

How many moles of C5H7Cl3 do we have if we have 37.32 g of C5H7Cl3?

Molar mass definition: 1 mole of C5H7Cl3 = 173.5 g C5H7Cl3

173.5 g C5H7Cl3

= 0.2152 mol C5H7Cl3}Molar mass (shown in conversion factor form converting from grams to moles)

45

( )0.725 mol C5H7Cl3 1 mole C5H7Cl3( )

How many grams of C5H7Cl3 do we have if we have 0.725 mol of C5H7Cl3?

173.5 g C5H7Cl3= 126 g C5H7Cl3}

Molar mass (shown in conversion factor form converting from moles to grams)

Molar mass definition: 1 mole of C5H7Cl3 = 173.5 g C5H7Cl3

46

( )0.725 mol C5H7Cl3 1 mole C5H7Cl3( )173.5 g C5H7Cl3= 126 g C5H7Cl3}

Molar mass (shown in conversion factor form converting from moles to grams)

( )37.32 g C5H7Cl3 ( )1 mole C5H7Cl3

173.5 g C5H7Cl3

= 0.2152 mol C5H7Cl3}

Molar mass (shown in conversion factor form converting from grams to moles)

Look at the similarities and differences in how the same molar mass can be used.

47

( )1 mole Ag2CrO4

331.8 g Ag2CrO4

= 16.8 g Ag+1 ions

A 25.8 g sample of Ag2CrO4 is used in this problem.

a) How many grams of Ag+1 ions are in the sample?

The thought process is: mass of sample to moles of sample to moles of silver ions to mass of silver ions.

In order to do this we need three equivalence statements:

2 mole of Ag = 1mole of Ag2CrO4

1 mole of Ag2CrO4 = 331.8 g of Ag2CrO4 this was calculated from

masses in the periodic table

( )25.8 g Ag2CrO4 ( )2 mole Ag+

1 mole Ag2CrO4( )1 mole Ag+1

107.9 g Ag+1

1 mole of Ag+ = 107.9 g Ag+

48

Percent Composition: The mass percent of each element present in a compound is called the percent composition.

Percent composition can be determined from the formula of a compound as in the example for water on the previous “slide” or it can be measured by analytical chemists in a laboratory as in the example below.

A chemist analyzes 0.25000 g of a salt and obtains the following results: the sample contained 0.09835 g Na and 0.15163 g Cl

What is the percent composition of the salt?

0.09835 g Na% Na = X 100 = 39.34% Na

0.25000 g sample( )0.15163 g Cl

% Cl = X 100 = 60.652% Cl0.25000 g sample( )

49

Percent Composition based on a chemical formula: H2O

2H 2X(1.008 g) = 2.016 g H

O 1X(16.00 g) = 16.00 g O

2.016 g H

18.02 g H2O% H = ( ) X 100 = 11.19 % H

1 mole H2O = 18.02 g

16.00 g O% O = X 100 = 88.79 % O

18.02 g H2O( )Mass of element in 1 mole of compound

Mass of 1 mole of compound( ) X 100 = % by mass of the element

50

Empirical Formula: the formula of a compound that uses the smallest whole number mole ratios of each element present in the compound.

Molecular Formula: the formula of a compound that uses the actual mole ratios of each element present in the compound.

Compare and Contrast these two ideas

Propene: one molecule of propene has 3 carbon atoms and 6 hydrogen atoms, so the molecular formula is: C3H6

but the smallest whole number ratio of carbon to hydrogen is 2 hydrogen atoms for every 1 carbon atom so the empirical formula is CH2

C3

3

H 6

3= CH2

51

If the percent composition is known for a compound, it can be used to calculate the empirical formula.

Step 1: Assume you have 100.00 g of the compound.

Step 2: Multiply the mass percents by 100.00 g to find the mass of each element present in 100.00 g of the compound.

Step 3: Convert the mass of each element into moles of each element.

Step 4: Calculate mole ratios (divide the larger number of moles by the smallest number of moles whenever possible).

Step 5: Write the formula using the mole ratios calculated. If a mole ratio is not a whole number, multiply each mole ratio by whatever number will convert them into whole numbers.

52

A compound known to contain C, H and Cl is analyzed and the percent composition is determined to be: 17.74% C, 3.722 % H, and 78.54% Cl. What is the empirical formula of the compound?

Steps 1 and 2 (100.00g)(17.74% C) = 17.74 g C

(100.00g)(3.722% H) = 3.722 g H

(100.00g)(78.54% Cl) = 78.54 g Cl

Step 3 (17.74 g C) ( )1 mol C

12.01 g C = 1.477 mol C

(3.722 g H) ( )1 mol H

1.008 g H = 3.692 mol H

(78.54 g Cl) 1 mol Cl

35.45 g Cl = 2.216 mol Cl( )

Step 4 1.477/1.477 = 1.000 mol C/1 mol C

3.692/1.477 = 2.500 mol H/1 mol C

2.216/1.477 = 1.500 mol Cl/1 mol C

Step 5 CH2.5Cl1.5

C2H5Cl3

53

If the molar mass and empirical formula of a compound are known, then the molecular formula can be determined.

If the empirical formula is C2H5Cl3 and the molar mass is 406.2 g/mol,

what is the molecular formula for the compound?

C2H5Cl3

2C 2X(12.01 g) = 24.02 g C5H 5X(1.008 g) = 5.040 g H3Cl 3X(35.45 g) = 106.4 g Cl

135.4 g in 1 mol of C2H5Cl3

406.2 g

135.4 g= 3.000 Therefore: C(2*3)H(5*3)Cl(3*3) = C6H15Cl9