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