Chemical Composition
Chapter 6
Tro, 2nd ed.
DEFINITIONS OF VARIOUS MASSES
Formula or molecular mass = S of atomic masses in
the chemical formula
Molecular mass = mass in amu for a molecule, from
nonmetal elements forming covalent bonds
Molecule is a covalent compound’s smallest unit,
made of nonmetals in covalent bonds
Formula mass = mass in amu for a formula unit or for
all compounds
Formula unit is the smallest whole number ratio of an
ionic compound with metallic and nonmetallic
elements in it (usually)
Molecular and formula mass
NaNO3
1 Na * 22.99 = 22.99
1 N * 14.01 = 14.01
3 O * 16.00 = 48.00
85.00 amu
(NH4)2SO4
2 N * 14.01 = 28.02
8 H * 1.008 = 8.064
1 S * 32.07 = 32.07
4 O * 16.00 = 64.00
132.15
amu
Molecular and formula mass
YOU PRACTICE:
PCl5
HNO3
CHLORINE GAS
ATOMS, MOLES & AVOGADRO’S
NUMBER
The mass of a single atom is too small to measure on a balance.
mass of hydrogen atom = 1.673 x 10-24 g
Chemists require a unit for counting which can express large
numbers of atoms using simple numbers.
Chemists have chosen a unit for counting atoms.
That unit is the
ATOMS, MOLES & AVOGADRO’S
NUMBER
1 mole = 6.0221 x 1023 objects
6.0221 x 1023 objects/mole is Avogadro’s Number
6.0221 x 1023 is a very
LARGE
number
ATOMS, MOLES & AVOGADRO’S
NUMBER
1 mole of any element contains 6.0221 x 1023
particles of that substance.
The atomic mass in grams of any element contains 1
mole of atoms.
This is the same number of particles as there are in
exactly 12 grams of carbon-12.
1 mole of atoms = 6.0221 x 1023 atoms
1 mole of molecules = 6.0221 x 1023 molecules
1 mole of ions = 6.0221 x 1023 ions
1 mole of formula units = 6.0221 x 1023 formula units
AVOGADRO’S NUMBER
& MOLAR MASS
The molar mass of an element is its atomic
mass in grams INSTEAD of amu.
It contains 6.0221 x 1023 atoms (Avogadro’s
number) of the element.
The units are grams/mole.
Relating atomic mass and molar
mass & Avogadro’s Number:
Atomic mass
Molar mass
Number of
atoms
H
1.008 amu
1.008 g
6.0221x1023
Mg
24.31 amu
24.31 g
6.0221x1023
Na
22.99 amu
22.99 g
6.0221x1023
This is per atom.
This is per mole.
This is per mole.
Element
THE MOLE MAP
ATOMS, MOLECULES,
FORMULA UNITS, IONS
X NA
/NA
MOLES
XM
/M
MASS (grams)
/D
XD
VOLUME, mL
ATOMS, MOLECULES
& AVOGADRO’S NUMBER
Practice:
1. How many atoms in 0.100 moles of He?
2. How many moles are represented by 9.00 x 109 atoms of He?
1. 0.100 mol * 6.0221 x 1023 atoms/mol = 6.0221 x 1022 He atoms
2.
9.00x109atoms
= 1.50 x 10-14 mol of He
6.0221 x 1023 atoms/mol
OR
9.00x109atoms *
1 mole
= 1.50 x 10-14 mol He
6.0221 x 1023 atoms
MOLAR MASS – JUST LIKE
CALCULATING MOLECULAR MASS
1. Find molar mass of liquid bromine, Br2
2. Find molar mass of ethanol CH3CH2OH (organic structural
formula)
3. Find molar mass of ammonium phosphate.
1.
2 * 79.90 g/mol = 159.80 g/mol
2.
2 C * 12.01 = 24.02 g/mol
6 H * 1.008 = 6.048 g/mol
1 O * 16.00 = 16.00 g/mol
46.068
= 46.07 g/mol (sig figs)
3.
149.12 g/mol
MASS, MOLECULES & THE MOLE:
USING THE MOLE MAP
EXAMPLE: Use mole map to find number of
molecules in 0.4607 grams of ethanol.
First use molar mass to go from mass to moles:
0.4607 * 1 mol
= 0.01000 MOLES
46.07 g
Then use moles and Avogadro’s Number to find
number of molecules:
0.1000 mol * 6.0221 x 1023 molecules/mol
= 6.022 x 1021 molecules
MASS, MOLECULES & THE MOLE:
USING THE MOLE MAP
EXAMPLE: Use mole map to find number of
grams of lead in 9.00 x 1018 atoms of lead.
First use Avogadro’s Number to find moles:
9.OO X 1018 atoms *
1 mol
= 1.494 X 10-5 mol
6.0221 * 1023 atoms
Then use molar mass to find grams:
1.494 X 10-5 mol Pb * 207.2 g/mol = 3.10 X 10-3 g
(back to 3 sig figs)
COMPOUNDS, MOLAR MASS
& AVOGADRO’S NUMBER
Chemical formulas are written to tell us:
- the number of atoms of each element
- the type of compound (ionic, covalent, acid,
or organic)
- the structure of the compound
COMPOUNDS, MOLAR MASS
& AVOGADRO’S NUMBER
NaNO3
1 Na atom
1 N atom
3 O atoms
(NH4)2SO4
2 N atoms
8 H atoms
1 S atom
4 O atoms
PCl5
HNO3
1 P atom 1 H atom
5 Cl atoms 1 N atom
3 O atoms
ionic
formula
unit
ionic
formula
unit
covalent acid
molecule molecule
(until aq)
These relationships are present when
hydrogen combines with chlorine.
H
Cl
HCl
6.0221 x 1023 H
atoms
6.0221 x 1023 Cl
atoms
6.0221 x 1023 HCl
molecules
1 mol H atoms
1 mol Cl atoms
1 mol HCl
molecules
1.008 g H
35.45 g Cl
36.46 g HCl
1 molar mass H
atoms
1 molar mass Cl
atoms
1 molar mass HCl
molecules
How many oxygen atoms are present in 2.00 mol of
oxygen molecules?
Two conversion factors are needed:
 6.022 x 1023 molecules O 2 


1
mol
O


2
 2 atoms O 
 1 mol O 

2 
Conversion sequence:
moles O2 → molecules O2 → atoms O
 6.022 x 1023 molecules O 2   2 atoms O 
(2.00 mol O 2 ) 



1 mol O 2

  1 molecule O 2 
= 2.41 x1024 atoms O
How many nitrogen atoms are in 56.04 g of N2?
The molar mass of N2 is 28.02 g/mol.
Conversion sequence: g N2 → moles N2 → molecules N2
→ atoms N
 1 mol N 2   6.022 x 10 molecules N 2 
(56.04 g N 2 ) 


1
mol
N
28.02
g
N


2
2 
23
 2 atoms N 
 1 molecule N 

2 
= 2.409 x 10
24
atoms N
ELEMENTAL COMPOSITION
Elemental composition is the percent composition of a
compound, which is the mass percent of each element in the
compound
Example:
H2O is 11.19% H by mass & 88.81% O by mass
percent = part * 100%
whole
Elemental composition of sodium chloride, NaCl:
1 Na*22.99= 22.99 g/mol (22.99/58.44)*100 = 39.34%
1 Cl*35.45 = 35.45 g/mol (35.45/58.44)*100 = 60.66%
58.44 g/mol
SUM = 100.00%
Percent Composition From Formula
If the formula of a compound is known, a twostep process is needed to calculate the
percent composition.
Step 1 Calculate the molar mass of the
formula.
Step 2 Divide the total mass of each element in
the formula by the molar mass and multiply
by 100.
Percent Composition From Formula
You practice with ethanol: CH3CH2OH
2 C*12.01 = 24.02 g/mol (24.02/46.07)*100 = 52.14%
6 H*1.008 = 6.048 g/mol (6.048/46.07)*100 = 13.13%
1 O*16.00 = 16.00 g/mol (16.00/46.07)*100 = 34.73%
= 46.07 g/mol
SUM = 100.00%
Percent Composition
From Experimental Data
Percent composition can be calculated from
experimental data without knowing the
composition of the compound.
Step 1 Calculate the mass of the compound
formed.
Step 2 Divide the mass of each element by the
total mass of the compound and multiply by
100.
A compound containing nitrogen and oxygen is found
to contain 1.52 g of nitrogen and 3.47 g of oxygen.
Determine its percent composition.
Step 1 Calculate the total mass of the compound
1.52 g N
3.47 g O
4.99 g = total mass of compound
Continued:
Step 2 Divide the mass of each element
by the total mass of the
compound formed.
 1.52 g N 

 (100) = 30.5%
 4.99 g 
 3.47 g O

 4.99 g

 (100) = 69.5%

O
69.5%
N
30.5%
Mole Relationships in
Chemical Formulas
Since we count atoms and molecules in mole units,
we can find the number of moles of a constituent
element if we know the number of moles of the
compound
Moles of Compound
Moles of Constituents
1 mol NaCl
1 mol Na, 1 mol Cl
1 mol H2O
2 mol H, 1 mole O
1 mol CaCO3
1 mol Ca, 1 mol C, 3 mol O
1 mol C6H12O6
6 mol C, 12 mol H, 6 mol O
Mole Relationships in
Chemical Formulas
Class work: Skillbuilder 6.6
How many moles of oxygen are in 1.4 moles of
sulfuric acid? How many grams of oxygen is
this?
1.4 mol H2SO4 * 4 mol O/mol H2SO4 = 5.6 mol
of oxygen
5.6 mol O * 16.00 g/mol = 89.6 grams of O
Empirical Formula versus
Molecular Formula
The empirical formula or simplest formula gives the
smallest whole-number ratio of the atoms present in a
compound.
The molecular formula is the true formula of a compound.
The empirical formula gives the relative number of
atoms of each element present in the compound.
The molecular formula represents the total number of
atoms of each element present in one molecule of a
compound.
Notice that these are all covalent compounds or diatomic
elements. That’s because ionic formulas are by definition the
lowest whole number ratio, with only a few exceptions like mercury
(I) chloride, Hg2Cl2.
Calculating
Empirical Formulas
Step 1 Assume a definite starting quantity (usually 100.0
g) of the compound, if the actual amount is not given,
and express the mass of each element in grams.
Step 2 Convert the grams of each element into moles of
each element using each element’s molar mass.
Step 3 Divide the moles of atoms of each element by the
moles of atoms of the element that had the smallest
value.
– If the numbers obtained are whole numbers, use them
as subscripts and write the empirical formula.
– If the numbers obtained are not whole numbers, go on
to step 4.
Calculating
Empirical Formulas
Step 4 Multiply the values obtained in step 3 by the
smallest numbers that will convert them to whole
numbers
Use these whole numbers as the subscripts in the
empirical formula.
The results of calculations may differ from a whole
number.
If they differ ±0.05, round off to the next nearest whole
number.
Deviations greater than 0.05 unit from a whole number usually
mean that the calculated ratios have to be multiplied by a
whole number.
Some Common Fractions and Their Decimal Equivalents
Decimal
Resulting Whole
Equivalent
Number
Common Fraction
1
4
0.25
1
1
3
0.333…
1
0.666…
2
0.5
1
0.75
3
2
3
1
2
3
4
Multiply the
decimal equivalent
by the number in
the denominator of
the fraction to get
a whole number.
The analysis of a salt shows that it contains 56.58%
potassium (K); 8.68% carbon (C); and 34.73%
oxygen (O). Calculate the empirical formula for
this substance.
Step 1 Express each element in grams. Assume 100
grams of compound.
K = 56.58 g
C = 8.68 g
O = 34.73 g
Continued:
Step 2 Convert the grams of each element to moles.
 1 mol K atoms 
 1.447 mol K atoms
K:  56.58 g K  

 39.10 g K 
 1 mol C atoms 
C atoms
atoms
C: 8.68 g C  
 0.723 mol C

 12.01 g C  C has the smallest number
of moles
 1 mol O atoms 
 2.171 mol O atoms
O:  34.73 g O  

 16.00 g O 
Continued:
Step 3 Divide each number of moles by the smallest
value.
1.447 mol
0.723 mol
K=
= 2.00
C:
= 1.00
0.723 mol
0.723 mol
2.171 mol
O=
= 3.00
0.723 mol
The simplest ratio of K:C:O is 2:1:3
Empirical formula K2CO3
Now you try:
The percent composition of a compound is
25.94% nitrogen (N), and 74.06% oxygen
(O). Calculate the empirical formula for this
substance.
N2O5
Calculating the Molecular Formula from
the Empirical Formula
The molecular formula can be calculated from the
empirical formula if the molar mass is known.
The molecular formula will be equal to the
empirical formula or some multiple, n, of it.
To determine the molecular formula evaluate n.
n is the number of units of the empirical formula
contained in the molecular formula.
molar mass
n=
= number of empirical
formula units
mass of empirical formula
What is the molecular formula of a compound which
has an empirical formula of CH2 and a molar mass of
126.2 g?
Calculate the mass of each CH2 “mol”
1 C = 1(12.01 g/mol) = 12.01g/mol
2 H = 2(1.01 g/mol) = 2.02g/mol
14.03g/mol
126.2 g
n
 9 (empirical formula units)
14.03 g
The molecular formula is (CH2)9 = C9H18
Group Work:
What is the empirical formula and the molecular
formula of lactic acid, which is 40.00% C, 6.71% H
and 53.29% O, with a molar mass of 90.1 g?
C3H6O3 (or CH3CHOHCOOH)