Why is Knowledge of
Composition Important?
Because everything in nature is either chemically or
physically combined with other substances it is
necessary to have knowledge of chemical
composition. To know the amount of a material in a
sample, you need to know what fraction of the
sample it is.
•
Examples that show the need to know about
chemical composition:
– the amount of sodium in sodium chloride for
dietary need
– the amount of iron in iron ore for steel production
– the amount of hydrogen in water for hydrogen
fuel
– the amount of chlorine in Freon to estimate
ozone depletion
Counting Atoms by Moles
•
If we can find the mass of a particular number of atoms, we can use this
information to convert the mass of an element sample to the number of atoms in
the sample.
•
The number of atoms we will use is 6.022 x 1023 and we call this a mole
– 1 mole = 6.022 x 1023 things, like 1 dozen = 12 things
– The dozen and the mole make it easier to talk about large quantities
– A mole can refer to anything, but we usually use it to talk about atoms,
molecules, and ions
•
mole = number of particles equal to the number of atoms in 12 g of C-12
– 1 atom of C-12 weighs exactly 12 amu
– 1 mole of C-12 weighs exactly 12 g
•
The number of particles in 1 mole is called Avogadro’s Number = 6.0221421 x
1023 (remember Avogadro’s # to 4 Sig. Figs.)
– 1 mole of C atoms weighs 12.01 g and has 6.022 x 1023 atoms
• the average mass of a C atom is 12.01 amu
Mass Number is Not the Same as Atomic Mass
•
•
the atomic mass is an experimental number determined from all naturally
occurring isotopes
the mass number refers to the number of protons + neutrons in one isotope
– natural or man-made
Calculating Atomic Mass
Gallium has two naturally occurring isotopes: Ga-69 with mass
68.9256 amu and a natural abundance of 60.11% and Ga-71 with
mass 70.9247 amu and a natural abundance of 39.89%. Calculate
the atomic mass of gallium.
Solution:
Convert the percent natural abundance into decimal form.
Ga-69  0.6011
Ga-71  0.3989
Determine the Formula to Use
Atomic Mass = (abundance1)∙(mass 1) + (abundance2)∙(mass 2) + ...
Apply the Formula:
Atomic Mass = 0.6011 (68.9256 amu) + 0.3989 (70.9247 amu) = 69.72
amu
Relationship Between Moles and Mass
• The mass of one mole of atoms is called the molar
mass.
• The molar mass of an element, in grams, is
numerically equal to the element’s atomic mass, in
amu.
One mole
Converting Between Moles and Number of Atoms
A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in
the ring?
•
Write down the given quantity and its units.
Given: 1.1 x 1022 Ag atoms
•
Write down the quantity to find and/or its units.
Wanted: ? Moles
•
Collect Needed Conversion Factors:
1 mole Ag atoms = 6.022 x 1023 Ag atoms
1 mole Ag
1.110 Ag atoms 
 moles
23
6.022 10 Ag atoms
22
= 1.8266 x 10-2 moles Ag
= 1.8 x 10-2 moles Ag
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 mole Na, 1 mole 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
Masses of different substances that equal to one mole
Converting from Grams to
Moles and Moles to Grams
How do you calculate the number of
moles of sulfur in 57.8 g of sulfur?
1 mole S
57.8 g S 
 moles S
32.06 g S
= 1.80287 moles S
= 1.80 moles S
Converting Between
Grams and Number of Atoms
How many aluminum atoms are in an
aluminum can with a mass of 16.2 g?
1 mole Al 6.022 10 atoms Al
16.2 g Al 

26.98 g Al
1 mole Al
23
= 3.6159 x 1023 atoms Al
Sig. Figs. & Round: = 3.62 x 1023 atoms Al
Converting Between
Grams and Moles of Compound
Calculate the mass (in grams) of 1.75 mol of
water
18.02 g H 2 O
1.75 mol H 2 O 
 g H 2O
1 mol H 2 O
= 31.535 g H2O
Sig. Figs. & Round: = 31.5 g H2O
Converting Between Grams and
Number of Molecules
Find the mass of 4.78 x 1024 NO2
molecules?
4.78 10 24 molec NO2 
1 mole NO2
46.01 g NO2

6.022 10 23 molec NO2 1 mole NO2
= 365.21 g NO2
Sig. Figs. & Round: = 365 g NO2
Converting Between Grams of a
Compound and Grams of a Constituent
Element
L-Carvone, (C10H14O), is found in spearmint oil. It has a
pleasant odor and mint flavor. It is often added to chewing
gum, liquors, soaps and perfumes. Find the mass of
carbon in 55.4 g of carvone.
Molar Mass C10H14O = 10(atomic mass C) + 14(atomic
mass H) + 1(atomic mass O)
= 10(12.01) + 14(1.01) + (16.00) = 150.2 g/mo
1 mole C10H14O = 150.2 g C10H14O
1 mole C10H14O  10 mol C
1 mole C = 12.01 g C
Solution Strategy “converting units” :
g
C10H14O
mol
C10H14O
1 mol C10 H14O
150.2 g C10 H14O
10 mol C
1 mol C10 H14O
mol
C
g
C
12.01 g C
1 mol C
1 mole C10 H14O
10 mol C
12.01 g C
55.4 g C10 H14O 


150.2 g C10 H14O 1 mol C10 H14O 1 mole C
= 44.2979 g C
= 44.3 g C
Percent Composition
•
•
Percentage of each element in a compound by mass can be determined
from:
1.
the formula of the compound
2.
the experimental mass analysis of the compound
The percentages may not always total to 100% due to rounding
part
Percentage 
 100%
whole
The mass percent tells you the mass of a constituent element in 100 g of
the compound.
The fact that NaCl is 39% Na by mass means that 100 g of NaCl
contains 39 g Na.
This can be used as a conversion factor:
100 g NaCl  39 g Na
39 g Na
g NaCl 
 g Na
100 g NaCl
or
100 g NaCl
g Na 
 g NaCl
39 g Na
Example - Percent Composition from the Formula
C2H5OH
1.
Determine the mass of each element in 1 mole of the compound:
Mass of Carbon = 2 moles C x (12.01 g) = 24.02 g
Mass of Hydrogen = 6 moles H x (1.008 g) = 6.048 g
Mass of Oxygen = 1 mol O x (16.00 g) = 16.00 g
2.
Determine the molar mass of the compound by adding the masses of
the elements:
1 mole C2H5OH = 46.07 g
3.
Divide the mass of each element by the molar mass of the compound
and multiply by 100%:
24.02g
100%  52.14%C
46.07g
16.00g
 100%  34.73%O
46.07g
6.048g
100%  13.13%H
46.07g
Example – Percent Composition of Carvone if a 30.0 g
sample contains 24.0 g of C, 3.2 g O and the rest H?
1.
Determine the masses of all the elements in the sample
C = 24.0 g, O = 3.2 g
H = 30.0 g – (24.0 g + 3.2 g) = 2.8 g
2.
Divide the mass of each element by the total mass of the
sample then multiply by 100% to give its percentage
24.0g
 100%  80.0%C
30.0g
2.8g
 100%  9.3%H
30.0g
3.2g
 100%  11%O
30.0g
Hydrates
Hydrates are compounds that include water molecules as part
of their structure
Examples include:
Na2CO3 • 10 H2O
NiCl2 • 6 H2O
What does this have to do with percent composition?
Percent Composition of
Water
Let’s say we take 1.20 g of NiCl2 • x H2O (a hydrate)
With lots of heat from a Bunsen burner, we remove all the
water and are left with 0.65 g of NiCl2. What is the percent
composition of water in the hydrate?
(1.20 g of NiCl2 • 6 H2O) - (0.65 g of NiCl2) =
= 0.55 g H2O
% Comp. H2O = 0.55 g H2O (part) x 100% = 45%
1.20 g NiCl2 • x H2O (whole)
*We didn’t need to know the exact composition of the NiCl2
hydrate
% Composition and Formula
We find a compound and determine that it is composed of 31.2%
nitrogen and 68.8% oxygen. What is the chemical formula of the
compound?
For every 100. grams of material, there is 30.9 g N and 69.1 g O.
We first convert to moles.
30.9 g N
1 mole N
14.01 g
= 2.21 mole N
69.1 g O
1 mole O
16.0 g
= 4.32 mole O
Examples: Empirical Formulas
Hydrogen Peroxide
Molecular Formula = H2O2
Empirical Formula = HO
Benzene
Molecular Formula = C6H6
Empirical Formula = CH
Glucose
Molecular Formula = C6H12O6
Empirical Formula = CH2O
Finding an Empirical Formula
from Experimental Data
A laboratory analysis of aspirin determined the following
mass percent composition. Find the empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
Example:
Find the empirical formula of
aspirin with the given mass
percent composition.
•
Information
Given: 60.00 g C, 4.48 g H, 35.53 g O
Find: empirical formula, CxHyOz
Collect Needed Conversion Factors:
1 mole C = 12.01 g C
1 mole H = 1.01 g H
1 mole O = 16.00 g O
– calculate the moles of each element
60 .00 g C 
4.48 g H 
1 mol C
 4.996 mol C
12.01 g C
1 mol H
 4.44 mol H
1.01 g H
1 mol O
35.52 g O 
 2.220 mol O
16.00 g O
This gives the formula of
C4.996H4.44O2.220. But, the formula
must have the atoms in whole
number ratios!
Find the mole ratio by dividing by the smallest number
of moles!
C 4.996 H
2.220
4.44
2.220
O 2.220
2.220
C 2.25 H 2 O1
Multiplying the subscripts by 4 gives the final ratio!
C9H8O4
All these molecules have the same Empirical
Formula. How are the molecules different?
Name
Molecular
Formula
Empirical
Formula
glyceraldehyde
C3H6O3
CH2O
erythrose
C4H8O4
CH2O
arabinose
C5H10O5
CH2O
glucose
C6H12O6
CH2O
Molecular Formulas
•
•
The molecular formula is a multiple of the empirical formula
To determine the molecular formula you need to know the empirical
formula and the molar mass of the compound
Molar Massreal formula = factor used to multiply subscripts
Molar Massempirical formula
Determine the Molecular Formula of Cadinene if it has a molar mass of
204 g/mol and an empirical formula of C5H8
Determine the empirical formula.
Determine the molar mass of the empirical formula:
5 C = 60.05 g, 8 H = 8.064 g
C5H8 = 68.11 g
Divide the given molar mass of the compound by the molar mass of the
empirical formula and round to the nearest whole number:
204 g
3
68.11g
Multiply the empirical formula by the factor
above (3) to give the molecular formula:
(C5H8)3 = C15H24
Cadinenes: Sesquiterpenes
found in junipers and
cedars (oil of cade)