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Basilar chemistry laws
Dr.ssa Rossana Galassi
320 4381420
[email protected]
Atom structure:
protons number (in the
nucleus) equal to the
electrons number
(around the nucleus). In
the nucleus are located
also the neutrons.
These latters are
neutral
Electricity:
Two faced gold foils
connected with a
metal pome.
When electrized
ebane is put close
the gold foils
disclose to each
other
Radioactivity: alfa, beta particles
and gamma radiation
Atom contains subparticles
Electron discovery and
determination of its charge/mass.
• :
Cathod rays
Experiments to Determine the Properties of Cathode Rays
OBSERVATION
CONCLUSION
1. Ray bends in magnetic field.
2. Ray bends towards positive
plate in electric field.
consists of charged particles
3. Ray is identical for any cathode.
consists of negative particles
Determination of the electron
charge (Millikan)
Millikan used his findings to also calculate the mass of an
electron.
mass of electron =
mass
determined by J.J. Thomson and
others
X
charge
charge
= (-5.686x10-12 kg/C)
(-1.602x10-19C)
= 9.109x10-31kg = 9.109x10-28g
Scoperta del protone (esperienza
di Goldstein): raggi canale
Rutherford’s a-scattering experiment
and discovery of the atomic nucleus.
General features of the atom today.
•The atom is an electrically neutral, spherical entity composed of a positively
charged central nucleus surrounded by one or more negatively charge
electrons.
•The atomic nucleus consists of protons and neutrons.
Properties of the Three Key Subatomic Particles
Charge
Name(Symbol)
Mass
Location
Absolute(C)* Relative(amu)† Absolute(g) in the Atom
Proton (p+)
+1.60218x10-19
1.00727
Neutron (n0)
0
1.00866
Electron (e-)
-1.60218x10-19
0.00054858
1.67262x1024
1.67493x10-24 Nucleus
9.10939x10-28
* The coulomb (C) is the SI unit of charge.
†
Nucleus
The atomic mass unit (amu) equals 1.66054x10-24 g.
Outside
Nucleus
Atomic Symbols, Isotopes, Numbers
A
Z
X
X = Atomic symbol of the element
A = mass number; A = Z + N
Z = atomic number
(the number of protons in the nucleus)
N = number of neutrons in the nucleus
Isotope = atoms of an element with the same
number of protons, but a different number
of neutrons
Determining the Number of Subatomic
Particles in the Isotopes of an Element
PROBLEM: Silicon(Si) is essential to the computer industry as a major
component of semiconductor chips. It has three naturally
occurring isoltopes: 28Si, 29Si, and 30Si. Determine the number
of protons, neutrons, and electrons in each silicon isotope.
PLAN:
We have to use the atomic number and atomic masses.
SOLUTION: The atomic number of silicon is 14. Therefore
28Si
has 14p+, 14e- and 14n0 (28-14)
29Si
has 14p+, 14e- and 15n0 (29-14)
30Si
has 14p+, 14e- and 16n0 (30-14)
Calculating the Atomic Mass of an Element
PROBLEM: Silver(Ag: Z = 47) has 46 known isotopes, but only two occur
naturally, 107Ag and 109Ag. Given the following mass
spectrometric data, calculate the atomic mass of Ag:
PLAN:
Isotope
Mass(amu)
Abundance(%)
107Ag
106.90509
51.84
109Ag
108.90476
48.16
We have to find the weighted average of the isotopic masses,
so we multiply each isotopic mass by its fractional abundance
and then sum those isotopic portions.
SOLUTION:
mass(g) multiply
of each by fractional
portion of atomic mass
atomic mass
add isotopic portions
isotope abundance of each
from each isotope
isotope
mass portion from 107Ag =
106.90509amu x 0.5184 = 55.42amu
mass portion from 109Ag = 108.90476amu x 0.4816 = 52.45amu
atomic mass of Ag = 55.42amu + 52.45amu = 107.87amu
The periodic table
Elements are ordered in increasing
number of atomic number Z
-there vertical groups and horizontal
periods
-due formalisms for groups: 1-8 A main
groups o B groups or groups from 1 to
18
Three areas: metals, not metals and
metalloids
The periodic table
This table was built, first by Mendeleev, by
following some rules:
• Chemical elements are arranged in the
table accordingly to the increasing Atomic
Number
• Elements are arranged in the table by
considering some chemical properties.
Dimitri Mendeleev table (18341907)
Mendeleev developed the
periodic table based on
atomic weights because
the concept of atomic
numbers was not known
till the development of the
structure of the atom in the
early 20th century.
Dimitri Mendeleev table (1834-1907)
•
•
•
•
•
•
•
•
The world’s first view of Mendeleev’s
Periodic Table – an extract
•
from Zeitschrift fϋr Chemie, 1869.
•
Concerning the relation between the properties and
atomic weights of elements. By D. Mendeleev. Arranging
the elements in vertical columns with increasing atomic
weights, so that the horizontal rows contain similar elements,
again in increasing weight order, the following table is
obtained from which general predictions can be drawn
Elements show a periodicity of properties if listed in order of
size of atomic weights.
Elements with similar properties either have atomic weights
that are about the same (Pt, Ir, Os) or increase regularly (K,
Rb, Cs).
The arrangement of the elements corresponds to their
valency, and somewhat according to their chemical properties
(eg Li, Be, B, C, N, O, F).
The commonest elements have small atomic weights.
It continues ....
The value of the atomic weight determines the character of
the element.
There are unknown elements to discover eg elements similar
to Al and Si with atomic weights in range 65-75.
The atomic weights of some elements may be changed from
knowing the properties of neighbouring elements. Thus the
atomic weight of Te must be in range 123-126. It cannot be
128.
Some typical properties of an element can be predicted from
its atomic weight.
Dimitri Mendeleev table (1834-1907)
The greatness of Mendeleev was that not only did he leave spaces for elements that were not yet
discovered but he predicted properties of five of these elements and their compounds. How foolish he
would have seemed if these predictions had been incorrect but fortunately for him three of these
missing elements were discovered by others within 15 years (ie within his lifetime). The first of these
Mendeleev had called eka-aluminium because it was the one after aluminium (eka = 1 in Sanskrit) and
was identified in Paris (1875) by Paul Emile Lecoq de Boisbaudran who named it gallium after the
Latin name for France. Mendeleev was ecstatic when he heard of its properties which nearly matched
his eka-aluminium. However de Boisbaudran's value for gallium's density (4.9 g/cm3) differed from
Mendeleev's prediction. Mendeleev told the Frenchman, who re-measured the density to find
Mendeleev was right! It is interesting to speculate whether de Boisbaudran was pleased or irritated by
this. The table compares Mendeleev's predictions with de Boisbaudran's discovery.
Eka-aluminium (Ea)
Atomic weight
Density of solid
Melting point
Valency
Method of discovery
Oxide
About 68
3
6.0 g/cm
low
3
Probably from its spectrum
Formula Ea2O3, density
3
5.5 g/cm . Soluble in both acids
and alkalis.
Gallium (Ga)
69.72
3
5.9 g/cm
o
29.78 C
3
Spectroscopically
Formula Ga2O3, density 5.88
3
g/cm . Soluble in both acids and
alkalis.
Dimitri Mendeleev table (1834-1907)
The photograph shows a giant wall
Periodic Table erected in St
Petersburg, Russia, in 1934.
The
periodic
table at
Milan
EXPO
2015
pavilion
The modern periodic table.
hydrogen
H
from Hydrogenum
Phosphoros
P
from Phosphorum
Gold
Au
from Aurum
Sodium Na
from Natrium
10 elements are the most abundant:
O(47%), Si(26%), Al, Fe, Ca, Mg, Na, K, Ti, H
About 99,5% w/w in hydrosphere, atmosphere
and outer shell of lithosphere
http://www.tavolaperiodica.unicam.it/
http://www.rsc.org/periodic-table
http://www.ptable.com/?lang=it
http://www.chemicalelements.com/
Metals, not metals and metalloids
Metals, metalloids, and nonmetals.
Metals
They are solids (except for mercury), high heat
and electrical conducibility, ductility, malleability,
and alloys forming.
Not metals
Solids, liquids and gas, generally do not conduct
electricity (except graphite, C)
Metalloids
B, Si, Ge, As, Sb, Te, are elements that can
behave as metals or not metals
Alkali metals
Earth alkali metals
13° or 3 A group elements
allotropes
Silicon compounds
Gases in the periodic table:
diatomic molecules
Sulfur: S8 molecule
http://www.tavolaperiodica.unicam.it/
Sulfur: allotropes
Noble gases
http://www.tavolaperiodica.unicam.it/
Transition metals
Stretching from 2A to 3 A is a series of elements called the transition
metals, called also B-groups. These elements fill the 1 B to 8B in the fourth
till the seventh periods in the center of the periodic table.
Most occur naturally combined with other elements, but Cu, Ag, Au and Pt
are less reactive and they can be found in nature as pure elements.
They are very important for their commercial use, in the technological
applications (materials, paints, coins, batteries, catalytic converters ….),
biological involvments (hemoglobin, vitamins, enzymes…)
Rare earths
The rows at the bottom of the table accomodate the lanthanides (the
series of elements between lanthanium Z=57, and hafnium Z=72), and
the actinides (the series of elements between actinium Z=89, and
They have important applications in the field of nuclear reactors (U), in
the color television picture tubes (La, Gd) or as contrast agent in
diagnostic imaging.
Molecules, compounds and
formulas
A molecule is the smallest identifiable unit by
which a pure compound such as sugar or water
can be divided still retaining the composition
and the chemical properties of the substance.
Such substances are composed of identical
molecules consisting of atoms of two or more
elements firmly bound together
Molecules and atoms
The modern atomic theory is based on
ancient postulates
They give the basis of the understanding of
chemical reactions and compound
formations
Dalton (1766-1844 d. C.) defined that
elements are constitued by atoms putting
the basis of the modern theory. However,
Dalton postulated the wrong assumption that
atoms were not possible to divide
Joseph Proust
French chemist Joseph Proust proposed the law of
definite composition or proportions based on his
experiments conducted between 1798 and 1804 on
the elemental composition of water and copper
carbonate.
In 1806, Proust summarized his observations in what
is now called Proust's Law. It stated that chemical
compounds are formed of constant and
defined ratios of elements, as determined by mass.
For example, carbon dioxide is composed of one
carbon atom and two oxygen atoms. Therefore, by
mass, carbon dioxide can be described by the fixed
ratio of 12 (mass of carbon):32 (mass of oxygen), or
simplified as 3:8.
Dalton’s Atomic Theory
The Postulates
1. All matter consists of atoms.
2. Atoms of one element cannot be converted into
atoms of another element.
3. Atoms of an element are identical in mass and other
properties and are different from atoms of any other
element.
4. Compounds result from the chemical combination of
a specific ratio of atoms of different elements.
1. All matter is composed of atoms. The atom is the smallest body that
retains the unique identity of the element.
2. Atoms of one element cannot be converted into atoms of another
element in a chemical reaction. Elements can only be converted
into other elements in nuclear reactions.
3. All atoms of an element have the same number of protons and
electrons, which determines the chemical behavior of the element.
Isotopes of an element differ in the number of neutrons, and thus
in mass number. A sample of the element is treated as though its
atoms have an average mass.
4. Compounds are formed by the chemical combination of two or more
elements in specific ratios.
• When this theory was postulated the
subatomic particles were not known.
• However, this theory introduces the mass
ratio and the mass conservation concepts.
These latters are still valid and recognised
by modern theories.
Dalton’s Atomic Theory
explains the mass laws
Mass conservation
Atoms cannot be created or destroyed postulate 1
or converted into other types of atoms. postulate 2
𝐴 = 𝜋𝑟 2
Since every atom has a fixed mass, postulate 3
during a chemical reaction atoms are combined
differently and therefore there is no mass change
overall.
2H2 + O2 => 2H2O
Dalton’s Atomic Theory
explains the mass laws
Definite composition
Atoms are combined in compounds in postulate 3
specific ratios
and each atom has a specific mass.
postulate 4
So each element has a fixed fraction of the total mass
in a compound.
Dalton’s Atomic Theory
explains the mass laws
Multiple proportions
Atoms of an element have the same mass
postulate 3
and atoms are indivisible. postulate 1
So when different numbers of atoms of elements
combine, they must do so in ratios of small, whole
numbers.
Mass conservation law (Lavoisier
1743-1794)
The total mass of the substance involved
in a trasformation remains unchanged
By performing the following reaction above a balance, weighting before and after
the reaction
BaCl2 + CuSO4 => BaSO4  + CuCl2
The law of mass conservation:
mass remains constant during a chemical reaction.
Law of Mass Conservation:
The total mass of substances does not change during a chemical
reaction.
reactant 1
+
reactant 2
total mass
=
calcium oxide
+
carbon dioxide
CaO
+
CO2
56.08
+
product
44.00
total mass
calcium carbonate
CaCO3
100.08
Law of Definite (or Constant) Composition:
No matter the source, a particular compound is
composed of the same elements in the same parts
(fractions) by mass.
Calcium carbonate
Analysis by Mass
(grams/20.0g)
8.0 g calcium
2.4 g carbon
9.6 g oxygen
20.0 g
Mass Fraction
(parts/1.00 part)
Percent by Mass
(parts/100 parts)
0.40 calcium
0.12 carbon
0.48 oxygen
40% calcium
12% carbon
48% oxygen
1.00 part by mass
100% by mass
Calculating the Mass of an Element in a Compound
PROBLEM: Pitchblende is the most commercially important compound of
uranium. Analysis shows that 84.2 g of pitchblende contains
71.4 g of uranium, with oxygen as the only other element. How
many grams of uranium can be obtained from 102 kg of
pitchblende?
PLAN:
The mass ratio of uranium/pitchblende is the same no matter
the source. We can use the ratio to find the answer.
SOLUTION:
mass(kg) of pitchblende
mass(kg) of uranium
mass (kg) of uranium =
mass(kg) pitchblende x
mass(kg) uranium in pitchblende
mass(kg) pitchblende
71.4kg uranium
mass(g) of uranium
= 86.5 kg
= 102 kg pitchblende x
84.2kg pitchblende uranium
86.5 kg uranium x
1000g
kg
= 8.65 x 104g uranium
Law of Multiple Proportions:
If elements A and B react to form two compounds, the different
masses of B that combine with a fixed mass of A can be expressed
as a ratio of small whole numbers.
Example: Carbon Oxides A & B
Carbon Oxide I : 57.1% oxygen and 42.9% carbon
Carbon Oxide II : 72.7% oxygen and 27.3% carbon
Assume that you have 100g of each compound.
In 100 g of each compound: g O = 57.1 g for oxide I & 72.7 g for oxide II
g C = 42.9 g for oxide I & 27.3 g for oxide II
gO
gC
57.1
=
42.9
gO
72.7
gC
27.3
= 1.33
= 2.66
2.66 g O/g C in II
1.33 g O/g C in I
=
2
1
Law of Multiple Proportions:
If elements A and B react to form two compounds, the different
masses of B that combine with a fixed mass of A can be expressed
as a ratio of small whole numbers.
Given the following data for two compounds, SnO2 and SnO, what is the whole
number ratio of the oxygen of SnO to the oxygen of SnO2? Does this follow the
law of multiple proportions?
•1:1. Yes, it follows the law of multiple proportions.
•2:1. Yes, it follows the law of multiple proportions.
•1:2. Yes, it follows the law of multiple proportions.
Law of Multiple Proportions:
If elements A and B react to form two compounds, the different
masses of B that combine with a fixed mass of A can be expressed
as a ratio of small whole numbers.
Which of the following compounds CANNOT demonstrate the law of multiple
proportions?
1. NO, NO2
2. CO, CO2
3. H2O, H2O2
4. Na2S, NaF
Chemical reactions
Matter cannot be created neither
destroyed
In a reaction, matter undergoes a
transformation without mass loss.
C (solid) + O2 (gaseous) => CO2 (gas) + heat (energy)
12 amu + 32 amu => 44 amu (micro)
12 g + 32 g
=> 44 g (macro)
Reaction balancing
balancing
___ Al(s) + ___ Br2(l)  ___ Al2Br6(s)
2 Al + 3 Br2 =>1Al2Br6
6Cl2(g) + P4(s)
 4PCl3 (l)
Equation
intial (mol)
initial (g)
variation
6Cl2 (g)
6 mol
425 g
-6 mol
(-425 g)
+
+
+
+
P4(s)

1 mol

124 g

-1 mol
(-124 g)
Complete reaction
0 mol
+
0 mol
4PCl3(l)
0 mol
0 g
+4 mol


4 mol
(425+124=549g)
Cl2 (g) + P4 (s)  PCl3 (l)
The rapresentation of a chemical reaction is named chemical equation
• Reactants are written on the left of the
arrow whilst products are on the right
Cl2(g) + P4(s)
reactants

PCl3 (l)
products
letters (s), (g), and (l) represent the physical state of the reagents and the products
6Cl2(g) + 1 P4(s)

4PCl3 (l)
• The bold numbers in the chemical
equation are the stoichiometric coefficients
Conservation of the mass
6x2=12 atomi di Cl
6Cl2(g) + P4(s)
4 atomi di P
4x3=12 atomi di Cl

4PCl3 (l)
4 atomi di P
The relationship between the reagents and the
products amount is called stoichiometry and the
coefficients in the chemical equation are called
stoichiometric coefficients
6Cl2(g) + P4(s)
 4PCl3 (l)
The mole
Definition of mole(mol) - the amount of a substance that
contains the same number of entities as there are atoms in
exactly 12 g of carbon-12.
This amount is 6.022x1023. The number is called Avogadro’s
number and is abbreviated as N.
One mole (1 mol) contains 6.022x1023 entities (to four
significant figures)
602200000000000000000000 a huge number!
The mole
concept of mole(mol) – being atoms and molecules very very very
small to have an amount which is macroscopically measurable the
number of them should be huge, as the Avogadro number is.
As usually the term a couple of glasses, a dozen of eggs, a pair of
socks delimits a certain amoun of daily life objects, in chemistry is
used the mole of a substance to quantify the amount we refer.
The Molar Mass (MM) is the mass in grams
of a mole of that substance.
Measure unit: g/mol
The mole
1. How many molecules are present in 5 moles of C?
n atoms = 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 ′ 𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 Ɲ 𝑥 5 = 3.011 𝑥 10 24
2. How many mole are present in 9.011 𝑥 10 24 atoms of C?
n mole = 9.011 𝑥 10 24
𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 ′ 𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 Ɲ = 15
3. How many molecules are present in 10 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻2𝑂?
n molecules = 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 ′ 𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 Ɲ 𝑥 10 = 6.022 𝑥 10 24
The mole
12 g of C (AW = relative atomic weigh= 12 amu) contains Ɲ of atoms, and it
consists of 1 mole
18 g of water (MW= relative molecular weigh = 18 amu) contains Ɲ of molecules,
and it consists of 1 mole
Molar Mass is the mass in g of a mole of substance.
Number of mole n = 𝑀𝑀
𝑚𝑎𝑠𝑠 𝑔
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
How many moles for 180 g of water?
Number of mole n = 𝑀𝑀
𝑚𝑎𝑠𝑠 𝑔
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
= 180 g /18 g /mole = 10 moles
The mole
C (solid) + O2 (gas) => CO2 (gas) + cal (energy)
12 amu + 32 amu => 44 amu (micro)
12 g + 32 g
=> 44 g (macro)
The proportionality constant between the
mcroscopic units (amu) and macroscopic
units (grams) is the
Avogadro number Ɲ
One mole of
common
substances.
CaCO3
100.09 g
Oxygen
32.00 g
Water
18.02 g
Copper
63.55 g
Information Contained in the Chemical Formula of Glucose
C6H12O6 ( M = 180.16 g/mol)
Carbon (C)
Hydrogen (H)
Oxygen (O)
Atoms/molecule
of compound
6 atoms
12 atoms
6 atoms
Moles of atoms/
mole of compound
6 moles
of atoms
12 moles
of atoms
6 moles
of atoms
Atoms/mole of
compound
6(6.022 x 1023)
atoms
12(6.022 x 1023)
atoms
6(6.022 x 1023)
atoms
Mass/molecule
of compound
6(12.01 amu)
=72.06 amu
12(1.008 amu)
=12.10 amu
6(16.00 amu)
=96.00 amu
12.10 g
96.00 g
Mass/mole of
compound
72.06 g
Summary of Mass Terminology
Term
Definition
Unit
Isotopic mass
Mass of an isotope of an element
amu
Atomic mass
Average of the masses of the naturally
occurring isotopes of an element
weighted according to their abundance
amu
Sum of the atomic masses of the atoms
(or ions) in a molecule (or formula unit)
amu
(also called
atomic weight)
Molecular
(or formula) mass
(also called
molecular weight)
Molar mass (M)
Mass of 1 mole of chemical entities
(also called
(atoms, ions, molecules, formula units)
gram-molecular weight)
g/mol
Interconverting Moles, Mass, and Number of Chemical Entities
no. of grams
Mass (g) = no. of moles x
g
1 mol
No. of moles = mass (g) x
1 mol
no. of grams
No. of entities = no. of moles x
6.022x1023 entities
1 mol
1 mol
No. of moles = no. of entities x
6.022x1023 entities
M
MASS(g)
of element
Summary of the mass-molenumber relationships for
elements.
M (g/mol)
AMOUNT(mol)
of element
Avogadro’s
number
(atoms/mol)
ATOMS
of element
Calculating the Mass and the Number of Atoms in a
Given Number of Moles of an Element
PROBLEM:
(a) Silver (Ag) is used in jewelry and tableware but no longer in U.S.
coins. How many grams of Ag are in 0.0342mol of Ag?
(b) Iron (Fe), the main component of steel, is the most important
metal in industrial society. How many Fe atoms are in 95.8g of Fe?
PLAN:
(a) To convert mol of Ag to g we have to use
the #g Ag/mol Ag, the molar mass M.
SOLUTION: 0.0342mol Ag x 107.9 g Ag = 3.69g Ag
mol Ag
PLAN: (b) To convert g of Fe to atoms we first
have to find the #mols of Fe and then
convert mols to atoms.
SOLUTION: 95.8g Fe x mol Fe
= 1.72mol Fe
55.85g Fe
6.022x1023atoms Fe = 1.04x1024 atoms
1.72mol Fe x
mol Fe
Fe
amount(mol) of Ag
multiply by M of Ag
(107.9g/mol)
mass(g) of Ag
mass(g) of Fe
divide by M of Fe
(55.85g/mol)
amount(mol) of Fe
multiply by 6.022x1023
atoms/mol
atoms of Fe
MASS(g)
of compound
Summary of the mass-molenumber relationships for
compounds.
M (g/mol)
AMOUNT(mol)
of compound
chemical
formula
Avogadro’s
number
(molecules/mol)
MOLECULES
(or formula units)
of compound
AMOUNT(mol)
of elements in
compound
Calculating the Moles and Number of Formula Units
in a Given Mass of a Compound
PROBLEM:
Ammonium carbonate is white solid that decomposes with
warming. Among its many uses, it is a component of baking
powder, first extinguishers, and smelling salts. How many
formula unit are in 41.6 g of ammonium carbonate?
After writing the formula for the
compound, we find its M by adding the
masses of the elements. Convert the given
mass, 41.6 g to mols using M and then the
mols to formula units with Avogadro’s
number.
SOLUTION:
The formula is (NH4)2CO3.
mass(g) of (NH4)2CO3
divide by M
amount(mol) of (NH4)2CO3
multiply by 6.022x1023
formula units/mol
number of (NH4)2CO3 formula units
M = (2 x 14.01 g/mol N)+(8 x 1.008 g/mol H)
+(12.01 g/mol C)+(3 x 16.00 g/mol O)
mol (NH4)2CO3
41.6 g (NH4)2CO3 x
96.09 g (NH4)2CO3
6.022x1023 formula units (NH4)2CO3
mol (NH4)2CO3
2.61x1023 formula units (NH4)2CO3
Mass percent from the chemical formula
Mass % of element X =
atoms of X in formula x atomic mass of X (amu)
x 100
molecular (or formula) mass of compound(amu)
Mass % of element X =
moles of X in formula x molar mass of X (amu)
molecular (or formula) mass of compound (amu)
x 100
Calculating the Mass Percents and Masses of
Elements in a Sample of Compound
PROBLEM: Glucose (C6H12O6) is the most important nutrient in the living
cell for generating chemical potential energy.
(a) What is the mass percent of each element in glucose?
(b) How many grams of carbon are in 16.55g of glucose?
PLAN:
We have to find the total mass of
glucose and the masses of the
constituent elements in order to
relate them.
SOLUTION:
(a)
Per mole glucose there are
6 moles of C
12 moles H
6 moles O
amount(mol) of element
X in 1mol compound
multiply by M(g/mol) of X
mass(g) of X in 1mol of
compound
divide by mass(g) of
1mol of compound
mass fraction of X
multiply by 100
mass % X in compound
continued
6 mol C x
12.01 g C
= 72.06 g C
12 mol H x
16.00 g O
= 96.00 g O
= 12.096 g H
mol H
mol C
6 mol O x
1.008 g H
M = 180.16 g/mol
mol O
(b)
mass percent of C =
72.06 g C
180.16 g glucose
= 0.3999 x 100 = 39.99 mass %C
12.096 g H
mass percent of H =
mass percent of O =
180.16 g glucose
96.00 g O
180.16 g glucose
= 0.06714 x 100 = 6.714 mass %H
= 0.5329 x 100 = 53.29 mass %O
The Molar Mass (MM) is the mass in grams
of a mole of that substance.
Measure unit: g/mol