Seager_10e_Ch02_PowerPoint.pptx

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Seager, Slabaugh, Hansen, General, Organic, and Biochemistry, Nineth Edition. © 2022 Cengage. All Rights Reserved. May not be scanned,
copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1
Chapter 2
Atoms and Molecules
copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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Learning Objectives (1 of 2)
• Use symbols for chemical elements to write formulas for chemical compounds
• Identify the characteristics of protons, neutrons, and electrons
• Use the concepts of atomic number and mass number to determine the
number of subatomic particles in isotopes and to write correct symbols for
isotopes
• Use atomic weights of the elements to calculate molecular weights of
compounds
• Use isotope percent abundances and masses to calculate atomic weights of
elements

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Learning Objectives (2 of 2)
• Use the mole concept to obtain relationships between number of moles,
number of grams, and number of atoms for elements, and use those
relationships to obtain factors for use in factor-unit calculations
• Use the mole concept and molecular formulas to obtain relationships between
number of moles, number of grams, and number of atoms or molecules for
compounds, and use those relationships to obtain factors for use in factor-unit
calculations

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Symbols and Formulas
• A unique name and symbol is used to represent each element
• Based on elemental properties or are derived from names of famous scientists, places,
astronomical bodies, or mythological characters
• Elemental symbols: Based on the name of the element and consist of one capital letter or a
capital letter followed by a lowercase letter

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Table 2.1 - The Chemical Elements and Their
Symbols (1 of 2)
Ac actinium Es einsteinium Mt meitnerium Sb antimony (stibium)a
Ag silver (argentum)a Eu europium N nitrogen Sc scandium
Al aluminum F fluorine Na sodium (natrium)a Se selenium
Am americium Fe iron ( ferrum)a Nb niobium Sg seaborgium
Ar argon FI flerovium Nd neodymium Si silicon
As arsenic Fm fermium Ne neon Sm samarium
At astatine Fr francium Nh nihonium Sn tin (stannum)a
Au gold (aurum)a Ga gallium Ni nickel Sr strontium
B boron Gd gadolinium No nobelium Ta tantalum
Ba barium Ge germanium Np neptunium Tb terbium
Be beryllium H hydrogen O oxygen Tc technetium
Bh bohrium He helium Og oganesson Te tellurium
Bi bismuth Hf hafnium Os osmium Th thorium
Bk berkelium Hg mercury (hydrargyrum)a P phosphorus Ti titanium
Br bromine Ho holmium Pa protactinium TL thallium
C carbon Hs hassium Pb lead (plumbum)a Tm thulium
aElements with symbols not derived from their English names.

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Table 2.1 - The Chemical Elements and Their
Symbols (2 of 2)
Ca calcium I iodine Pd palladium Ts tennessine
Cd cadmium In indium Pm promethium U uranium
Ce cerium Ir iridium Po polonium V vanadium
Cf californium K potassium (kalium)a Pr praseodymium W tungsten (wolfram)a
Cl chlorine Kr krypton Pt platinum Xe xenon
Cm curium La lanthanum Pu plutonium Y yttrium
Cn copernicium Li lithium Ra radium Yb ytterbium
Co cobalt Lr lawrencium Rb rubidium Zn zinc
Cr chromium Lu lutetium Re rhenium Zr zirconium
Cs cesium Lv livermorium Rf rutherfordium
Cu copper (cuprum)a Mc moscovium Rg roentgenium
Db dubnium Md mendelevium Rh rhodium
Ds darmstadtium Mg magnesium Rn radon
Dy dysprosium Mn manganese Ru ruthenium
Er Erbium Mo Molybdenum S Sulfur
aElements with symbols not derived from their English names.

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Elements in the Human Body

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Compound Formula
• Consists of the symbols of the atoms found in the molecule
• Each elemental symbol represents one atom of the element
• If more than one atom is present in the compound, then a subscript follows the elemental
symbol

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Table 2.2 - Examples of Compound Formulas

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Example 2.1 - Writing Compound Formulas
• Write formulas for the following compounds:
a. Nitrogen dioxide: One nitrogen (N) atom and two oxygen (O) atoms
b. Sulfuric acid: Two hydrogen (H) atoms, one sulfur (S) atom, and four oxygen (O) atoms

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Example 2.1 - Solution
a. The single N atom will not have a subscript because ones are understood and
never written, and the two O atoms will be represented by writing a subscript 2
• The molecular formula is NO2
b. Using similar reasoning, the H atom will have a subscript 2, the S atom will
have no subscript, and the O atom will have a subscript 4
• The molecular formula is H2SO4

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Compound Formulas Practice
• The clear liquid is carbon disulfide
• It is composed of carbon (left) and sulfur (right)
• If carbon disulfide contains one atom of
carbon for every two atoms of sulfur, what
is the chemical formula for carbon
disulfide?
• Answer - CS2

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Atomic Structure
• Atoms are made up of three subatomic
particles
• Protons, neutrons, and electrons
• Protons and neutrons
• Tightly bound together to form the central
portion of an atom called the nucleus
• Electrons
• Located outside the nucleus
• Move rapidly throughout a relatively large
volume of space surrounding the nucleus
Electrons move rapidly around a massive
nucleus. This figure is not drawn to scale.
For a nucleus of the size shown, the closest
electrons would be at least 80 m away.

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Table 2.3 - Characteristics of the
Fundamental Subatomic Particles
Characterstics
Particle
Common
Symbols
Charge (±) Mass (g) Mass (u) Location
Electron e− 1− 9.07 × 10−28 1/1836 Outside nucleus
Proton p, p+, H+ 1+ 1.67 × 10−24 1 Inside nucleus
Neutron n 0 1.67 × 10−24 1 Inside nucleus

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Atomic Structure Review
• Which subatomic particles are represented by the pink spheres?
• Answer - Electrons
• Which subatomic particles are represented by the yellow and blue spheres?
• Answer - Protons and neutrons
• What structure is formed by the yellow and blue spheres?
• Answer - The nucleus

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Atomic and Mass Numbers
• Atomic number of an atom
• Equal to the number of protons in the nucleus and the number of electrons in an atom
• Symbolically represented by Z
• Mass number of an atom
• Equal to the sum of the number of protons and neutrons in the nucleus of an atom
• Symbolically represented by A

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Atomic and Mass Numbers Application
(1 of 2)
• Based on the information given below, what is the atomic number of fluorine?
• Answer - The atomic number of fluorine is 9
• In the periodic table, the atomic number is written as a whole number above the symbol F
• In the written description, fluorine is said to have 9 protons (the atomic number is the number
of protons)
• In the symbol, the number 9 is written in the atomic number or Z (lower left) position

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Atomic and Mass Numbers Application
(2 of 2)
• Based on the information given below, what is the mass number of fluorine?
• Answer - The mass number of fluorine is 19
• In the written description, fluorine is said to have 9 protons and 10 neutrons (the mass
number is the sum of the numbers of protons and neutrons)
• In the symbol, the number 19 in written in the mass number or A (upper left) position
• Note - The periodic table does not show the mass number for an individual atom
• It lists an average mass number for a collection of atoms

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Isotopes
• Atoms that have the same number of protons in the nucleus but different
numbers of neutrons
• Same atomic number but different mass numbers
• All isotopes of the same element have the:
• Same number of electrons outside the nucleus
• Same number of protons in the nucleus

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Ways to Represent Isotopes
• Represented by the following symbol:
• Z is the atomic number
• A is the mass number
• E is the elemental symbol
• Example -
• This symbol represents an isotope of hydrogen that contains 11 protons in the nucleus
• Represented by the elemental name, which is followed by the mass number
• Example - Hydrogen-2 is an isotope of hydrogen

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Example 2.2 - Using the Periodic Table
• Use the periodic table to answer the following questions about isotopes:
a. What are the mass number, atomic number, and isotope symbol for an atom that contains 7
protons and 8 neutrons?
b. How many neutrons are contained in an atom of nickel-60?
c. How many protons and how many neutrons are contained in an atom with a mass number of
26 and the symbol Mg?

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Example 2.2 - Solution (a)
• The mass number, A, equals the sum of the number of protons and the number
of neutrons
• A = 7 + 8 = 15
• The atomic number, Z, equals the number of protons
• Z = 7
• According to the periodic table, the element with an atomic number of 7 is
nitrogen, with the symbol N
• The isotope symbol is

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Example 2.2 - Solution (b)
• According to the periodic table, nickel has the symbol Ni, and an atomic
number, Z, of 28
• The mass number, 60, is equal to the sum of number of protons and the
number of neutrons
• The number of protons is equal to the atomic number, 28
• Therefore, the number of neutrons is 60 – 28 = 32
• The atom contains 32 neutrons

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Example 2.2 - Solution (c)
• According to the periodic table, the element with the symbol Mg is magnesium,
which has an atomic number of 12
• Therefore, the atom contains 12 protons
• Since A, the number of protons plus neutrons is equal to 26, the number of
neutrons is 26 – 12, or 14
• The atom contains 14 neutrons

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Relative Masses
• Numbers that are given beneath the symbol and name for each element in the
periodic table
• Provide simple means of comparing the masses of atoms

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Atomic Mass Unit (u)
• Used to express the relative masses of atoms
• 1 u = 1/12 the mass of a carbon-12 atom
• One carbon-12 atom has a relative mass of 12 u
• Atom with a mass equal to 1/12 the mass of a carbon-12 atom would have a relative mass of
1 u
• Atom with a mass equal to twice the mass of a carbon-12 atom would have a relative mass
of 24 u

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Atomic Weight
• Relative mass of an average atom of an
element expressed in atomic mass units
• Example - According to the periodic table,
the atomic weight of N is 14.0 u and Si is
28.1 u
• It also means that two N atoms have a total mass
very close to the mass of a single Si atom

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Electron Shell Occupancy
• Relative mass of a molecule expressed in atomic mass units
• Calculated by adding together the atomic weights of the atoms in the molecule
• Example - The formula for a molecule of water is H2O
• This means one molecule of water contains two atoms of hydrogen, H, and one atom of
oxygen, O
• Molecular weight of water is then the sum of two atomic weights of H and one atomic weight
of O

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Example 2.4 - Atomic Weights and Molecular
Weights
• Use atomic weights from the periodic table to determine the molecular weight
of urea, CH4N2O, the chemical form in which much nitrogenous body waste is
excreted in the urine

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• According to the formula given, a urea molecule contains one carbon atom, C,
four hydrogen atoms, H, two nitrogen atoms, N, and one oxygen atom, O
• The molecular weight is calculated as follows:
• Rounded to four significant figures, the correct answer is 60.06 u

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Molecular Weight Practice
• The clear liquid is carbon disulfide, CS2
• It is composed of carbon (left) and sulfur (right)
• What is the molecular weight for carbon disulfide?
• Answer

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Isotopes and Atomic Weights
• Atomic weight of elements that occur as mixtures of isotopes is the average
relative mass of the atoms in the isotope mixture
• Average mass of each particle in a group of atoms is obtained by dividing the
total mass of the group by the number of particles in the group
• Practical way of determining the average mass of a group of isotopes is to use an imaginary
sample of an element containing 100 atoms
• Use the percentage of each isotope to represent the number of atoms of each isotope in the group
• Total mass equals the sum of masses contributed by each isotope

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Isotopes and Atomic Weights - Example
• A specific example is shown below for the element boron that consists of
19.78% boron-10 with a mass of 10.01 u and 80.22% boron-11 with a mass of
11.01 u
• This calculated value matches the value given in the periodic table

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Example 2.5 - Isotope and Atomic Weight
Relationships
• Calculate the atomic weight of chlorine, given that the naturally occurring
element consists of 75.53% chlorine-35 (mass = 34.97 u) and 24.47% chlorine-
37 (mass = 36.97 u)

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• This result is slightly different from the periodic table atomic weight value of
35.45 because of slight errors introduced in rounding the isotope masses to
four significant figures

36
Avogadro’s Number and the Mole Concept
• Avogadro’s number - Number of atoms or molecules in a specific sample of an
element or compound
• Mole (mol): Number of particles contained in a sample of an element or compound with a
mass in grams equal to the atomic or molecular weight, respectively
• 1 mol = 6.022×1023
• Example - 1 mol S atoms = 6.02×1023 S atoms = 32.1 g S
• Following factors can be generated for use in factor-unit calculations:
• 1 mol S atoms = 6.02×1023 particles S atoms
• 6.02×1023 S atoms = 32.1 g S
• 1 mol S atoms = 32.1 g S

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The Mole and Chemical Calculations
• Mole concept can be used to obtain factors that are useful in chemical
calculations involving both elements and compounds

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Mole Calculation Example (1)
• Calculate the number of moles of Ca contained in a 15.84 g sample of Ca
• Solution:
• Notice that the g Ca units in the denominator of the factor cancel the g Ca units in the given
quantity, leaving the correct units of mole Ca for the answer

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Example 2.7 - Factor-Unit Calculations for
Sulfur
• Determine the following using the factor-unit method of calculation and factors
obtained from the preceding three relationships given for sulfur (S):
a. The mass in grams of 1.35 mol of S
b. The number of moles of S atoms in 98.6 g of S
c. The number of S atoms in 98.6 g of S
d. The mass in grams of one atom of S

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Example 2.7 – Solution (1 of 2)
a. Known quantity is 1.35 mol of S, and the unit of the unknown quantity is grams
of S
• Factor comes from the relationship 1 mol S atoms = 32.1 g S
b. Known quantity is 98.6 g of S, and the unit of the unknown quantity is moles of
S atoms
• Factor comes from the same relationship used in (a)

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c. Known quantity is 98.6 g of S, and the unit of the unknown quantity is the
number of S atoms
• Factor comes from the relationship 6.02×1023 S atoms = 32.1 g S
d. Known quantity is one S atom, and the unit of the unknown is grams of S
• Factor comes from the same relationship used in (c), 6.02×1023 S atoms = 32.1 g S

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• Note that the factor is the inverse of the one used in (c) even though both came
from the same relationship
• Thus, we see that each relationship provides two factors

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The Mole Concept Applied to Compounds
• One mole of any compound is a sample of the compound with a mass in grams
equal to the molecular weight of the compound
• 1 mol CO2 molecules = 6.02 ×1023 CO2 molecules = 44.0 g CO2
• Following relationships can be used to generate factors for use in factor-unit calculations:
• 1 mol CO2 molecules = 6.02 ×1023 CO2 molecules
• 6.02 ×1023 CO2 molecules = 44.0 g CO2
• 1 mol CO2 molecules = 44.0 g CO2

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• How many moles of O atoms are contained in 11.57 g of CO2?
• Solution
• Note that the factor used was obtained from two of the six quantities given on the previous
slide

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• How many CO2 molecules are needed to contain 50.00 g of C?
• Solution:
• Note that the factor used was obtained from two of the six quantities given on a previous slide

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• What is the mass percentage of C in CO2?
• Solution:
• Mass percentage is calculated using the following equation:
• If a sample consisting of 1 mole of CO2 is used, the mole-based relationships given earlier
show that 1 mole CO2 = 44.01 g CO2 = 12.01 g C + 32.00 g O
• Thus, the mass of C in a specific mass of CO2 is known and the problem is solved as follows:

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Mole Calculation Example (5) (1 of 2)
• What is the mass percentage of oxygen in CO2?
• Solution:
• Mass percentage is calculated using the following equation:
• Once again, a sample consisting of 1 mole of CO2 is used to take advantage of the
mole-based relationships given earlier where:
• 1 mole CO2 = 44.01g CO2 = 12.01 g C + 32.00g O

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Mole Calculation Example (5) (2 of 2)
• Thus, the mass of O in a specific mass of CO2 is known and the problem is
solved as follows:
• Notice that the % C + % O = 27.29% + 72.71% = 100%, which should be the case
because C and O are the only elements present in CO2

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Example 2.8 - Factor-Unit Calculations for
Carbon Dioxide
• Determine the following using the factor-unit method of calculation and factors
obtained from the preceding three relationships given for carbon dioxide, CO2:
a. The mass in grams of 1.62 mol of CO2
b. The number of moles of CO2 molecules in 63.9 g of CO2
c. The number of CO2 molecules in 63.9 g of CO2
d. The mass in grams of one molecule of CO2

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a. The known quantity is 1.62 mol of CO2, and the unit of the unknown quantity is
g CO2
• The factor comes from the relationship 1 mol CO2 molecules = 44.0 g CO2
b. The known quantity is 63.9 g of CO2, and the unit of the unknown quantity is
mol of CO2 molecules
• The factor comes from the same relationship used in (a)

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c. The known quantity is, again, 63.9 g of CO2, and the unit of the unknown is the
number of CO2 molecules
• The factor comes from the relationship 6.02×1023 CO2 molecules = 44.0 g CO2
d. The known quantity is 63.9 g of CO2, and the unit of the unknown quantity is
mol of CO2 molecules
• The factor comes from the same relationship used in (c), but the factor is the inverse of the
one used in (c)

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The Mole and Chemical Formulas
• Chemical formulas represent the numerical relationships that exist among
atoms in a compound
• Example - H2O represents a 2:1 fixed ratio of hydrogen atoms to oxygen atoms
in a water molecule
• Following relationships can be derived:
• 6.02×1023 H2O molecules contain 12.04×1023 H atoms and 6.02×1023 O atoms
• 1 mol of H2O molecules contains 2 mol of H atoms and 1 mol of O atoms
• 18.0 g of water contains 2.0 g of H and 16.0 g of O

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Mole Calculations: Steps

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Example 2.11 - Mass Percentage Calculations
• Ammonia (NH3) and ammonium nitrate (NH4NO3) are commonly used
agricultural fertilizers
• Which one of the two contains the higher mass percentage of nitrogen (N)?

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Example 2.11 - Solution (1 of 2)
• In each case, the mass percentage of N is given by
• We will use 1 mol of each compound as a sample because the mass in grams of
1 mol of compound and the mass in grams of N in the 1 mol of compound are
readily determined
• One mol of NH3 weighs 17.0 g and contains 1 mol of N atoms, which weighs 14.0 g

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Example 2.11 - Solution (2 of 2)
• Similarly, 1 mol of NH4NO3 weighs 80.0 g and contains 2 mol of N atoms,
which weigh 28.0 g

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Laboratory Application (1) (1 of 3)
• The mass of an empty crucible is 20.057 g
• Copper sulfate pentahydrate is added to the crucible
• The combined mass is 45.551 g
• What is the mass of the copper sulfate pentahydrate?
• Solution

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• The crucible is heated and allowed to cool
• The mass of the crucible and anhydrous solid after heating is 38.547 g
• What is the mass of the anhydrous solid?
• What is the mass of the water that was removed by heating?

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• How many moles of water (H2O) were removed by heating?
• How many moles of anhydrous copper sulfate (CuSO4) remained after heating?
• What is the ratio of moles anhydrous copper sulfate to moles of water?

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• The formula for a hydrate is written as CuSO4·x H2O, where x is the number of
moles of water associated with each mole of CuSO4
• Based on the calculated mole ratio, what is the experimentally determined formula for
copper sulfate pentahydrate?
• Solution
• CuSO4·3.355 H2O

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• The actual formula for copper sulfate pentahydrate is CuSO4·5 H2O
• Why is there a difference between the actual formula and the experimentally determined
formula?
• Answer
• The copper sulfate pentahydrate was not heated thoroughly enough
• Some of the water remained in the “anhydrous ” copper sulfate, which caused the number of
moles of water removed by heating to be smaller than the actual value

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• What is the mass percentage of H2O in CuSO4·5 H2O?
• What mass of water could have been removed from the 25.494 g copper sulfate
pentahydrate if the sample had been heated completely?

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• What percentage of the total amount of water was removed during the
experiment?
• Had all the water been removed, what would the mass of the anhydrous copper
sulfate been after heating? Would this have been the mass displayed on the
balance? Why or why not?
• Answer: No, the mass displayed on the balance would have been 36.353 g because the
anhydrous copper sulfate was in the 20.057 g crucible

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