Building the World

Quantifying Chemical Compounds

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Quantifying Chemical Compounds Theory

The Law of Multiple Proportions

Elements combine in different ways to form substances whose mass ratios are small whole number multiples of each other.

The Law of Definite Proportions

All samples of a pure chemical substance always contain the same proportion of elements by mass regardless of their origin.

By mass, Carbon dioxide ($CO_2$) is: 73% Oxygen and 27% Carbon

All carbon dioxide is chemically the same in every location.

pure chemical substances are chemically the same

Return of the Mole!

We know:

One mole is equal to $6.022$x$10^{23}$ objects.
One mole of any element equals its molar mass.

1 mole of carbon (C) has a mass of: 12.01 g

Amedeo Avogadro - Inventor of the mole.

Molecules and the Mole:

One mole is equal to $6.022$x$10^{23}$ objects.
One mole of any compound is equal to the sum of the molar masses of all elements in the compound.

1 mole of carbon dioxide $CO_{2}$ a mass of : $12.01\,g + 16.00\,g + 16.00\,g = 44.01\,g$

Numbers of moles IS A DIRECT COMPARISON of the relative number of atoms or molecules!

The mole is used extensively in chemical problems because numbers of atoms can be compared!

Comparing relative numbers of atoms has a specific notation. We use an empirical or molecular formula to describe the relative number of atoms in each compound!

Empirical Formula Vs. Molecular Formula

Empirical Formula

The lowest whole integer numbers representing an atomic ratio of a molecule using a chemical description.

Empirical Formula of hydrogen peroxide: HO

Molecular Formula:

A chemical description of the actual complete molecule.

Molecular Formula of hydrogen peroxide: $H_2O_2$

NOTE

Some compounds have empirical and molecular formulas that are the same.

Empirical Formula of water: $H_{2}O$
Molecular Formula of water: $H_{2}O$

Periodic Trends

We know:

The periodic table is organized to help predict the properties of elements.
Elements down a column have similar chemical properties.

Periodic Trends:

The periodic table is organized to help us determine useful information about elements. For example: atomic radius, electronegativity, and ionization energy. Learning periodic trends can help us understand why certain elements have the properties we observe.

Electronegativity

The ability of a bonded atom to attract electrons

Moving down a column on the periodic table electronegativity decreases

Moving across a row on the periodic table electronegativity increases

The difference in electronegativity determines bond type

Electronegativity differences (ΔEN) for bonded atoms can be calculated by subtracting the least electronegative atom from the atom with the highest electronegativity.

For hydrochloric acid (HCl):
Electronegativity of Cl=3.0
Electronegativity of H=2.1
∆𝐄𝐍=𝟑.𝟎-𝟐.𝟏=𝟎.𝟗

Atomic Radius

Is the bonding radius determined from averaging measurements of many compounds and molecules

The bond length of a two bonded atoms is determined by adding their bond radii

Moving down a column on the periodic table the principle quantum number (n) of the outermost electrons increases, orbitals are larger, therefore the atomic radii are larger.

Moving across a row the electrons in the outermost shell feel the charge from the nucleus more strongly. Electrons are closer to the nucleus, orbitals are smaller, therefore the radii are smaller

On the periodic table bond radius increases down a column.

On the periodic table bond radius decreases across a row.

Atomic Radius and Ions:

An atom that has lost an electron (cation) will have a smaller radius than the neutral atom. Fewer electrons = smaller electron cloud = smaller ionic radius.

An atom that has gained an election (anion) will have a larger radius than the neutral atom. More electrons = larger electron cloud = larger ionic radius.

Cations have a smaller ionic radius than the neutral atom.

Anions have a larger ionic radius than the neutral atom.

Ionization Energy

The amount of energy required to remove one electron from an atom (or ion) in a gaseous state.

Moving down a column on the periodic table the principle quantum number (n) of the outermost electrons increase, orbitals are larger, therefore it takes less energy to remove an electron because they are farther from the nucleus.

Moving across a row the electrons in the outermost shell feel the charge from the nucleus more strongly, orbitals are smaller, therefore it takes more energy to remove an electron because they are closer to the nucleus.

On the periodic table ionization energy decreases down a column.

On the periodic table ionization energy increases across a row.

Examples

Quantifying Chemical Compounds Example Problem 1

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Problem:

What is the molar mass of copper sulfate $CuSO_{4}$?

What is the problem asking?

Molar mass of $CuSO_{4}$ = g/mol = ?

What do you have?

The formula for copper sulfate = $CuSO_{4}$

What do you need?

The molar masses of Cu, S, and O from the periodic table. Cu = 63.55 g/mol S = 32.07 g/mol O = 16.00 g/mol

Set up the problem

Molar mass of $CuSO_{4}$
= (Cu) + (S) + 4(O)
= $(63.55\frac{g}{mol}) + (32.07\frac{g}{mol}) + 4(16.00\frac{g}{mol}) $

Calculate your answer

Molar mass of $CuSO_{4}$
= $(63.55\frac{g}{mol}) + (32.07\frac{g}{mol}) + 4(16.00\frac{g}{mol}) $
= $159.62\frac{g}{mol}$
The molar mass of copper sulfate is $159.62\frac{g}{mol}$

Check your answer.

The molar mass of copper sulfate is $159.62\frac{g}{mol}$

Quantifying Chemical Compounds Example Problem 2

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Problem:

How many molecules of copper sulfate ( $CuSO_4)$ are in a 200 mg sample?

What is the problem asking?

Number of molecules of $CuSO_{4}$ = # of molecules = ?

What do you have?

Mass of $CuSO_{4}$ = 200 mg
Molar mass of $CuSO_{4}$ from previous question = 159.62 g/mol

What do you need?

Avogadro’s number:

$= 6.022$x$10^{23}\frac{\text{molecules}}{\text{mol}}$

Conversion factor:

$\frac{\text{1 g}}{\text{1000 mg}}$

Set up the problem

$200\,mg\, CuSO_{4}$x$\frac{1\,g}{1000\,mg}$x$\frac{1\,mol\,CuSO_{4}}{159.62\,g\, CuSO_{4}}$x$\frac{6.022\text{x}10^{23}\,molecules\,CuSO_{4}}{mol\,CuSO_{4}}$

Calculate your answer

$200\,mg\, CuSO_{4}$x$\frac{1\,g}{1000\,mg}$x$\frac{1\,mol\,CuSO_{4}}{159.62\,g\, CuSO_{4}}$x$\frac{6.022\text{x}10^{23}\,molecules\,CuSO_{4}}{mol\,CuSO_{4}}$
$= 7.55$x$10^{20}$ molecules $CuSO_{4}$

There are $7.55$x$10^{20}$ molecules of $CuSO_{4}$ in a 200 mg sample.

Check your answer.

There are $7.55$x$10^{20}$ molecules of $CuSO_{4}$ in a 200 mg sample.

Reasoning:

By definition of the mole there are $= 6.022$x$10^{23}\frac{\text{molecules}}{\text{mol}}$ of $CuSO_{4}$ in a 159.62 gram sample.
159.62 grams of $CuSO_{4}$ = 159 620 mg of $CuSO_{4}$ = $= 6.022$x$10^{23}$ molecules of $CuSO_{4}$
A 200 mg sample of $CuSO_{4}$ has $7.55$x$10^{20}$ molecules of $CuSO_{4}$.
159 620 mg is 3 orders of magnitude larger than 200 mg.
It is obvious that 200 mg would have fewer molecules than 159 620 mg, the answer makes numerical and logical sense.

Quantifying Chemical Compounds Example Problem 3

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The following problem is broken down into steps that can be seen by clicking on the tabs below.

Problem:

How many grams of copper are in a kilogram of copper sulfate ($CuSO_4$)?

What is the problem asking?

Mass of Cu in grams in one kilogram of $CuSO_{4}$ = g of Cu = ?

What do you have?

Mass of $CuSO_{4}$ = 1 kg
Molar mass of $CuSO_{4}$ from previous question = 159.62 g/mol
Molar mass of Cu from previous question = Cu = 63.55 g/mol
The relationship between Cu and $CuSO_{4} = \frac{1\,mol\,Cu}{1\,mol\,CuSO_{4}}$

What do you need?

Conversion factor:

$\frac{\text{1000 g}}{\text{1 kg}}$

Set up the problem

$1\,kg\,CuSO_{4}\,$x$\frac{1000\,g}{1\,kg}$ x $\frac{1\,mol\,CuSO_{4}}{159.62\,g\,CuSO_{4}}$x $\frac{1\,mol\,Cu}{1\,mol\,CuSO_{4}}$ x $\frac{63.55\,g\,Cu}{1\,mol\,Cu}$

Calculate your answer

$1\,kg\,CuSO_{4}\,$x$\frac{1000\,g}{1\,kg}$ x $\frac{1\,mol\,CuSO_{4}}{159.62\,g\,CuSO_{4}}$x $\frac{1\,mol\,Cu}{1\,mol\,CuSO_{4}}$ x $\frac{63.55\,g\,Cu}{1\,mol\,Cu}$
$= 398.13 \,g\, Cu$

There are 398.13 g Cu in one kilogram of $CuSO_{4}$

Check your answer.

There are 398.13 g Cu in one kilogram of $CuSO_{4}$

Reasoning:

One kilogram has 1000 g, copper has a molar mass of 63.55 g/mol.
In copper sulfate, copper has a molar mass that is approximately two times greater than sulfur and four times greater than oxygen; with four oxygen atoms and one sulfur atom.
It is reasonable that that copper would be approximately forty percent of the mass of one kilogram of copper sulfate.

Quantifying Chemical Compounds Example Problem 4

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The following problem is broken down into steps that can be seen by clicking on the tabs below.

Problem:

What is the empirical formula of a compound containing 94% oxygen and 6% hydrogen by mass

What is the problem asking?

The empirical formula = the smallest whole number ratio of H to O = $0_{?}$ $H_{?}$

What do you have?

The compound contains 94% oxygen and 6% hydrogen
100 g of the compound has 94 g of oxygen and 6 g of hydrogen.

What do you need?

The molar mass of oxygen and hydrogen from the periodic table.
O = 16.00 g/mol
H = 1.008 g/mol

Set up the problem

$94.0\,g \, O \,$ x $\frac{1\,mole\,O}{16.0\,g\,O} =$ 5.9 moles oxygen
$6.0\,g \, H \,$ x $\frac{1\,mole\,H}{1.0\,g\,H} =$ 6.0 moles hydrogen

Ratio = 5.9 moles oxygen : 6.0 moles hydrogen
To get the smallest whole number ratio divide by the smallest molar quantity and round to nearest whole number.

Calculate your answer

$= \frac{5.9}{5.9} \text{moles oxygen:} \frac{6.0}{5.9} \text{moles hydrogen}$
=1:1.01
=1:1
$0_{1}H_{1} = OH$

The empirical formula of a compound containing 94% oxygen and 6% hydrogen by mass is OH.

Check your answer.

The empirical formula of a compound containing 94% oxygen and 6% hydrogen by mass is OH.

Reasoning:

It may seem like the empirical formula should have a larger amount of oxygen than hydrogen, but remember the percentage is based on the mass of the element.
Since one mole of oxygen atoms has a mass approximately sixteen times larger than one mole of hydrogen atoms this answer makes logical and numerical sense.

Quantifying Chemical Compounds Example Problem 5

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The following problem is broken down into steps that can be seen by clicking on the tabs below.

Problem:

A compound with the empirical formula OH is found to have a molar mass of $34.0 \frac{g}{mol}$ , what is the molecular formula?

What is the problem asking?

What is the molecular formula?
$0_{?}H_{?}$

What do you have?

Molar Mass of molecular compound = 34.0 g/mol

What do you need?

Molar Mass of empirical compound OH =
(molar mass O) + (molar mass of H) =
16.00 g/mol + 1.01 g/mol = 17.01 g/mol =OH Molar Mass

Set up the problem

Molecular formula molar mass = $x $(Molar Mass of OH) $= 34.0 \frac{g}{mol}$
Solve for $x$
$x =\frac{34.0 \frac{g}{mol}}{17.0 \frac{g}{mol}}$

Calculate your answer

$x = \frac{34.0 \frac{g}{mol}}{17.0 \frac{g}{mol}}$
$= 2$
$x$(OH) = molecular formula
= 2(OH) = $H_{2}O_{2}$

The molecular formula of a compound with an empirical formula of OH and a molar mass of 34.0 g/mol is $H_{2}O_{2}$.

Check your answer.

The molecular formula of a compound with an empirical formula of OH and a molar mass of 34.0 g/mol is $H_{2}O_{2}$.

Problem Set

Quantifying Chemical Compounds Problem Set

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Practice Problems

Quantifying Chemical Compounds problem set

Problem Solutions

Quantifying Chemical Compounds problem set solutions

Quiz

Building the World

Chemical Bonding

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Chemical Bonding Theory

Chemical Bonding Terminology

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Chemical Bonding

Classes of Chemical Bonding

Ionic

Electrostatic attraction between positive and negative ions.

Bonding atoms are generally metals and non-metals.

Electron is transferred from one element to another.

Ionic Bonding in Sodium Chloride (NaCl)

Covalent

Electron sharing between atoms.

Bonding atoms are nonmetals.

Electrons are equally shared by atoms.

Covalent Bonding in Hydrogen ($H_2$)

Polar Covalent

Electron sharing between atoms.

Bonding atoms are nonmetals.

Electrons are unevenly shared between atoms.

Polar Covalent Bonding in Hydrogen Fluoride (HF)

Ionic Bonding

We know:

Metals have tendency to lose electrons, forming cations.
Nonmetals have a tendency to gain electrons, forming anions.
Metals and nonmetals want to attain the nearest noble gas configuration.

Ionic bonding:

When metal and nonmetal atoms approach each other an electron transfer takes place.
The metal transfers its electron(s) to the nonmetal so both elements reach the nearest noble gas configuration.
The charged ions are then attracted to each other due to electrostatic forces.

Electrostatic Attraction Between Ions = Ionic Bond

Covalent Bonding

We know:

Atoms will try to attain a noble gas configuration.
Valence electrons are the electrons in the outermost shell of an atom.

Covalent Bonding:

Nonmetals will share valence electrons to satisfy a noble gas configuration.

Nonmetals Sharing Electrons = Covalent Bond

Electron Sharing in a Hydrogen Molecule
Electron are shared equally between the two atoms

Polar Covalent Bonding

We know:

An atom’s electronegativity is a measure of how well it attracts electrons to itself while in a chemical bond.
The difference in electronegativity between two bonded atoms can be calculated.

Polar Covalent Bonding:

A covalent bond between atoms that have a significant difference in their electro-negativities. Polar covalent bonds have unequal electron sharing between atoms.

Nonmetals Sharing Electrons Unequally = Polar Covalent Bond

Molecules that have unequal electron sharing have partial charge separation.

The atom in a molecule that is more electronegative will have a larger electron density and a slight net negative charge delta negative(δ-).

The atom in a molecule that is less electronegative will have a smaller electron density and a slight net positive charge delta positive(δ+).

Electronegativity and Bond Types

The electronegativity difference (ΔEN) between bonded atoms can be used to identify which type of bond exists in a molecule

Electronegativity Differences and Bond Type

(ΔEN)	Bond Type	Example
0.0-0.4	Covalent (no charge on atoms)	$O_{2}$ Oxygen
0.4-2.0	Polar Covalent (partial charge on atoms)	CO Carbon Monoxide
2.0 +	Ionic (full charge on ions)	KI Potassium Iodide

Valence Electrons and Bonding

We know:

Valence electrons are the electrons in the outermost shell of an atom.

Valence Electrons:

The valence electrons are the electrons involved in chemical bonding between atoms.
The number of valence electrons an element has determines the chemical properties of that element.
Elements in a column (group) have the same number of valence electrons, this is why elements in a group have similar chemistry.

Valence Electrons Continued:

The valence electrons of a main group element are located in the outermost shell.
The valence electrons of a transition metal are located in the outermost d orbitals, as well as the outermost shell.

Valence electrons are responsible for the chemical properties of an atom.

Valence Electrons and Lewis Structures

Lewis Structures:

Representing the valence electrons of a main group element using dots surrounding the chemical symbol.

Lewis Theory:

Chemical bonding is the attainment of a stable electron configuration through the sharing or transfer of electrons.
Lewis Theory uses the octet rule to predict bonding.
Lewis structures can be used to demonstrate bonding in molecules by showing atoms in a molecule sharing electrons to attain a full octet.

Octet:

A full outer shell containing eight electrons.

The Octet Rule

A stable electron configuration can be attained with eight electrons in the outermost shell.
Bonding atoms will transfer or share electrons to satisfy the octet rule, each atom will have access to eight electrons in its outermost shell.
The octet rule can be used to predict how atoms bond.

This rule only works for the second period (row) of the periodic table.
Elements beyond the second row can access d or f orbitals, elements are larger and have more room for bonding.

Hyper-coordination: Elements beyond the second row (after Neon) can have expanded octets.
The octet rule is a useful tool for predicting bonding in molecules especially in organic chemistry.

Lewis Structures Terminology

Formal Charge:

The charge on an atom in a Lewis structure, assuming all electrons are shared equally. Formal charge is calculated with the following formula:

(valence $e^{-}$) - ($\frac{1}{2}$bonding $e^{-}$) - (lone electrons) = Formal Charge

Formal charge is used to determine the best Lewis structure for a compound that has more than one choice. The lowest possible charge indicates the most energetically favorable structure.

Resonance Structures:

When more than one Lewis structure is allowed for a compound drawing both structures with a double headed arrow between them is the correct way to indicate resonance.

Lewis Structures

First Row:

Second Row and the Octet Rule:

Lewis Structures and Bonding:

Lewis Structures for Molecules Method

1.

Determine the total number of valence electrons for the entire molecule by adding the amount of valence electrons for each molecule, for anions add an extra electron, for cations subtract an electron.

2.

Write the correct structure for the molecule, use a line to connect the bonding atoms, each line represents two electrons. Subtract bonding electrons from total electrons.
Hydrogen will always be terminal.
Generally the least electronegative atom will be in the center.

3.

Distribute remaining electrons (as pairs) first to the terminal atoms, then to the central atom. Subtract from your total as you fill octets.
Hydrogen has a full shell with two electrons. Do not assign electrons (other than bonding) to hydrogen.

4.

If there are any atoms without a full octet, move electrons to form double or triple bonds, use arrows to indicate electron movement. Redraw your structure.

Examples

Problem Set

Chemical Bonding Problem Set

Below are two documents. One is practice problems, the second is the same problems with solutions.
They can be downloaded and changed to suit your needs.

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Practice Problems

Problem Solutions

Quiz

Building the World

Quantifying Chemical Compounds

Apply your Knowledge

The Law of Multiple Proportions

Elements combine in different ways to form substances whose mass ratios are small whole number multiples of each other.

The Law of Definite Proportions

All samples of a pure chemical substance always contain the same proportion of elements by mass regardless of their origin.

By mass, Carbon dioxide (\(CO_2\)) is: 73% Oxygen and 27% Carbon

All carbon dioxide is chemically the same in every location.

Return of the Mole!

We know:

Molecules and the Mole:

1 mole of carbon dioxide \(CO_{2}\) a mass of : \(12.01\,g + 16.00\,g + 16.00\,g = 44.01\,g\)

Numbers of moles IS A DIRECT COMPARISON of the relative number of atoms or molecules!

The mole is used extensively in chemical problems because numbers of atoms can be compared!

Empirical Formula Vs. Molecular Formula

Empirical Formula

Molecular Formula:

NOTE

Periodic Trends

We know:

Periodic Trends:

Electronegativity

Atomic Radius

On the periodic table bond radius increases down a column.

On the periodic table bond radius decreases across a row.

Atomic Radius and Ions:

Cations have a smaller ionic radius than the neutral atom. Anions have a larger ionic radius than the neutral atom.

Ionization Energy

On the periodic table ionization energy decreases down a column.

On the periodic table ionization energy increases across a row.

Quantifying Chemical Compounds Example Problem 1

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Problem:

What is the molar mass of copper sulfate \(CuSO_{4}\)?

Quantifying Chemical Compounds Example Problem 2

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Problem:

How many molecules of copper sulfate (CuSO4) C u S O 4 ) CuSO_4) are in a 200 mg sample?

Quantifying Chemical Compounds Example Problem 3

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Problem:

How many grams of copper are in a kilogram of copper sulfate (\(CuSO_4\))?

Quantifying Chemical Compounds Example Problem 4

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Problem:

What is the empirical formula of a compound containing 94% oxygen and 6% hydrogen by mass

Quantifying Chemical Compounds Example Problem 5

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Problem:

A compound with the empirical formula OH is found to have a molar mass of \(34.0 \frac{g}{mol}\) , what is the molecular formula?

Quantifying Chemical Compounds Problem Set

Below are two documents. One is practice problems, the second is the same problems with solutions. They can be downloaded and changed to suit your needs.

Click or tap the images below to view.

Practice Problems

Problem Solutions

What is the molar mass of Sr\(_{3}\)(PO\(_{4}\))\(_{2}\)?

A)

B)

C)

D)

What is the mass of one mole of CaCO\(_{3}\)?

A)

B)

C)

D)

The formula of a compound is Cu\(_{2}\)O. This is a(n):

Empirical formula

Molecular formula

A formula, but not enough information is given to specify type.

The atomic radius of H, Li, K and Rb are 37, 152, 227, and 248 pm respectively. Use the position of sodium in the periodic table and the given information to choose the most correct atomic radius of chlorine.

A)

B)

C)

D)

Predict which element has the largest ionization energy using the periodic table.

A)

B)

C)

D)

Cations have a smaller ionic radius than the neutral atom.

Anions have a larger ionic radius than the neutral atom.

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Click or tap the grey boxes to the left and right
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How many molecules of copper sulfate ( $CuSO_4)$ are in a 200 mg sample?

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to scroll through the examples.

Click or tap the grey boxes to the left and right
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Click or tap the grey boxes to the left and right
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Below are two documents. One is practice problems, the second is the same problems with solutions.
They can be downloaded and changed to suit your needs.

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Electron Sharing in a Hydrogen Molecule
Electron are shared equally between the two atoms

Click or tap the grey boxes to the left and right
to scroll through the examples.

Click or tap the grey boxes to the left and right
to scroll through the examples.

Below are two documents. One is practice problems, the second is the same problems with solutions.
They can be downloaded and changed to suit your needs.