Introduction to Chemical Nomenclature¶

This notebook will outline and explain the following concepts:

Section 1: Review (Periodic Table, ions, and atomic structure)
Section 2: How Elements Form Compounds (ionic bonds versus covalent bonds)
Section 3: Naming Binary Ionic Compounds
Section 4: Naming Ionic Compounds with Polyatomic Ions
Section 5: Writing Formulas for Ionic Compounds
Section 6: Writing Formulas for Molecular Compounds
Section 7: Naming Molecular Compounds

Section 1: Review¶

Make reference to the following periodic table throughout this notebook:

Note: The content covered in the following multiple choice is review material (concepts that you should already be familiar with).

Question 1: The elements on the Periodic Table are

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "In electronically neutral form." 
option_2 = "All metals or metal alloys."
option_3 = "All charged with either a positive or negative charge."
option_4 = "Unorganized and cannot be grouped together in any way."

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mUnorganized and cannot be grouped together in any way.
[1m(B) [0;0mAll metals or metal alloys.
[1m(C) [0;0mIn electronically neutral form.
[1m(D) [0;0mAll charged with either a positive or negative charge.

Question 2: The subatomic particle that is invovled in the bonding of atoms is called the

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Electron" 
option_2 = "Proton"
option_3 = "Neutron"
option_4 = "Nucleus"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mProton
[1m(B) [0;0mElectron
[1m(C) [0;0mNucleus
[1m(D) [0;0mNeutron

Question 3: What is the name of the atomic model below?

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Bohr Model" 
option_2 = "Quantum Mechanical Model"
option_3 = "Plum Pudding Model"
option_4 = "Solar System Model"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mSolar System Model
[1m(B) [0;0mQuantum Mechanical Model
[1m(C) [0;0mBohr Model
[1m(D) [0;0mPlum Pudding Model

Question 4: What is the symbol and charge for each subatomic particle?

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Protons (p{}), electrons (e{}), neutrons (n{}).".format(chr(0x207A), chr(0x207B), chr(0x2070)) 
option_2 = "Protons (p{}), electrons (e{}), neutrons (n{}).".format(chr(0x207B), chr(0x207A), chr(0x2070))
option_3 = "Protons (p{}), electrons (e{}), neutrons (n{}).".format(chr(0x207A), chr(0x2070), chr(0x207B))
option_4 = "None of the options are correct."

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mNone of the options are correct.
[1m(B) [0;0mProtons (p⁻), electrons (e⁺), neutrons (n⁰).
[1m(C) [0;0mProtons (p⁺), electrons (e⁰), neutrons (n⁻).
[1m(D) [0;0mProtons (p⁺), electrons (e⁻), neutrons (n⁰).

Question 5: A cation is a _______ charged ion. An anion is a _______ charged ion.

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "positively; negatively" 
option_2 = "negatively; positively"
option_3 = "neutral; negatively"
option_4 = "positively; neutral"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mneutral; negatively
[1m(B) [0;0mpositively; negatively
[1m(C) [0;0mnegatively; positively
[1m(D) [0;0mpositively; neutral

Section 2: How Elements Form Compounds¶

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Much of chemical activity involves the transfer or sharing of electrons.

Ionic compounds are combinations of cations and anions, which are generally (but not always) metals and nonmetals. Their compounds are held together by electrostatic attractions between oppositely charged ions (similar to a magnetic attraction). This is called an ionic bond.

Covalent (or molecular) compounds are generally composed of nonmetals only. Electrons are shared between atoms. This is called a covalent bond.

Section 3: Naming Binary Ionic Compounds¶

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Steps for naming binary ionic compounds:

Name the cation (often a metal) first.
Name the anion (often a nonmetal) second. For the anion, drop the ending and add “ide”.
Use Roman numerals in parentheses to indicate which charge of cation is used. This is only used for metals that make more than one charge. Below is a table of the Roman numerals from 1 to 10:

Number	Roman Numeral
1	I
2	II
3	III
4	IV
5	V
6	VI
7	VII
8	VIII
9	IX
10	X

Examples¶

Name the following ionic compounds:

Ca\(_3\)N\(_2\)
Mg\(_3\)P\(_2\)
Al\(_2\)O\(_3\)
CuCl\(_2\)
FeBr\(_3\)
CoN

#----------

#import ipywidgets as widgets
#from ipywidgets import Output, VBox, Layout
#from IPython.display import clear_output, display, HTML

#----------

out1 = Output()
button1_step1 = widgets.Button(description="Answers 1-3", layout=Layout(width='20%', height='100%'), button_style='primary')
count1 = 1

text1_1 = widgets.HTMLMath(value=r"1. Calcium nitride")
text1_2 = widgets.HTMLMath(value=r"2. Magnesium phosphide")
text1_3 = widgets.HTMLMath(value=r"3. Aluminum oxide")

def on_button1_step1_clicked(b):
    global count1
    count1 += 1
    with out1:
        clear_output()
        if count1 % 2 == 0:
            display(text1_1, text1_2, text1_3)
            
display(VBox([button1_step1, out1]))
button1_step1.on_click(on_button1_step1_clicked)

#----------

#import ipywidgets as widgets
#from ipywidgets import Output, VBox, Layout
#from IPython.display import clear_output, display, HTML

#----------

out2 = Output()
button1_step2 = widgets.Button(description="Answers 4-6", layout=Layout(width='20%', height='100%'), button_style='primary')
count2 = 1

text1_4 = widgets.HTMLMath(value=r"4. Reference the periodic table. Notice how copper makes more than one charge (+2 or +1). Recall how compounds are electronically neutral. Since chlorine always makes a -1 charge (and there are two chlorine atoms), this implies that copper must be taking a +2 charge in this situation.")
text1_4_2 = widgets.HTMLMath(value=r" Answer: Copper (II) chloride")
text1_5 = widgets.HTMLMath(value=r"5. Iron (III) bromide (similar to #4)")
text1_6 = widgets.HTMLMath(value=r"6. N takes a -3 charge and we have one nitrogen and one cobalt atom. Cobalt must be taking a +3 charge to make the compound neutral.")
text1_6_2 = widgets.HTMLMath(value=r"Answer: Cobalt (III) nitride")
def on_button1_step2_clicked(b):
    global count2
    count2 += 1
    with out2:
        clear_output()
        if count2 % 2 == 0:
            display(text1_4, text1_4_2, text1_5, text1_6, text1_6_2)
            
display(VBox([button1_step2, out2]))
button1_step2.on_click(on_button1_step2_clicked)

Practice (Multiple Choice)¶

Choose the option with the correct name for the following ionic compounds.

Question 1: CaCl\(_2\)

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Calcium chloride" 
option_2 = "Calcium chlorine"
option_3 = "Chlorine Calcium"
option_4 = "Calcium (II) chloride"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mChlorine Calcium
[1m(B) [0;0mCalcium chlorine
[1m(C) [0;0mCalcium (II) chloride
[1m(D) [0;0mCalcium chloride

Question 2: MgBr\(_2\)

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Magnesium bromide" 
option_2 = "Magnesium bromine"
option_3 = "Magnesium (II) bromide"
option_4 = "Magnesium (II) bromine"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mMagnesium (II) bromine
[1m(B) [0;0mMagnesium bromine
[1m(C) [0;0mMagnesium bromide
[1m(D) [0;0mMagnesium (II) bromide

Question 3: FeI\(_3\)

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Iron (III) iodide" 
option_2 = "Iron iodine"
option_3 = "Iron iodide"
option_4 = "Iron (III) iodinide"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mIron iodide
[1m(B) [0;0mIron (III) iodinide
[1m(C) [0;0mIron (III) iodide
[1m(D) [0;0mIron iodine

Question 4: CuCl

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Copper (I) chloride" 
option_2 = "Copper chloride"
option_3 = "Copper chlorine"
option_4 = "Calcium (I) chlorine"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mCopper chlorine
[1m(B) [0;0mCalcium (I) chlorine
[1m(C) [0;0mCopper chloride
[1m(D) [0;0mCopper (I) chloride

Question 5: Cr\(_3\)N\(_2\)

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Chromium (II) nitride" 
option_2 = "Chromium (VI) nitride"
option_3 = "Chromium (III) nitride"
option_4 = "Chromium nitride"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mChromium (II) nitride
[1m(B) [0;0mChromium (III) nitride
[1m(C) [0;0mChromium (VI) nitride
[1m(D) [0;0mChromium nitride

Practice (Fill-in-the-blank)¶

Write the names for the following ionic compounds. Spelling, capitalization, and spacing matters! (Match the format of the examples above). Note that the first letter of each name is capitalized and roman numerals are capitalized.

blank1 = widgets.Text(
    value='',
    placeholder='Type name here & press ENTER',
    description="LiCl",
    disabled=False
)


display(blank1)

           
def callback1(wdgt):
    if (wdgt.value == "Lithium chloride") or (wdgt.value == "lithium chloride"):
        print("Correct! 👏                                                                                      ", end='\r')
        blank1.on_submit(callback1)
    else:
        print("Hint: Name the metal 1st and the nonmetal 2nd. Remember to change the nonmetal's ending to 'ide'!", end='\r')
        blank1.on_submit(callback1)

blank1.on_submit(callback1)

blank2 = widgets.Text(
    value='',
    placeholder='Type name here & press ENTER',
    description="Sr{}P{}".format(chr(0x2083), chr(0x2082)),
    disabled=False
)


display(blank2)

           
def callback2(wdgt):
    if (wdgt.value == "Strontium phosphide") or (wdgt.value == "strontium phosphide"):
        print("Correct! 👏                                                                                      ", end='\r')
        blank2.on_submit(callback2)
    else:
        print("Hint: Name the metal 1st and the nonmetal 2nd. Remember to change the nonmetal's ending to 'ide'!", end='\r')
        blank2.on_submit(callback2)

blank2.on_submit(callback2)

blank3 = widgets.Text(
    value='',
    placeholder='Type name here & press ENTER',
    description="Ag{}S".format(chr(0x2082)),
    disabled=False
)


display(blank3)

           
def callback3(wdgt):
    if (wdgt.value == "Silver sulfide") or (wdgt.value == "silver sulfide"):
        print("Correct! 👏                                                                                      ", end='\r')
        blank3.on_submit(callback3)
    else:
        print("Hint: Name the metal 1st and the nonmetal 2nd. Remember to change the nonmetal's ending to 'ide'!", end='\r')
        blank3.on_submit(callback3)

blank3.on_submit(callback3)

blank4 = widgets.Text(
    value='',
    placeholder='Type name here & press ENTER',
    description="SnI{}".format(chr(0x2084)),
    disabled=False
)


display(blank4)

           
def callback4(wdgt):
    if (wdgt.value == "Tin (IV) iodide") or (wdgt.value == "tin (iv) iodide"):
        print("Correct! 👏                                                         ", end='\r')
        blank4.on_submit(callback4)
    else:
        print("Hint: Remember to reference if the metal makes more than one charge!", end='\r')
        blank4.on_submit(callback4)

blank4.on_submit(callback4)

blank5 = widgets.Text(
    value='',
    placeholder='Type name here & press ENTER',
    description="PdS{}".format(chr(0x2082)),
    disabled=False
)


display(blank5)

           
def callback5(wdgt):
    if (wdgt.value == "Palladium (IV) sulfide") or (wdgt.value == "palladium (iv) sulfide"):
        print("Correct! 👏                                                         ", end='\r')
        blank5.on_submit(callback5)
    else:
        print("Hint: Remember to reference if the metal makes more than one charge!", end='\r')
        blank5.on_submit(callback5)

blank5.on_submit(callback5)

Section 4: Naming Ionic Compounds with Polyatomic Ions¶

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Polyatomic ions are groups of atoms that tend to stay together and carry an overall ionic charge. Reference the following table of polyatomic ions throughout the rest of this notebook:

Name	Formula
Ammonium	NH\(_4\) \(^{+1}\)
Bicarbonate	HCO\(_3\) \(^{-1}\)
Carbonate	CO\(_3\) \(^{-2}\)
Carbonite	CO\(_2\) \(^{-2}\)
Chlorate	ClO\(_3\) \(^{-1}\)
Chlorite	ClO\(_2\) \(^{-1}\)
Chromate	CrO\(_4\) \(^{-2}\)
Cyanide	CN \(^{-1}\)
Hydroxide	OH \(^{-1}\)
Iodate	IO\(_3\) \(^{-1}\)
Iodite	IO\(_2\) \(^{-1}\)
Nitrate	NO\(_3\) \(^{-1}\)
Nitrite	NO\(_2\) \(^{-1}\)
Phosphate	PO\(_4\) \(^{-3}\)
Phosphite	PO\(_3\) \(^{-3}\)
Sulfate	SO\(_4\) \(^{-2}\)
Sulfite	SO\(_3\) \(^{-2}\)

Section 2: Steps for naming ionic compounds with polyatomic ions.

For polyatomic compounds, name the cation (metal) first followed by the name of the polyatomic ion. The endings do not change.
Like before, use Roman numerals in parentheses to indicate which charge of cation is used. This is only used for metals that make more than one charge.

Examples¶

Name the following ionic compounds:

Al(OH)\(_3\)
NaHCO\(_3\)
Sn(NO\(_3\))\(_4\)

#----------

#import ipywidgets as widgets
#from ipywidgets import Output, VBox, Layout
#from IPython.display import clear_output, display, HTML

#----------

out3 = Output()
button2_step1 = widgets.Button(description="Answers", layout=Layout(width='20%', height='100%'), button_style='primary')
count3 = 1

text2_1 = widgets.HTMLMath(value=r"1. Answer: Aluminum hydroxide. OH is a compound called 'hydroxide' (see the table of polyatomic ions above). Also, aluminum makes only one charge.")
text2_2 = widgets.HTMLMath(value=r"2. Answer: Sodium bicarbonate. NaHCO$_3$ is a compound called 'bicarbonate'. Also, sodium makes only one charge.")
text2_3 = widgets.HTMLMath(value=r"3. Answer: Tin (IV) nitrite. Sn(NO$_3$)$_4$ is a compound called 'nitrite'. Notice that Sn (tin) takes two charges (+2 or +4); therefore, we need to use Roman numerals. Nitrite takes a -1 charge, so tin is taking a +4 charge in this situation.")

def on_button2_step1_clicked(b):
    global count3
    count3 += 1
    with out3:
        clear_output()
        if count3 % 2 == 0:
            display(text2_1, text2_2, text2_3)
            
display(VBox([button2_step1, out3]))
button2_step1.on_click(on_button2_step1_clicked)

Practice (Multiple Choice)¶

Choose the option with the correct name for the following ionic compounds involving polyatomic ions.

Question 1: KNO\(_3\)

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Potassium nitrate" 
option_2 = "Potassium nitride"
option_3 = "Potassium (II) nitrate"
option_4 = "Potassium (II) nitride"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mPotassium (II) nitride
[1m(B) [0;0mPotassium nitride
[1m(C) [0;0mPotassium (II) nitrate
[1m(D) [0;0mPotassium nitrate

Question 2: Mg\(_3\)(PO\(_4\))\(_2\)

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Magnesium phosphate" 
option_2 = "Magnesium (II) phosphate"
option_3 = "Magnesium phosphide"
option_4 = "Magnesium (II) phosphide"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mMagnesium phosphide
[1m(B) [0;0mMagnesium (II) phosphate
[1m(C) [0;0mMagnesium phosphate
[1m(D) [0;0mMagnesium (II) phosphide

Practice (Fill-in-the-blank)¶

Write the names for the following ionic compounds involving polyatomic ions. As before, spelling, capitalization, and spacing matters! Note that the first letter of each name is capitalized and Roman numerals are capitalized.

blank6 = widgets.Text(
    value='',
    placeholder='Type name here & press ENTER',
    description="NaOH",
    disabled=False
)


display(blank6)

           
def callback6(wdgt):
    if (wdgt.value == "Sodium hydroxide") or (wdgt.value == "sodium hydroxide"):
        print("Correct! 👏                                                    ", end='\r')
        blank6.on_submit(callback6)
    else:
        print("Hint: Remember to reference the table of polyatomic ions above!", end='\r')
        blank6.on_submit(callback6)

blank6.on_submit(callback6)

blank7 = widgets.Text(
    value='',
    placeholder='Type name here & press ENTER',
    description="Pb(NO{}){}".format(chr(0x2082), chr(0x2084)),
    disabled=False
)


display(blank7)

           
def callback7(wdgt):
    if (wdgt.value == "Lead (IV) nitrite") or (wdgt.value == "lead (iv) nitrite"):
        print("Correct! 👏                                                    ", end='\r')
        blank7.on_submit(callback7)
    else:
        print("Hint: Remember to reference the table of polyatomic ions above!", end='\r')
        blank7.on_submit(callback7)

blank7.on_submit(callback7)

Section 5: Writing Formulas for Ionic Compounds¶

Back to Top

We now consider how to convert a compound name to a formula. Since each ionic compound is electronically neutral, we have the following steps:

Note the ionic charges and the symbols for both the cation and anion.
Determine the lowest number of each element that would make the entire compound electronically neutral.

Note:

Some metals can make more than one kind of ion. Roman numerals in brackets indicate the charge of the ion.
If you require more than one polyatomic ion, remember to put the entire polyatomic ion in brackets along with a subscript indicating the number required.

Examples¶

Write the formulas for the following ionic compounds:

Calcium iodide
Sodium phosphide
Iron (III) oxide
Iron (II) oxide
Sodium carbonite
Aluminum sulfate

#----------

#import ipywidgets as widgets
#from ipywidgets import Output, VBox, Layout
#from IPython.display import clear_output, display, HTML

#----------

out4 = Output()
button3_step1 = widgets.Button(description="Answers", layout=Layout(width='20%', height='100%'), button_style='primary')
count4 = 1

text3_1 = widgets.HTMLMath(value=r"1. Answer: CaI$_2$. The charges involved are Ca$^{+2}$ and I$^{-1}$. We need exactly 2 iodine atoms to cancel calcium's +2 charge.")
text3_2 = widgets.HTMLMath(value=r"2. Answer: Na$_3$P. We have Na$^+$ and P$^{-3}$, so we need 3 sodium atoms to cancel phosphorus' -3 charge.")
text3_3 = widgets.HTMLMath(value=r"3. Answer: Fe$_2$O$_3$. The Roman numeral III tells us that we have Fe$^{+3}$. Since oxygen takes a -2 charge, we need exactly 2 Fe atoms and 3 O atoms to arrive at an electronically neutral compound.")
text3_4 = widgets.HTMLMath(value=r"4. Answer: FeO. We have Fe$^{+2}$ and O$^{-2}$, so we need exactly 1 Fe atom and 1 O atom to arrive at an electronically neutral compound.")
text3_5 = widgets.HTMLMath(value=r"5. Answer: Na$_2$CO$_2$. We have Na$^+$ and CO$_2^{-2}$, so we need 2 Na atoms and 1 CO$_2$ molecule.")
text3_6 = widgets.HTMLMath(value=r"6. Al$^{+3}$ and SO$_4^{-2}$ $\implies$ Al$_2$(SO$_4$)$_3$.")

def on_button3_step1_clicked(b):
    global count4
    count4 += 1
    with out4:
        clear_output()
        if count4 % 2 == 0:
            display(text3_1, text3_2, text3_3, text3_4, text3_5, text3_6)
            
display(VBox([button3_step1, out4]))
button3_step1.on_click(on_button3_step1_clicked)

Practice (Multiple Choice)¶

Choose the option with the correct formula for the following ionic compounds.

Question 1: Sodium bromide

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "NaBr" 
option_2 = "BrNa"
option_3 = "SBr{}".format(chr(0x2082))
option_4 = "SBr"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mBrNa
[1m(B) [0;0mNaBr
[1m(C) [0;0mSBr₂
[1m(D) [0;0mSBr

Question 2: Zinc iodide

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "ZnI{}".format(chr(0x2082)) 
option_2 = "ZnI"
option_3 = "Zn{}I{}".format(chr(0x2082), chr(0x2082))
option_4 = "ZnIO{}".format(chr(0x2083))

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mZn₂I₂
[1m(B) [0;0mZnIO₃
[1m(C) [0;0mZnI₂
[1m(D) [0;0mZnI

Question 3: Cobalt (II) chloride

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "CoCl{}".format(chr(0x2082))
option_2 = "CoClO{}".format(chr(0x2083))
option_3 = "CoCl"
option_4 = "Co{}Cl".format(chr(0x2082))

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mCo₂Cl
[1m(B) [0;0mCoClO₃
[1m(C) [0;0mCoCl₂
[1m(D) [0;0mCoCl

Question 4: Lead (IV) selenide

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "PbSe{}".format(chr(0x2082))
option_2 = "Pb{}Se{}".format(chr(0x2082), chr(0x2084))
option_3 = "PbSe"
option_4 = "PbSe".format(chr(0x2084))

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mPbSe
[1m(B) [0;0mPbSe
[1m(C) [0;0mPb₂Se₄
[1m(D) [0;0mPbSe₂

Question 5: Silver nitrate

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "AgNO{}".format(chr(0x2083))
option_2 = "Ag(NO){}".format(chr(0x2083))
option_3 = "Ag{}N".format(chr(0x2083))
option_4 = "AgNO{}".format(chr(0x2082))

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mAg₃N
[1m(B) [0;0mAgNO₃
[1m(C) [0;0mAg(NO)₃
[1m(D) [0;0mAgNO₂

Question 6: Iron (II) phosphate

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "Fe{}(PO{}){}".format(chr(0x2083), chr(0x2084), chr(0x2082))
option_2 = "FePO{}".format(chr(0x2084))
option_3 = "Fe{}PO{}".format(chr(0x2083), chr(0x2084))
option_4 = "Fe{}(PO{}){}".format(chr(0x2082), chr(0x2084), chr(0x2083))

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mFe₃PO₄
[1m(B) [0;0mFePO₄
[1m(C) [0;0mFe₃(PO₄)₂
[1m(D) [0;0mFe₂(PO₄)₃

Section 6: Writing Formulas for Molecular Compounds¶

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Recall that a binary molecular compound is composed of two nonmetals and that they are share electrons via covalent bonds (ex. CO\(_2\)). There are two main steps to writing formulas for molecular compounds:

Write the symbols of each nonmetal.
Write the appropriate subscripts for each nonmetal that correspond to the prefixes used.

Note: The prefix mono is never used on the first nonmetal.

Here is a list of prefixes used for molecular compounds:

Number	Prefix
1	mono
2	di
3	tri
4	tetra
5	penta
6	hexa
7	hepta
8	octa
9	nona
10	deca

Examples¶

Write the formulas for the following molecular compounds.

Carbon dioxide
Dinitrogen trioxide
Carbon tetrafluoride

#----------

#import ipywidgets as widgets
#from ipywidgets import Output, VBox, Layout
#from IPython.display import clear_output, display, HTML

#----------

out5 = Output()
button4_step1 = widgets.Button(description="Answers", layout=Layout(width='20%', height='100%'), button_style='primary')
count5 = 1

text4_1 = widgets.HTMLMath(value=r"1. Answer: CO$_2$. No prefix in front of carbon implies that there is one carbon. The prefix 'di' implies that there are two oxygen atoms.")
text4_2 = widgets.HTMLMath(value=r"2. Answer: N$_2$O$_3$")
text4_3 = widgets.HTMLMath(value=r"3. Answer: CF$_4$")

def on_button4_step1_clicked(b):
    global count5
    count5 += 1
    with out5:
        clear_output()
        if count5 % 2 == 0:
            display(text4_1, text4_2, text4_3)
            
display(VBox([button4_step1, out5]))
button4_step1.on_click(on_button4_step1_clicked)

Practice (Multiple Choice)¶

Choose the option with the correct formula for the following molecular compounds.

Question 1: Sulfur trioxide

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "SO{}".format(chr(0x2083))
option_2 = "S{}O".format(chr(0x2083))
option_3 = "S{}O{}".format(chr(0x2083), chr(0x2083))
option_4 = "Not possible"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mS₃O
[1m(B) [0;0mS₃O₃
[1m(C) [0;0mNot possible
[1m(D) [0;0mSO₃

Question 2: Diboron trioxide

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "B{}O{}".format(chr(0x2082), chr(0x2083))
option_2 = "B{}O{}".format(chr(0x2083), chr(0x2082))
option_3 = "2B3O"
option_4 = "3B2O"

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0m2B3O
[1m(B) [0;0mB₃O₂
[1m(C) [0;0m3B2O
[1m(D) [0;0mB₂O₃

Question 3: Phosphorus trihydride

#import ipywidgets as widgets

#Assign each multiple choice to these four variables
#Option_1 contains the answer
option_1 = "PH{}".format(chr(0x2083))
option_2 = "P{}H".format(chr(0x2083))
option_3 = "P(OH){}".format(chr(0x2083))
option_4 = "P{}(OH)".format(chr(0x2083))

multiple_choice(option_1, option_2, option_3, option_4)

[1m(A) [0;0mP(OH)₃
[1m(B) [0;0mP₃H
[1m(C) [0;0mPH₃
[1m(D) [0;0mP₃(OH)

Section 7: Naming Molecular Compounds¶

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Steps:

Write the name of the first nonmetal and then the second.
Place the appropriate prefixes in front of each of the names. The prefixes correspond to the subscripts of each element.
Drop the ending of the last element named and add ide.

Note: Sometimes the prefixes are shortened when the ending vowel of the prefix “conflicts” with a starting vowel in the compound. This makes the name easier to pronounce. For example, “tetraoxide” is written as “tetroxide” and “monooxide” is shortened to “monoxide”.

Recall the following list of prefixes:

Number	Prefix
1	mono
2	di
3	tri
4	tetra
5	penta
6	hexa
7	hepta
8	octa
9	nona
10	deca

Examples¶

Name the following molecular compounds.

CS\(_2\)
N\(_2\)O\(_4\)
P\(_4\)S\(_{10}\)

#----------

#import ipywidgets as widgets
#from ipywidgets import Output, VBox, Layout
#from IPython.display import clear_output, display, HTML

#----------

out6 = Output()
button5_step1 = widgets.Button(description="Answers", layout=Layout(width='20%', height='100%'), button_style='primary')
count6 = 1

text5_1 = widgets.HTMLMath(value=r"1. Answer: Carbon disulfide. The prefix 'mono' is never used for the first element.")
text5_2 = widgets.HTMLMath(value=r"2. Answer: Dinitrogen tetroxide. Recall that 'tetraoxide' is shortened to 'tetroxide'. This avoids two vowels being next to each other.")
text5_3 = widgets.HTMLMath(value=r"3. Answer: Tetraphosphorus decasulfide")

def on_button5_step1_clicked(b):
    global count6
    count6 += 1
    with out6:
        clear_output()
        if count6 % 2 == 0:
            display(text5_1, text5_2, text5_3)
            
display(VBox([button5_step1, out6]))
button5_step1.on_click(on_button5_step1_clicked)

Practice (Fill-in-the-blank)¶

Write the names for the following molecular compounds. Remember that spelling and spacing matters.

blank8 = widgets.Text(
    value='',
    placeholder='Type name here & press ENTER',
    description="CO",
    disabled=False
)


display(blank8)

           
def callback8(wdgt):
    if (wdgt.value == "Carbon monoxide") or (wdgt.value == "carbon monoxide"):
        print("Correct! 👏                                    ", end='\r')
        blank8.on_submit(callback8)
    else:
        print("Hint: Remember to drop the vowel in the prefix!", end='\r')
        blank8.on_submit(callback8)

blank8.on_submit(callback8)

blank9 = widgets.Text(
    value='',
    placeholder='Type name here & press ENTER',
    description="CF{}".format(chr(0x2084)),
    disabled=False
)


display(blank9)

           
def callback9(wdgt):
    if (wdgt.value == "Carbon tetrafluoride") or (wdgt.value == "carbon tetrafluoride"):
        print("Correct! 👏                                                                  ", end='\r')
        blank9.on_submit(callback9)
    else:
        print("Hint: Remember that the prefix 'mono' is never used for the first element!", end='\r')
        blank9.on_submit(callback9)

blank9.on_submit(callback9)

blank10 = widgets.Text(
    value='',
    placeholder='Type name here & press ENTER',
    description="N{}O{}".format(chr(0x2082), chr(0x2085)),
    disabled=False
)


display(blank10)

           
def callback10(wdgt):
    if (wdgt.value == "Dinitrogen pentoxide") or (wdgt.value == "dinitrogen pentoxide"):
        print("Correct! 👏                                    ", end='\r')
        blank10.on_submit(callback10)
    else:
        print("Hint: Remember to drop the vowel in the prefix!", end='\r')
        blank10.on_submit(callback10)

blank10.on_submit(callback10)

Summary¶

There are generally two types of compounds: ionic and covalent.
Ionic compounds are composed of a cation (usually a metal) and an anion (usually a nonmetal).
Covalent compounds are usually composed of nonmetals only.
There are two separate naming systems for both ionic and covalent compounds.
It is critical to become fluent in both naming systems in order to achieve success and understanding in chemistry.

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Callysto Curriculum Notebooks

Introduction to Chemical Nomenclature¶

Section 1: Review¶

Section 2: How Elements Form Compounds¶

Section 3: Naming Binary Ionic Compounds¶

Examples¶

Practice (Multiple Choice)¶

Practice (Fill-in-the-blank)¶

Section 4: Naming Ionic Compounds with Polyatomic Ions¶

Examples¶

Practice (Multiple Choice)¶

Practice (Fill-in-the-blank)¶

Section 5: Writing Formulas for Ionic Compounds¶

Examples¶

Practice (Multiple Choice)¶

Section 6: Writing Formulas for Molecular Compounds¶

Examples¶

Practice (Multiple Choice)¶

Section 7: Naming Molecular Compounds¶

Examples¶

Practice (Fill-in-the-blank)¶

Summary¶