Research Guides: CHEM 1411 Course Guide: Module 2

Dimensional Analysis

Dimensional analysis: A method to convert a given result from one system of units to another.

The video (by Tyler DeWitt) below is an introduction to dimensional analysis. It discusses the reasons why & how it can be used in every day life.

Exploring more of the video discussion, let's discuss another conversion factor. Here is an equivalent statement: 2.54 cm = 1 in

If you divide both sides of that equation by 2.54 cm, you get:

1 in
1 = ------------
2.54 cm

This expression is called a unit factor. Since 1 inch and 2.54 cm are exactly equivalent, multiplying any expression by this unit factor will not change its value.

Problem:

A pin has a length of 2.85 cm. What is the length in inches (in)?

1 in 2.85 in
2.85 cm X ------------ = ------------ = 1.12 in
2.54 cm 2.54 in

We start off with our given, which is 2.85 cm and just like in the video, the centimeter units cancel out to give inches for the result. If you notice, the answer also has 3 significant figures just like our given 2.85. Remember the conversion factor are exact numbers and would not be considered for significant figures due to having infinite numbers.

The next video (by Tyler DeWitt) goes more into detail in the explanation and instruction of how to do dimensional analysis.

For a different explanation of Dimensional Analysis, view The Factor-Label Method video by Bozeman Science.

Length
SI unit: meter (m) 1 meter = 1.0936 yards 1 centimeter = 0.39370 inch 1 inch = 2.54 centimeters (exactly) 1 kilometer = 0.62137 mile 1 mile = 5280 feet 1 mile = 1.6093 kilometers 1 angstrom = 10-10 meter 1 angstrom = 100 picometers

Mass
SI unit: kilogram (kg) 1 kilogram = 1000 grams 1 kilogram = 2.2046 pounds 1 pound = 453.59 grams 1 pound = 0.45359 kilograms 1 pound = 16 ounces 1 ton = 2000 pounds 1 ton = 907.185 kilograms 1 metric ton = 1000 kilograms 1 metric ton = 2204.6 pounds 1 atomic mass unit = 1.66056 x 10-27 kilograms

Volume
SI Unit: cubic meter (m3) 1 liter = 10-3 m3 1 liter = 1dm3 1 liter = 1.0567 quarts 1 gallon = 4 quarts 1 gallon = 8 pints 1 gallon = 3.7854 liters 1 quart = 32 fluid ounces 1 quart = 0.94633 liter

Temperature
SI Unit: kelvin (K) 0 K = –273.15 °C 0 K = –459.67 °F K = °C + 279.15 5 °C = --------(°F — 32) 9 9 °F = --------(°C + 32) 5

Energy
SI Unit: joule (j) 1 joule = 1kg • m2/s2 1 joule = 0.23901 calorie 1 joule = 9.4781 x 10-4 btu (British thermal unit) 1 calorie = 4.184 joules 1 calorie = 3.965 x 10-3 btu 1 btu = 1055.06 joules 1 btu = 252.2 calories

Pressure
SI Unit: pascal (Pa) 1 pascal = 1 N/m2 1 pascal = 1 kg/m • s2 1 atomsphere = 101.325 kilopascals 1 atmosphere = 760 torr (mm Hg) 1 atmosphere = 14.70 pounds per square inch 1 bar = 105 pascals

Sometimes, dimensional analysis is not a one-step process and can take multiple steps to solve a single problem.

The video (by Tyler DeWitt) below discusses, explains, and instructs how to do multi-step unit conversion problems.

Let's work out a multiple conversion problem going step-by-step!

A student has entered a 10.0 km run. How long is the run in miles?

First, we know that we have to convert kilometers to miles, and then we know that our given is 10.0 km.

Next, we need to know our equivalence statements, which are:

1 km = 1000 m
1 m = 1.094 yd
1760 yd = 1 mi

Now that we know our equivalence statements, let's figure out the strategy. We can follow this process:

kilometers → meters → yards → miles

Alright! Let's try to follow this process going step by step:

Kilometers to Meters
1000 m
10.0 km X ------------ = 1.00 x 10⁴ m Note the km unit cancel out to give the meters result.
1 km

Meters to Yards
1.094 yd Note the m unit cancel out to give the yd result. Also, there should be
1.00 x 10⁴ m X ------------ = 1.094 x 10⁴ yd 3 significant figures in the answer, but this is an intermediate step and the
1 m FINAL result should round off to 3 significant figures.

Yards to Miles
1 mi Note in this case that 1 mi equals exactly 1760 yd by designation, and therefore,
1.094 x 10⁴ yd X ------------ = 6.216 mi making 1760 and exact number.
1760 yd

Since the given is 10.0 km, the result can only have 3 significant figures and should be rounded to 6.22 mi making the final answer:

10.0 km = 6.22 mi

Now if we combine the steps, it should look like this:

1000 m 1.094 yd 1 mi
10.0 km X ------------ X ------------ X ------------ = 6.22 mi
1 km 1 m 1760 yd

Helpful Tip:

In doing chemistry problems, you should ALWAYS include the units for the quantities used. ALWAYS check to see that the units cancel to give the correct units for the final result.

How do you do dimensional analysis problems with numbers that have a top and a bottom?

The video (by Tyler DeWitt) below will help explain and instruct on how to work out problems with numbers that have both a top and bottom unit.

Let's try out a problem of our own!

A car is advertised as having a gas mileage of 15 km/L. Convert to miles per gallon.

We already know how to do the initial part from changing km to mi, so let's start with doing that:

First, we need our equivalence statements:

1 km = 1000 m
1 m = 1.094 yd
1760 yd = 1 mi
1 L = 1.06 qt
4 qt = 1 gal

Now, let's do part one of our conversion problem:

15 km 1000 m 1.094 yd 1 mi Note that we were able to cancel out the km unit to mi, but we still
------------ X ------------ X ------------ X ------------ = 9.3238636 mi/L have yet to convert L to gallon.
L 1 km 1 m 1760 yd

We had to make sure that we canceled out the units by putting the same unit on the bottom of the next part of the equation (e.g., 15 km is on the top of the first step and 1 km is on the bottom of the second, 1000 m is on the top of the second step and 1 m is on the bottom of the third step).

Now, we are going to reverse the process. We have 1 L on the bottom of the first step so we need to make sure that the next L unit is on the top of the next step.

9.3238636 mi 1 L 4 qt Remember that we have to round the final result to obtain the
--------------------- X ------------ X ------------ = 35.18439 mi/gal = 35 mi/gal correct number of significant figures which should be 2 for
L 1.06 qt 1 gal this problem because our given is 15.

Here is what it would look like if it is done all together:

15 km 1000 m 1.094 yd 1 mi 1 L 4 qt
------------ X ------------ X ------------ X ------------ X ------------ X ------------ = 35.18439 mi/gal = 35 mi/gal
L 1 km 1 m 1760 yd 1.06 qt 1 gal

Everything that we did in the first step is done in purple while everything that we did in the second step is done in green.

Again, it helps when you write down your units and are able to cross them out as you go. This helps check your process to make sure that you are doing it correctly. ALWAYS INCLUDE UNITS!

The answers are located in the boxes: Highlight the text inside the box to reveal answer

Here are some conversion factors that you should probably know:

1 km = 1000 m	1 cm = 10 mm	1 L = 1000 mL	1 g = 1000 mg
1 m = 1000 mm	1 m = 100 cm	1 kg = 1000 g

Here are other conversion factors:

1 mi = 1.61 km	1 L = 1.06 qt	1 oz = 28.3 g	1 mL = 20 drops	1 cm3 = 1 mL
1 qt = 0.946 L	1 lb = 454 g	1 ton = 2000 lbs	1 gal = 4 quarts	1 in = 2.54 cm
1 kg = 2.2 lbs	1 mi = 5280 ft	365 days = 1 year

Convert 25.0 g to kg.

1 kg
25.0 g X ------------ = 0.0250 kg
1000 g

The answer should contain 3 significant figures because of the given 25.0. The leading zeros in the answer do not count as significant figures and the trailing zero does count because of the decimal point in the result.

How many minutes are there in 1.6 years?

365 days 24 hr 60 min
1.6 yr X ------------ X ------------ X ------------ = 840,960 min --> Corrected answers: 840,000 min OR 8.4 x 10⁵ min
1 yr 1 day 1 hr

Because our given has 2 significant figures, our answer needs to have 2 as well. In this case, we can write the answer two different ways. The numbers that are underlined are the only numbers that count as significant figures. The trailing zeros in the first answer do not count because there is no decimal point in the result.

If you are going 55 mph, what is your speed in m/s?

55 mi 1.61 km 1000 m 1 hr 1 min
------------ X ------------ X ------------ X ------------ X ------------ = 25 m/s
1 hr 1 mi 1 km 60 min 60 s

These questions were obtained from Hudson High School Honors Course worksheet.

An interesting characteristic of an atom is its ability to combine with other atoms to form compounds. Atoms have electrons that are responsible for bonding one atom to another.

Chemical bonds - Forces that hold atoms together in compounds.

Covalent bonds - Atoms share electrons and make molecules.

Examples: H2, CO2, H2O, NH3, O2, CH4

Molecule - A bonded collection of two or more atoms of the same or different elements.

Chemical formula - The representation of a molecule in which the symbols for the elements are used to indicate the types of atoms present and subscripts are used to show the relative numbers of atoms.

Example: Carbon dioxide is CO2; This molecule contains one atom of carbon and two atoms of oxygen.

Structural formula - The representation of a molecule in which the relative positions of the atoms are shown and the bonds are indicated by lines.

Structural formulas may or may not indicate the actual shape of the molecule.
Example:

Water can be represented as

H — O — H OR

Ion - An atom or group of atoms that has a net positive or negative charge.

Cation - A positive ion; often metals since they lose electrons to become positively charged

Helpful Hint: Cat = "paws"itive

Anion - A negative ion; often nonmetals since they gain electrons to become negatively charged

Helpful Hint: Think of the first "n" in anion = negative

Cations and anions have opposite charges and attract each other. This force of attraction between oppositely charged ions is called ionic bonding.

Polyatomic ion - An ion containing a number of atoms.

Helpful Hint: When dealing with polyatomic ions, you should memorize the formula and the charge.

Ionic solid - A solid containing cations and anions that dissolves in water to give a solution containing the separated ions, which are mobile and thus free to conduct an electric current.

The Periodic Table

Download a PDF version of the periodic table from the American Chemical Society.

An interactive version of the periodic table is available on the Royal Society of Chemistry website.

Atomic number - The number of protons and the number located above each element symbol.

Examples: Carbon (C) has atomic number 6, while lead (Pb) has atomic number 82.

Symbols for the Elements That Are Based on the Original Names
Current Name	Original Name	Symbol
Antimony	Stibium	Sb
Copper	Cuprum	Cu
Iron	Ferrum	Fe
Lead	Plumbum	Pb
Mercury	Hydrargyrum	Hg
Potassium	Kalium	K
Silver	Argentum	Ag
Sodium	Natrium	Na
Tin	Stannum	Sn
Tungsten	Wolfram	W

Metals - Efficient conductors of heat and electricity, malleable (can be hammered into thin sheets), ductile (can be pulled into wires), and often lustrous (although tarnishes readily). Metals tend to lose electrons to form POSITIVE ions (cations).

Nonmetals - Appear in the upper-right corner of the table except for Hydrogen which resides in the upper-left corner. Lack the physical properties that characterize the metals. Nonmetals tend to gain electrons to form NEGATIVE ions (anions).

Groups or families - The vertical columns in the periodic table and have similar chemical properties.

Group A - Representative elements

Group 1A (alkali metals): lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), Francium (Fr); Very active elements that readily form ions with a 1+ charge when they react with nonmetals.
Group 2A (alkaline earth metals): beryllium (Be), magnesium (Mg), calcium (Ca), Strontium (Sr), barium (Ba), radium (Ra); All form ions with a 2+ charge when they react with nonmetals.
Group 7A (halogens): fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At); They form diatomic molecules and F, Cl, Br, & I react with metals to form salts containing ions with a 1- charge.
Group 8A (noble gases): helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn); All exist under normal conditions as monatomic (single-atom) gases and have little chemical reactivity.

Group B - Transition elements; They are all metals and have numerous oxidation/valence states.

Periods - The horizontal rows of elements in the periodic table.

First period (Horizontal row 1): hydrogen (H) and helium (He)
Second period (Horizontal row 2): lithium (Li), beryllium (Be), boron (B), carbon (C), nitrogen (N), oxygen (O), fluorine (F), neon (Ne)
Third period and so on

Naming Compounds

Examples of the rules for naming binary ionic compounds
Compound	Ions Present	Name
NaCl	Na+, Cl-	Sodium chloride
KI	K+, I-	Potassium Iodide
CaS	Ca2+, S2-	Calcium sulfide
Li₃N	Li+, N3-	Lithium nitride
CsBr	Cs+, Br-	Cesium bromide
MgO	Mg2+, O2-	Magnesium oxide

Binary compounds: Compounds composed of two elements.

The first type we will consider is ionic compounds.
Binary ionic compounds: Contain a cation (+ ion) that is ALWAYS written first in the formula and an anion (- ion).

Ionic compound: Metal + Nonmetal(s)

Rules for Naming Type I Binary Compounds

The cation is ALWAYS named first and the anion second.
A monatomic cation takes its name from the name of the element (e.g., Na+ = sodium for compounds containing this ion).
A monatomic anion is named by taking the root of the element name and adding -ide (e.g., Cl- = chloride).

Helpful Tip: The illustration below shows the general trend for ionic charges

Common Monatomic Cations and Anions
Cation	Name	Anion	Name
H⁺	Hydrogen	H^-	Hydride
Li⁺	Lithium	F^-	Fluoride
Na⁺	Sodium	Cl^-	Chloride
K⁺	Potassium	Br^-	Bromide
Cs⁺	Cesium	I^-	Iodide
Be²⁺	Beryllium	O^2-	Oxide
Mg²⁺	Magnesium	S^2-	Sulfide
Ca²⁺	Calcium	N^3-	Nitride
Ba²⁺	Barium	P^3-	Phosphide
Al³⁺	Aluminum

Formulas From Names:

The compound needs to be neutral. This means that you need to get the charges to balance out.

Let's look at the above example of Lithium nitride.

The ions are: Li+ and N3-

The nitrogen ion contains a 3- charge while lithium has a 1+ charge. The three from the charge of nitrogen is going to become the subscript of lithium and lithium's 1 is going to become the subscript to nitrogen. This is called the criss-cross method and the final product will look like this:

Li3N
Now, you can see that there are 3 lithiums to the 1 nitrogen to balance out the charges and make the compound neutral. Since the nitrogen shows in the formula, the 1 is not needed and that is why you don't see it.

What about magnesium oxide (MgO)?

Both of those ions contain 2: Mg2+, O2-

But from the above information, the formula should look like this: Mg₂O₂
The lowest ratio is needed, and for this example, the lowest ratio is 1:1. The twos will cancel out to leave 1 and give us the final product of:

MgO

This video by Tyler DeWitt gives more detailed instructions on how to write ionic formulas

Practice: Highlight the text inside the box to reveal answer
Problems	Answers (Highlight boxes)	Reasons (Highlight boxes)
Sodium bromide	NaBr	Sodium = Na+ and bromine = Br- By using the criss-cross method, the 1 from Na will become the subscript of Br The 1 from the Br will become the subscript of Na.
GaAs	Gallium arsenide	Ga = Gallium (stays the same) As = arsenic Remove -ic and add -ide for arsenide
KCl	Potassium chloride	K = Potassium (stays the same) Cl = Chlorine Remove -ine and add -ide for chloride
Aluminum sulfide	Al2S3	Aluminum = Al3+ and sulfur = S2- The 3 from the Al will become the subscript of S The 2 from S will become the subscript of Br You must have 2 Al and 3 S to balance the charge and make the compound neutral
Mg₃P₂	Magnesium phosphide	Mg = Magnesium (stays the same) P = Phosph~~orous~~ Remove -orous and add -ide
Lithium oxide	Li2O	Lithium = Li+ and Oxygen = O2- The 1 from Li will become the subscript of O The 2 from O will become the subscript of Li You must have 2 Li to balance the charge of O
CaBr	Calcium bromide	Ca = Calcium (stays the same) Br = Brom~~ine~~ Remove the -ine and add -ide
Barium sulfide	BaS	Barium = Ba2+ and Sulfur = S2- After using the criss-cross method, you are left with Ba2S2 The 2's are the same and become the lower ratio which is 1:1 Final answer is BaS
NaF	Sodium fluoride	Na = Sodium (stays the same) F = Fluor~~ine~~ Remove the -ine and add -ide
MgCl₂	Magnesium chloride	Mg = Magnesium (stays the same) Cl = Chlor~~ine~~ Remove -ine and add -ide
Sodium nitride	Na3N	Sodium = Na+ and Nitrogen = N3- The 1 from Na is going to become the subscript of N The 3 from N is going to become the subscript of Na You must have 3 Na to balance out the charge of N.

Common Type II Cations
Ion	Systematic Name
Fe3+	Iron(III)
Fe2+	Iron(II)
Cu2+	Copper(II)
Cu+	Copper(I)
Co3+	Cobalt(III)
Co2+	Cobalt(II)
Sn4+	Tin(IV)
Sn2+	Tin(II)
Pb4+	Lead(IV)
Pb2+	Lead(II)
Hg2+	Mercury(II)
Hg₂²⁺	Mercury(I)
Ag+	Silver
Zn2+	Zinc
Cd2+	Cadmium
The mercury(I) ions ALWAYS occur bound together to form Hg22+ ions. Although silver, zinc, and cadmium are transition metals, they form only one type of ion and a Roman numeral is not used.

In this section, you will come to find out that there are metals that have more than one type of positive ion and can form more than one type of ionic compound with a given anion.

Iron is a good example. Iron can be: Fe2+ or Fe3+
These will be written as: iron(II) and iron(III)

Helpful Tip:
Roman numerals are required to help specify which ion is being used. Elements that form only one cation DO NOT need to be identified by a Roman numeral.

FUN FACT: There is another system for naming these ionic compounds. The lower charged ion will have the ending -ous while the higher charged ion will have the ending -ic . So, iron will be written as: ferrous ion and ferric ion

This video by Tyler DeWitt teaches you how to name ionic compounds that have transition metals in them.

Table for Naming Binary Ionic Compounds
Does the compound contain Type I or Type II cations?
Type I	Type II
Name the cation using the element name.	Using the principle of charge balance, determine the cation charge.
	Include in the cation name a Roman numeral indicating the charge.

Helpful Tip:
Silver, zinc, and cadmium only have ONE type of ion even though they are transition metals. This means that they don't need Roman numerals when their name is being written.

Practice Problems: Highlight boxes for answer
Problems	Answers: Highlight boxes to reveal	Reasons: Highlight boxes to reveal
Titanium(III) bromide	TiBr₃	Titanium = Ti3+ and Bromine = Br- Use the criss-cross method, the 3 from Ti will become the subscript for Br The 1 from Br will become the subscript for Ti You need 3 Br to balance the charge of Ti
Tin(IV) selenide	SnSe2	Tin= Sn4+ and Selenium = Se2- The 4 from Sn will become the subscript for Se The 2 from Se will become the subscript for Sn Sn2Se4 is not the lowest ratio. If you divide by two, the final answer is SnSe2
Co2O3	Cobalt(III) oxide	You can tell that the subscripts are the lowest ratio because the 2 subscript came from O which is O2- (the number didn't change) So, when you reverse the criss-cross method, the 3 subscript from O returns as Co charge giving you Cobalt(III) Ox~~ygen~~ Remove -ygen and add -ide
V2S5	Vanadium(V) sulfide	Sulfur = S2- By looking at Vanadium's subscript, you can see that the 2 is still there. When you reverse the criss-cross method, vanadium will have a charge of 5 and you will have Vanadium(V) in your answer Sulfur Remove -ur and add -ide
Copper(I) oxide	Cu2O	Copper = Cu+ and Oxygen = O2- The criss-cross method - The 1 from Cu becomes the subscript for O The 2 from O becomes the subscript for Cu
Pb3N2	Lead(II) nitride	Nitrogen = N3- Pb's subscript is 3 which means that the numbers haven't changed Reverse the criss-cross method and the 2 subscript becomes the charge of Pb to give you Lead(II) in your answer Nitr~~ogen~~ Remove -ogen and add -ide
AgBr	Silver Bromide	Silver only has one type of ion and does not need Roman numerals Brom~~ine~~ Remove -ine and add -ide
Titanium (II) selenide	TiSe	Titanium = Ti2+ and Selenium = Se2- Criss-cross to give Ti2Se2 Divide by 2 to obtain the lowest ratio and answer is TiSe
SnS2	Tin(IV) sulfide	Sulfur = S2- Notice that Sn does not have a subscript. It should be 2 since S has a charge of 2 This means that the lowest ratio is being used. If you multiply each subscript by 2, then the S subscript will give you 4 and you will have Sn2S4. Reverse criss-cross and you will have Tin(IV) as your answer Sulfur Remove -ur and add -ide

For more practice problems and a video tutorial, view Tyler DeWitt's Naming Ionic Compounds with Transition Metals Practice Problems

For even further help, visit the Name-inator website.

Common Polyatomic Ions
Ion	Name	Ion	Name
Hg22+	Mercury(I)	NCS- or SCN-	Thiocyanate
NH4+	Ammonium	CO32-	Carbonate
NO2-	Nitrite	HCO3-	Hydrogen Carbonate (Bicarbonate is a widely used common name)
NO3-	Nitrate	ClO- or OCl-	Hypochlorite
SO32-	Sulfite	ClO2-	Chlorite
SO42-	Sulfate	ClO3-	Chlorate
HSO4-	Hydrogen Sulfate (bisulfate a widely used common name)	ClO4-	Perchlorate
OH-	Hydroxide	C2H3O2-	Acetate
CN-	Cyanide	MnO4-	Permanganate
PO43-	Phosphate	Cr2O72-	Dichromate
HPO42-	Hydrogen Phosphate	CrO42-	Chromate
H2PO4-	Dihydrogen Phosphate	O22-	Peroxide
		C2O42-	Oxalate
		S2O32-	Thiosulfate

Polyatomic Ion: An ion containing a number of atoms

Oxyanions: Anions containing an atom of a given element and different numbers of oxygen atoms

When there are two members in a series:
- The name with the smaller number of oxygen atoms will end in -ite
- The name with the larger number of oxygen atoms will end in -ate

Example:
SO32- = sulfite and SO42- = sulfate

When there are more than two members in a series:
- The fewest oxygen atoms will have the prefix hypo-
- The most oxygen atoms will have the prefix per-

Example:
ClO- = hypochlorite and ClO4- = perchlorate

Helpful Tip:
MEMORIZE THE POLYATOMIC IONS

This video by Tyler DeWitt teaches you how to write formulas for ionic compounds that contain polyatomic ions.

Practice Problems: Highlight boxes to reveal answers
Problems	Answers: Highlight boxes	Reasons: Highlight boxes
Ammonium chloride	NH4Cl	Ammonium = NH4+ and Chlorine = Cl- Use the criss-cross method, The 1 from NH4 becomes the subscript for Cl The 1 from Cl becomes the subscript for NH4
Iron(III) nitrate	Fe(NO3)3	Iron(III) = Fe3+ and Nitrate = NO3- Use the criss-cross method The 3 from Fe becomes the subscript of NO3 Notice that NO3 is in parentheses, you're not adding the subscripts. T hey are separate. You have 3 oxygens for the nitrate and now 3 nitrates for the compound. The 1 from nitrate becomes the subscript for Fe
Cr(PO4)2	Chromium(VI) phosphate	Phosphate has a charge of 3- (PO43-) Cr has no subscript from PO43- which means that the formula now is on the lowest ratio Multiply the subscripts by 3 to get the regular subscripts back giving Cr a subscript of 3 and PO4 a subscript of 6 Reverse criss-cross - Cr has a charge of 6 to write Chromium(IV)
V(CO3)2	Vanadium(IV) carbonate	Carbonate has a charge of 2- (CO32-) V has no subscript and that means that the formula is on the lowest ratio Multiply the subscripts by 2 to get the regular subscripts back giving V a subscript of 2 and CO3 a subscript of 4 Reverse criss-cross - V has a charge of 4 to write Vanadium(IV)
Scandium(III) hydroxide	Sc(OH)3	Scandium(III) = Sc3+ and Hydroxide = OH- Use the criss-cross method The 1 from OH becomes the subscript of Sc The 3 from Sc becomes the subscript of OH Remember to put OH in parenthesis because this shows you have 3 OH instead of just 3 H
Strontium acetate	Sr(C2H3O2)2	Strontium = Sr2+ and Acetate = C2H3O2- Use the criss-cross methodd The 1 from C2H3O2 becomes the subscript of Sr The 2 from Sr becomes the subscript of C2H3O2 Remember to use parenthesis on C2H3O2 to show that you have 2
Ag3PO4	Silver phosphate	Silver only has one type of ion and does not need a Roman numeral. Ag = Silver PO4 = Phosphate
KMnO4	Potassium permanganate	K = Potassium MnO4 = Permanganate
Cadmium sulfite	CdSO3	Cadmium = Cd2+ and Sulfite = SO32- Use criss-cross method The 2 from SO3 becomes the subscript of Cd The 2 from Cd becomes the subscript of SO3 You should have Cd2(SO3)2 but you can make the subscript smaller Divide by 2 and obtain the lowest ratio to get final answer of CdSO3

For more practice problems on naming compounds and writing formulas, visit Mr. Carman's Blog Chemical Compounds Practice Quiz.

Prefixes Used to Indicate Number in Chemical Names
Prefix	Number
mono-	1
di-	2
tri-	3
tetra-	4
penta-	5
hexa-	6
hepta-	7
octa-	8
nona-	9
deca-	10

Binary Covalent Compounds: Formed by two nonmetals

These compounds do not contain ions

Naming Binary Covalent Compounds

The first element in the formula is named using the full element name.
The second element is named as if it were an anion.
The prefixes are used to indicate the numbers of atoms present.
The prefix mono- is never used for naming the first element.

Example:
CO = Carbon monoxide NOT monocarbon monoxide

Helpful Tip:
If the element begins with a vowel, the final o or a of the prefix is dropped. Carbon monoxide NOT carbon monooxide. Dinitrogen tetroxide NOT dinitrogen tetraoxide.

Examples of naming covalent compounds
Compound	Systematic Name	Common Name
N2O	Dinitrogen monoxide	Nitrous oxide
NO	Nitrogen monoxide	Nitric oxide
NO2	Nitrogen dioxide
N2O3	Dinitrogen trioxide
N2O4	Dinitrogen tetroxide
N2O5	Dinitrogen pentoxide

Some compounds are just always referred to by their common name.

Examples:

H2O = Water
NH3 = Ammonia
H2O2 = Hydrogen Peroxide

This video by Tyler Dewitt teaches you how to write names for Covalent Molecular Compounds.

Practice Problems: Highlight boxes to reveal answers.
Problems	Answers
CO2	Carbon dioxide
SCl2	Sulfur dioxide
diphosphorus pentoxide	P2O5
CCl4	Carbon tetrachloride
tricarbon tetranitride	C3N4
selenium difluoride	SeF2
B2H6	diboron hexahydride
nitrogen tribromide	NBr3
S3N2	trisulfur dinitride
Se3S6	triselenium hexasulfide

For more practice problems on naming compounds and writing formulas, visit Mr. Carman's Blog Chemical Compounds Practice Quiz.

Acid: A molecule in which one or more H ions are attached to an anion.

The rules for naming acids depend on whether the anion contains oxygen.
If the name of the anion ends in -ide, the acid is named with the prefix hydro- and the suffix -ic.

Example:
HCl = Hydrochloric acid

When the anion contains oxygen, the acidic name is formed from the root name of the anion with a suffix of -ic or -ous.

If the anion name ends in -ate the suffix -ic is added to the root name.
- H2SO4 = contains the sulfate anion (SO42-) and is called sulfuric acid
- HC2H3O2 = contains the acetate ion (C2H3O2-) and is called acetic acid
If the anion name ends in -ite, then -ous is used.
- H2SO3 = contains the sulfite anion (SO32-) and is called sulfurous acid
- HNO2 = contains the nitrite anion (NO2-) and is called nitrous acid

Application of Acid Rules
Acid	Anion	Name
HClO4	Perchlorate	Perchloric acid
HClO3	Chlorate	Chloric acid
HClO2	Chlorite	Chlorous acid
HClO	Hypochlorite	Hypochlorous acid

This video by Tyler DeWitt will teach you how to look at the chemical formula for an acid and then write its name.

Names of Non-Oxygen-Containing Acids (aqueous solutions)
Acid	Name
HF	Hydrofluoric acid
HCl	Hydrochloric acid
HBr	Hydrobromic acid
HI	Hydroiodic acid
HCN	Hydrocyanic acid
H2S	Hydrosulfuric acid

Name of Some Oxygen-Containing Acids
Acid	Name
HNO3	Nitric acid
HNO2	Nitrous acid
H2SO4	Sulfuric acid
H2SO3	Sulfurous acid
H3PO4	Phosphoric acid
HC2H3O2	Acetic acid

For more practice problems on naming compounds and writing formulas, visit Mr. Carman's Blog Chemical Compounds Practice Quiz.
For more practice problems and a video tutorial, view Tyler DeWitt's Naming Acids Practice Problems.