Finding Molarity

General Chemistry Index

Where are we going with this? This page will assist in developing the ability to perform calculations to determine the composition of a compound or mixture when given the necessary information and to apply lab data to determine the empirical and molecular formula of a compound.

Finding Molarity
This sounds like something someone just made up as a joke…

Seems like a definition should come now. Good news! This is really very, very simple.

Molarity: the number of moles of something (the solute) in a liter of a solution.

https://en.wikipedia.org/wiki/Molar_concentration

AKA: Molar concentration, because… two names are better than one?

That's it. If you put one mole of something into water, then top it off to exactly one liter, you have something that has a molarity of one.

If you put two moles of something into water, then top it off to exactly one liter, you have something that has a molarity of two.

Hang on for this one! If you put FOUR moles of something into water, then top it off to exactly TWO liters, you have something that has a molarity of two.

Seems like there would be a formula!

Where…

M is molarity,

n is number of moles, and

V is volume…

M = n/V

There's a trick though! Of course there is!

V is the volume you end up with! So you mix stuff such that you end up with some volume. Then you divide. Hence the "top it off to exactly" wording in the examples above.

So… this seems really easy…

It can be slightly more difficult. Suppose you are given grams of solute, not moles? OR! Suppose you are supposed to find the grams needed for some given molarity! Yeah, that! So… Let's do some examples…

 

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EXAMPLE 1

What is the molarity of a 2 liter solution containing 3.4 moles of CuSO4?

Find M where

V = 2 liters
n = 3.4 moles

M = n/V

M = 3.4 moles / 2 liters

M = 1.7 M

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EXAMPLE 2

How many grams of K2SO4 are needed to make 2 liters of a 4 Molar solution?

That escalated fast! Let's do it in two steps:

1. How many moles of K2SO4 are needed to make 2 liters of a 4 Molar solution?

2. How much does that many moles of K2SO4 weigh in grams?

Hmm… seems legit.

1. How many moles of K2SO4 are needed to make 2 liters of a 4 Molar solution?

Find n where

V = 2 liters
M = 4 

M = n/V

4 moles/liter = n / 2 liters

Multiply both sides of the equation by 2 and handle the unit to clear the fraction on the right, because math.

2 • 4 moles = n

8 moles = n

So we need 8 moles of K2SO4. 

2. How much does that many moles of K2SO4 weigh in grams?

We know how many moles we need (8 moles), so we need to convert that to grams.

Moles to grams you multiply… grams to moles divide…

grams = n • atomic mass

Having the number of moles, we now need to find the molecular mass of K2SO4. Using a periodic table (and rounding) we can find the atomic masses of the elements:

K = 39.1 g/mole

S = 32.1 g/mole

O = 16.0 g/mole

So, the molecular mass of K2SO4 is the sum of all those atom masses:

2 • K = 78.2

1 • S = 32.1

4 • O = 64.0

78.2 + 32.1 + 64 = 174.2 g/mole

 

Okay… so, one mole weighs 174.2 grams. How much does… what was that number again?

So we need 8 moles of K2SO4. 

Moles to grams you multiply… grams to moles divide…

grams = moles • g/mole 

grams = 8 moles • 174.2 g/mole  <----- moles cancel

grams = 1393.6 g 

So, the answer is… 

You need 1393.6 grams of K2SO4 to make 2 liters of a 4 Molar solution?

_________________________________

EXAMPLE 3

What would be the molarity of a .5 liter solution in which was 18.65 grams of KCl?

Looks like we have two steps again…

1. How many moles are in 18.65 grams of KCl?

2. What is the molarity of a .5 liter solution in which are that many moles of KCl?

Let's do this!

1. How many moles are in 18.65 grams of KCl?

Moles to grams you multiply… grams to moles, divide…

Where n is number of moles, g is mass in grams, and atomic mass is the atomic mass of the substance, then…

n = g / atomic mass

Atomic mass of KCl is

1 • K = 39.1

1 • Cl = 35.5

Add those up…

Atomic Mass = 39.1 + 35.5

Atomic Mass = 74.6 g / mole

Now, we can find how many moles there are…

n = g / atomic mass

n = 18.65 / 74.6 moles <-- the units reduce to moles because math

n = .25 moles

So, 18.65 grams of KCl is .25 moles.

2. What is the molarity of a .5 liter solution in which are that many moles (.25 moles) of KCl?

Find M where

V = .5 liters
n = .25 moles

 

M = n/V

M = .25 moles / .5 liters

M = .5 M

The molarity of a .5 liter solution in which are 18.65 grams of KCl is 0.5 M.

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So, there you go… Essentially, we are taking our understanding of moles and grams and all that, then adding… the solvent. 

• Find out how many moles you have and divide by the volume.
• Or flip it around and find out how many moles you need and convert moles to grams.

Not too bad!

ADDITIONAL COMMENTS:

• In EXAMPLE 2, we made a LOT of a 4 molar solution. Most situations will call for much smaller quantities.
  
  
• Solutes actually have a MAXIMUM amount that will dissolve in in a solvent, and when that happens, the solution is said to be saturated.

RESOURCE:

Here's a link to a kinda cool spreadsheet you can copy that will do a lot of the math for you!

 

Finding Empirical Formulas

General Chemistry Index

Where are we going with this? This page will assist in developing the ability to perform calculations to determine the composition of a compound or mixture when given the necessary information and to apply lab data to determine the empirical and molecular formula of a compound.

Finding Empirical Formulas
This does NOT sound fun…

We probably should start off with some definitions and examples, eh?

Empirical Formula: The integer ratio of elements in a compound reduced to lowest multiples.

Examples:

H2O but not H4O2

CO2  but not C2O4

https://en.wikipedia.org/wiki/Empirical_formula

It is important to note that more than one compound can have the same empirical formula. What? This is because it represents the ratio of parts. It does not specify exactly how the parts are combined, as a molecular formula does.

"Glucose (C6H12O6), ribose (C5H10O5), acetic acid (C2H4O2), and formaldehyde (CH2O) all have different molecular formulas but the same empirical formula: CH2O. This is the actual molecular formula for formaldehyde, but acetic acid has double the number of atoms, ribose has five times the number of atoms, and glucose has six times the number of atoms" (Source).

Given any percent composition, it's not terribly hard to find the empirical formula. Really.

How does it work?

In addition to the examples on the linked page above, Google will provide many, many examples. We should add one more! Yeah, let's do that!

EXAMLE:

A compound is made up of 74.20% sodium (Na) and 25.80% oxygen (O). What is its empirical formula. I'm taking a stab at it and guessing this is sodium oxide…

Step 1: Assume there is 100 grams of the substance AND (This means that 74.20% of 100 grams is the mass of the Na… etc.)  convert the percents into masses.

Na = 74.20 grams

O = 25.80 grams.

Step 2: Convert grams to moles of atoms. Moles to grams you multiply, grams to moles divide.

Moles of Na = 74.20 grams • (1 mol / 22.99 g/mol) = 3.2275 moles

Moles of O = 25.80 grams • (1 / 15.99 g/mol) = 1.6135 moles

Now, you have the ratio of moles, albeit rendered in very unwieldy decimals. Not to fear!

Step 3: Divide the very unwieldy decimals by the smallest value to get something… better:

1.6135 moles of O / 1.6135 = 1        <--- THIS was the plan. Get this guy to 1

3.2275 moles Na / 1.6135 = 2      <-- Rounding this off from 2.00030, because… rounding.

Step 4: SOMETIMES you have to multiply the results by 2 in order to get integers.

For instance in C3H6O2, you will end up with:

1.5 moles C

3 moles H

1 mole O

Multiply by 2 to get rid of the 1.5:

3 moles C

6 moles H

2 moles O

The empirical formula is the atomic symbols and the calculated number of moles. For the two examples above in Step 3 and Step 4:

Na2O

C3H6O2

___________________________

If you are given the composition in terms of the masses, the work is even easier.  

The process is to find out how many moles you have of each thing, then use the smallest of those to convert to an integer ratio.

Let's look at the math in isolation…

Say you have something that is made up of .2 moles of X and .4 moles of Y.

You can probably see that you have twice as many moles of Y as you do X, so the answer would be 

XY2

Probably… 

So, what's the math on that?

You have, based on the masses, a ratio of

.2 : .4

To make that into an integer ratio, divide everything by the smallest number…

.2  : .4  ÷ .2 = 1 : 2

Now, back to the chemistry…

Find the empirical formula given a 7.315 gram sample containing 0.815 grams of hydrogen and 6.500 grams of oxygen.

Use these atomic masses:

O = 16.0

H = 1.0

Step 1: Find the mole ratio present in the sample.

Step 1a: How many moles of each thing do you have?
Moles to grams you multiply, grams to moles divide!

Divide grams given by atomic masses;

Moles O = 6.50 grams ÷ 16.0 g/mole

Moles O = 0.40625

Moles H = 0.815 grams ÷ 1.0 g/mole

Moles H = 0.815

So, knowing how many moles you have, you now… have the mole ratio!

The ratio of O to H is

0.40625 : 0.815

Step 2: Divide that ratio by the smallest number to get an Integer ratio

0.40625 : 0.815 ÷ 0.40625 = 1 : 2.006

Step 3: The integer ratio (rounded off) becomes the subscripts for the empirical formula:

H2O   <--- the subscript on the O would be 1, but we don't write that.

Last thought… Sometimes, you will have to multiply by 2 (just like before, above) to get a whole number, but that's not a big deal.

___________________________

That's it! Wasn't so bad, afterall!

Finding Percent Composition From Molecular Formulas

General Chemistry Index

Where are we going with this? This page will assist in developing the ability to perform calculations to determine the composition of a compound or mixture when given the necessary information and to apply lab data to determine the empirical and molecular formula of a compound.

Finding Percent Composition From Molecular Formulas
Well, now… That's a whole lot of words!

We probably should start off with some definitions and examples, eh?

Molecular Formula: The integer ratio of elements in a compound that represents the numbers of each type of atom in the compound.

https://www.google.com/search?q=chemistry+molecular+formula

This is not new!

Working with molecular formulas goes way back. We should have a good sense about this.

So, what about that percent composition, then? What's that?

Percent Composition: The percent of the mass of a compound made up by the mass of each of the constituent elements.

https://www.google.com/search?q=chemistry+percent+composition

Another way to look at it is to think of it this way… What percent of the mass of XXXX is made up by YYYYY? What percent of water is made up by oxygen? What percent of sucrose is made up by carbon?

We can make it look more like chemistry this way…

Find the percent composition of each element in CuBr2.

So, percent… in this case, you'll be doing 

Part ÷ Whole X 100%          <--------  Look at the bottom of this page for more explaining!

Using the masses from a periodic table:

Cu = 63.5

Br = 79.9

we find the molecular mass (the whole) of CuBr2

CuBr2 = 63.6 + (2 X 79.9) 

CuBr2 = 223.3

Now, we have everything we need! Recap? Sure?

Cu = 63.5

Br = 79.9

CuBr2 = 223.3

What percent is the Cu? Let's do that Part ÷ Whole X 100% thing.

Percent Cu = 63.5 / 223.3 X 100%
Percent Cu = 28.44%

What percent is the Br? Let's do that Part ÷ Whole X 100% thing. BUT!! 

There are TWO Br in the molecule. So, the mass of the Br is not 79.9, but rather 159.8. 

Percent Br = 159.8 / 223.3 X 100%
Percent Br = 71.56%

Let's try another, but follow some steps… Steps are good, right?

Step 1: Find all the masses of the elements.

Step 2: Find the mass of the compound.

Step 3: Find the "part" masses of the compound.

Step 4: Do that Part ÷ Whole X 100% thing

Find the percent composition of each element in C2H4O2.

Step 1:

C = 12.01

H = 1.01

O = 15.99

Step 2:

C2H4O2 = 60.04

Step 3

C2 = 24.02

H4 = 4.04

O2 = 31.98

Step 4:

Percent C = 24.02 / 60.04 X 100%
Percent C = 40.01 %

Percent H = 4.04 / 60.04 X 100%
Percent H = 6.73%

Percent O = 31.98/60.04 X 100%
Percent O = 53.26%

Not so bad! That's it. 

https://youtu.be/iEpkBXfPFto

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What is this Part ÷ Whole X 100% thing?

Math… If you want to know what percent of something you have, you… do that.

A classroom has 10 red desks and 30 blue desks. What percent of the desks are red?

The "part" you are looking at is the red desks. The "whole" is ALL of the desks.

Percent red desks =  Part ÷ Whole X 100%

Percent red desks = 10 ÷ 40 X 100%

Percent red desks = .25 X 100%

Percent red desks = 25%

It works like this for all "what percent of" questions. Math…

How Double Replacement Products Form

 General Chemistry Index

Where are we going with this? This page will assist in developing the ability to predict products of simple reactions as listed in of reactions: synthesis (i.e., combination), decomposition, single displacement, double displacement, acid/base, and combustion.

How Double Replacement Products Form 

These reactions look like this…

AB + CD --> AD + BC

The double replacement reaction looks more complex than synthesis or single replacement. However (ironically), the possible products are a lot easier to predict. 

Remember that opposites attract thing? Also, that chemistry grammar thing of putting positives first? Well, for double replacement, those two things add up to making this really easy.

So, look at this…

MX + NY --> ???

Grammar says M and N are positive and X and Y are negative. (But you should actually look it up!)

Presuming whoever wrote the question followed the grammar rules, since opposites attract, the only thing that might be possible is for the X and Y to swap places. 

Example? Sure. And how about those polyatomic ions? Okay, that too.

AgNO3 + K2SO4 --> ????

The trick is to realize that SO4 and NO3 are polyatomic ions. How about some coloring?

AgNO3 + K2SO4 --> ????

Blue is positive. Red is negative (don't hate… just a rough life)

Opposites attract. Blue cannot be with blue. Red cannot be with red.

There's only one possible product. Swap the negative parts: 

AgNO3 + K2SO4 = Ag2SO4 + KNO3

The only thing left to do is figure out IF IT ACTUALLY WILL react. That… That's a whole 'nother thing. (The reaction above does, by the way.) 

There's a lot of rules related to this, and then one of the rules has rules. Yeah… I know.

How Compounds Form

General Chemistry Index

Where are we going with this? This page will assist in developing the ability to predict products of simple reactions as listed in of reactions: synthesis (i.e., combination), decomposition, single displacement, double displacement, acid/base, and combustion.

How Compounds Form 
What happens if I mix this with that?

Not everything will combine with each other. There are certain and specific rules regarding what will combine with what.

In chemistry, with regard to "charge" opposites attract. Something with a negative charge will be attracted to (and could theoretically) combine with a different thing having a positive charge.

For example, Na has a charge of +1. O has a charge of -2. Therefore, these two elements will combine. 

Likewise, the polyatomic molecule, OH has a charge of -1. It will also combine with something having a positive charge, such as K which is +1

Figuring out if something will actually combine is the first step in finding the subscripts that produce a neutrally charged molecule.

Let's go through this a little…

______________________

Synthesis / Combination Reactions

So in this kind of reaction you have the model form

A + B --> AB

If this is going to work A has to be positive and B has to be negative. (Chemistry grammar prefers that the positives are written first. Generally, this is also going left to right across the periodic table.)

So, let's mess with this!

Can Na combine chemically with K?

Finding the charge of Na on a chart or on the periodic table tells us that it is +1.

Finding the charge of K on a chart or on the periodic table tells us that it is +1.

Well… remember that "opposites attract" rule? So NO. No reaction.

How about Mg and Cl?

Chart… Periodic table… Mg is +2. Cl is -1

YES! These two will combine!

Okay, synthesis is easy!

______________________

Single Replacement / Displacement

These reactions look like this…

Single replacement

AB + C --> AC + B

or

AB + C --> CB + A

Think about that opposites attract thing. A is positive (chemistry grammar). B is negative. C can be either positive or negative. The compound that might form¹ will be the one where C replaces which ever part of AB has the same charge.

Let's say this…

A is +1

B is -1

C is +2

C and A CANNOT combine because they are both positively charged. Therefore C might replace A and combine with B.

Real example…

K2O + Li -->

What could Li replace? Chart… Periodic table… charges…

K is +1

O is -2

Li is +1

Therefore, the Li might² replace the K. It could combine with the O.

SUMMARY

So, in single replacement, the single thing can only replace the thing in the compound that has the same charge.

--------

¹ Whether or not it will form depends on if C is more highly active than the thing it is trying to replace.

² Have to look it up in the activity series table…

______________________

Double Replacement / Displacement

These reactions look like this…

AB + CD --> AD + BC

The double replacement reaction looks more complex than synthesis or single replacement. However (ironically), the possible products are a lot easier to predict. 

Remember that opposites attract thing? Also, that chemistry grammar thing of putting positives first? Well, for double replacement, those two things add up to making this really easy.

So, look at this…

MX + NY --> ???

Grammar says M and N are positive and X and Y are negative. (But you should actually look it up!)

Presuming whoever wrote the question followed the grammar rules, since opposites attract, the only thing that might be possible is for the X and Y to swap places. 

Example? Sure. And how about those polyatomic ions? Okay, that too.

AgNO3 + K2SO4 --> ????

The trick is to realize that SO4 and NO3 are polyatomic ions. How about some coloring?

AgNO3 + K2SO4 --> ????

Blue is positive. Red is negative (don't hate… just a rough life)

Opposites attract. Blue cannot be with blue. Red cannot be with red.

There's only one possible product. Swap the negative parts: 

AgNO3 + K2SO4 = Ag2SO4 + KNO3

The only thing left to do is figure out IF IT ACTUALLY WILL react. That… That's a whole 'nother thing. (The reaction above does, by the way.) 

There's a lot of rules related to this, and then one of the rules has rules. Yeah… I know.

SUMMARY

So, in double replacement, the negatives swap places. 

Solubility and Double Replacement

 General Chemistry Index

Where are we going with this? This page will assist in developing the ability to describe, classify, and give examples of various kinds of reactions: synthesis (i.e., combination), decomposition, single displacement, double displacement, acid/base, and combustion.

Solubility and Double Replacement

Part of predicting products of chemical reactions is determining if it will even react. One of the rules regarding the potential of a double replacement reaction to react is solubility of the products.

The complete set of predictors for if something will react in a double replacement situation is…

• one of the products is water (H₂O), 
• one of the products is a gas, or an unstable product that breaks down to form a gas (for example, H₂S, H₂CO₃ → H₂O + CO₂, H₂SO₃ → H₂O + SO₂, or NH₄OH → H₂O + NH₃)
• one of the products is an insoluble solid (an example could be BaSO4)

There it is… Solubility… Right smack in the rules! Understanding solubility is fundamental to one of the rules that is part of figuring out if a double replacement reaction will occur!

• So, what is solubility in the context of chemistry?
  
  
• Solubility is the property of a solid, liquid or gaseous chemical substance called solute to dissolve in a solid, liquid or gaseous solvent. It is the state of a substance of being soluble(being able to dissolve)
  
  
• Solubility refers to how well a substance dissolves in water. If a substance when placed in water dissolves then that substance is soluble. If it does not dissolve then it is not soluble.  
  
  
• Solubility is the ability of a substance or the solute to dissolve in the solvent.
  
  
• Solubility is the measure of how much a substance can dissolve in a gas, liquid, or solid substance. The thing that is being dissolved is the solute and the thing it dissolves in is the solvent. 

The concept of solubility should be fairly familiar to anyone. There are examples of things that dissolve and things that don’t around every home.

For instance, these things will dissolve in water:

• Sugar
• Salt 
• Alka-Seltzer
• Honey 

Of course, not everything will dissolve in water. Just to list a few things that won't…

• Coffee grounds
• Flour
• Tea leaves
• Pepper
• Sand 
• Paraffin wax
• Oil 

There is a word for not being able to dissolve: insoluble. Just to complete the thought, there ought to be a definition.

Insoluble: the property of a substance meaning that very little of it dissolves in a particular solvent under the given conditions.

An insoluble substance will not dissolve to an appreciable extent in the solvent, even after mixing.

Before we get back to predicting reaction products, we should make sure we know all the fancy words! Talking about soluble and insoluble falls into the context of solutions, so let’s get those defined.

Solution: A special type of homogeneous mixture composed of two or more substances.

Solute: A substance that is dissolved in a solution 

Solvent: substance, ordinarily a liquid, in which other materials dissolve to form a solution

Insoluble Solids and Predicting Products of
Double Replacement / Displacement Reactions

Okay, now that we know what we are talking about we can get back to that prediction of products thing…

First you have to realize you are dealing with a double replacement reaction. That’s pretty easy. Typically, double replacement reactions start off with two compounds and end up with two different compounds 

Double Replacement/Displacement reactions fit into “normal” form…

AB + CD → AD + CB

or

AY + BX → AX + BY  

For instance, here are some examples of double replacement reactions:

AgNO3 + NaCl → AgCl + NaNO3 

BaCl2 + Na2SO4 --> BaSO4 + 2NaCl 

So, figuring out what kind of reaction it might be is done. The next question is “Will it actually react?” 

Answering that requires looking at the potential products and figuring out if they fit the rules. And one of those rules (described above) has to do with solubility:

• one of the products is is an insoluble solid (an example could be BaSO4)

That means that, somehow, you have to know if something is soluble or not. That means, not only must you realize it is a double replacement reaction, but that you also must figure out if the product (if not water and not a gas) is soluble. There should be rules or something? Surely?

 

1. Salts containing Group I elements (Li+, Na+, K+, Cs+, Rb+) are soluble. There are few exceptions to this rule. Salts containing the ammonium ion (NH4+) are also soluble.

2. Salts containing nitrate ion (NO3-) are generally soluble.

3. Salts containing Cl -, Br -, or I - are generally soluble. Important exceptions to this rule are halide salts of Ag+, Pb2+, and (Hg2)2+. Thus, AgCl, PbBr2, and Hg2Cl2 are insoluble.

4. Most silver salts are insoluble. AgNO3 and Ag(C2H3O2) are common soluble salts of silver; virtually all others are insoluble.

5. Most sulfate salts are soluble. Important exceptions to this rule include CaSO4, BaSO4, PbSO4, Ag2SO4 and SrSO4 .

6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al3+ are insoluble. Thus, Fe(OH)3, Al(OH)3, Co(OH)2 are insoluble.

7. Most sulfides of transition metals are highly insoluble, including CdS, FeS, ZnS, and Ag2S. Arsenic, antimony, bismuth, and lead sulfides are also insoluble.

8. Carbonates are frequently insoluble. Group II carbonates (CaCO3, SrCO3, and BaCO​3) are insoluble, as are FeCO3 and PbCO3.

9. Chromates are frequently insoluble. Examples include PbCrO4 and BaCrO4.

10. Phosphates such as Ca3(PO4)2 and Ag3PO4 are frequently insoluble.

11. Fluorides such as BaF2, MgF2, and PbF2 are frequently insoluble.

SOURCE

You’re probably thinking… “That’s a lot of stuff to sift through! Someone should create a chart!”

The following list indicates some common compounds and whether or not they are soluble–and also which of the rules apply.

Compound	Soluble or Insoluble	Reason
NaCl	Soluble	Salts containing Group 1 ions such as Na⁺ are soluble.
FeS	Insoluble	Most transition-metal sulfides are insoluble.
BaSO4	Insoluble	Most sulfates are soluble, but BaSO₄ is an important insoluble exception.
Ag2CO3	Insoluble	Carbonates are insoluble.
AgBF4	Soluble	Most transition-metal sulfides are insoluble.
NaHCO3	Soluble	Compounds containing Na⁺ are soluble because sodium is a Group 1 element.
NO3	Soluble	Salts containing the nitrate ion are generally soluble in water.
AgCl	Insoluble	Many ionic solids do not dissolve in water
Ca(OH)2	Soluble	Most hydroxides have low solubility; Ca(OH)₂ is one of the more soluble Group 2 hydroxides.

Okay, what are you supposed to do with this information?

Since formation of an insoluble compound is one of the rules to predict if a double replacement/displacement reaction will occur…

(Remember those three rules?)

• one of the products is water (H2O), 
• one of the products is a common gas (H2S, H2CO3, H2SO3 and NH4OH), 
• one of the products is is an insoluble solid (an example could be BaSO4)

…you can use the information presented and the chart to see if the reaction satisfies the rule about insoluble solids.

For example, these two reactions will occur because one of the products is insoluble in water.

Na2SO4 + Ba(NO3)2 → BaSO4 + 2 NaNO3 (the BaSO4 is insoluble)

2NaOH(aq) + FeCl2(aq) → 2NaCl(aq) + Fe(OH)2(s) 

On the other hand, this reaction will not

Sr(NO3)2+KCl → NR

Okay… the point should be made! Right? 

Double replacement / displacement reactions are plentiful, and it’s super easy to come up with combinations of compounds that won’t react.

However…

The rules on this page should go a long way in helping predict if they will or not. Of course, chemistry has plenty of exceptions, but the information here is a solid starting point.

______________

CREDITS

This document was written collaboratively by the Fourth Period Lowell High School 2020 Honors Chemistry Class and compiled by the site operator

Brian Bright - Lowell High School, 2023

Bill Snodgrass - University of Memphis, 1985, 1991, 2000

Sophia Emery - Lowell High School, 2023

Logan Ritter- Lowell High School, 2022

Walter Kotlin - Lowell High School, 2023

Madelyn Logan - Lowell High School, 2023

Isabelle McPheron - Lowell High School, 2022

Kayanna Seely - Lowell High School,  2023

Payton Gard- Lowell high school, 2022

Jakob Gricus - Lowell High School, 2023

Shelby Hilliard - Lowell High School, 2022

Liv Tully- Lowell High School, 2022

Josh Babin- Lowell High School, 2023

Neutral Molecules

General Chemistry Index

Where are we going with this? This page will assist in developing the ability to predict products of simple reactions as listed in of reactions: synthesis (i.e., combination), decomposition, single displacement, double displacement, acid/base, and combustion.

Neutral Molecules 
What happens if I mix this with that?

Predicting products of chemical reactions is a process by which potential reactants are scrutinized to determine if they will react and if so what product(s) will be formed.

The products that are formed must be "neutrally" charged. Since some of the elements have positive charges only not everything can combine into a compound with everything else. To get a "neutrally charged" molecule, some part of the molecule has to be positive and the other part negative.

MORE…

Once you figure out that it WILL react, the next thing to do is to find the subscripts that will result in a neutrally charged molecule. With ionic compounds, its fairly easy…

The valences will sum to zero. In other types of compounds, there's a similar thing going on at the conceptual level with covalent bonds. There are oxidation numbers, and they have to end up… let's just say they also add up to zero.

But what does that even mean!!!!

Let's go with explanation by example…

Let's take some Al and combine it with Cl.

So, we are trying to end up with a molecule that is neutrally charged. That is to say that the sum of the positive charges and the sum of the negative charges adds up to zero.

Think…
 

Suppose you have 3 things with a charge of +2. That means you have a charge of +6.

 

You need a charge of -6 to even that out. If the negative thing you are dealing with has a charge of -3, then you need 2 of them to get to -6.

In chemistry, you can specify how many of a molecule's parts you have with subscript. (The coefficient tells how many molecules you have.)

 

Okay, if we are going to combine Al and Cl we'll end up with some number of Al atoms and some number of Cl atoms so that the sum of the charges is zero.

Thus… we need to find out what the charges of one atom is. We get to go find the charges on these two things. Luckily, there is a chart. And also the periodic table.

So, here we go:

Al has a charge of +3

That means it has 3 electrons it will give up.

Cl has a charge of -1

That means it will accept 1 electron.

We can write it like this.

Al⁺³

Cl^-1

So, the goal is to get a zero charge and all we have to work with is Al and Cl…

Al⁺³ = +3      

Cl^-1 = -1    

AlCl  is this:

Al⁺³ Cl^-1= +2         <--Not 0

So one Al and one Cl does NOT become neutral. Bring on the subscripts!

Al2Cl  is this:     Al⁺³ Al⁺³ Cl^-1 = +5         <--Not 0  That made it worse!

AlCl2  is this:     Al⁺³ Cl^-1 Cl^-1 = +1         <--Not 0  But closer!

AlCl3  is this:     Al⁺³ Cl^-1  Cl^-1 Cl^-1 = 0        <--THIS IS 0  It takes 3 -1 Cl to become neutral with 1 +3 Al!

It works the same way with polyatomic ions.

If you have:

Fe+3

OH-1

You need 3 minus ones to neutralize 1 plus 3.

Hence, you get this:

Fe(OH)3

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There is a "trick" called the criss-cross method:

1. Take the charge number without the sign of the negative ion and use it as the subscript of the positive ion.
   
   
2. Take the charge number without the sign of the positive ion. and use it as the subscript of the negative ion.
   
   
3. Reduce to "lowest terms."

Tricks are nice. Let's try that!

N is -3

Mg is +2

So let's see… step 1… step 2… 

Mg3N2

Step 3… reduce… not needed in this case. You might see occasionally a 4 and 2… 

Woo! Easy!

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Let's try another one… just for fun?

Be and O

We can write it like this.

Be⁺²

O^-2

So, the goal is to get a zero charge and all we have to work with is Be and O…

Be⁺² O^-2= +2         <--THIS IS 0  It takes 1 -2 O to become neutral with 1 +2  Be

So you would write it as…

BeO

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If you use the criss-cross method, you would take the 2 from O and put it on the Be and then the 2 from Be and put it on the O.

That would be:

Be2O2

In this case, you have the SAME subscripts on both elements, so you reduce to "lowest terms" and end up with the same thing as above.

BeO