Wednesday, December 16, 2020

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:

H2but not H4O2

CO 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

CuBr= 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. 



_______________________________________________

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…







Monday, November 30, 2020

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 SOand 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 blueRed 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.


Wednesday, November 18, 2020

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 SOand 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. 



Monday, November 16, 2020

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 common gas (H2S, H2CO3, H2SO3 and NH4OH), 
  • one of the products is 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 (solid) meaning that will not dissolve in a solvent even after mixing.

An insoluble substance won't dissolve in a solvent after letting it sit and mix

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 = AC + BD 

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 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

Covalent bonds in water are stronger than ionic bonds in salt 

FeS8

Insoluble

No rules satisfied

BaSO4

Insoluble

Ba ion and Sulphate ion are large in size

Ag2CO3

Insoluble 

Carbonates are insoluble.

AgBF4

Soluble 

 

NaHCO3

Soluble

The reaction is exothermic in that heat and Carbonic acid are produced. Carbonic acid is unstable so breaks up into carbon dioxide and water.

NO3

Soluble 

Salts containing this anion are called nitrates.  Almost all nitrates are soluble in water. 

AgCl 

Insoluble 

Many ionic solids do not dissolve in water

Ca(OH)2

Soluble

Exception to the sulfide rule, calcium is the exception. 


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 redacted 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


Wednesday, November 11, 2020

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.


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!

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

AlCl 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



_______________________

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!

_______________________

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


_______________________

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

Wednesday, November 4, 2020

Moles to Grams to Moles

General Chemistry Index

Where are we going with this? This page will give the ability to demonstrate an understanding of the law of conservation of mass through the use of particle diagrams and mathematical models.


Moles to Grams to Moles
Moles to grams you multiply. Grams to moles divide.

This is so important to understand, I made a cringy video. Really.



Did you catch that? 

Moles to grams you multiply. Grams to moles, you divide!

Divide? Multiply? What??? Multiply what!

How Does This Thing Work?

Okay, back to that periodic table… Back to that atomic number… 

The atomic number gives the mass in grams of one mole of atoms of the element.

If you have a molecule of atoms, you can find the molecular mass by adding up all of the atomic masses of all of the atoms in the molecule. Really.

If you know the molecular mass of a substance, you know how much one mole of that substance weighs in grams. 

Ah ha!!!

Moles to grams you multiply. Grams to moles, you divide!

 Moles = Grams ÷ Molar mass

Grams = Moles X Molar mass


EXAMPLES:

Water is H20.

Two atoms of Hydrogen. One atom of oxygen.  Thus, adding up the masses of the atoms, we find that one molecule of water has the molecular mass of 31.998 grams / mole.

Suppose you have 5.5 grams of water… How many moles do you have?

Moles = Grams ÷ Molar mass 

Moles = 5.5 ÷ 31.998

Moles = 0.17198


Suppose you have .5 moles of water… How many grams do you have?

Grams = Moles X Molar mass

Grams = .5 X 31.998

Grams = 15.999


Moles to grams you multiply. Grams to moles, you divide!

Sunday, October 25, 2020

Predicting Reactions: A System

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.


Predicting Reactions: A System 
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.

What happens if I mix this baking soda with vinegar?
What happens if I let this spilled gasoline sit on the painted garage floor?
What happens if I pour bleach directly onto my clothes?

Predicting chemical reactions does not only take place in the lab, but actually is a part of everyday life! However, in the lab, we can be more specific and isolate the this and the that more.

So… I told you this would be long! Maybe another soda or cup of coffee?

Recall: the goal is to end up with neutral molecules no matter what you have to do to the subscripts on the PRODUCT side.

Let's just say we are starting with "real" molecules on the reactant side, to begin with, okay? Coming up with a system to predict the products is, at best, a starting place. 

Because of the vast scope of chemistry, there will be variations that this system will not cover (probably). And, there are always exceptions!

Back to the original question, but flipping it: If I mix this with that, what happens?

The system I'm going to offer expands on that question thusly:

What do you have to start with?

This step identifies what kind of reaction you could be looking at. Is it synthesis? Decomposition? Etc. 

 
What does it become? 
  1. Will it even react? 
  2. What combination of atoms are needed in the product such that the compound molecule is neutrally charged? (AKA what are the subscripts?) 
  3. What is the FINAL balanced equation?
 

So, here is a "decision tree" for what to do to predict the products of a chemical reaction. 



Predicting the Products of Chemical Reactions Decision Tree



What do you have to start with?

• Two elements: goto Synthesis Reaction below (Click) 

• One compound: goto Decomposition Reaction below (Click)

• An element and a compound: goto Single Displacement Reaction below (Click) 

• Two compounds: goto Double Displacement Reaction below (Click) 





What does it become?

Synthesis Reaction (background)
 
First off, will they react? 
For two elements to react, they need to have opposite valence charges (one plus, one minus = one has four or less electrons in its valence orbitals, one has four or more)

Secondly, if they will react, how many of each are needed to get a neutrally charged compound molecule in the product?

The subscripts denote how many of each thing will be present in the neutrally charged molecule. You can "criss-cross" the charges of two elements (then reduce mathematically) to find the right numbers.

For instance, take carbon and oxygen.

Oxygen has a valence of -2.

Carbon has a charge of ±4. (Since oxygen is negative, we'll use the +4)

C + O2 --> ??

So, it has to be C?O?

Criss-cross the charges:  C2O4 

Reduce the subscripts mathematically: C2O4   becomes  CO2

C + O2 --> CO2

Thirdly, balance the equation such that the same number of each type of atom is present on both sides:

C + O2 --> CO(was already in balance)


______________________________

Decomposition Reaction (background)
 
First off, will they react? 
Not everything will break apart easily. Somethings only break apart at high temperatures.

It seems fair to presume that, if given a predicting products exercise, the compound will, by some means react and decompose into the parts.

Secondly, if… Well, this one is pretty easy. Whatever you start with breaks apart. But… into how many pieces! (Probably two.) Usually, it will look like this:

AB --> A + B 

Both the A and the B have to be neutral molecules or elements. And don't forget about those diatomic elements

Thirdly, balance the equation such that the same number of each type of atom is present on both sides:

2H2O → 2 H2 + O2

It could be tricky, though!

Na2CO3 → Na2O + CO2


______________________________

Single Displacement Reaction (background)

First off, will they react? 
For one thing to replace another… Let's say it like this… For A to replace B

A + BC --> ?? + ????

A has to be more highly reactive than B. How would anyone know that? There is a chart!

So, sodium won't replace potassium. Etc.!
 
Also, either B or C could be a polyatomic ion! That makes it harder to figure out what is being replaced. Look into the BC part and match one of them to the A with regards to location on the periodic table. There's a good chance that A will be in a family/group that is near the family or group of B or C. (You'll have to be openminded about this claim when dealing with transition elements.)
 
Secondly, if they will react, how many of each are needed to get a neutrally charged compound in the product?

You know what you are starting with, so the reactant side is done. Let's do aluminum and HCl as an example…

Al + HCl --> ?? + ?

Since Al will, in fact, replace H, the product side will be:

Al + HCl --> AlCl + H

Product side subscript time:
This should be fun!

1. H is diatomic, so it will be H2

2. The AlCl has to become neutrally charged. 

The subscripts denote how many of each thing will be present in the neutrally charged molecule. You can "criss-cross" the charges of two elements (then reduce mathematically) to find the right numbers.

For instance, take carbon and oxygen.

Al has a valence of +3.

Cl has a charge of -1.

The balanced molecule will be AlCl3


Thirdly, balance the equation such that the same number of each type of atom is present on both sides:
2Al + 6HCl --> 2AlCl3 + 3H2


______________________________


Double Displacement Reaction (background)

First off, will they react? 
For one thing to replace another… Let's say it like this… For A to replace B

AB + CD --> ???? + ????

Deciding what is A and what is B can be hard when polyatomic ions are involved. Really, the only way to get good at this is to do it a lot.

Look at A and C first. Are they both metals? Hmmm… What about B and D? Both polyatomic ions? If so, do they both have a positive or negative charge (There is a chart!)? If both B and D are negative and both A and C are positive (which is kinda the normal way of writing molecules) then you can find your potential "swaps."

At this point, you have to answer the question! Will they react?

Answering this question is complex, since the reactivity of each of the four parts is in play. How is it done? (CLICK HERE)

So… 

1. Does it form water? 

2. Does it form a gas? 

3. And then that insolubility thingeasy


Secondly, if they will react, how many of each part is needed to get a neutrally charged compound in the product?

The work done in the first step should have resulted in you knowing what the AB and CD parts are.  The product will become AD and BC

You know what you are starting with, so the reactant side is done. Let's do the follow as an example…

Fe2(SO4)3 + KOH → 
So, there's some AB and CD up there? That looks like a sentence in a foreign language!

So this… 
 
AB          +   CD  → AD + CB
Fe2(SO4)3 + KOH

(When you predict the swapped product, put in the parentheses to start with, at least in your head.)

Fe2(SO4)3 + K(OH)1

Fe2(SO4)3 + K(OH)1 → Fe?(OH)? + K?(SO4)?


Product side subscript time:
This should be… never mind!

1. Do that thing with the charges to get balanced molecules. For instance…

K (from periodic table) has a charge of + 1

SO4 has a charge of -2

Criss-cross the charges to get K2(SO4)1
Chemistry "grammar" says we don't write ones and if the subscript is one don't use parentheses.

Thus, we get:

K2SO4 

Using the the example from above, working through the process of getting both products to a neutral charge, we get the unbalanced (but each part is a neutrally charged molecule) equation:

?Fe2(SO4)3 + ?KOH → ?K2SO4 + ?Fe(OH)3


Thirdly, balance the equation such that the same number of each type of atom is present on both sides.

So for the above example…

Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3