How About Calculating Some Moles and Grams?

 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.

How About Calculating Some Moles and Grams?
Let's take another look at this stoichiometry thing… 

The essence here is that we are dealing with the quantities of stuff that goes into a chemical reaction. How much of one thing reacts with another thing and how much of what is produced?

Let's get some numbers that are nice and round (and that you can't Google). 

Substance	Mass
Ft - Fictionite	10
M - Madeupnium	30
Lg - Legendite	15
Hc - Howcomeium	5
N - Nowayium	12
NHc (Polyatomic ion)

And… a problem to work through… Let's do grams given…

If 5.0 grams of FtM2 is completely reacted with LgNHc without leftovers, how many grams of the second substance is needed?



Copy this page onto paper, filling in the boxes and doing the calculations…

STEP 1: Balance the reaction by finding the coefficients that result in the conservation of matter. Type the coefficients in the boxes provided.

 FtM₂ +  LgNHc -->  Ft(NHc)2 +   LgM

Step 2: Find the expected molar ratio from the coefficients. (You'll use this in step 4, below).

The coefficients from Step 1 represent a molar ratio of how the compound will combine. It can be thought of as a "recipe" of moles. 


Step 3: Molar Masses

You will need the molar mass of what is given, so might as well just get the molar masses of everything.

Calculate Molar Masses

FtM₂  =  : LgNHc =   --> Ft(NHc)2. =   : LgM =  

Step 4: Find the actual moles of whatever is given:

Recall that moles can be found several ways; See Step 3 here.

In this problem, use the grams given with the molar mass of the substance to find moles:

Moles = grams / molar mass

Moles present of given substance = grams given / molar mass

Moles present of given substance = 

Step 5: Find what proportion of the expected moles is actually given

• So, the coefficients in Step 1 are the moles expected.

• The actual number of moles present was calculated in Step 3.

Some proportion of the "recipe" is present.

Ratio Multiplier = moles present / moles expected of the given

Ratio Multiplier = 


Step 6: Find the molar ratio of what you actually are working with based on what was given.

Now, we take the expected ratio (coefficients from Step 2) and multiply it by the ratio multiplier to find the ratio we are actually working with (present ratio or working ratio):

(The numbers below are the SAME as the coefficients above in step 1).

 :  -->  :   Expected Molar Ratio (Coefficients from Step 2)

X  Ratio Multiplier

________________________________________

 :  -->  :   Present, Working Molar Ratio




Step 7: Find  whatever you are supposed to find from the working molar ratio.

To find masses…

Multiply the present, working molar ratio (Step 6) by the molar masses (Step 3) to get grams present / needed / produced.

 :  -->  :   Grams present / needed / produced.

To find gas volume at STP…

Multiply the present, working molar ratio (Step 6) by 22.4 to get liters of gas at STP.

 :  -->  :   liters of gas at STP.

If working with Molarity in a solution…

Where M is molarity, n is number of moles, and V is volume in liters

M = n/v

So,

M =  present, working molar ratio (Step 6) ÷ V

What would that look like on paper with a real equation? (As if you can read the handwriting…)

Starting with .5 grams of hydrogen gas, how much oxygen gas would be needed and how much water would be produced?

LINK TO TEMPLATE—Click Here

________________________

Spoiler alert! Answers to the walkthrough follow!

STEP 1: Balance the reaction by finding the coefficients that result in the conservation of matter (The "1"s in green would not normally be written).

1FtM₂ +2LgNHc --> 1Ft(NHc)2 + 2 LgM

Step 2: Find the expected molar ratio from the coefficients. (You'll use this in step 4, below).

The coefficients represent a molar ratio of how the compound will combine. It can be thought of as a "recipe" of moles. 

1: 2 --> 1:2

Step 3: Molar Masses

You will need the molar mass of what is given, so might as well just get the molar masses of everything.

Calculate Molar Masses

FtM₂  = 70 : LgNHc = 32 --> Ft(NHc)2. = 44 : LgM = 45 

Step 4: Find the actual moles of whatever is given:

Recall that moles can be found several ways; See Step 3 here.

In this problem, use the grams given with the molar mass of the substance to find moles:

Moles = grams / molar mass

Moles present of given substance = grams given / molar mass

Moles present of given substance = 5 / 70

Moles present of given substance = 0.071

 

Step 5: Find what proportion of the expected moles is actually given

• So, the coefficients in Step 1 are the moles expected.

1: 2 --> 1:2

• The actual number of moles present was calculated in Step 3.

Some proportion of the "recipe" is present.

Ratio Multiplier = moles of given present / moles if given expected

Ratio Multiplier = 0.071 / 1

Ratio Multiplier = 0.071

Step 6: Find the molar ratio of what you actually are working with based on what was given.

Now, we take the expected ratio (coefficients from Step 2) and multiply it by the ratio multiplier (Step 5) to find the ratio we are actually working with (present ratio or working ratio):

(The numbers below are the SAME as the coefficients above in step 1).

1       :       2          -->    1         :         2               Expected Molar Ratio (Coefficients from Step 2) 

 

                                                  X  0.071 

 

0.071  :   0.142.     -->   0.071   :   0.142            Present, Working Molar Ratio

Step 7: Find  whatever you are supposed to find from the working molar ratio.

To find masses…

Multiply the present, working molar ratio (Step 6) by the molar masses (Step 3) to get grams present / needed / produced.

FtM₂  = 70 : LgNHc = 32        -->   Ft(NHc)2  = 44     :    LgM = 45  Molar Masses

 

0.071          :    0.142.               -->       0.071                :      0.142     Present, Working Molar Ratio

FtM₂  = 5.   :  LgNHc = 4.57   -->   Ft(NHc)2. = 3.14   :  LgM = 6.43   Grams present / needed / produced.

____________________________

To find gas volume at STP…

Multiply the present, working molar ratio (Step 6) by 22.4 to get liters of gas at STP.

0.071              :    0.142            -->       0.071              :     0.142     Present, Working Molar Ratio

 

                                                                                        X 22.4 

 

FtM₂  = 1.6    :    LgNHc = 3.2   --> Ft(NHc)2. = 1.6   :   LgM = 3.2  liters of gas at STP.

____________________________

If working with Molarity in a solution…

Where M is molarity, n is number of moles, and V is volume in liters

M = n/v

So, 

M =  present, working molar ratio (Step 6) ÷ V

Parts Per Million

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.

Parts Per Million
This doesn't sound bad, does it?

https://www.epa.gov/dwreginfo/drinking-water-regulations

You might have heard of his one. Like, how many parts of something is allowed in something.

The definition is pretty obvious… Parts per million is how many times (by mass) that something appears within a million parts of the whole thing. It is abbreviated as PPM. 

A more formal definition: 

Parts Per Million (ppm) is the number of units of mass of a substance per million units of total mass.

Because of the relationship between grams and water, it turns out that PPM in water turns out to be:

PPM = mg / l

where mg is mass of the "added in thing" and l is volume of water in liters.

PPM is often used to discuss contaminants in something. For instance, EPA states that drinking water needs to not have more than 1.3 parts per million of copper in it. Most utility agencies are required to report how their water tests, and often it will include PPM data.

Besides PPM, concentrations can be reported at even lower concentrations. PPB is parts per billion. Both are analogous to percent, witch is parts per one hundred.

2021-01-21 Air Quality,
Lowell, Indiana

Air Quality is also discussed in terms of what part of the whole contains contaminants. The amounts present are often reported in PPM or PPB. 

Okay, that's nice… Hmm… Seems like there would be some math here, right?

So, yeah… Math…

Let's think about this… You know… logic… If something is half and half, it would be 500,000 parts per million, right?

So, if you take the mass of the part divided by the mass of the whole, then multiply by 1,000,000, you'll have PPM, right?

Let's try!

Mass of whole = 1,000,000

Mass of part = 500,000

Part / Whole • 1,000,000 = PPM

500,000 / 1,000,000 • 1,000,00 = PPM

.5 • 1,000,000 = PPM

500,000 = PPM

Well, logic and formula converged nicely! That's always a good thing!

https://sciencing.com/calculate-concentration-ppm-6935286.html

So, let's say that the formulas for PPM are…

PPM = Mass of Part / Mass of Whole • 1,000,000

Mass of Part = PPM * Mass of Whole /1,000,000

Mass of Whole = Mass of Part • 1,000,000 / PPM

because algebra…

Let's keep going! Let's say that ThingX has a concentration of 2 parts per million in a mixture with ThingY.

So, if you have 8 million grams of ThingY, what is the mass of ThingX?

Mass of Whole • PPM /1,000,000 = Mass of Part 

 

8,000,000 gr • 2 /1,000,00 = Mass of Part

16 gr = Mass of Part

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SO WHAT?

It seems like there should be something useful about this? Is there?

• PPM is a great way to discuss the presence of small amounts of things within something else. It's more "tidy" than saying that there is 0.0003% of "this" in "that." Saying that it 3 PPM is a lot more tidy. I suppose that's a matter of opinion, huh?
  
  
• If you are talking about contamination, LOW PPM is good.

Net Ionic Equations

 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.

Net Ionic Equations 

Balancing Net Ionic Equations is really a subset of balancing equations, in general. So, what's special? Why does this get a special level of consideration?

Good question!

In the net ionic equation, you are looking at some stuff in solution mixed with some more stuff in solution. When a compound is in solution, it can break apart into the constituent ions. So, NaCl in water results in there being Na⁺ and Cl^- ions hanging around the polar H2O molecules.  If we mix other things in, it's possible to get other positive and negative ions all mixed up in the water. 

Then, suppose the positive ion from one compound joins with the negative ion from the other to form an insoluble solid? That solid will precipitate out of the solution leaving all the other ions floating around… just watching. 

So, you end up starting with a reaction like :this (Wait? Can you end up starting? Does that even make sense?)  :

NaCl + AgNO3 → NaNO3 + AgCl

Dealing with net ionic reactions, the idea is that the reactants are dissolved in water, so the parts come… apart. The equation is then written with even more subscripts!

aq means it is "aqueous" which means the ions are just mixed up in the water.

s means it is a not-dissolved solid. (This will usually be on the product side!)

l means liquid. 

So, we put the stuff into the liquid (which isn't really part of the reaction, but is the—if you will—container for the reaction). The ions come apart (we said that already) and so we rewrite the equations as…

Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO3⁻³(aq )→

Now, on the product side where we had…

→ NaNO3 + AgCl

But, we now need to do that aq - s - l thing… meaning we need to know what will and what won't dissolve.  Skipping to the outcome, turns out that AgCl is insoluble. So…

Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO3⁻(aq) → Na⁺(aq)  +  NO3⁻(aq)  + AgCl(s)

Comparing the reactant side to the product side, we can see some things that don't change… they were ions before and ended up ions after… They just "watched" the reaction. We call these spectator ions (really, no joking).

Spectator ions: ions in a chemical reaction that do not change during a chemical reaction.

Na⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO3⁻(aq) → Na⁺(aq)  +  NO3⁻(aq)  + AgCl(s)

Now for the "net" part of this whole thing! In business, net profit is the money left over after the expenses are taken out. The part left over… 

So, we just cancel out the things that don't change. As if they aren't even there. As if they are just watching…

Cl⁻(aq) + Ag⁺(aq) →  + AgCl(s)

A net ionic equation shows only the ions found in the larger balanced chemical equation that are directly involved in the chemical reaction.

That's it. 

Stoichiometry Walkthrough

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.

Stoichiometry Walkthrough
Let's take another look at this stoichiometry thing… 

The essence here is that we are dealing with the quantities of stuff that goes into a chemical reaction. How much of one thing reacts with another thing and how much of what is produced?

In a different article, steps were developed around the quantities being measured in grams. In this one, we'll build on that and go a little more into details.

Let's get going…

Since we are dealing with stuff in a chemical reaction, the first part of the process is to figure out the reaction. 

STEP 1: Begin with a balanced chemical reaction.
Which isn't always that easy, right? All the predicting products and stuff! Finding coefficients… All that!

STEP 2: From the balanced chemical reaction, determine the mole ratio of the reactants and products.

Those coefficients again! And all that math…  

Let's look at a simple reaction as an example (adding the "1" coefficient that chemistry grammar says we don't actually write/type):

2Na + 1Cl2 --> 2NaCl

So, based on that balanced equation, we know that the reactants will combine in a set, fixed molar ratio of 2 to 1 and produce 2 moles of product.

Lost to some students is that the 2:1-->2 ratio is ALWAYS present in the balanced equation. However, it can be any fraction or multiple, so long as the ENTIRE ratio changes. So, all of the following ratios of moles are valid for that reaction.

1 X — 2 : 1 --> 2

10 X —20 : 10 --> 20

.1 X — .2 : .1 --> .2

1/2 X — 1 : .5 --> 1

7 X — 14 : 7 --> 14

.134 X — .268 : .134 --> .268

There are literally infinite sets of numbers that will fit the 2 : 1 --> 2 pattern. Just multiply the whole ratio by any number and the results will fit the pattern.

Once you have the mole ratio from the balanced equation, you are ready to use that with the actual amounts present, needed, or produced.

STEP 3: Find out how many moles of one thing you are working with.

The number of moles (n) is found several ways.

If you are given the mass of something, moles can be found (where n is moles) using…

 n = mass in grams / molar mass

If dealing with gases, then where P is pressure, V is volume, R is a constant, and T is temperature Kelvin, then the ideal gas law can be used:

PV = nRT

Many times, the ideal gas law is estimated to be:

n = V/22.4 liters

If you are working with a solution, then you can use the molarity (M) of the solution to find moles (n). Where V is volume in liters then

M = n/v

M • V = n

so…

n = M • V

Whatever is given, by some means, you figure out how many moles are present of something in the reaction.

STEP 4: Use the number of moles present with the molar ratio to find out how many moles of everything else is needed or will be produced.

Because reactions occur in fixed ratios of moles, and because one mole of any substance has a set mass, it is pretty easy easy to figure everything out.

______________________

EXAMPLE

Starting with 2.3 grams of Na, how much Cl2 will be needed to completely react without any excess?

STEP 1: Balanced Equation

2Na + 1Cl2 --> 2NaCl

STEP 2: Ratio (molar) of coefficients

2 : 1 --> 2

STEP 3: Actual moles present

 Let's use these atomic weights:

Na = 23.0

Cl = 35.5

Convert the grams of Na present to moles present:

moles = grams / molar mass

n = 2.3 / 23

n = .1 moles 

 

STEP 4 : Needed and produced moles

In this step, you are first thinking this: How many moles do I have as a percentage of how many moles I was supposed to have?

So, divide what is present by the amount of that thing needed in the ratio given (STEP 2) by the coefficients of the balanced equation.

decimal amount given  = given moles / needed needed

decimal amount given = .1 / 2

decimal amount given = .05

Then you are thinking this: If I start with (some decimal amount given) of the one thing, then I need that same decimal amount of everything else.

So, multiply the whole molar ratio (STEP 2) by the decimal amount given .

2 : 1 --> 2

      X  .05 

________

.1 : .05 --> .1

Hence, starting with  .1 moles of Na, you need  .05 moles of Cl2 and you will produce .1 moles of NaCl.

It is easy to convert further into grams/liters of Cl2 and/or NaCl.

grams = moles • molar mass

grams Cl2 = .05 X 71          <-- Each Cl has a mass of 35.5 and Cl2 has 2 of them so…

grams Cl2 = .05 X 3.55

Finding grams of NaCl will be left as an exercise for the reader.*

n = V/22.4 liters

.05 = V/22.4 liters

.05 • 22.4 litres = V

1.12 liters = V

SUMMARY:

Stoichiometry is much harder to spell than it is to do. Using the four steps provided, it really is just a matter of multiplying and dividing.

Begin with a Balanced equation

Find the Ratio from the coefficients

Calculate the Actual number of moles present

Calculate the Needed moles for the other reactants and products.

BRAN is good, right?

And, just what does this look like on paper?

__________________

* This footnote has nothing to do with this topic. But…

When in college a zillion years ago, I had this text book that would go through some vastly complex calculation, then skip like 50 steps that were replaced by something like:

Converting from the above relationship to the following is intuitively obvious to the most casual observer.

Nah… never.

Then, it would end up and add something that amounted to this:

Creating ongoing world peace will be left as an exercise to the reader.

Finding Molality

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 Molality
Do we really need this? Some people do!

https://www.google.com/search?q=molality

Yet, again, we should whip up a definition!

Molality: the number of moles of something (the solute) in a kilogram of a solvent.

That… seems a lot like molarity. Indeed! But this is moles per kilogram, not moles per liter. And it is the mass of the solvent, not the mass of the solution.

Yet again, there should be a formula, right?

Where m is molality, mol is number of moles of solute, and kg is mass in kg of the solution, then…

m = mol/kg

Note: it is lowercase m for molality.

So, that doesn't look too hard… We should do an example!

_________________

EXAMPLE:

Find the molality of a solution containing 5 moles of NaCl dissolved into 2 kg of water.

m = ?

mol = 5

kg = 2

m = mol/kg           <--- write the formula

m = 5 / 2            <--- plug in the numbers given, then do math

m = 2.5 mol/kg

_________________

Well… that wasn't too bad!

Why is this even a thing?

https://www.google.com/search?q=where+is+molality+used

Working with concentrations when temperature is involved introduces changes in volume (because things contract and expand with decreases and increases in temperature). If you want to talk about the effect a concentration of something has, and don't want to discuss temperature, then you can use molality.

 

Walkthrough: Calculating Molar Mass

 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.

Walkthrough: Calculating Molar Mass
This rings a bell… But… a second look is always a good idea, right?

Calculating molar mass s a part of numerous chemistry concepts; anytime you are dealing with moles and grams, such as in doing stoicheomitry, finding percent composition, and calculating molarity. Let's take a look!

What exactly is the molar mass? Molar mass is the mass in grams of one mole of the substance.

Molar mass us very useful in things like…

This reaction needs 2 moles of iron. How much does two moles of iron weigh?

What is the molarity of a 2 liter solution that contains 284 grams of K2SO4?

The process is pretty simple, actually. Let's check it out!

The setup here is… something something something, you need to know how much one mole of ____ weighs in grams (see useful things above).

_______________________

EXAMPLE: 

What is the molar mass of MgCl2? So, we need to find out how much one mole of that compound weighs in grams…

Step 1: Use the molecular formula to identify ALL the TYPES of atoms in the compound.

Let's see… we got Mg and Cl…

Step 2: Use a periodic table to find the atomic masses of each of the elements in the compound.

Mg = 24.3 g/mole

Cl = 35.5 g/mole

Step 3: Use the subscripts from the molecular formula to figure out how many of each type of atom you have in the compound.

Remember if no subscript is written, it is assumed to be 1, so… Looks like we have…

Mg1Cl2

 

Step 4: Math

4a. Multiply the atomic mass by the subscript to find the total mass of each element present in the compound:

Mass of Mg: 24.3  X 1 = 24.3

Mass of Cl2:  35.5 X 2 = 71.0

4b. Add up the mass of each element in the compound to find the total mass of the compound.

Molecular Mass of MgCl2 = Mass of Mg + Mass of Cl2

Molecular Mass of MgCl2 = 24.3 + 71.0

Molecular Mass of MgCl2 = 95.3 g/mole

Step 5: You are done. That's the answer.

Here's what that looks like written by hand:

Looks like it was written by a 3rd grader…