In this unit, you have heard about many different types of stoichiometry problems. This discussion gives you an opportunity to share your insights about solving stoichiometry problems with your classmates. Generally, if you are having trouble with a particular topic, someone else in the class is having the same problems. More importantly, it is likely that someone in the class can explain the problem in a way that you can understand. Use this opportunity to get help on the subjects you find difficult and give advice to others about problems they are having.
- Talk about how to convert from grams to moles, and
- how to determine the percent composition by mass of elements in compounds using both laboratory data and molecular formulas.
- Tell your classmates how you go about determining the empirical and molecular formula of a compound given combustion data and the molar mass of the compound.
- Discuss the simplest way to relate the amount of reactants and products in a chemical equation by looking at mole ratios.
- Solution stoichiometry is often confusing at first, so take a minute to talk about how to make a solution of a given molarity and the how to make dilutions from it.
- Discuss how to determine the limiting reagent in a reaction and how to find the theoretical and percent yield.
- Gravimetric and volumetric analysis can both be used to determine the amount of a substance dissolved in a solution. Talk about why you might choose one method over the other.
- Discuss how a titration works and what types of things need to be considered before doing a titration.
- Finally, remember to discuss how to calculate how many particles there are in a mole.
Josh Dos Santos
ReplyDelete1)Converting from grams to moles is all about ratios.
In the same way that you convert radians to degrees in math, you can convert grams to moles and moles to grams.
pi radians=180 degrees
so Xdegrees * (pi/180 degrees) will yield (X/180)pi
and Xpi * (180 degrees/pi) will yield 180X
In the same way...
Xgrams can be changed into YMoles.
One can do this with molar mass. The periodic table states that One mole of an element will weigh Z grams...
So..
X grams * (1mol/Z grams) will yeild (X/Z)mol
where z is the number of grams in one mole and x is the number of grams in the sample.
So..
Ymol * (Z grams/1 mol) which will give YZgrams
In essence it is the number of moles given multiplied by the number of grams in one mole, the number on the periodic table.
A shortcut is to remember that if Ymol of Xgrams will always be (X/Z mol) and z is the molar mass, then you always just divide the number of grams given by the number provided in the periodic table, without having to write an equation.
In the same way, if you remember that X grams is YZmol and z is the molar mass, then you just multiply the number of moles given by the number in the periodic table.
If you want moles you divide by the number in the periodic table, molar mass.
If its grams you want, you multiply the number you have by the number in the periodic table, the molar mass of the element.
Hope this helps,
Josh Dos Santos
9. Finally, remember to discuss how to calculate how many particles there are in a mole.
ReplyDeleteWoah it's like déjà vu from the last discussion about moles! Since we all had sooo much fun with that last one, how about a little more "mole talk" :) haha ok ok here it is:
First, to define a mole. One mole of a substance will give huge amount of that substance (like we saw), so to make those numbers more manageable we use Avogadro's Number. For every substance 1 mol = 6.022x10^23 particles. Every substance has the same number of particles/mole.
Therefore, to find the number of particles in a substance you simply multiply the number of moles in the substance by 6.022x10^23.
If you are given the amount of a substance in grams all you need to do is convert it to moles by dividing the number of grams by the amount of grams/mole of that substance.
EX: How many particles are there in 14.2g of gold :)
First: Convert grams to moles.
14.2gAu X 1 mol Au/196.67gAu
=0.0722 mol Au
*this may seem like a small amount but wait 'till we multiply this by Avogadro's Number, that 10^23 will make this number explode!
Second: Multiply number of moles by Avogadro's Number (6.022 x 10^23).
.0722 x (6.22x10^23)
=4.348 x 10^22
=43,480,000,000,000,000,000,000
*"'aye you, that's a lot of gold!" haha I know, bad joke.
I hope this helps keep our minds nice and fresh :)
Sincerely,
Kelsey Sloper
Josh,
ReplyDeleteHoley Moley. At first my mind was all jumbled up with all the x's and y's and the example with pi/radians reminded me of PreCal last year...But I did find your examples useful. After my headache from past memories went away haha.
I like how you put the info into little chunks. That definitely made it more manageable. Thanks!
-Kels
This comment has been removed by the author.
ReplyDelete2) Percent composition is basically how much of an element is in a certain compound. For this example, I will use water (H2O). First, we need to find the molecular mass of H2O.
ReplyDelete2(1.008) + 16 = 18.016 g/mol
Now,in order to find the percentage of hydrogen in H2O, we need to take the molecular mass of hydrogen and divide it by the molecular mass of H2O. Then we need to multiply the quotient by 100% because we are trying to find the percentage.
2.016/18.016 = .1119 x 100% = 11.19%
This means that in every mole of water, there is around 11.19% of hydrogen in each mol. Now we have to do the same for oxygen.
16/18.016 = .8880 x 100% = 88.80%
This shows that there is around 88.80% of oxygen in every mole of water. With this, we have found the percent composition of water, which is 88.80% oxygen and 11.19% hydrogen.
There are two ways to check if the percent composition that was found is correct. The addition of the percentages should yield 100% or around that number. so...
88.80 + 11.19 = 99.99
This is very close to 100 so this is okay. The second way is to take the molecular mass of an element in the compound and divide it by the percent. The answer should come out to be the molecular mass of the compound, which H2O for us. In the case of hydrogen...
2.016/.1119 = 18.016
The answer is 18.016 so our percent composition of H2O being 88.80% oxygen and 11.19% hydrogen is correct.
Now I'm thirsty...
@Kelsey
ReplyDeleteIt was smart to refer back to our last discussion to illustrate Avogadro's number. It is such a big number and if there is just one mistake the the answer can be entirely different than what it should be.
Also, great job in defining the mole first before going into any calculations. That can save a lot of questions of where you got 6.022 x 10^23 from.
One more thing. Half of that gold is mine. haha
6)Discuss how to determine the limiting reagent in a reaction and how to find the theoretical and percent yield.
ReplyDelete- The limition agent is which you have a limited amount of compound or element.To find a theoretical yield, you can simply calculate it. To get a percent yield, you can do actual yield / theoretical yield.
In chemical formular, C+O2=CO2, let's say that we have 14g of carbon. Carbon now became the limiting reagent. Oxigen is excessive.
14/14(molar mass of carbon) = 1
we have 1 mole of carbon. Thus, we have 1 mole of carbon dioxide.
that molar mass of carbon dioxide is 46.
we have 46 * 1 = 46 g of carbon dioxide.
that is theoratical yield
Let's say that you've done the experiment and got 45g of CO2
that is actual yield.
45(actual yield)/46(theoratical yield) x 100 =
(percent yield)
@Andrew
It was easy to understand that you have given examples on each step. I can easily know what you are trying to say.
number 3) was mine... Moon Kyu Kim
ReplyDeleteMoon Kyu Kim
ReplyDeleteI mean.. 6)
josh Dos Santos
ReplyDeleteKels, like how you used a sample question..smart.
I also think it was good to remind us of how big a mole is so we don't freak out when we get a small number in our calculations, since when multiplied by avogadro's number, it will be much bigger.
P.S.
Holey MOLEY haha
o and did you get that "aye you" joke from that john tucker movie?
P.P.S.
Thanks:) i try:)
Oh dang, I'm late -_-;;;;
ReplyDeleteI'm sorry for the delay. So i'll take number 7 because it is LUCKY SEVEN!!!!! MUHAHAHAHA.....
Anyway, to get serious, there is two way to find out how many moles of certain substance is in the solution given. You can use either gravimetric or volumetric analysis to find the moles of solution. However, i prefer to use Gravimetric analysis than volumetric analysis for certain reasons.
one of the main reason why i prefer Gravimetric analysis is that volumetric analysis requires you to balance a difficult Reduction-oxidation reaction, which is just mind boggling. For example, here is an reduction oxidation reaction for volumetric analysis that you need to balance:
Fe(2+)+ MnO4- + H+ => Fe(3+) + Mn2+ + H2O
Now you break this up into two distinct group of Oxidation and Reduction
Oxidation:
Fe2+ -> Fe3+
Reduction:
MnO4- -> Mn2+
Now, since this is taking place in Acidic solution, you add H+ in order to balance out each sides, which makes the equation look like:
Oxidation:
5 Fe2+ -> 5 Fe3+ + 5e-
Reduction:
MnO4- + 8 H+ + 5 e- -> Mn2+ + 4 H2O
This two equation is now combined again to give one equation of:
MnO4- + 8 H+ + 5 e- + 5 Fe2+ -> 5 Fe3+ + 5 e- + Mn2+ + 4 H2O
Now just take a glance at this. How beastly is it? It is just unbearable.....
On the other hand, gravimetric analysis makes it easier because what we do is a precipitation reaction. and because it is precipitation reaction, it is simple more ration and mole converting that is required. This makes the whole equation much easier.
Also, easy calculation lowers the possibility of having the wrong, which is also good.
Of course, I acknowledge that the gravimetric formula won't always be easy equations. However, Gravimetric analysis is easier to me, and i recommend gravimetric analysis for you guys.....
P.S. Whew this was long. Anyway, i tried to make Volumetric analysis look not bad. And its a mere suggestion, so it might be wrong people. Please don't criticize me. Well, i don't even think anybody will read this anyway in the first place...... ToT
Andrew
ReplyDeleteNice job on #2. You organized very well and made it easy for me to understand better. However, just to make a comment, if you have had rounded the numbers up more precisely, couldn't you have gotten a precise, exact 100%?
I mean, i'm not trying to criticize or attack or assault or anything, but i just thought that being accurate was important.
Anyway, Nice job.
Also, nice job. Josh.
ReplyDeletejust by substituting all those crazy numbers with that alphabets made me easier to understand. And also you are the first one to post a comment on here. How impressive.
Allison VanOtterdyk
ReplyDelete3) A molecular formula for a compound is basically a a list of elements and the number of particles that each element possesses to neutralize the compound. For example C6H14 or Hexane. Now the empirical formula basically just divides the number of particles by the greatest common denominator, in this case 2 thus the empirical formula would be C3H7.
I'm late on this too so I look bad...pooh.
I must congratulate you Victor, it was long, but very understandable, after all, sometimes a longer explanation is better than one that is short, you allow more detail and room to explain.
ReplyDeleteBy: Megan Dickson
ReplyDelete2) Determining the percent composition can seem overwhelming at first, but it is easy when you know what to do and you follow a basic procedure.
When you are working with data in a lab, you can determine the percent composition using the information you have from your laboratory procedure. For example, you may have found that you have 5.3 grams of silver in a 15.9 gram sample of the compound of silver nitrate, which is AgNO3. You can easily find the percent composition of silver in the compound by making a fraction of the two values that you have. You would but the amount of silver as the numerator in the fraction and the total amount of silver nitrate as the denomitor in the reaction, and you could use your calculator to divide.
5.3 grams of silver/ 15.9 grams of silver nitrate = 0.3333
You would then multiply the number you get by 100 to obtain a percent value.
0.3333 x 100 = 33.33% silver
I often find writing the numbers and formulas together to be confusing, so I write out the names of the elements and formulas. I find that easier to manage.
When working with molecular formulas, the process in essentially the same. However, you must use the periodic table and your knowledge of molar mass to solve a percent composition problem.
You begin by analyzing the formula of a compoud. Let's look at the compound water H2O.
First, we want to find the molar mass of the entire compound. We do this by adding the molar masses of the individual elements that make up the compound. In water, we have 2 hydrogrens, which have a molar mass of 1.01 g/mol each, and oxygen, which has a molar mass of about 16.0 g/mol.
(1.01 g/mol x 2) + 16.0 g/mol = 18.02 g/mol
Now, to find the percent composition of hydrogen, we place the molar mass of hydrogen present over the total molar mass of the compound.
2.02 g/mol of hydrogen/ 18.02 g/mol of water = 0.1121 g/mol
We again multiply by 100 to get a percent amount.
0.1121 x 100= 11.21% hydrogen
I hope that I was able to help anyone struggling with this concept!
Megan Dickson
ReplyDeleteKelsey, that was very helpful :) I like how you related concepts so well. It helped me to understand a unfamiliar concept of Chemistry by relating it to a concept that I am more familiar with.
Thanks!
Megan
Ellen Cho
ReplyDelete4.Discuss the simplest way to relate the amount of reactants and products in a chemical equation by looking at mole ratios.
To find amount of reactants or products in a equation, it is able to find with mole ratios if an amount of a reactant or product is given.We have to use mole of given amount of reactant or product, the mole ratio, and conversion of mole to grams.
For example, let's say an equation is
2 Al(s) + Fe2O3(s) => Al2O3(s) + 2 Fe(l) and 130 grams of Fe2O3 in reantant is given. If the question is to find the amount of Al2O3, the mole ratio is needed.
First, find a mole of Fe2O3. It gives about 0.814 moles. Then, use mole ratio to find how much of mole is producted in Al2O3. Since the mole ratio is 1:1, the mol of Al2O3 is also 0.814. [(0.814mol Fe2O3)*(1mol Al2O3/Fe2O3)=0.814mol Al2O3]
After finding mol of Al2O3, just convert it into grams which is about 83 grams. It tells that 130 grams of Fe2O3 is reacted to produce about 83 grams of Al2O3.
Ellen Cho
ReplyDeleteI didn't know what Gravimetric and volumetric analysis are, but now I learned what they are and how to use them. But I don't think I can use them in equations by my own. It's so confusing to me.
Haily Oh
ReplyDeleteTo me,finding the theoretical and percent yield were difficult. But make those in simple, memorizing the equation is the easiest way actual mass/ predicted mass x 100.(percent yield)
For example, C(s) + O2(g) --- > CO2(g)
If you burn 12 grams of carbon to make CO2, then amount of carbon dioxide expected is one mol of CO2 or 44 grams of CO2.
Sadly the amount you will get will probably be less than 44 grams and more like 34 grams of CO2. The problem is a competing reaction that happens. Some carbon reacts to make CO.
2 C(s) + O2(g) --- > 2 CO(g)
The carbon participating in this "side" reaction will not be able to make CO2. The reaction will not yield 100% of the expected CO2.
The amount of carbon dioxide produced, 34 grams of CO2 is only 77% and not 100 % of the expected 44 grams.
Thanks for Josh's explaining on Converting from grams to moles is all about ratios is helpful to understand its concept. I always try to use the book when I need to solve that kind of problem and now I dont have to do it anymore!
ReplyDeleteEsther Lee
ReplyDelete6) limiting reagent is completely consumed in a reaction and limits the amounts of products formed.
For example, if there is 6 mols of A and 5 mols of B reacts to produce C, with following chemical equation:
3A+ B -> 3C
Here, B can be used five times but A only can be used two times thereby limiting the use of reactants.
theoretical yield is calculated amount of product
to calculate the theoretical yield, balance the equation and get the moles of each reactant. Then find the limiting reagent. Convert that limiting reagent into grams.
actual yield is the measured amount of product
percent yield:(actual yield(grams)/theoretical yield(grams)) x 100
Josh, nice comparison!
ReplyDeleteIt was interesting that you compared converting from moles to grams with converting from radians to degrees.
Esther Lee
Mrs. Eastman
ReplyDeleteGreat job on the explanations! You guys could teach these skills =) Also, Victor: the Gravimetric Analysis lab is only 3 labs away! Woohoo!
Keep up the great work!
1)
ReplyDeleteYou get the substance gram and divide with the gram/mol of that substance.
For example, you have 12 grams of P than 12/30.97 will give you the mol you have.
2)
In lab you can use what you gain and what you had and divide these two and multiply 100 to get %. For example, if you had 6 grams of Ag and you gained 16 grams of AgNO3 than you do 6/16*100.
3)
You should get all the substances to the mol and divide the substances with the smallest mol substance that you have than the whole number that you gain from this process will be the numbers of the substances in the empirical formula.
And now you should change the empirical formula to the gram and compare with the original gram you gain and time the ratio till you get the same gram.
4)
Using ratio will give the ratios of the reactants and products. After you know how many mols you need you just need to convert the mol to the grams.
5)
Just have to remember that it is quantitative relationships of reactants and products. Morality is Mol/L so it is the density of the mol in L, so to make diluted you just have to add more water in it.
6)
Limiting agent is the first reactant in a chemical reaction to run out from the reaction. Theoretical yield is the product that can be obtained from given amounts of reactants in a chemical reaction, so it represents the maximum amount. To get percent yield = Theoretical yield/ Actual yield * 100.
7)
I will chose gravimetric analysis if I can get the precipitate, because it is more precise. The reason is Volumetric analysis is just adding the reactants until the person can see the change visually, which is not accurate.
8)
Should be slowly added to the solution. Volume should be precisely measured. Lastly, you should know what is in inside of titration.
9)
As having the final and starting volumes for titration, get the molar of titration and change it to the mol of the product.
Thank you
Sincerely,
Dwayne Hahm
I didn't know much about stoichiometry and it confused me alot. but Megan's examples gave me better view on this. Thank you.
ReplyDeleteDwayne Hahm