For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

### Related Articles:

### Stoichiometry

- All Stoichiometric calculations pass through mols, the basic unit of Chemistry
- There is always a path connecting data you are given and data requested. Express this path pictorially, then convert to mathematics.

### Stoichiometry

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Stoichiometry 0:25
- Introduction to Stoichiometry
- Example 1
- Example 2
- Example 3
- Example 4
- Example 5: Questions
- Example 5: Part A - Limiting Reactant
- Example 5: Part B
- Example 5: Part C

### AP Chemistry Online Prep Course

### Transcription: Stoichiometry

*Hello, and welcome back to Educator.com*0000

*Today, we are going to continue our quick review of the basics of chemistry before we jump into some solid AP work.*0002

*We're going to go over stoichiometry today.*0010

*I'm going to talk about several varieties of stoichiometry problems.*0013

*Stoichiometry, essentially, is about amounts; that is what it is.*0018

*In chemistry, the basic unit of amount is the mole.*0026

*This is something that I know you all have heard before, so really quickly: a mole is 6.02x10 ^{23} particles.*0031

*"Particles," because in chemistry, we can talk about atoms; we can talk about molecules; we can talk about ions, electrons, protons...whatever.*0042

*I tend to use just the generic term "particles"; and the problem itself will specify what particle we're talking about.*0049

*In one mole (so this is a conversion factor) of something, we have 6.02x10 ^{23} particles.*0057

*Now, the periodic table is arranged based on the mole; a molar mass, that you see on the periodic table--that is one mole of oxygen atoms, one mole of magnesium atoms, one mole of iron atoms--those weigh those particular numbers.*0065

*So, the mole is the standard unit that we use in chemistry--again, because we can't count individual atoms (they're too tiny), so we have to choose a unit, kind of like a dozen or a century--things like that.*0079

*Stoichiometry basically concerns conversion factors: setting up a bunch of conversion factors based on what you know versus what you want.*0093

*There is a series of steps to that.*0103

*So, I'm going to introduce this notion of how to deal with stoichiometry problems.*0106

*Because the mole is always what we're talking about, the mole is the central hub; you start from the mole and you go to the mole, or you pass through the mole getting to where you want.*0111

*One way or the other, you are going to have to use a mole.*0120

*The reason is: the equations in chemistry (like, for example, 2H _{2}+O_{2}→ 2H_{2}O, the reaction of hydrogen and oxygen gas to produce water)--well, these numbers here--the coefficients--the 2, 1, and the 2--they represent moles.*0122

*What this says is, "Two moles of hydrogen gas need to react with one mole of oxygen gas in order to produce two moles of H _{2}O."*0143

*So, all equations--balanced equations--represent the standard that we use to work from.*0150

*It's all in moles.*0158

*Now, in any stoichiometry problem, any species that you're dealing with always has (this is the best way to think about it) moles (you know what, I'm going to actually draw this up here) of X.*0160

*Every species that is mentioned in a stoichiometry problem--something like that; from the mole, we can go to grams of X, or we can go to the number of particles of X.*0179

*Again, X might be a molecule; it might be an atom; it might be an ion; it depends.*0198

*Later on, we might need to go to liters of X.*0202

*But, they all have to pass through the mole.*0209

*Again, every species under discussion has one of these little things associated with it.*0213

*If a stoichiometry problem has two species that are mentioned, and there is some relationship among the two species (which we will get to in a minute), that species is also going to have something like this.*0218

*Let's call it mol y, and again, you can go to grams of y; number of particles of y; or maybe liters of y.*0231

*Now, here is what is interesting: the reason that I say that each species--you should write something like this: when you do something like this, this will give you a pathway to actually find out how to write the problem, stoichiometrically; how to solve it, mathematically.*0250

*This gives you a solution path.*0264

*Any time you are moving from one species to another--like, for example, in a minute we're going to sort of choose a couple of reactants and a couple of products and stoichiometric problems--movement this way or that way is based on the mole ratio, and these mole ratios are precisely the coefficients in the balanced reaction.*0267

*Now, this will make more sense in a minute, when we actually do a problem, and I'll show you the way to actually do this.*0286

*Hopefully, you are already good at this; but if not, this will be a good review and a little bit of practice.*0293

*Having said that, let's just jump into a problem, and I think it will make sense.*0299

*OK, so example number 1: How many Cl atoms are there in 18.50 grams of phosphorus pentachloride?*0304

*OK, let's see: so now, let's identify the species: one of the species is chlorine; one of the species is PCl _{5}; and so, I'm going to write that.*0333

*I'm going to write mol PCl _{5}, and I'm going to write mol Cl; those are the two species.*0344

*Now, what is it that I'm interested in?--I'm actually interested in the number of atoms, so I need the number of Cl atoms.*0354

*OK, what am I given?--I'm given grams of PCl _{5}.*0365

*I'm given grams of PCl _{5}; I'm supposed to find the number of chlorine atoms.*0375

*Here is how you do it: you go...this is the path that you follow, and each arrow represents a conversion factor.*0380

*I need to go from grams of PCl _{5} to moles of PCl_{5}.*0390

*From moles of PCl _{5}, I need to go to moles of Cl; this is going to be the mole ratio.*0394

*From moles of Cl, I can go to the number of chlorine atoms.*0400

*Here is the setup (actually, let me draw this a little bit higher, so that I can make some room): grams PCl _{5}...*0405

*This is why we say each species has this little setup here: when you set it up like this, you can actually go ahead and draw the pathway; so here, I'm going from grams of y, to moles of y, to moles of x, to number of particles of x.*0418

*That's it; I'm following a pathway, and I can go anywhere I want, but I'm always passing through moles; there is a mole ratio.*0434

*OK, you always write what you are given: 18.50 grams of PCl _{5}.*0441

*This is one conversion factor, two conversion factors, three conversion factors: I know I'm going to have three conversion factors--I don't even need to fill in the numbers yet.*0451

*I can just automatically do this.*0461

*Now, the thing that you are coming from is going to be the unit on the bottom; the thing that you are going to is going to be the unit on the top.*0463

*So, I'm going to have grams of PCl _{5}, moles of PCl_{5}; well, how many grams in a mole of PCl_{5}?*0471

*Well, the molar mass is 208.22 in one mole of PCl _{5}.*0481

*Now, my second conversion: what is the relationship between moles of PCl _{5} and moles of Cl?*0487

*One mole of PCl _{5} releases 5 moles of Cl; that is just based on the molecular formula--it is PCl_{5}.*0493

*All right, PCl _{5}: one of these produces five of these.*0501

*So, again, the thing that you are going to is on top--mole of Cl; the thing you are coming from is down at the bottom--mole of PCl _{5}.*0507

*Well, the relationship is: one mole of this produces five moles of that; that is the mole ratio; that is this step.*0524

*Now, I'm looking for number of Cl atoms in one mole of Cl.*0532

*Well, one mole of Cl--one mole of anything contains 6.02x10 ^{23} (let me actually write "atoms").*0544

*So again, here, the particle we're talking about is atoms.*0558

*When I do this multiplication, I end up with (OK, I can write it down here) 2.67x10 ^{23} atoms of Cl; that is my final answer.*0561

*Again, it's just based on this relationship; every species has a relationship; from mole, you can go to grams, to particles, to liters.*0578

*Then, among the species, there is a mole ratio--a relationship.*0588

*In this one, it was 1 mole of this produces 5 moles of that.*0592

*That is how you proceed through this network.*0595

*You have something that you are given; you have another thing that you want; you find a path through there.*0598

*If you're given this and you want that, you find a path through there, and it always passes through moles, because the mole is the basic unit of chemistry.*0603

*OK, we'll get to some more stoichiometry in just a minute.*0614

*The next part of the review: I'm going to cover an empirical formula problem.*0618

*Oftentimes, you are given a certain amount of data, and you need to find the empirical formula of a compound.*0623

*The empirical formula is the formula that gives the lowest number ratio of those things.*0630

*So, for example, if I had something like C _{6}H_{12}O_{6}, which is glucose; and let's say I had another sugar, C_{12}H_{24}O_{12}; well, these are two different sugars.*0636

*But notice, the base formula--I can actually divide it by 6 here; I end up with CH _{2}O, and I end up with CH_{2}O, right?*0656

*That is the lowest number ratio of atoms to each other.*0666

*In other words, all sugars have a ratio of one carbon to two hydrogens to one oxygen.*0670

*That is the empirical formula; it's a general umbrella formula that covers all of the molecules of a given compound.*0676

*Each individual compound has its own molecular formula.*0684

*These are molecular formulas; they are specific to the compound.*0687

*But, an empirical formula covers both; that's it.*0690

*So, let's go ahead and do an empirical formula problem.*0695

*Let's see...let's write this as example 2: all right, I have a compound that is 24.5% sodium, and 14.9% silicon, and 60.6% fluorine.*0698

*I want to know what the empirical formula is.*0722

*OK, here is how you handle it: because you are given percentages...percentage is based on 100; so, since they give us the percentages, let's just presume that we have 100 grams of something.*0724

*So, we can convert these percentages: 24.5% of 100 grams is 24.5 grams.*0735

*What I'm going to do is: I'm going to convert these to moles.*0741

*So, sodium: I take 24.5 grams, and I convert to moles: well, one mole of that is about 23 grams, so I get 1.07 moles of sodium.*0744

*Now, I do the same for silicon: silicon--I have 14.9%, which is 14.9 grams if we take a 100-gram sample; we're just doing this so we can make our math easy.*0760

*Well, one mole of silicon is 32.07 grams, so I get 0.46 moles of silicon.*0771

*Then, when we do fluorine, we have 60.6 grams of fluorine; one mole of fluorine is 19 grams, and I end up with 3.2 moles of that.*0782

*Well, now that I have it in moles, I divide by the lowest number of moles; so I divide all of them by 0.46, 0.46, 0.46, and when I do that, I end up with 2.3 here; I end up with 1 here, of course, and here I end up with 7.*0796

*I need whole-number ratios; so 2.3--in order to make this a whole number, I have to multiply by 3, which means I have to multiply everything else by 3.*0822

*This becomes 21; this becomes 3; and this becomes 7.*0832

*So, this is my empirical formula: I have: Na _{7}Si_{3}F_{21}.*0837

*That is my empirical formula.*0851

*It could be Na _{14}Si_{6}F_{42}, or any other multiple, but the basic relationship, the ratio, of sodium to silicon to fluorine is 7:3:21.*0853

*This procedure always allows you to find the empirical formula.*0866

*You change the percentage to gram, gram to mole for all of them; you divide by the lowest number of moles to standardize, so that you have at least one of them with a 1; and then you multiply by an appropriate constant to make sure that all of them give you whole numbers, because we can't leave it as 2.3; you can't have a fraction of an atom.*0872

*That is a standard empirical formula problem.*0891

*OK, so now we're going to take the next step, and I'm going to show you a molecular formula problem.*0894

*Once we find the empirical formula, if we happen to have the molar mass of a particular compound, we can find the actual molecular formula.*0901

*Let's do that one.*0908

*Let's go ahead and make this example 3: Menthol is composed of carbon, hydrogen, and oxygen.*0913

*1.005 grams is combusted to produce 2.829 grams of CO _{2} and 1.159 grams of H_{2}O.*0932

*If menthol is 156 grams per mole, what is its molecular formula?*0969

*Here, we're going to end up having to find the mass of the individual components--the carbon, the hydrogen, the oxygen.*0994

*We're going to use the empirical formula, and then, from there, we're going to use the fact that they give us the molar mass.*1003

*We're going to divide that molar mass, the 156, by the mass of the empirical formula to see how many empirical units actually go into the whole formula.*1010

*That is how we get the molecular formula.*1018

*Let's just jump right on in.*1021

*Notice, this didn't give you the percentages straight on; this is a slightly different problem--slightly more complicated.*1023

*Let's see: 2.89 grams of CO _{2}; let's go ahead and...let me write this equation down: menthol, when you combust it, it's with O_{2}, and it's going to produce CO_{2} + H_{2}O.*1029

*All right, now: 2 CO _{2}...how many grams of carbon...OK, so now we have 2.829 grams of CO_{2}.*1054

*One mole of CO _{2} weighs 44 grams; now we're going to do...so we know that all of the carbon ends up in CO_{2}; all of the hydrogen in menthol ends up in H_{2}O; it's the oxygen that is split between the two.*1077

*We need to find the number of grams of carbon, the number of grams of hydrogen; they give us the total number of grams of menthol that were combusted, so the balance of that is going to be the oxygen.*1092

*Then I can start my problem.*1104

*2.829 grams of CO _{2}) times...again, this is a mole ratio: 1 mole of CO_{2} produces one mole of carbon, right?*1107

*One mole of CO _{2} produces one mole of carbon, times 12 grams of carbon per mole, and I end up with 0.7715 grams of carbon.*1119

*Now, I have 1.159 grams of water times...1 mole of water is 18 grams: 1 mole of H _{2}O produces 2 moles of hydrogen, right?*1135

*Erase that: 2 moles of hydrogen--1 mole of this produces 2 moles of hydrogen.*1159

*Hydrogen is 1 gram per mole, so I get 0.1288 grams of (oops, this isn't carbon; this is) hydrogen.*1165

*Now, if I want to find the mass of oxygen, I take the 1.005 grams of menthol, and I subtract the amount of carbon, 0.7715, plus 0.1288, and I end up with a mass of 0.1047 grams of oxygen.*1182

*Now I have my masses of each, based on a combustion analysis.*1211

*I hope you understand how it is that I came up with that.*1216

*Now, I can start my empirical formula problem.*1218

*I have 0.7715 grams of carbon, times 1 mole over 12 grams; that gives me 0.06429--often, with these empirical formula problems, you want to carry out the decimals as far as you can, to get a good number.*1222

*Then, if I do hydrogen, I have 0.1288 grams, times 1 mole, which is 1 gram, and I get 0.1288--this is in moles.*1242

*Oxygen: I have 0.1047 grams; 1 mole of that is 16 grams of oxygen (again, we'll use the oxygen atom), and this is going to equal 0.0065438 moles.*1258

*Now that we have moles, we need to divide by the smallest, and it looks like the smallest here is this one.*1284

*When we divide by that, when we divide by 0.0065438, and we divide by 0.0065438, we end up with 10 here; we end up with 20 here; and we end up with 1 here.*1289

*So, the empirical formula for menthol is C _{10}H_{20}O_{1}.*1312

*Now, (OK, so here is where we have to...) what we do is: we take the molar mass of the compound, the 156 (let me actually write this out), and we divide it by the molar mass of the empirical formula.*1324

*When we do that, again, the empirical formula is the lowest number ratio, so in this particular case, we had C _{10}H_{20}O_{1}; that means the carbon to hydrogen to oxygen ration is 10:20:1.*1363

*Well, it might be C _{20}H_{40}O_{2}; we don't know.*1380

*So, we divide the mass of the empirical formula into the molar mass of the compound, which we do know.*1385

*This ends up being...they said it was 156 grams per mole, and the molar mass of the empirical happens to also be 156 grams per mole; therefore, the ratio is 1.*1391

*Therefore, menthol is C _{10}H_{20}O_{1}.*1405

*If this number were 3, it would be C _{30}H_{60}O_{3}; that's it.*1411

*This is the actual...not only empirical formula, but it happens to be the molecular formula for menthol.*1416

*This process will always work for molecular formula.*1422

*You run through it; you find the empirical formula; and then you divide the molar mass of the compound by the molar mass of the empirical formula, and that will give you the ratio--the number by which to multiply all of these little subscripts in order to get your final molecular formula.*1425

*Let's do a stoichiometric calculation based on an equation.*1443

*The fermentation of glucose produces ethyl alcohol and carbon dioxide according to the following equation: C _{6}H_{12}O_{6}, and fermentation (so no oxygen involved), goes to 2C_{2}H_{5}OH + 2 CO_{2}.*1450

*The question is, "How many grams of glucose are required to produce 5.2 grams of ethanol, C _{2}H_{5}OH?*1474

*So, how many grams of glucose do I need in order to produce 5.2 grams of C _{2}H_{5}OH?*1502

*OK, so let's take a look: the two species that we are talking about are glucose and C _{2}H_{5}OH; we need to pass through moles.*1508

*They want...we are given grams of C _{2}H_{5}OH that we want; we want to find the number of grams of glucose.*1519

*Here is the process: based on this equation (I'll write it like this), I need grams of glucose.*1526

*Grams of glucose comes from moles of glucose.*1538

*Moles of glucose I can get from moles of ethanol.*1543

*Moles of ethanol comes from grams of ethanol.*1551

*That is the process: I am given grams of this--I am given 5.2 grams of this; I have to go to the moles--from grams to moles, mole to mole (I use mole ratio--that one), and then from mole back to gram.*1559

*This is my path: I go here to here to here to here; that's it--I have 1, 2, 3 arrows; that is three conversion factors.*1574

*Let's go ahead and write this down; we always start with what we're given.*1586

*5.2 grams of EtOH (EtOH is just a shorthand for ethanol); that is 1, that is 2, that is 3; grams of EtOH here; I'm going to moles.*1590

*Moles of EtOH--notice, I'm not going to do the numbers until afterward--I have moles of EtOH here, because, again, they have to cancel, and it's also the thing that I'm coming from.*1605

*I'm going to moles of glucose, and this is moles of glucose; grams of glucose...*1617

*5.2 grams of ethanol; ethanol happens to be 46 grams per mole; the mole ratio--how many moles of ethanol to moles of glucose?*1629

*Moles of ethanol to moles of glucose: 1; that is why you need a balanced equation.*1638

*Glucose is 180 grams per mole; therefore, my final answer is 10.17 grams of glucose.*1645

*10.17 grams of glucose will produce 5.2 grams of ethanol--standard stoichiometry.*1655

*The two species were glucose and CH _{3}OH; there is my 1 mole of glucose; mole of ethanol; these are the relationships: I need grams of glucose coming from grams of ethanol; this is my path.*1664

*If you can't just do it automatically, draw it out; see the solution path, and the path will tell you what the mathematics looks like.*1681

*Let's finish off this lesson with a limiting reactant problem.*1692

*Limiting reactant problems are going to come up very, very often, so basically, it's just...a balanced equation is such that it says, "All of this will react with all of that; nothing is left over."*1697

*But in real life, it doesn't work like that; often, when we run a reaction, one of the reactants is going to be left over; one of them is going to run out.*1709

*The thing that runs out--when it runs out, the reaction stops.*1717

*It's a limiting reactant--it limits the outcome of the reaction.*1720

*It controls how much product you actually get, because when it runs out, again, the reaction stops.*1725

*Let's take a look at this equation here: 2 aluminum hydroxide, plus 3 moles of sulfuric acid, produce 1 mole of aluminum sulfate, plus 6 waters.*1732

*I'm given 0.350 moles of aluminum hydroxide, and I'm given 0.450 moles of sulfuric acid.*1756

*My questions are: a) which is the limiting reactant? part b) how many grams of aluminum sulfate (Al _{2}(SO_{4})_{3}) can be recovered, theoretically?*1769

*That is what we are calculating; we are calculating a theoretical yield, as if everything went right, theoretically.*1805

*And part c) is: How many moles of excess reactant are left over?*1811

*Let's do part a: what is the limiting reactant?*1831

*We need to find out which one of these runs out first.*1834

*You can choose either one to start with: basically, what you're going to do is...here is what it looks like: I'm going to go ahead and choose the aluminum hydroxide.*1837

*0.350 moles of Al(OH) _{3}; the mole ratio between this and this is 2:3.*1847

*Moles of Al(OH) _{3} (this is the problem with chemistry--there is just so much writing as far as these symbols are concerned--the names!) to moles of H_{2}SO_{4}; I need to do a mole ratio calculation.*1863

*It is 2 moles of this to 3 moles of that; this actually tells me that I require 0.525 moles of H _{2}SO_{4}.*1885

*In other words, if I have .350 moles of aluminum hydroxide, this calculation tells me that I actually need .525 moles of H _{2}SO_{4} to completely react with it.*1897

*Well, this is required; the question is: I require .525--do I have .525? I do not (let me do this in red).*1908

*I don't have .525--I only have .50; therefore, H _{2}SO_{4} is limiting.*1921

*Because it's limiting, it controls the rest of the reaction; this is the number that I use to find out how much of this I'm going to get; not this.*1932

*When this runs out, the reaction stops; there is going to be excess of this.*1940

*Therefore, this is what controls how much I have.*1944

*Let's go ahead and do part b, which is, "How many grams of aluminum sulfate can be recovered?"*1947

*All right, well, since this is my limiting reactant, that is the one that I take: 0.450 moles of H _{2}SO_{4} times...well, what is the mole ratio here?--it's 3:1; 3 moles of H_{2}SO_{4}...this is not going to work...*1954

*3 moles of H _{2}SO_{4}...oh, these crazy lines; OK, let me try this one more time.*1987

*Let's start from the top: 1 mole of aluminum sulfate comes from 3 moles of H _{2}SO_{4}, and then we have 342.7 grams of aluminum sulfate per mole.*2002

*There we go; this cancels that; that cancels this; my final answer is going to be in grams, and it's going to be =51.32 grams of Al _{2}(SO_{4})_{3} (it's tedious writing all of those symbols, isn't it?).*2035

*There you go; no, I'm not going to have...I do not want those lines to show up; therefore, I'm going to write this one more time.*2061

*=51.32 grams of Al _{2}(SO_{4})_{3}; there we go!*2073

*Based on .450 moles of H _{2}SO_{4}, the most aluminum sulfate I can recover is 51.32 grams.*2088

*The reason is: because that is what controls the reaction.*2096

*OK, now: this final problems says, "How many moles of excess reactant will be left over?"*2101

*This is our limiting, so this is our excess: we need to find out how much is used, and then subtract it from the amount that we started with, and that will give us the amount left over.*2107

*The amount used is, again, going to be based on the limiting reactant; so, part C: 0.450 moles of H _{2}SO_{4}...*2117

*Well, I have that 3 moles of H _{2}SO_{4} requires 2 moles of aluminum hydroxide.*2132

*I end up with 0.30 moles of aluminum hydroxide used.*2145

*0.350 moles-0.30 moles used (let me actually write these out: moles of Al(OH) _{3} minus 0.03 mol used up) leaves me with 0.05 moles of aluminum hydroxide to end up with.*2156

*You see what we did: once you know the limiting reactant, that controls the rest of the reaction.*2191

*.450 moles of H _{2}SO_{4} uses up .3 moles of aluminum hydroxide.*2196

*I started with .350; I subtract, and I'm left with .05 moles of aluminum hydroxide; that is my excess reactant.*2201

*OK, so in this lesson, we have done some basic stoichiometry coverage.*2212

*We have talked about empirical formula; we have talked about molecular formula; and we finished off with a limiting reactant problem.*2215

*That, and a previous lesson with naming, should provide sort of a good, general foundation for the rest.*2222

*Starting with the next lesson, we're going to jump straight into some good, solid AP chemistry.*2230

*Until then, thank you for joining us here at Educator.com; we'll see you next time; goodbye.*2234

1 answer

Last reply by: Professor Hovasapian

Fri Apr 7, 2017 6:28 PM

Post by Xiao Liu on April 7 at 05:18:37 PM

Should we be concerned with significant digits on the AP test?

1 answer

Last reply by: Professor Hovasapian

Fri Jan 13, 2017 7:30 PM

Post by William Cole on December 17, 2016

Hi Professor!

Around 14:40-15:00 in the video you say we multiply by 3 to find the correct empirical formula, as it gives us a whole number. In my intro to chem class last year I learned it is okay to just round the number to the nearest whole. So the 2.3 you multiplied by 3 to 7, could have been rounded down to just 2. Do you know if this is accepted in the AP test and also generally correct?

Thank you in advance!

0 answers

Post by Areez Khaki on December 8, 2016

Hi Professor, i have a question about question b in the limiting reactants section, how did u come up with 342.7 grams of aluminium sulfate

1 answer

Last reply by: Professor Hovasapian

Thu May 7, 2015 4:12 AM

Post by R Abdullah on May 7, 2015

Hello Professor Hovasapian

Before I ask my question, I'd like to thank you for providing us with these amazing lessons. They have helped me so much in my academic career and I believe I've even developed a passion for mathematics from watching your lessons on Linear Algebra and Multivariable Calculus.

I took the AP Chemistry exam on May 5th and on the free-response section, there was a simple question asking for the percent yield of a reaction. I solved it with the actual yield over theoretical yield being in grams. However, most of my peers solved the percent yield through mols of actual yield over mols of percent yield. I'm not sure which is correct and which one AP considers correct.

Once again, thank you for all that you have done.

Rasheed Abdullah

1 answer

Last reply by: Professor Hovasapian

Thu Jan 8, 2015 1:48 AM

Post by Lyngage Tan on January 5, 2015

hi professor in example 2 i believe the molar mass of silicon is 28.09 g/mol. and going through the solution i ended up with this ratios Na= 1.06568731/0.530437878 = 2 Si= 0.530437878/0.530437878= 1 and F 3.189473684/0.530437878 = 6. so the empirical formula is Na2SiF6.

1 answer

Last reply by: Professor Hovasapian

Fri Nov 7, 2014 9:15 PM

Post by Noorhan A. on November 6, 2014

Hello Prof. Hovasapian

Thank you so much for your detailed videos which explain some difficult concepts very thoroughly. Your videos helped me go from a C to an A+ on my Chemistry tests! Currently, I am taking Honors Chemistry, which is a weighted course and is very rigorous but does not cover some concepts from AP Chemistry. My school does not offer AP Chemistry but I would still like to take the exam in May. Do you have any suggestions for how I can self-study? Also, do sig figs matter in the AP Chem test? My teacher takes points off our tests if we do not use the correct number of sig figs.

0 answers

Post by Milan Ray on September 29, 2014

haha, pen really was acting up!

1 answer

Last reply by: Professor Hovasapian

Sun Sep 28, 2014 11:23 PM

Post by Quazi Niloy on September 28, 2014

what happens wlhen the empirical formula does not equal to 100%? Is this just a matter of error or is there a way to go about doing a problem like it?

1 answer

Last reply by: Professor Hovasapian

Tue Aug 5, 2014 6:04 PM

Post by Jason Kim on August 5, 2014

So I am assuming that at 14:38 you rounded 2.3*3 which is 6.9 to just 7 professor?

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Last reply by: Professor Hovasapian

Tue Jun 10, 2014 8:55 PM

Post by Alice Rochette on June 10, 2014

for example 2 you said 1 mol of Si = 32.07g, how did you find that? because in my periodic table it says that the atomic weight for silicon is 28.08g... anyway thank you!!

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Fri Mar 14, 2014 7:50 PM

Post by sadia sarwar on February 26, 2014

so how did you get the last answer for example 4?? it was great tutorial though, thanks!!

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Thu Dec 26, 2013 3:14 PM

Post by Nada Al Bedwawi on December 25, 2013

I didn't quite understand how in example 3 while converting from mol of CO^2 to mole of C.I thought that could only be done if by molar ratio(as in a chemical reaction)

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Last reply by: Professor Hovasapian

Tue Aug 27, 2013 9:15 PM

Post by Marian Iskandar on August 27, 2013

Excellent tutorial! Regardless of the tiny errors made, the solutions were laid out so clearly...left no room for ambiguity. Thank you again, Professor! :)

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Thu Aug 22, 2013 5:04 PM

Post by Venugopal Ghanta on August 22, 2013

you look and sound a lot like albert einstein

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Tue Jul 2, 2013 6:00 PM

Post by Sarawut Chaiyadech on July 2, 2013

You are Very good teacher thank you so much Professor

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Fri Jun 21, 2013 6:15 PM

Post by Angela Patrick on June 21, 2013

Is the answer to example four in significant figures?

I got the same answer but thought that there were only two significant figures and got the answer 10.

If the answer 10.17 is in significant figures can you explain why there are four significant figures?

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Last reply by: Professor Hovasapian

Sat Dec 1, 2012 5:47 PM

Post by Andrew Stewart on December 1, 2012

I do not understand how to obtain the final answer of 2.67 x 10^23.?

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Sat Oct 27, 2012 4:25 PM

Post by Max Mayo on October 26, 2012

Thanks for the help. Has anyone ever told you that you look a lot like Albert Einstein?

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Last reply by: sadia sarwar

Wed Feb 19, 2014 2:50 AM

Post by Revanth Guptha on September 8, 2012

so what is the correct answer for Example 2?

thanks for the great tutorial!! :D :)

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Last reply by: Professor Hovasapian

Mon Sep 3, 2012 6:26 PM

Post by Anuradha Kumar on September 3, 2012

Didn't he use the atomic mass of sulfur instead of silicon? Great tutorial nonetheless! :)

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Sun Aug 19, 2012 8:59 PM

Post by Arshin jain on August 17, 2012

It's not necessary to take general chemistry before AP chemistry, as long as you understand everything, right?

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Sun Aug 12, 2012 3:47 PM

Post by Suresh Sundarraj on August 12, 2012

that was a great lecture, thanks a lot!

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Last reply by: Professor Hovasapian

Sun Sep 1, 2013 10:47 PM

Post by noha nasser on August 5, 2012

and i think that there is something wrong in the final answer of example 4 too... tried to calculate it many times and i still couldn't get it :)

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Post by mateusz marciniak on May 5, 2012

i agree that was the atomic mass of sulfur not silicon, great tutorial nonetheless

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Last reply by: Professor Hovasapian

Mon Jul 16, 2012 12:42 AM

Post by chenyu liu on April 25, 2012

i think you made a mistake in14:04 by using sulfur's atomic number for silicon.

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Post by Trace Shapiro on March 18, 2012

Visually defining (mapping) the solution path through the stoichiometric problems is absolutely genius! I can't believe I have never been taught this before! Thank you so much!!!