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As a young boy, I remember staring in wonder at a pot of boiling water. Searching for an explanation for the bubbles that formed, I believed for a time that the motion of the hot water drew air down into the pot, which then bubbled back to the surface. Little did I know that what was happening was even more magical than I imagined - the bubbles were not air, but actually water in the form of a gas.
The different states of matter have long confused people. The ancient Greeks were the first to identify three classes (what we now call states) of matter based on their observations of water. But these same Greeks, in particular the philosopher Thales (624 - 545 BCE), incorrectly suggested that since water could exist as a solid, liquid, or even a gas under natural conditions, it must be the single principal element in the universe from which all other substances are made. We now know that water is not the fundamental substance of the universe; in fact, it is not even an element.
To understand the different states in which matter can exist, we need to understand something called the Kinetic Molecular Theory of Matter. Kinetic Molecular Theory has many parts, but we will introduce just a few here. One of the basic concepts of the theory states that atoms and molecules possess an energy of motion that we perceive as temperature. In other words, atoms and molecules are constantly moving, and we measure the energy of these movements as the temperature of the substance. The more energy a substance has, the more molecular movement there will be, and the higher the perceived temperature will be. An important point that follows this is that the amount of energy that atoms and molecules have (and thus the amount of movement) influences their interaction with each other. Unlike simple billiard balls, many atoms and molecules are attracted to each other as a result of various intermolecular forces such as hydrogen bonds, van der Waals forces, and others. Atoms and molecules that have relatively small amounts of energy (and movement) will interact strongly with each other, while those that have relatively high energy will interact only slightly, if even at all, with others.
How does this produce different states of matter? Atoms that have low energy interact strongly and tend to “lock” in place with respect to other atoms. Thus, collectively, these atoms form a hard substance, what we call a solid. Atoms that possess high energy will move past each other freely, flying about a room, and forming what we call a gas. As it turns out, there are several known states of matter; a few of them are detailed below.
Solids are formed when the attractive forces between individual molecules are greater than the energy causing them to move apart. Individual molecules are locked in position near each other, and cannot move past one another. The atoms or molecules of solids remain in motion. However, that motion is limited to vibrational energy; individual molecules stay fixed in place and vibrate next to each other. As the temperature of a solid is increased, the amount of vibration increases, but the solid retains its shape and volume because the molecules are locked in place relative to each other. To view an example of this, click on the animation below which shows the molecular structure of ice crystals.
Liquids are formed when the energy (usually in the form of heat) of a system is increased and the rigid structure of the solid state is broken down. In liquids, molecules can move past one another and bump into other molecules; however, they remain relatively close to each other like solids. Often in liquids, intermolecular forces (such as the hydrogen bonds shown in the animation below) pull molecules together and are quickly broken. As the temperature of a liquid is increased, the amount of movement of individual molecules increases. As a result, liquids can “flow” to take the shape of their container but they cannot be easily compressed because the molecules are already close together. Thus liquids have an undefined shape, but a defined volume. In the example animation below we see that liquid water is made up of molecules that can freely move past one another, yet remain relatively close in distance to each other.
Gases are formed when the energy in the system exceeds all of the attractive forces between molecules. Thus gas molecules have little interaction with each other beyond occasionally bumping into one another. In the gas state, molecules move quickly and are free to move in any direction, spreading out long distances. As the temperature of a gas increases, the amount of movement of individual molecules increases. Gases expand to fill their containers and have low density. Because individual molecules are widely separated and can move around easily in the gas state, gases can be compressed easily and they have an undefined shape.
Solids, liquids, and gases are the most common states of matter that exist on our planet. If you would like to compare the three states to one another, click on the comparison animation below. Note the differences in molecular motion of water molecules in these three states.
Plasmas are hot, ionized gases. Plasmas are formed under conditions of extremely high energy, so high, in fact, that molecules are ripped apart and only free atoms exist. More astounding, plasmas have so much energy that the outer electrons are actually ripped off of individual atoms, thus forming a gas of highly energetic, charged ions. Because the atoms in plasma exist as charged ions, plasmas behave differently than gases, thus representing a fourth state of matter. Plasmas can be commonly seen simply by looking upward; the high energy conditions that exist in stars such as our sun force individual atoms into the plasma state.
As we have seen, increasing energy leads to more molecular motion. Conversely, decreasing energy results in less molecular motion. As a result, one prediction of Kinetic Molecular Theory is that if we continue to decrease the energy (measured as temperature) of a substance, we will reach a point at which all molecular motion stops. The temperature at which molecular motion stops is called absolute zero and has been calculated to be -273.15 degrees Celsius. While scientists have cooled substances to temperatures close to absolute zero, they have never actually reached absolute zero. The difficulty with observing a substance at absolute zero is that to “see” the substance, light is needed, and light itself transfers energy to the substance, thus raising the temperature. Despite these challenges, scientists have recently observed a fifth state of matter that only exists at temperatures very close to absolute zero. Bose-Einstein Condensates represent a fifth state of matter only seen for the first time in 1995. The state is named after Satyendra Nath Bose and Albert Einstein who predicted its existence in the 1920’s. B-E condensates are gaseous superfluids cooled to temperatures very near absolute zero. In this weird state, all the atoms of the condensate attain the same quantum-mechanical state and can flow past one another without friction. Even more strangely, B-E condensates can actually “trap” light, releasing it when the state breaks down.
Several other less common states of matter have also either been described or actually seen. Some of these states include liquid crystals, fermionic condensates, superfluids, supersolids and the aptly named strange matter. To read more about these phases, visit the Phase page of Wikipedia, linked to below in the Further Exploration section.
Phase transitions
The transformation of one state of matter into another state is called a phase transition. The more common phase transitions even have names; for example, the terms melting and freezing describe phase transitions between the solid and liquid state, and the terms evaporation and condensation describe transitions between the liquid and gas state. Phase transitions occur at very precise points, when the energy (measured as temperature) of a substance in a given state exceeds that allowed in the state. For example, liquid water can exist at a range of temperatures. Cold drinking water may be around 4ºC. Hot shower water has more energy and thus may be around 40ºC. However, at 100°C under normal conditions, water will begin to undergo a phase transition into the gas phase. At this point, energy introduced into the liquid will not go into increasing the temperature; it will be used to send molecules of water into the gas state. Thus, no matter how high the flame is on the stove, a pot of boiling water will remain at 100ºC until all of the water has undergone transition to the gas phase. The excess energy introduced by a high flame will accelerate the liquid-to-gas transition; it will not change the temperature. The heat curve below illustrates the corresponding changes in energy (shown in calories) and temperature of water as it undergoes a phase transition between the liquid and gas states.
As can be seen in the graph above, as we move from left to right, the temperature of liquid water increases as energy (heat) is introduced. At 100ºC, water begins to undergo a phase transition and the temperature remains constant even as energy is added (the flat part of the graph). The energy that is introduced during this period goes toward breaking intermolecular forces so that individual water molecules can “escape” into the gas state. Finally, once the transition is complete, if further energy is added to the system, the heat of the gaseous water, or steam, will increase.
This same process can be seen in reverse if we simply look at the graph above starting on the right side and moving left. As steam is cooled, the movement of gaseous water molecules and thus temperature will decrease. When the gas reaches 100ºC, more energy will be lost from the system as the attractive forces between molecules reform; however the temperature remains constant during the transition (the flat part of the graph). Finally, when condensation is complete, the temperature of the liquid will begin to fall as energy is withdrawn.
Phase transitions are an important part of the world around us. For example, the energy withdrawn when perspiration evaporates from the surface of your skin allows your body to correctly regulate its temperature during hot days. Phase transitions play an important part in geology, influencing mineral formation and possibly even earthquakes. And who can ignore the phase transition that occurs at about -3ºC, when cream, perhaps with a few strawberries or chocolate chunks, begins to form solid ice cream.
Now we understand what is happening in a pot of boiling water. The energy (heat) introduced at the bottom of the pot causes a localized phase transition of liquid water to the gaseous state. Because gases are less dense than liquids, these localized phase transitions form pockets (or bubbles) of gas, which rise to the surface of the pot and burst. But nature is often more magical than our imaginations. Despite all that we know about the states of matter and phase transitions, we still cannot predict where the individual bubbles will form in a pot of boiling water.
Stoich is short for stoichiometry. Don't be scared off by that name; I remembered in ninth grade when I heard that I thought I wasn't going to get through the year. It's a 13-letter name for something that's very easy, yet important part to becoming successful in chemistry. This is where the stuff becomes important to learn, because it's directly on the test.
The Mole and Molar Calculations
I am surprised to hear that this is the point where some chemistry students lose all hope; trying to figure out what a mole is. I admit, I was confused too, but keeping this in mind made life a lot easier: a mole is simply the measure of how many. A similar unit is a dozen, and since everyone knows what this is, I'll use this first.
Let's say you have a dozen eggs. What does this mean? It just means you have twelve of something. It doesn't say how much it weighs, or how much space it occupies, just how much. A dozen siege tanks doesn't have the same mass nor volume to a dozen specks of dust. There's just twelve of each.
Moles are the same concept. One mole of something has 6.022 x 1023 objects in it. It's a lot, but atoms are small, so it's a good unit in chem. 6.022 x 1023 is NA, called Avogadro's number. It's a pretty big number. One mole of seconds is 4 million times as long as the age of the earth. One mole of meters is over 600 times the diameter of the Milky Way. For our purposes, it's pretty good.
It's important to understand the relation of amu's and moles. One mole of something that weighs one amu is one gram. In other words, 1 g = 6.022 x 1023 amus. So one mole of hydrogen (H2) would weigh 1.0079 x 2 = 2.0158 grams. One mole of carbon is exactly 12 grams; that's how they derived Avogadro's number.
Often it's important to find out how many moles is some given quantity of mass. You can use dimensional analysis (DA) for this. Let's say you want to know how many moles is 36 grams of water, and also how many molecules are in it. First you must find the molar mass of water (how many amus it weighs). Water is H2O, that means 2 of hydrogen and one of oxygen. So the adding of those masses is:
1.008 x 2 + 16.00 = 18.00 amu.
Now comes the DA part.
So it's 2 moles. And since 1 mole is NA molecules, there are 2 x NA = 1.2 x 1024 molecules.
The key to solving these problems is getting the g/mol (molar mass, or amus) of the substance, and then using that in a DA setup to convert back and forth between grams and moles. One amu is one g/mol.
Example:
How many moles are in 4.5 grams of glucose (C6H12O6)?AnswerFirst you must find out the molar mass of glucose. Adding the weights of the individual atoms, you get:
6 x 12.00 amu + 12 x 1.008 amu + 6 + 16.00 amu = 180.1 amus = 180.1 g/mol
Then you use DA:
The answer is .025 moles. Be careful to use DA correctly, and that you will have to flip it so that the units in the top and bottom cancel out.
Example:
You have a sample of glucose (C6H12O6), that weighs 2.5 grams. How many carbon molecules are in it?AnswerFind the number of moles of glucose (we already found molar mass of glucose in last problem, 180.1.)
In each glucose molecule there are six carbons. So if you have .014 moles of glucose, you must have 6 times as many carbons in it. .014 moles x 6 = .084 moles C. Just multiply it by Avogadro's number to get the answer.
.084 moles x 6.022 x 1023=5.1 x 1022 carbon molecules.
Percent Mass
This is just another way to show how much of something is in a compound. It's how much of the substance is made of one particular element, the percent by mass (I bet you could figure that out by the title). So if there is 60 g of something, and in that compound there is 12 g of carbon, then you would say the percent mass is 20% (12 g / 60 g x 100%).
Note that this number can't be directly obtained from the moles. Percent moles is not the percent mass too. You must convert or die.
Example:
Find the mass percent of each element of penicillin (C14H20N2SO4).Here's a hint: if they don't give you how much there is, assume one mole.
So if there is one mole of penicillin, there must be 14 moles of carbon, 20 moles of hydrogen, and so on.
To find the mass of each element in this mole of penicillin, just do the DA using the g/mol conversion.
Mass of carbon = 14 moles C x 12.01 g / mol = 168.4 g.
Mass of hydrogen = 20 moles H x 1.008 g / mol = 20.16 g.
Mass of nitrogen = 2 moles N x 14.01 g / mol = 28.02 g.
Mass of sulfur = 1 mole S x 32.07 g / mol = 32.07 g.
Mass of oxygen = 4 moles O x 16.00 g / mol = 64.00 g.
The total mass can be found just by adding:
168.4g + 20.16g + 28.02g + 32.07g + 64.00g = 312.6 g
Percent mass is mass of one thing divided by the total mass, times hundred percent:
Percent mass carbon = 168.4 g / 312.6 g x 100% = 53.87%
Percent mass hydrogen = 20.16 g / 312.6 g x 100% = 6.449%
Percent mass nitrogen = 28.02 g / 312.6 g x 100% = 8.964%
Percent mass sulfur = 32.07 g / 312.6 g x 100% = 10.24%
Percent mass oxygen = 64.00 g / 312.6 g x 100% = 20.47%
And the percents add up to 99.993%, which is close enough to 100%.
Empirical and Molecular Formulas
The molecular formula is the type that we've always been using: H2O, C6H12O6, C2H4, and all the like. It's how many atoms of each element are present in one molecule of the substance. The empirical formula is the reduced form of this. It's the ratio of atoms. For H2O, there's 2 hydrogens for every oxygen. So the empirical formula is the same. However in C6H12O6, there is a ratio of 6 : 12 : 6. That can be reduced, so the ratio is 1 : 2 : 1. Therefore the empirical formula is CH2O. Likewise, the ratio 2 : 4 in the last one can be reduced to 1 : 2. The empirical for it is is CH2.
Example:
The empirical formula of a compound is found to be C2H3O4. The molar mass was found to be 273.1 g / mol. Find the molecular formula.Answer
Let's find the molecular mass of the empirical formula.
12.01 g/mol x 2 + 1.008 g/mol x 3 + 16.00 g/mol x 4 = 91.04 g / mol.
And since the molecular formula is just a multiple of the empirical, let's see if 273.1 g / mol is a multiple of 91.04 g / mol.
273.1 g/mol / 91.04 g/mol = 3.000.
Sure enough, the number goes divides almost evenly three times. That means the molecular must have three times the number of particles of the empirical. Multiplying each by three gives the molecular formula to be C6H9O12.
It's important to remember that the molecular formula is a multiple of the empirical formula in both subscripts and molar masses.
Chemical Equations and Balancing Them
_________________________________________________________________________________
A chemical reaction is when one or more molecules rearrange their atoms to form new molecules. Here is an example:
CH4 + O2 ---> CO2 + H2O
You can see this visually below:
+
->
Methane
Oxygen
+
Carbon Dioxide
Water
The things on the left side are called the reactants, and the things on the right are called the products. The reactants are generally transformed into the products until one of the reactants runs out, or the system reaches equilibrium (a balance).
As another example, if you wanted to say that two molecules combine to form one, just use the coefficients. Just like in algebra, you can put numbers in front of the molecule to show how many of them are in the equation. Let's say 2 of Molecule A and 3 of Molecule B are needed to make 4 of Molecule C. The equation would look like:
2A + 3B ---> 4C
Another thing that's important to right down is what state each reactant and product is, by little subscript letters. The symbols are:
Solid: (s)
Liquid: (l) (When you write it by hand, it's a lowercase cursive l.)
Gas: (g)
Aqueous (means dissolved in water): (aq)
So the first equation should have really been written as:
CH4 (g) + O2 (g) ---> CO2 (g) + H2O (g)
It's funny how they teach you to do that, but on the AP test on the Equation Writing section, they will not look for those symbols. But your teacher most certainly will, so get into the habit!
That equation is better, but it still is missing a key point in chemical reactions. Atoms cannot be created nor destroyed; they can only be put in different orders. If you look above, there are two oxygens (green) on the left and three oxygens on the right. Where did that extra oxygen come from? The key is to realize that these aren't one to one reactions. That is, they all don't have coefficients of one. So you have to balance.
Let's start with hydrogen. There's four on the left and two on the right, so multiply the molecule on the right by two:
CH4 (g) + O2 (g) ---> CO2 (g) + 2H2O (g)
Now the hydrogens are balanced, but the oxygens aren't. Now we have 4 on right and 2 on left. Multiply atom with oxygen on left by two:
CH4 (g) + 2O2 (g) ---> CO2 (g) + 2H2O (g)
If you count them, everything adds up. There are the same number of atoms on the left as on the right. That is the correct chemical equation.
Here is The Art of Balancing Instruction Manual, by none other than Takalah:
Before doing anything, check if it all adds up.If the same number of atoms exist on both sides, then it's already balanced and you're done.If not, start your balancing with the atom that is combined with another on both sides, and the one that has the most atoms. You really do want to start with the most complex one. For instance, in C2H5OH + O2 CO2 + H2O, you would start with H first, because it's in a molecule with other atoms on both sides, and the numbers of atoms (5 and 2) are greater than the other one that fits the first criteria (C).
Leave the atoms that stand by themselves for last. Like in the equation in the last rule, you would save O2 for last. Since changing the coefficient for this only affects one atom, you can use it as a "finishing touch", balancing the oxygens after all the other ones are done, but without messing another atom up. Don't change the molecular formula!!!As frustrating as some of these problems can be at first, don't be tempted to change one little number in some molecular formula just to make something balance. Remember that changing the coefficient will just change the number of particles that are involved in the reaction. Changing the subscripts will change the whole molecule into something else. The molecule is a unit. So don't do it!
Example:
Write balanced equations for the following:
Glucose (C6H12O6) reacts with oxygen gas to form carbon dioxide and water.
Indium reacts with oxygen to make indium (III) oxide.
The unbalanced reaction is: C6H12O6 (s) + O2 (g) ---> CO2 (g) + H2O (g)The most complicated atom is H, since it's in glucose and water, and that the subscripts are the largest. Let's see; there are 12 on the left and 2 on the right, so let's try multiplying right by 6:C6H12O6 (s) + O2 (g) ---> CO2 (g) + 6H2O (g)
Now you move on to the next complicated atom; C, since it's in both carbon dioxide and glucose. There are six on left, and one on right; try multiplying right one by 6. You get:
C6H12O6 (s) + O2 (g) ---> 6CO2 (g) + 6H2O (g)
Everything is basically done now, you got have to finish it by balancing oxygens using the O2 molecule. There's 6 on left; and 6x2+6 = 18 on right. (Remember that you have to multiply coefficient by subscript to get number of atoms.) So there must be 12 atoms on right to balance it. Twelve oxygen atoms is 6 oxygen molecules, so the final answer is:
C6H12O6 (s) + 6O2 (g) ---> 6CO2 (g) + 6H2O (g)
You have to remember the stuff you learned in the last part to do this. Indium (III) oxide is In2O3. So the equation is:In (s) + O2 (g) ---> In2O3 (g)Start with the most complicated one, which is oxygen. Multiplying left by three and right by two gives
In (s) + 3O2 (g) ---> 2In2O3 (g)
Now to balance In's, there are four total In's on right. So there must be 4 In's in left. Answer is:
4In (s) + 3O2 (g) ---> 2In2O3 (g)
AgNO3 (aq) + H2SO4 (aq) ---> Ag2SO4 (s) + HNO3 (aq)This is the hardest one yet. Oxygen is too hard to do right now, since it's in all four. Let's start with Ag, by getting two on left:2AgNO3 (aq) + H2SO4 (aq) ---> Ag2SO4 (s) + HNO3 (aq)
That changes the N to 2N. Looking at the right, we have only one N. Let's multiply that one by 2 too.
2AgNO3 (aq) + H2SO4 (aq) ---> Ag2SO4 (s) + 2HNO3 (aq)
Now we get 2 H's on the right, but that's favorable, since we have 2 H's on left. In fact, if you find out how many oxygen's are on each side, you gots ten, and that's your answer. Note that solving these can be like a chain reaction; solving for one gets you how many of another, then you balance that, and it goes on.
Stoichiometric Calculations
One thing to realize when doing any calculations is that moles and coefficients are interchangeable. Both mean number of particles, or multiples thereof. So in 2A + 3B ---> 4C, it either means 2 particles of A combine with 3 particles of B to make 4 particles of C, or 2 moles reacts with 3 moles to make 4 moles.
With that in mind, here is a simple problem.
Example:
The formation of water from hydrogen and oxygen gas is: 2H2 (g) + O2 (g) ---> 2H2O (l). What mass of water will form from 12.0 grams of hydrogen and excess oxygen (assuming the reaction goes to completion?)AnswerFirst what you must do in any of these problems is get all given masses into moles. You are given 12.0 grams of hydrogen, let's see how many moles that is:
12.0 g H2 x 1 mol / 2.02 g = 5.94 mol H2
So how much water is formed? According to the equation, for every 2 moles of hydrogen, 2 moles of water are produced, or in other words, a 1 to 1 ratio. So 5.94 moles of water will be formed. The question asks for what mass, so we're not quite done yet.
5.94 mol H2O x 18.0 g / mol = 107. grams of H2O.
The "oxygen in excess" is important. We're assuming that there's tons of oxygen, enough to supply whatever the hydrogen needs. If there was only .0000000000000000000001 moles of oxygen, 5.94 moles of water obviously can't form.
Let's try one more of this type before moving on to LRs.
Example:
The newly discovered element Takalahium (symbol Tak; molecular mass = 411 g/mol) combines with oxygen to form Takalahium Oxide. The unbalanced equation is:
Tak + O2 ---> Tak2O3
How many grams of Tak Oxide are formed when burning 8.00 kilograms of Tak? Answer
First and foremost, the balanced equation is needed. That would be:
4Tak + 3O2 ---> 2Tak2O3
Then convert all given masses to moles:
8.00 kg Tak = 8000 g x 1 mol / 411 g = 19.5 moles.
Since there are 2 Tak's for every 1 Tak Oxide, there must be half as many moles of Tak Oxide, or 9.50 moles. You can also use DA to do the same mole ratios.
Before you can get grams, you must first find the molar mass of Tak Oxide, which is no problem:
Mass = 2 x 411 g + 3 x 16.0 g = 870 g/mol.
Then you find the mass:
9.50 moles x 870 g / 1 mol = 8260 grams = 8.26 kilograms.
In these last two problems, we always assumed that one of the reactants was in excess. This simplified things, but it's often not the case in many reactions. You have given amounts of each reactant. So how do you do this? Simple, just find out which out runs out first molewise. The one that does is called the limiting reactant, or just LR. One quick example:
Example:
Tak + Asdf ---> TakAsdf
You are given 45.0 grams of Takalahium (411 g/mol) and 500. grams of Asdfur (5620.5 g/mol). Find the limiting reactant in the formation of Takalahium Asdfite. Answer
The limiting reactant isn't necessarily the one with the smallest mass. Remember that reactions look at number of particles, not at mass. So let's get both of these values into moles:
45.0 g Tak x 1 mol / 411 g = .109 moles.
500.g Asdf x 1 mol / 5620.5 g = .0890 moles.
As both are being consumed, you'll see that Asdf is used up first. Therefore it's the limiting reactant, even though there is over ten times as much of it by mass.
Note that mole ratios are important. Let's say 2 Tak's become 1 Asdf. Then it becomes a little bit more than just finding the bigger number, but not very harder. You start with one, say, Tak. Assume that it is the LR. Then, according to the mole ratio, you would need at least half as much Asdf for Tak to become the LR. Do you have a half? Well, one-half of .109 moles is about .05 something, and you do have that much. So if Asdf isn't your LR, then Tak must be. (Of course if it was exactly equal to half, then you could use either one, there would be no LR, and everything gets used up.)
What happens if you started with looking at Asdf? Well you need twice as much Asdf's. Twice .0890 is .17 something. You don't have that much Tak, therefore Tak is your LR, no matter how you look at it.
Once you find the LR, you can go ahead and finish the problem just like the others. So what's the difference? Once you find the LR, you use that amount and throw away the other amount. So like in the above example, you use the .0890 value, and forget about the .109. Of course, if they ask you to refer to the original amount, such as how much reactant is left over, then you need to use it. But not for stoich.
Example:
In the previous example, how much Takalahium Asdfide is produced by the reaction? And how much of each element is left over?AnswerSince the Asdf was the limiting reactant, with .0890 moles, you use this value for all remaining stoich problems. .109 moles is not how much Tak reacts, it will be less than that. Since it's a one-to-one ratio, .0890 moles of Tak will react with .0890 moles of Asdf to form .0890 moles of TakAsdf. The question asks what mass was produced. That's easy:
Molar mass = 411 g/mol + 5620.5 g/mol = 6031.5 g/mol,
.0890 mol TakAsdf x 6031.5 g / mol = 537 grams of TakAsdf.
As to the second prompt, you know that .109 moles Tak was initially present, and .0890 moles or Tak was consumed in reaction. A simple subtraction and conversion, and there's your answer:
Moles left over = Moles initially there - Moles used up = .109 mol - .0890 mol = .020 mol.
.020 mol Tak x 411 g / mol = 8.22 g
As to the Asdf, all of it was used up, so there's 0 g left.
Percent Yield
One little thing. You can test the above stuff in a lab, like measuring out the reactants, and see if you get the calculated amount of product. You probably won't get the exact number, but rather a number that's below it. There could be many reasons for it, such as maybe the reaction doesn't go to completion, or maybe there's impurities or something. The theoretical yield is how much you should get, according to stoich. The actual yield is how much you actually get, determined in a lab. The percent yield is the ratio of these two. The equation:
Yields are masswise. That is, how much you get is measured in kilograms, grams, mass.
Note that the percent yield will always be under 100%. The theoretical yield is the max you can get. (You're not going to get more substance than calculations predict.)
Now it's time for the big example, that ties together balancing equations, stoich, LRs, and percent yield.
Example:
Here's the combustion of ethane (not balanced):
C2H6 + O2 ---> CO2 + H2O
32.0 g of ethane was burned with 15.0 grams of oxygen gas, and 10.8 grams of carbon dioxide was formed. Calculate the percent yield of carbon dioxide. Answer
Before doing anything, get the equation balanced!
2C2H6 + 7O2 ---> 4CO2 + 6H2O
(If you didn't know how to do that, review the rules of balancing. I do have a 20 megabyte limit here!)
Now to convert all given masses into moles:
32.0 g ethane x 1 mol / 30.0 g = 1.07 mol
15.0 g oxygen x 1 mol / 32.0 g = .469 mol
Which one is the LR? Let's look at ethane first. If ethane is indeed the LR, then the moles oxygen reacted would be:
1.07 mol ethane x 7 mol oxygen / 2 mol ethane = 3.74 moles oxygen needed.
Since the oxygen will have run out first, oxygen is the limiting reactant. Disregard the 1.07 mol.
Now to find the theoretical yield. 4 moles of carbon dioxide will form from 7 moles of oxygen:
.469 mol oxygen x 4 mol CO2 / 7 mol oxygen = .268 moles.
Now to get the mass:
.268 moles x 44.0 g / mol = 11.8 grams.
That's how much you should get. But experiment shows that you get only 10.8 grams. Finding theoretical yield:
10.8 grams / 11.8 grams x 100 % = 91.5%
