Chapter 3 Mass Relationships in Chemical Reactions
3.1 Atomic mass
3.2 Avogadro's Number and the Molar Mass of an Element
the slight variation in neutrons accounts for the average atomic mass of 12.01 on the periodic table
3.3 Molecular Mass
3.4 The Mass Spectrometer
3.5 Percent Composition of Elements
3.6 Experimental Determination of Empirical Formulas
3.7 Chemical Reactions and Chemical Equations
3.8 Amounts of Reactants and Products
3.9 Limiting Reagents
3.10 reaction Yield
3.1 Atomic Mass
Because of the extremely small size of atoms finding their exact mass is impossible, but it is feasible to find the mass of an atom relative to another. The atomic mass is the mass of an atom in atomic mass units. Scientists have established a standard unit of measure (or the atomic mass unit) when finding atomic mass, which is equal to 1/12 the mass of one carbon 12 atom. For example, it is found that one hydrogen atom is 8.4 percent as large as a carbon 12 atom, so the atomic mass is .084 x 12 amu = 1.008 amu, which is the mass found on the periodic table! One may notice that on the periodic table the atomic mass of carbon is 12.01 instead of 12, but there is a reason for this slight variation. Since most naturally occurring elements have more than one isotope, when we find the atomic mass of an element we generally settle for the average mass of the isotopes. When finding the average, the original calculation of the common form of carbon (98.9 %) is multiplied by 12, and then the isotope of carbon 13 (1.1 %) is multiplied by 13, or the atomic mass. The resulting calculation is .989 x 12 + .011 x 13 to give approximately 12.01.
3.2 Avogadro's Number and the Molar mass of an Element
Lets face it, no scale can be created to find the masses of groups of atoms in atomic mass units. A solution to this problem was created in the form of another scientific unit, or the mole, abbreviated mol. One mole is equivalent to the amount of a substance that contains as many atoms, molecules or various other particles as there are in 12 grams of the carbon-12 isotope. From this amount the molar mass can be determined, namely the mass in kilograms of one mole of units. The actual number of atoms in one carbon-12 isotope is extremely important. It is known as Avogadro's number, or about 6.022 x 10^23, named after Italian scientist Amadeus Avogadro. This is the number of atoms in one mole of a substance.
-Note that molar mass = atomic mass in amu For example, the molar mass of sodium is 22.99 g = atomic mass, 22.99 amu
Amadeus Avogadro, born August 9, 1776 in Italy
Another example is Titanium, molar mass = 47.88 g = atomic mass, 47.88 amu
Important conversion factors!
12 grams of carbon-12 = 1 mol carbon-12 atom (equality)
12 g carbon-12 atoms / 1 mol carbon-12 atoms
1 mol carbon-12 atoms / 6.022 x 10^23 carbon-12 atoms
Avogadro's number can thus be used to convert relationships between a) the mass and the moles of atoms b) the number of atoms and mass.
(1 mol x / molar mass of x) and (1 mol x / 6.022 x 10^23 x atoms). (x denotes an element)
3.3 Molecular Mass
If the masses of component atoms in a molecule are known, we can find the molecular mass of the molecule. The molecular mass is the sum of the atomic masses in the molecule! To do this, we multiply the atomic mass of each element by the respective numbers of each atom.
For example,
the molecular mass of CO2= 44 amu because
the mass of C = 12 amu because there is only one carbon atom
and the mass of O2 is 32 amu because there are two oxygen atoms
Thus, 12 + 32 = 44 amu
With the molecular mass, the molar mass of a compound can be found and the molecular mass of a compound in amu is equal to the molar mass in grams. Using the equations in this section we can find the quantity of moles or atoms in a given compound. 3.4 The Mass Spectrometer
Of course scientists wouldn't settle for so few methods of finding molecular and atomic masses, so they devised a new method called mass spectrometry, using a device
Mass spectrometer in action, showing the atomic masses of carbon dioxide
appropriately named the spectrometer.
"Okay-who put my lunch through the mass spectrometer..?"
This tool is used to measure properties of light over a section of the electromagnetic spectrum. A gaseous stream is hit with high energy electrons. Positive ions are created when these electrons touch each the gaseous atoms or molecules. The ions are accelerated by oppositely charged plates, which are curved into circular paths by a magnet. If the charge of each ion is e and the mass is m, the radius of the path of the ions is dependant on e/m (or the charge-to-mass ratio). The smaller the e/m ratio, the wider the curve which separates ions with the same e and distinguishes them by mass. Then the ions reach a detector which shows the current type for each which will be proportional to the number of ions. This allows scientists to discover previously unknown isotopes.
-The first mass spectrometer was created by F. W. Aston in the 1920's.
-The device provides concrete evidence of isotopes such as neon-20. Recently, there has been a great interest in a new virtual lab that is being developed by Carnegie Mellon University and the University of Pittsburgh. This virtual mass spectrometry lab will be a great way for students to study this topic in depth through the internet. ---> VMSL 3.5 Percent Composition of Compounds Lesson Walkthrough
As if there aren't enough scientific calculations in this lesson, this section touches upon another. This time we want to find out the percent composition by mass of a compound. Percent composition by mass is the actual percent by mass of every element in a compound. To find % by mass, divide the mass of each element in one mole of compound by the molar mass of the compound and then multiply by 100. For example, in CO2the molar mass is 44 (12 from carbon and 32 from oxygen). 12 / 44 = 27.3 % carbon, and 2 x 16 / 44 = 72.7 % oxygen, the 16 is multiplied by two because there are two moles of oxygen, the same applies for any element and amount of moles for each compound when calculating percent composition by mass. This process can also be reversed given the percent composition by mass, the emperical formula can be found. Scientists can thus find the quality of a substance and what mass of each element is present.
Percent Composition by Mass Ibuprofen (C13H18O2):
Ibuprofen
75.69% C, 8.80% H, 15.51% O
As you can see, the percents accurately reflect the chemical formula and picture shown of an ibuprofen molecule.
An example problem: More examples
find the percent composition of H2SO4
2x1.008 g H / 98 g H2SO4 x 100 = 2.06 % hydrogen
32 g S / 98 g H2SO4x 100 = 32.7 % sulfur
16 x 4 g O / 98 g H2SO4x 100 = 65.3 % oxygen
3.6 Experimental Determination of Molecular Formulas
Once we obtain the empirical formula of a compound given the percent composition allows us to identify compounds experimentally. The basic procedure behind this is to put a sample of a compound into an apparatus which is heated. Absorbers in connecting tubes will collect hydrogen, oxygen and carbon molecules which will reveal what elements are in the compound.
In calculating the actual empirical formula mathematically, each respective mass of every element in a compound must be converted into moles. Each number of moles are set as the subscript for each respective element (see example below to see how this works). Then, check for the lowest number of moles in the resulting equation, and divide that by each number in the equation, which will produce reduced numbers for each coefficient. These reduced numbers will generally be very close to a whole number and can thus be rounded to make many of the basic chemical equations commonly seen in chemistry, such as H2O.
-Example: Find the empirical formula of methane if 6.0 grams of methane can be decomposed into 4.5 grams of carbon and 1.5 grams of hydrogen.
Any formulas calculated from percent by mass is always an empirical formula because it is reduced to have the lowest coefficients possible. To calculate the molecular formula of a compound, we must know both the empirical formula and the approximate molar mass. First, calculate the mass of the empirical formula. Then, divide the molar mass of the compound by the mass of the empirical formula. The resulting number is a ratio of the molecular to the empirical formula. Then multiply this number by the coefficients of each element in the compound to get the molecular formula.
-This is an example of how to calculate the molecular formula given three masses of carbon, hydrogen, and oxygen.
Lactic acid
3.7 Chemical Reactions and Chemical Equations
A chemical reaction is a process in which one or more substances is changed into one or more new substances. These reactions range from lighting a match to mixing dangerous chemicals in laboratories you may often see on TV; they are all over the place and very important in the world of chemistry. Math has always a problem for many people, but when it comes to chemistry equations are just as important as mathematical ones! You may be wondering, what is a chemical equation?? Well, a chemical equation is a written representation of a chemical reaction. This was devised by scientists to show how equations work, what will form and what it takes for the result to be formed. They consist of symbols both mathematical and chemical.
Two examples of chemical equations are as follows:
2HgO(s) --> 2Hg(l) + O2(g)
2H2(g) + O2(g) --> 2H2O(l)
For these two reactions, many of the basic ideas involving chemical equations can be explained. Each equation has a product(s) and reactant(s) side. Each side is on either side of an arrow (-->) which represents a "yields" sign. Whichever side the arrow points to is the products side. The first sample equation with HgO is a decomposition reaction in which a compound is broken down to form multiple products. The second reaction (that forms water) is a synthesis reaction in which two or more reactants forms one or more products. A "+" sign denotes that reactants/products are being added or form together, respectively. These equations follow the law of conservation of mass, and there is thus the same amount of products as there are reactants in any given equation. Usually equations must be balanced to ensure the mass law is in effect. The subscripts represent the number of molecules present for each element or compound and the coefficients show how many moles are present. These coefficients are added after the equation is written to balance it. Note the letters l, g, and s in parenthesis for each sample reaction. These letters represent what phase state (solid liquid or gas) each reactant or product is. If there is an "aq" in the parenthesis it stands for "aqueous," or that it's dissolved in water.
Balancing equations:
In order to balance a chemical equation, it must be written out first to include what reactants form what products. For example,
an unbalanced chemical equation
the reaction between sodium and chlorine generates commonly used table salt. The unbalanced equation is: Na + Cl2 --> NaCl. Since there are two molecules of chlorine on the reactants side, there must be two molecules on the right. Placing a two in front of NaCl will allow the chlorine on each side to be in equilibrium, but now there is one more mole of sodium on the products side than the reactants side. The next step would be to add a two coefficient in front of the sodium on the reactants side so the element is equal on either side. The balanced equation is thus 2Na + Cl2 --> 2NaCl. In many situations, trial and error is the best way to balance an equation, placing coefficients in front of each item until each element is the same on both sides. Make sure to note that when a two is put in front of a compound like NaCl, it applies to each element it consists of. This basic example shows how to balance even the most complex equations.
3.8 Amounts of Reactants and Products
Now that chemical equations have been explained, this section delves into how much reactant/produce is actually present in a reaction. To do this, we must use concepts like the mole, as explained in section 3.2, and other quantitative terms/ concepts such as molar mass. The quantitative study of reactants and products in a chemical reaction is known as stoichiometry. The units for reactants can be liters, grams, ect...but to find the amount of product formed moles are used (regardless of the unit given). Using moles like this is known as the mole method, in which the coefficients of reactants/products denote the number of moles on either side of a reaction. There are many variations of questions that can be asked in regards to calculating amounts in reactions, but this sample can show the basic bread and butter of solving stoichiometric problems. Example: Let's say 4.8 moles of CO react with oxygen to form carbon dioxide. Since we want to find the moles of CO2 produced, we convert using CO in the denominator as follows:
moles of carbon dioxide produced= 4.8 mol CO x (1 mol CO2 /1 mol CO)
This information can also be manipulated and converted to show how many grams this is, simply by multiplying by the molar mass of carbon dioxide.
Try this one: If 10.7 grams of CO are the only given information in the same reaction, how many grams of carbon dioxide will be formed? The steps to solve this are to convert: grams of CO, to moles of CO, to moles of CO2, to grams of CO2.
Hint! When converting for each step, make sure the units of the first number are in the denominator of the second, so the units are converted to will be correct. Answer: 16 grams of carbon dioxide.
There are three main types of calculations involving the mole method in stoichiometry that can all be completed using general conversions:
1. moles of reactant --> moles of product
2. mass of reactant --> moles of reactant --> moles of product
3. mass of reactant --> moles of reactant -->moles of product --> mass of product 3.9-3.10 Limiting Reagentsand Reaction Yield
During the course of a chemical reaction, the amounts of reactants used are generally not in proportion with a balanced chemical equation. For example a scientist may use one mole more of a substance than the equation calls for. This is not a problem, because of limiting and excess reagents. When one reactant is totally used up and can't react any further, a reaction has reached completion. The reactant that limits how far the reaction can go is thus the limiting reagent. And who could have guessed excess reagents are the opposite? Excess reagents are the reactants in a reaction that are in abundance in proportion to the halting nature of the limiting reagents. Click here for an animated explanation of limiting reagents! An interesting fact is that chemists often use the more expensive chemical as the limiting reagent so that all of it will be consumed in their test experiments.
The amount of the limiting reagent of a reaction allows a theoretical yield to be calculated. This is the amount of product that would form if the entire limiting reagent is used up. The actual yield is appropriately the amount of product that actually forms when the reaction is carried out. So why not use just the actual yield? Since some reactions don't go through fully many times and situational factors may affect the yield, the theoretical yield is thus a good gauge on what the products should form. A calculation has been devised to check how accurate the actual yield is in relation to the theoretical. What is known as percent yield is this very computation. Temperature and pressure (see chapter 5 gas laws) can both affect the yield.
percent yield = (actual yield/ theoretical yield) x 100
LAB
The lab about to be explained was performed in order to show how to find a limiting reagent in a chemical reaction. It entails a reaction between potassium iodide and lead nitrate, and is a simple procedure. The pictures below depict the reaction in this lab, note the vibrant color the solution turns! This lab ties together both sections 3.9 and 3.10. Goggles, gloves, and an apron are important! Chemicals in this lab are toxic!
These are the materials necessary:
-250 mL beaker
-distilled water
-second beaker to collect filtrate
-8 grams of potassium iodide
-hot plate/ oven
-1 gram of lead nitrate
-funnel
-filter paper
-stir rod
To perform this lab, first measure out the necessary chemicals and record their masses. Add approximately 75 mL of the distilled water to the 250 mL beaker. Add the potassium iodide to the beaker and stir until it has all dissolved. Then add the lead nitrate and stir again until the reaction is fully complete in a few minutes. The beaker will instantly turn a vibrant orange/yellow color, it looks just like orange juice...so it's recommended not to do this lab in the morning. Then record the mass of the filter paper (for use later). Insert the filter paper into the funnel (fold if necessary so the filter will not leak any desired compound) and filter the solution. The object is now to get the mass of the product left in the filter. To do this, either heat the product in the filter paper in an oven or let it sit and dry, for the water must be evaporated in order to get the accurate mass. When it is dry, find the mass and subtract that of the filter paper, to find the mass of the precipitate alone. Questions:
-What is a limiting reagent, and what is the limiting reagent in this lab (also see section 3.9)?
A limiting reagent, which is commonly known as a limiting reactant, determines how much the chemicals will react and how long the reaction can go. Since there is a limited amount of the reagent, it will eventually be used up and the other reactant can't react with it any further. In this particular lab, the lead nitrate is the limiting reagent because the potassium iodide is much more abundant in the solution, and the lead nitrate will only react with a certain amount of that KI before it is used up.
-How can you determine the amount of the non-limiting reagent (potassium iodide) leftover from the reaction?
If the solution remaining after the reaction is heated, the water will evaporate and the potassium iodide will remain, then take the mass of this.
-How is it possible to have a percent yield lower than 100 percent?
In my particular trial of this lab, I added a little distilled water to the original beaker after filtering the solution to make sure the KI stuck to the sides was all filtered, but too much water was added, causing some of the filtrate to slip through the filter paper. This indicates that the filter paper wasn't secure enough in the funnel (maybe a piece of tape can solve this for future trials) and also that there was some potassium iodide left over in the actual filtrate, which can throw off results. This trial ended up with a yield of approximately two grams of lead (II) iodide. Chapter Over! Here are some review questions:
1. What is the molar mass of NH4F?
Answer: masses of each element are added together, N=14 grams, H=4 grams, and F= 19 grams.
14 + 4 + 19 = 38 grams.
2. Calculate the mass in grams of a single carbon atom.
Answer: series of simple conversions must be completed to solve this question. 12.01 grams are in one mole of carbon. One mole of carbon has 6.022 x 10^23 atoms (Avogadro's number, see section 3.2). Then set up a simple ratio to solve algebraically.
Mass = 12.01 x 6.022 x 10 ^ -23 grams
The mass of one carbon atom is 1.994 x 10^ -23 g .
3. Balance the following equation: Al + O2 --> Al2O3
Answer: 4Al + 3O2 --> 2Al2O3
4. Lets say that in a reaction 32 grams of carbon dioxide are formed when the theoretical yield in a reaction is actually 44 grams. What is the percent yield for this reaction?
Answer: (32 grams/ 44 grams) x 100 = approximately 73 percent yield. Acknowledgments the following pictures used in this wikispace are non original and all the property of the respective owners, linked below. Text included is resourced from the last source, the text book.
3.1 Atomic mass
3.2 Avogadro's Number and the Molar Mass of an Element
3.3 Molecular Mass
3.4 The Mass Spectrometer
3.5 Percent Composition of Elements
3.6 Experimental Determination of Empirical Formulas
3.7 Chemical Reactions and Chemical Equations
3.8 Amounts of Reactants and Products
3.9 Limiting Reagents
3.10 reaction Yield
3.1 Atomic Mass
Because of the extremely small size of atoms finding their exact mass is impossible, but it is feasible to find the mass of an atom relative to another. The atomic mass is the mass of an atom in atomic mass units. Scientists have established a standard unit of measure (or the atomic mass unit) when finding atomic mass, which is equal to 1/12 the mass of one carbon 12 atom. For example, it is found that one hydrogen atom is 8.4 percent as large as a carbon 12 atom, so the atomic mass is .084 x 12 amu = 1.008 amu, which is the mass found on the periodic table! One may notice that on the periodic table the atomic mass of carbon is 12.01 instead of 12, but there is a reason for this slight variation. Since most naturally occurring elements have more than one isotope, when we find the atomic mass of an element we generally settle for the average mass of the isotopes. When finding the average, the original calculation of the common form of carbon (98.9 %) is multiplied by 12, and then the isotope of carbon 13 (1.1 %) is multiplied by 13, or the atomic mass. The resulting calculation is .989 x 12 + .011 x 13 to give approximately 12.01.
3.2 Avogadro's Number and the Molar mass of an Element
Lets face it, no scale can be created to find the masses of groups of atoms in atomic mass units. A solution to this problem was created in the form of another scientific unit, or the mole, abbreviated mol. One mole is equivalent to the amount of a substance that contains as many atoms, molecules or various other particles as there are in 12 grams of the carbon-12 isotope. From this amount the molar mass can be determined, namely the mass in kilograms of one mole of units. The actual number of atoms in one carbon-12 isotope is extremely important. It is known as Avogadro's number, or about 6.022 x 10^23, named after Italian scientist Amadeus Avogadro. This is the number of atoms in one mole of a substance.
-Note that molar mass = atomic mass in amu For example, the molar mass of sodium is 22.99 g = atomic mass, 22.99 amu
Another example is Titanium, molar mass = 47.88 g = atomic mass, 47.88 amu
Important conversion factors!
12 grams of carbon-12 = 1 mol carbon-12 atom (equality)
12 g carbon-12 atoms / 1 mol carbon-12 atoms
1 mol carbon-12 atoms / 6.022 x 10^23 carbon-12 atoms
Avogadro's number can thus be used to convert relationships between a) the mass and the moles of atoms b) the number of atoms and mass.
(1 mol x / molar mass of x) and (1 mol x / 6.022 x 10^23 x atoms). (x denotes an element)
3.3 Molecular Mass
If the masses of component atoms in a molecule are known, we can find the molecular mass of the molecule. The molecular mass is the sum of the atomic masses in the molecule! To do this, we multiply the atomic mass of each element by the respective numbers of each atom.
For example,
the molecular mass of CO2= 44 amu because
the mass of C = 12 amu because there is only one carbon atom
and the mass of O2 is 32 amu because there are two oxygen atoms
Thus, 12 + 32 = 44 amu
With the molecular mass, the molar mass of a compound can be found and the molecular mass of a compound in amu is equal to the molar mass in grams. Using the equations in this section we can find the quantity of moles or atoms in a given compound.
3.4 The Mass Spectrometer
Of course scientists wouldn't settle for so few methods of finding molecular and atomic masses, so they devised a new method called mass spectrometry, using a device
-The first mass spectrometer was created by F. W. Aston in the 1920's.
-The device provides concrete evidence of isotopes such as neon-20. Recently, there has been a great interest in a new virtual lab that is being developed by Carnegie Mellon University and the University of Pittsburgh. This virtual mass spectrometry lab will be a great way for students to study this topic in depth through the internet. ---> VMSL
3.5 Percent Composition of Compounds Lesson Walkthrough
As if there aren't enough scientific calculations in this lesson, this section touches upon another. This time we want to find out the percent composition by mass of a compound. Percent composition by mass is the actual percent by mass of every element in a compound. To find % by mass, divide the mass of each element in one mole of compound by the molar mass of the compound and then multiply by 100. For example, in CO2the molar mass is 44 (12 from carbon and 32 from oxygen). 12 / 44 = 27.3 % carbon, and 2 x 16 / 44 = 72.7 % oxygen, the 16 is multiplied by two because there are two moles of oxygen, the same applies for any element and amount of moles for each compound when calculating percent composition by mass. This process can also be reversed given the percent composition by mass, the emperical formula can be found. Scientists can thus find the quality of a substance and what mass of each element is present.
Percent Composition by Mass Ibuprofen (C13H18O2):
75.69% C, 8.80% H, 15.51% O
As you can see, the percents accurately reflect the chemical formula and picture shown of an ibuprofen molecule.
An example problem: More examples
find the percent composition of H2SO4
2x1.008 g H / 98 g H2SO4 x 100 = 2.06 % hydrogen
32 g S / 98 g H2SO4x 100 = 32.7 % sulfur
16 x 4 g O / 98 g H2SO4x 100 = 65.3 % oxygen
3.6 Experimental Determination of Molecular Formulas
Once we obtain the empirical formula of a compound given the percent composition allows us to identify compounds experimentally. The basic procedure behind this is to put a sample of a compound into an apparatus which is heated. Absorbers in connecting tubes will collect hydrogen, oxygen and carbon molecules which will reveal what elements are in the compound.
In calculating the actual empirical formula mathematically, each respective mass of every element in a compound must be converted into moles. Each number of moles are set as the subscript for each respective element (see example below to see how this works). Then, check for the lowest number of moles in the resulting equation, and divide that by each number in the equation, which will produce reduced numbers for each coefficient. These reduced numbers will generally be very close to a whole number and can thus be rounded to make many of the basic chemical equations commonly seen in chemistry, such as H2O.
-Example: Find the empirical formula of methane if 6.0 grams of methane can be decomposed into 4.5 grams of carbon and 1.5 grams of hydrogen.
12.0 g
1.0 g
C=
0.375/.375=
1.0=
1
Here's another sample problem, empirical formula problem.
Any formulas calculated from percent by mass is always an empirical formula because it is reduced to have the lowest coefficients possible. To calculate the molecular formula of a compound, we must know both the empirical formula and the approximate molar mass. First, calculate the mass of the empirical formula. Then, divide the molar mass of the compound by the mass of the empirical formula. The resulting number is a ratio of the molecular to the empirical formula. Then multiply this number by the coefficients of each element in the compound to get the molecular formula.
-This is an example of how to calculate the molecular formula given three masses of carbon, hydrogen, and oxygen.
3.7 Chemical Reactions and Chemical Equations
A chemical reaction is a process in which one or more substances is changed into one or more new substances. These reactions range from lighting a match to mixing dangerous chemicals in laboratories you may often see on TV; they are all over the place and very important in the world of chemistry. Math has always a problem for many people, but when it comes to chemistry equations are just as important as mathematical ones! You may be wondering, what is a chemical equation?? Well, a chemical equation is a written representation of a chemical reaction. This was devised by scientists to show how equations work, what will form and what it takes for the result to be formed. They consist of symbols both mathematical and chemical.
Two examples of chemical equations are as follows:
2HgO(s) --> 2Hg(l) + O2(g)
2H2(g) + O2(g) --> 2H2O(l)
For these two reactions, many of the basic ideas involving chemical equations can be explained. Each equation has a product(s) and reactant(s) side. Each side is on either side of an arrow (-->) which represents a "yields" sign. Whichever side the arrow points to is the products side. The first sample equation with HgO is a decomposition reaction in which a compound is broken down to form multiple products. The second reaction (that forms water) is a synthesis reaction in which two or more reactants forms one or more products. A "+" sign denotes that reactants/products are being added or form together, respectively. These equations follow the law of conservation of mass, and there is thus the same amount of products as there are reactants in any given equation. Usually equations must be balanced to ensure the mass law is in effect. The subscripts represent the number of molecules present for each element or compound and the coefficients show how many moles are present. These coefficients are added after the equation is written to balance it. Note the letters l, g, and s in parenthesis for each sample reaction. These letters represent what phase state (solid liquid or gas) each reactant or product is. If there is an "aq" in the parenthesis it stands for "aqueous," or that it's dissolved in water.
Balancing equations:
In order to balance a chemical equation, it must be written out first to include what reactants form what products. For example,
3.8 Amounts of Reactants and Products
Now that chemical equations have been explained, this section delves into how much reactant/produce is actually present in a reaction. To do this, we must use concepts like the mole, as explained in section 3.2, and other quantitative terms/ concepts such as molar mass. The quantitative study of reactants and products in a chemical reaction is known as stoichiometry. The units for reactants can be liters, grams, ect...but to find the amount of product formed moles are used (regardless of the unit given). Using moles like this is known as the mole method, in which the coefficients of reactants/products denote the number of moles on either side of a reaction. There are many variations of questions that can be asked in regards to calculating amounts in reactions, but this sample can show the basic bread and butter of solving stoichiometric problems.
Example: Let's say 4.8 moles of CO react with oxygen to form carbon dioxide. Since we want to find the moles of CO2 produced, we convert using CO in the denominator as follows:
moles of carbon dioxide produced= 4.8 mol CO x (1 mol CO2 /1 mol CO)
This information can also be manipulated and converted to show how many grams this is, simply by multiplying by the molar mass of carbon dioxide.
Try this one: If 10.7 grams of CO are the only given information in the same reaction, how many grams of carbon dioxide will be formed? The steps to solve this are to convert: grams of CO, to moles of CO, to moles of CO2, to grams of CO2.
Hint! When converting for each step, make sure the units of the first number are in the denominator of the second, so the units are converted to will be correct.
Answer: 16 grams of carbon dioxide.
There are three main types of calculations involving the mole method in stoichiometry that can all be completed using general conversions:
1. moles of reactant --> moles of product
2. mass of reactant --> moles of reactant --> moles of product
3. mass of reactant --> moles of reactant -->moles of product --> mass of product
3.9-3.10 Limiting Reagents and Reaction Yield
During the course of a chemical reaction, the amounts of reactants used are generally not in proportion with a balanced chemical equation. For example a scientist may use one mole more of a substance than the equation calls for. This is not a problem, because of limiting and excess reagents. When one reactant is totally used up and can't react any further, a reaction has reached completion. The reactant that limits how far the reaction can go is thus the limiting reagent. And who could have guessed excess reagents are the opposite? Excess reagents are the reactants in a reaction that are in abundance in proportion to the halting nature of the limiting reagents. Click here for an animated explanation of limiting reagents! An interesting fact is that chemists often use the more expensive chemical as the limiting reagent so that all of it will be consumed in their test experiments.
The amount of the limiting reagent of a reaction allows a theoretical yield to be calculated. This is the amount of product that would form if the entire limiting reagent is used up. The actual yield is appropriately the amount of product that actually forms when the reaction is carried out. So why not use just the actual yield? Since some reactions don't go through fully many times and situational factors may affect the yield, the theoretical yield is thus a good gauge on what the products should form. A calculation has been devised to check how accurate the actual yield is in relation to the theoretical. What is known as percent yield is this very computation. Temperature and pressure (see chapter 5 gas laws) can both affect the yield.
percent yield = (actual yield/ theoretical yield) x 100
LAB
The lab about to be explained was performed in order to show how to find a limiting reagent in a chemical reaction. It entails a reaction between potassium iodide and lead nitrate, and is a simple procedure.
These are the materials necessary:
-250 mL beaker
-distilled water
-second beaker to collect filtrate
-8 grams of potassium iodide
-hot plate/ oven
-1 gram of lead nitrate
-funnel
-filter paper
-stir rod
To perform this lab, first measure out the necessary chemicals and record their masses. Add approximately 75 mL of the distilled water to the 250 mL beaker. Add the potassium iodide to the beaker and stir until it has all dissolved. Then add the lead nitrate and stir again until the reaction is fully complete in a few minutes. The beaker will instantly turn a vibrant orange/yellow color, it looks just like orange juice...so it's recommended not to do this lab in the morning. Then record the mass of the filter paper (for use later). Insert the filter paper into the funnel (fold if necessary so the filter will not leak any desired compound) and filter the solution. The object is now to get the mass of the product left in the filter. To do this, either heat the product in the filter paper in an oven or let it sit and dry, for the water must be evaporated in order to get the accurate mass. When it is dry, find the mass and subtract that of the filter paper, to find the mass of the precipitate alone.
Questions:
-What is a limiting reagent, and what is the limiting reagent in this lab (also see section 3.9)?
A limiting reagent, which is commonly known as a limiting reactant, determines how much the chemicals will react and how long the reaction can go. Since there is a limited amount of the reagent, it will eventually be used up and the other reactant can't react with it any further. In this particular lab, the lead nitrate is the limiting reagent because the potassium iodide is much more abundant in the solution, and the lead nitrate will only react with a certain amount of that KI before it is used up.
-How can you determine the amount of the non-limiting reagent (potassium iodide) leftover from the reaction?
If the solution remaining after the reaction is heated, the water will evaporate and the potassium iodide will remain, then take the mass of this.
-How is it possible to have a percent yield lower than 100 percent?
In my particular trial of this lab, I added a little distilled water to the original beaker after filtering the solution to make sure the KI stuck to the sides was all filtered, but too much water was added, causing some of the filtrate to slip through the filter paper. This indicates that the filter paper wasn't secure enough in the funnel (maybe a piece of tape can solve this for future trials) and also that there was some potassium iodide left over in the actual filtrate, which can throw off results. This trial ended up with a yield of approximately two grams of lead (II) iodide.
Chapter Over! Here are some review questions:
1. What is the molar mass of NH4F?
Answer: masses of each element are added together, N=14 grams, H=4 grams, and F= 19 grams.
14 + 4 + 19 = 38 grams.
2. Calculate the mass in grams of a single carbon atom.
Answer: series of simple conversions must be completed to solve this question. 12.01 grams are in one mole of carbon. One mole of carbon has 6.022 x 10^23 atoms (Avogadro's number, see section 3.2). Then set up a simple ratio to solve algebraically.
Mass = 12.01 x 6.022 x 10 ^ -23 grams
The mass of one carbon atom is 1.994 x 10^ -23 g .
3. Balance the following equation: Al + O2 --> Al2O3
Answer: 4Al + 3O2 --> 2Al2O3
4. Lets say that in a reaction 32 grams of carbon dioxide are formed when the theoretical yield in a reaction is actually 44 grams. What is the percent yield for this reaction?
Answer: (32 grams/ 44 grams) x 100 = approximately 73 percent yield.
Acknowledgments the following pictures used in this wikispace are non original and all the property of the respective owners, linked below. Text included is resourced from the last source, the text book.
http://portermason.com/johnny/tag/chemistry/
http://www.cartoonstock.com/directory/F/FOOD.asp
http://www.emc.maricopa.edu/faculty/farabee/BIOBK/BioBookCHEM1.html
http://www.windows.ucar.edu/physical_science/chemistry/co2_molecule_big_gif_image.html
http://www.calzim.com/online/online3_1/class_material/unit1/unit1.htm
http://serc.carleton.edu/research_education/geochemsheets/techniques/gassourcemassspec.html
https://reich-chemistry.wikispaces.com/file/view/double_displacement.jpg
Chang, Chemistry, Eighth Edition