USING ENGLISH TO CALCULATE
Stoichiometry
Let's start with how to say this word. Five syllables: STOY-KEE-AHM-EH-TREE. It's a big word that describes a simple idea. Stoichiometry is the part of chemistry that studies amounts of substances that are involved in reactions. You might be looking at the amounts of substances before the reaction. You might be looking at the amount of material that is produced by the reaction. Stoichiometry is all about the numbers.
All reactions are dependent on how much stuff you have. Stoichiometry helps you figure out how much of a compound you will need, or maybe how much you started with. We want to take the time to explain that reactions depend on the compounds involved and how much of each compound is needed.
What do you measure? It could be anything. When you're doing problems in stoichiometry, you might look at...
- Mass of Reactants (chemicals before the reaction)
- Mass of Products (chemicals after the reaction)
- Chemical Equations
- Molecular Weights of Reactants and Products
- Formulas of Various Compounds
Now, an example. Let's start with something simple like sodium chloride (NaCl). You start with two ions and wind up with an ionic/electrovalent compound. When you look at the equation, you see that it takes one sodium ion (Na+) to combine with one chlorine ion (Cl-) to make the salt. When you use stoichiometry, you can determine amounts of substances needed to fulfill the requirements of the reaction. Stoichiometry will tell you that, if you have ten million atoms of sodium and only one atom of chlorine, you can only make one molecule of sodium chloride. Nothing you can do will change that. It's like this:
Let's bump it up a level. When you mix hydrogen gas (H2) and oxygen gas (O2), nothing much happens. When you add a spark to the mixture, all of the molecules combine and eventually form water (H2O). You would write it like this:
What does stoichiometry look at here? First, look at the equation. Four hydrogen atoms and two oxygen atoms are on each side of the equation. It's an important idea to see that you need twice as many hydrogen atoms as you do oxygen atoms. The number of atoms in the equation will help you figure out how much of each substance you will need to make the reaction happen. If you make this an extreme example and fill a sealed container with one million hydrogen molecules and only one oxygen molecule, the spark won't make an explosion. There is no monster reaction to be created when there is only one oxygen molecule around. You will make two water molecules and be done.
What do you measure? It could be anything. When you're doing problems in stoichiometry, you might look at...- Mass of Reactants (chemicals before the reaction)
- Mass of Products (chemicals after the reaction)
- Chemical Equations
- Molecular Weights of Reactants and Products
- Formulas of Various Compounds
Now, an example. Let's start with something simple like sodium chloride (NaCl). You start with two ions and wind up with an ionic/electrovalent compound. When you look at the equation, you see that it takes one sodium ion (Na+) to combine with one chlorine ion (Cl-) to make the salt. When you use stoichiometry, you can determine amounts of substances needed to fulfill the requirements of the reaction. Stoichiometry will tell you that, if you have ten million atoms of sodium and only one atom of chlorine, you can only make one molecule of sodium chloride. Nothing you can do will change that. It's like this:
10,000,000 Na + 1 Cl --> NaCl + 9,999,999 Na
Let's bump it up a level. When you mix hydrogen gas (H2) and oxygen gas (O2), nothing much happens. When you add a spark to the mixture, all of the molecules combine and eventually form water (H2O). You would write it like this:
2H2 + O2 --> 2H2O
What does stoichiometry look at here? First, look at the equation. Four hydrogen atoms and two oxygen atoms are on each side of the equation. It's an important idea to see that you need twice as many hydrogen atoms as you do oxygen atoms. The number of atoms in the equation will help you figure out how much of each substance you will need to make the reaction happen. If you make this an extreme example and fill a sealed container with one million hydrogen molecules and only one oxygen molecule, the spark won't make an explosion. There is no monster reaction to be created when there is only one oxygen molecule around. You will make two water molecules and be done.
Are your students likely to recoil at the very mention of stoichiometry problems - those sticky exercises in which students are asked to find the weight of a product given the weight of a reactant in a chemical equation? In an effort to lessen student anxiety over such problems, we in the chemistry department at Columbia-Greene Community College developed a model called The Bridge Method of Stoichiometry. The method can be applied to any problem involving the relationships between constituents in a chemical reaction, and even students with little prior knowledge of chemistry can use the model successfully.
Bridging the Stoichiometry Gap
The Bridge Method uses a map model consisting of an imaginery bridge and several roads leading to and away from the bridge (see Figure 1). Students use the map as a guide to the steps they must follow in order to solve a stoichiometry problem. On the left side of the bridge is Known Country, the territory of the chemical for which information is given in the problem. On the right side of the bridge is Unknown Country, the land of the chemical about which information is sought. 
The sample problem illustrates how the bridge is used.
How many grams of nitrogen must react to form 51 grams of ammonia by the reaction:
N2 + 3 H2 --> 2 NH3 ?
The first step is to identify on the map both the starting point and the destination. The starting point would be 51 grams of ammonia on the left side of the bridge, and the destination ( or answer to the problem) would be grams of nitrogen on the right. First, starting with grams of ammonia, follow Formula Weight Road, to get to moles of ammonia. Second, cross over the Coefficients Bridge to reach moles of nitrogen. Third, turn on Formula Weight Road and proceed to grams of nitrogen.
The road names are clues to the calculations involved in the problem. To convert 51 grams of ammonia to moles of ammonia along Formula Weight Road, use the formula weight of ammonia, 17 grams per mole. Dimensional analysis (factor label method) shows that the 51 grams of ammoia is divided by 17 grams per mole:
51 g of NH3 X 1 mole of NH3
17 g of NH3
The manipulation carried this far would give an answer in moles of ammonia. Next, cross Coefficients Bridge. As the name suggests, the coefficients in the balanced equation is used in this step. The balanced equation shows that 2 moles of NH3 are chemically equivalent to1 mole of N2. In order for the units to cancel correctly, the conversion factor should be 1 mole of N2 / 2 moles of NH3. So far, this is the setup:
51 g of NH3 X 1 mole of NH3 X 1 mole of N2
17 g of NH3 2 moles of NH3
and the answer would be in moles of nitrogen. Finally, Formula Weight Road is used to reach grams of nitrogen. (The formula weight of nitrogen is 28 grams per mole.) The entire setup is therefore:
51 g of NH3 X 1 mole of NH3 X 1 mole of N2 X 28 g of N2
17 g of NH3 2 moles of NH3 1 mole of N2
Computing the answer gives you 42 grams of nitrogen.
Solving other problems
The bridge model can also be used to solve problems in converting grams of one substance into molecules or atoms of another. For example: How many molecules of hydrogen would be released if 10 grams of magnesium react with unlimited hydrochloric acid? Also, how many atoms of hydrogen is this? The balanced equation is:
Mg + 2 HCl -----> MgCl2 + H2
In this problem, the path to follow is from grams of magnesium to moles of magnesium via Formula Weight Road, then over the Coefficients Bridge to moles of hydrogen and along Route 6.02 X 1023 to molecules of hydrogen. After stopping at molecules of hydrogen, the journey continues down Subscript Street to atoms of hydrogen. Before the Bridge, use the formula weight of magnesium (24 grams) to get from grams of magnesium to moles. The first part of the setup is:
10 g of Mg X 1 mole of Mg
24 g of Mg
Next, Coefficients Bridge must be crossed. The balanced equation shows that 1 mole of magnesium is chemically equivalent to 1 mole of hydrogen, so 1 mole of H2 / 1 mole of Mg is now included in the setup:
10 g of Mg X 1 mole of Mg X 1 mole H2
24 g of Mg 1 mole Mg
Next, the map suggests that moles are converted into molecules via Route 6.02 X 1023 (a reminder that there are 6.02 X 1023 molecules per mole). Dimensional analysis dictates that this conversion factor be included in the form:
6.02 X 1023 molecules NH3
1 mole NH3
The calculation thus becomes:
10 g of Mg X 1 mole of Mg X 1 mole H2 X 6.02 X 1023 molecules H2
24 g of Mg 1 mole Mg 1 mole H2
and the answer is calculated to be 2.5 X 1023 molecules of hydrogen. The second part of the problem asks for the number of atoms of hydrogen. Subscript Street leads to atoms, meaning that the hydrogen will be used to convert from molecules to atoms of hydrogen. The formula H2 denotes that there are two atoms of hydrogen to every molecule. Thus,
2.5 X 1023 molecules H2 X 2 atoms of H2
1 molecules H2
yields 5 X 1023 atoms of hydrogen.
The Bridge Method can be used to solve many other problem variations. As long as one substance in a balanced equation is expressed in moles, grams, molecules, or atoms, any other substance in the equation can be so converted. The appeal of the method is that every stoichiometry problem is handled the same way.
Once your students are familiar with the territory, you can also expand the map to more complicated problems (see Figure 2). For example, if you start with the volume of a liquid sample, take Density Avenue to grams and then continue as before: Volume in millilitres X Density (in grams/millilitre) = grams, would be the first part of the setup. Or if the molarity of a solution and its volume are given, follow Litres Lane to get from molarity to moles. The setup would begin with Molarity (moles/ litres) X volume in litres = moles.
In the fall of 1980, 65 percent of students at Columbia-Greene who were tested on stoichiometry problems using the Bridge Method achieved a grade of A; only 8 percent failed. Following remedial work, those who failed went on to demonstrate that they understood stoichiometry problems in a subsequent exam.
As already stated, the Bridge Method is a confidence builder for students first starting out with stoichiometry problems. It dosen't take long, however, for the astute student to realize that the method is more than a gimmick, and that he or she is always converting to moles of known and then crossing the bridge to moles of unknown. At that point, the student is no longer dependant upon drawing the map, and is able to solve subsequent problems using the dimensional analysis (factor label) method.
USEFUL REFERENCE MATERIALS
http://www2.ucdsb.on.ca/tiss/stretton/chem1/stoicgap.html
http://www2.ucdsb.on.ca/tiss/stretton/chem1/stoicgap.html
http://www.chem4kids.com/files/react_stoichio.html
What is relationship between mole and volume?
BalasHapusthanks for visit bun. Mole Relations with Volume
Hapus"At the same temperature and pressure, the same amount of gas with the same number of moles, from this temperature STP (0 ° C 1 atm) 1 mole per volume of 22.4 liters is called the volume of molar gas. as we know the formula PV = nRT The formula explains that the mole and volume ratio is proportional. In other words the greater the mole then the volume is also getting bigger.
What is the difference between the volume in normal circumstances with the volume in the STP state?
BalasHapusthanks for visit novi. The volume in the standard state means the measurement of gas with 0 ° C and 1 atm pressure is called the standard state and is expressed by STP (Standard Temperature and Pressure).
HapusWhile the ordinary volume is the volume of gas measured when the gas measurement conditions at 25 ° C and 1 atm pressure is called the state of the room and expressed by RTP (Room Temperature and Pressure).
Hello hudia, can you explain about concept of avogardro's number with your own words?
BalasHapusthanks for visit ma. The avogadro number is a number representing how many atoms or molecules in a mole. A scientist named avogadro looks for avogadro numbers through C-12 atoms. The mass of 1 C-12 atom is 1,992 × 10-23 gram, then the number of atoms in 1 mol (12 gram) C-12 with the result of calculation 6,02. 10 ^ 23 and named as avogadro number. As well as to easily recognize units of numbers such as 1 dozen it is 12, then so is the avogadro number as a benchmark for finding the number of other atomic molecules.
HapusHai hudia can you explain whats the conclusion of this materi ?
BalasHapusThe conclusion stoichiometri is the study of the calculation of the quantity of a substance be it moles, mass, volume and so forth.
HapusWhat are the differences between empirical formula and molecular formula?
BalasHapus
HapusEmpirical formulas show the simplest whole number ratio of these atoms.
Molecular formulas show the actual ratio of atoms in the compound.
Here is an example:
Glucose is the a simple sugar whose molecular formula is C6H12O6
Its empirical formula would be CH2O which would be its molecular formula divided by the smallest whole number
What is the meaning of the molecular weight of the compound?
BalasHapusThanks for visit wahyu. The molecular weight or so-called relative molecule (Mr) is the weight of a molecule in the atom mass unit (sma). The molecular weight can be calculated by summing the weight of all the atoms that make up it. The molecular weight is an important variable because it is directly related to the physical properties of the polymer.
HapusWhat all emprical formula of compound including C and H
BalasHapus
HapusNotice also how it really doesn't make much of a difference. The trick is to know when to do that and it comes only via experience. Generally speaking, in empirical formula problems, C = 12, H = 1, O = 16 and S = 32 are sufficient.
There are times when using 12.011 or 1.008 will be necessary. If you hit a problem that just doesn't seem to be working out, go back and re-calculate with more precise atomic weights. These problems, however, are fairly uncommon.
For what it is worth, one piece of advice on rounding: don't round off on the moles if you see something like 2.33 or 4.665. That first one can be rendered as two and one-third (or seven thirds) and the second one as four and two-thirds (or fourteen thirds). In a situation like that, you would multiply by three to reach the smallest whole-number ratio rather than dividing by the smallest.
Can you explain to me how to know the MR of a substance.?
BalasHapusWe can look at the periodic table. If for compoun, we The molecular mass or molecular weight is the total mass of a compound. It is equal to the sum of the individual atomic masses of each atom in the molecule. It is easy to find the molecular mass of a compound with these steps.
HapusDetermine the molecular formula of the molecule.
Use the periodic table to determine the atomic mass of each element in the molecule.
Multiply each element's atomic mass by the number of atoms of that element in the molecule. This number is represented by the subscript next to the element symbol in the molecular formula.
Add these values together for each different atom in the molecule.
The total will be the molecular mass of the compound.
What is goal we learn stochimetry?
BalasHapus
HapusThanks for visis. Stoichiometry calculations are based on the fact that atoms are conserved. They cannot be destroyed or created. Numbers and kinds of atoms before and after the reactions are always the same. This is the basic law of nature.
So stoichiometry is very important.
Give an example for an equivalent reaction to stoichiometry ?
BalasHapus
HapusThanks for visit. For example Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.
Pb(OH)4+H2SO4→Pb(SO4)2+H2O
Pb(OH)4+H2SO4→Pb(SO4)2+H2O
SOLUTION
Start by counting the number of atoms of each element.
The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a coefficient of 2 should be added in front of H2SO4H2SO4 to have an equal number of sulfur on both sides of the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4 coefficient should be added in front of H2OH2O where there is a deficiency of oxygen. Count the number of elements now present on either side of the equation. Since the numbers are the same, the equation is now balanced.
Pb(OH)4+2H2SO4→Pb(SO4)2+4H2O
Please explain what is an imaginary bridge?
BalasHapus
HapusThanks for visit. Bridging the Stoichiometry Gap
The Bridge Method uses a map model consisting of an imaginery bridge and several roads leading to and away from the bridge (see Figure 1). Students use the map as a guide to the steps they must follow in order to solve a stoichiometry problem. On the left side of the bridge is Known Country, the territory of the chemical for which information is given in the problem. On the right side of the bridge is Unknown Country, the land of the chemical about which information is sought.
What is the difference between the basic laws of chemistry on the law of proust and the laws of dalton, and please give us an example?
BalasHapus
Hapuslaw of definite proportions (Joseph Proust, shortly after Lavoisier): Proust studied metal compounds, including metal oxides, carbonates and sulfides. From the work of Robert Boyle in the 17th century, it was understood that substances that could be broken down into more fundamental components were mixtures or compounds. Substances that could not be further broken down were referred to as elements. Thus, Proust deduced the so-called law of definite proportions:
In a given chemical compound, the proportion by mass of the elements that compose it are fixed, independent of the origin of the compound or its mode of preparation.
This is basically saying that sodium chloride, for example, is always NaCl, no matter how it is obtained, made, or prepared. There are no ``intermediate'' compounds.
The law of multiple proportions: (John Dalton, shortly after Proust): Studied gases and gaseous mixtures under different external conditions. Building on Proust's work, he noted that mathematically discrete manner in which elements combined to form different compounds. For example, in carbon monoxide (CO), the mass ratio of oxygen to carbon $m_{\rm O}/m_{\rm C} = 1.33$ and in carbon dioxide (CO$_2$) $m_{\rm O}/m_{\rm C} = 2.66$. Thus, since the amount of carbon is fixed in each compound, we can look at how the amount of oxygen varies, and we find that $m_{\rm O}({\rm in\ CO}_2)/m_{\rm O}({\rm in\ CO}) = 2$. The generalization of this idea is the law of multiple proportions:
When two elements form a series of compounds, the masses of one that combine with a fixed mass of the other are in the ratio of (small) integers to each other.
This law is obeyed by all gaseous compounds, which is what Dalton studied. Certain solids are exceptions to both this rule and the law of definite proportions. An example is the solid wüstite, which can range from Fe$_{0.95}$O to Fe$_{0.85}$O, depending on the mode of preparation. These formulae express the incommensurate compositions possible in this solid. But, since atoms are essentially indestructible (we have to work hard to get them to fission!), this explains why we do not find compounds such as C$_{13/7}$H$_{5/3}$ in nature. Dalton's observations led him to propose the notion of the atom as the fundamental and indestructible building blocks of matter.
how to Bridging the Stoichiometry Gap
BalasHapus
HapusThe Bridge Method can be used to solve many other problem variations. As long as one substance in a balanced equation is expressed in moles, grams, molecules, or atoms, any other substance in the equation can be so converted. The appeal of the method is that every stoichiometry problem is handled the same way.
Once your students are familiar with the territory, you can also expand the map to more complicated problems (see Figure 2). For example, if you start with the volume of a liquid sample, take Density Avenue to grams and then continue as before: Volume in millilitres X Density (in grams/millilitre) = grams, would be the first part of the setup. Or if the molarity of a solution and its volume are given, follow Litres Lane to get from molarity to moles. The setup would begin with Molarity (moles/ litres) X volume in litres = moles.