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fandomhigh2006-02-16 09:27 am
equalsmcsquared.livejournal.com) wrote in fandomhigh2006-02-16 09:27 am
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Chemistry (4th Period)
"Today, we're going to discuss limiting reagents. I know this stuff seems tedious but it is vital when doing lab work. You don't want unbalanced solutions. Trust me when I say that there is no need to live through an explosion."
Sometimes when reactions occur between two or more substances, one reactant runs out before the other. That is called the "limiting reagent". Often, it is necessary to identify the limiting reagent in a problem.
Example: A chemist only has 6.0 grams of C2H2 and an unlimitted supply of oxygen and desires to produce as much CO2 as possible. If she uses the equation below, how much oxygen should she add to the reaction?
2C2H2(g) + 5O2(g) ---> 4CO2(g) + 2 H2O(l)
To solve this problem, it is necessary to determine how much oxygen should be added if all of the reactants were used up (this is the way to produce the maximum amount of CO2).
First, we calculate the number of moles of C2H2 in 6.0 grams of C2H2. To be able to calculate the moles we need to look at a periodic table and see that 1 mole of C weighs 12.0 grams and H weighs 1.0 gram. Therefore we know that 1 mole of C2H2 weighs 26 grams (2*12 grams + 2*1 gram).
Then, because there are five (5) molecules of oxygen to every two (2) molecules of C2H2, we need to multiply the result by 5/2 to get the total molecules of oxygen. Then we convert to grams to find the amount of oxygen that needs to be added:
While the empirical formula is the simplest form of a compound, the molecular formula is the form of the term as it would appear in a chemical equation. The empirical formula and the molecular formula can be the same, or the molecular formula can be any positive integer multiple of the empirical formula. Examples of empirical formulas: AgBr, Na2S, C6H10O5. Examples of molecular formulas: P2, C2O4, C6H14S2, H2, C3H9.
One can calculate the empirical formula from the masses or percentage composition of any compound. We have already discussed percent composition in the section above. If we only have mass, all we are doing is essentially eliminating the step of converting from percentage to mass.
Example: Calculate the empirical formula for a compound that has 43.7 g P (phosphorus) and 56.3 grams of oxygen. First we convert to moles:
Next we divide the moles to try to get a even ratio.
When we divide, we did not get whole numbers so we must multiply by two (2). The answer=P2O5
Calculating the molecular formula once we have the empirical formula is easy. If we know the empirical formula of a compound, all we need to do is divide the molecular mass of the compound by the mass of the empirical formula. It is also possible to do this with one of the elements in the formula; simply divide the mass of that element in one mole of compound by the mass of that element in the empirical formula. The result should always be a natural number.
Example: if we know that the empirical formula of a compound is HCN and we are told that a 2.016 grams of hydrogen are necesary to make the compound, what is the molecular formula? In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by 1.008 we see that the amount of hydrogen needed is twice as much. Therefore the empirical formula needs to be increased by a factor of two (2). The answer is:
H2C2N2.
"There will be a test next Tuesday. Study."
Sometimes when reactions occur between two or more substances, one reactant runs out before the other. That is called the "limiting reagent". Often, it is necessary to identify the limiting reagent in a problem.
Example: A chemist only has 6.0 grams of C2H2 and an unlimitted supply of oxygen and desires to produce as much CO2 as possible. If she uses the equation below, how much oxygen should she add to the reaction?
2C2H2(g) + 5O2(g) ---> 4CO2(g) + 2 H2O(l)
To solve this problem, it is necessary to determine how much oxygen should be added if all of the reactants were used up (this is the way to produce the maximum amount of CO2).
First, we calculate the number of moles of C2H2 in 6.0 grams of C2H2. To be able to calculate the moles we need to look at a periodic table and see that 1 mole of C weighs 12.0 grams and H weighs 1.0 gram. Therefore we know that 1 mole of C2H2 weighs 26 grams (2*12 grams + 2*1 gram).
Then, because there are five (5) molecules of oxygen to every two (2) molecules of C2H2, we need to multiply the result by 5/2 to get the total molecules of oxygen. Then we convert to grams to find the amount of oxygen that needs to be added:
While the empirical formula is the simplest form of a compound, the molecular formula is the form of the term as it would appear in a chemical equation. The empirical formula and the molecular formula can be the same, or the molecular formula can be any positive integer multiple of the empirical formula. Examples of empirical formulas: AgBr, Na2S, C6H10O5. Examples of molecular formulas: P2, C2O4, C6H14S2, H2, C3H9.
One can calculate the empirical formula from the masses or percentage composition of any compound. We have already discussed percent composition in the section above. If we only have mass, all we are doing is essentially eliminating the step of converting from percentage to mass.
Example: Calculate the empirical formula for a compound that has 43.7 g P (phosphorus) and 56.3 grams of oxygen. First we convert to moles:
Next we divide the moles to try to get a even ratio.
When we divide, we did not get whole numbers so we must multiply by two (2). The answer=P2O5
Calculating the molecular formula once we have the empirical formula is easy. If we know the empirical formula of a compound, all we need to do is divide the molecular mass of the compound by the mass of the empirical formula. It is also possible to do this with one of the elements in the formula; simply divide the mass of that element in one mole of compound by the mass of that element in the empirical formula. The result should always be a natural number.
Example: if we know that the empirical formula of a compound is HCN and we are told that a 2.016 grams of hydrogen are necesary to make the compound, what is the molecular formula? In the empirical formula hydrogen weighs 1.008 grams. Dividing 2.016 by 1.008 we see that the amount of hydrogen needed is twice as much. Therefore the empirical formula needs to be increased by a factor of two (2). The answer is:
H2C2N2.
"There will be a test next Tuesday. Study."