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 unlimited supply of oxygen and he 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 g 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 g and H weighs 1.0 g. Therefore we know that 1 mole of C2H2 weighs 26 g (2 × 12 grams + 2 × 1 gram).
|6.0 g C2H2 x|
1 mol C2
(24.0 + 2.0)g C2
|= 0.25 mol C2H2|
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:
|0.25 mol C2H2 x|
5 mol O2
2 mol C2H2
32.0 g O2
1 mol O2
|= 20 g O2|
It is possible to calculate the mole ratios (also called mole fractions) between terms in a chemical equation when given the percent by mass of products or reactants.
percentage by mass = mass of part/ mass of wholeThere are two types of percent composition problems-- problems in which you are given the formula (or the weight of each part) and asked to calculate the percentage of each element and problems in which you are given the percentages and asked to calculate the formula.
In percent composition problems, there are many possible solutions. It is always possible to double the answer. For example, CH and C2H2 have the same proportions, but they are different compounds. It is standard to give compounds in their simplest form, where the ratio between the elements is as reduced as it can be-- called the empirical formula. When calculating the empirical formula from percent composition, one can convert the percentages to grams. For example, it is usually the easiest to assume you have 100 g so 54.3% would become 54.3 g. Then we can convert the masses to moles; this gives us mole ratios. It is necessary to reduce to whole numbers. A good technique is to divide all the terms by the smallest number of moles. Then the ratio of the moles can be transferred to write the empirical formula.
Example: If a compound is 47.3% C (carbon), 10.6% H (hydrogen) and 42.0% S (sulfur), what is its empirical formula?
To do this problem we need to transfer all of our percents to masses. We assume that we have 100 g of this substance. Then we convert to moles:
|= 3.94 moles|
|= 10.52 moles|
|= 1.310 moles|
Now we try to get an even ratio between the elements so we divide by the number of moles of sulfur, because it is the smallest number:
So we have: C3H8 S
Example: Figure out the percentage by mass of hydrogen sulfate, H2SO4.
In this problem we need to first calculate the total mass of the compound by looking at the periodic table. This gives us:
2(1.008) + 32.07 + 4(16.00) g/mol = 98.09 g/mol
Now, we need to take the weight fraction of each element over the total mass (which we just found) and multiply by 100 to get a percentage.
|= 0.0206 ∗ 100 = 2.06%|
|= 0.327 ∗ 100 = 32.7%|
|= 0.652 ∗ 100 = 65.2%|
Now, we can check that the percentages add up to 100%
65.2 + 2.06 + 32.7 = 99.96This is essentially 100 so we know that everything has worked, and we probably have not made any careless errors. So the answer is that H2SO4 is made up of 2.06% H, 32.7% S, and 65.2% O by mass.
Empirical Formula and Molecular Formula
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:
43.7 grams P
|= 1.41 moles|
56.3 grams O
|= 3.52 moles|
Next we divide the moles to try to get an 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 necessary 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: