In Chemistry, the Empirical formula of a chemical compound is defined as the simplest positive integer ratio of atoms present in a compound.
For instance, talking about the empirical formula to molecular formula, the empirical formula of SO (Sulfur Monoxide) and SO (Disulfur Dioxide), both compounds of sulfur and oxygen, have the same empirical formula. However, in terms of molecular formulas, the number of atoms in each molecule of a chemical compound is not the same. This means that the empirical formula does not consider the arrangement or number of atoms.
On this page, we will learn to calculate the empirical formula of compounds like the empirical formula of glucose and go through the empirical formula statistics as well.
Empirical and Molecular Formula
A chemical compound n-hexane has a molecular formula of CH3CH2CH2CH2CH2CH3 or simply, we can write it as. Here, we notice that there are 6 C-atoms arranged in a chain with 14 hydrogen atoms.
Though the formula of n-hexane is C6H14, its empirical formula is C3H7. By empirical formula, the C: H ratio is 3: 7.
From the above empirical formula and molecular formula, we understand the basic concept. Now, let us understand the concept of the empirical formula rule by calculating the empirical formula.
Empirical Formula Examples
Let us consider the molecular formula of Glucose, which is C6H12O6. Here, this formula has 2 moles of hydrogen for every mole of carbon and oxygen. Therefore, the empirical formula of glucose becomes CH2O.
Similarly, talking about Ribose whose molecular formula is C5H10O5. Here, for one mole of carbon and oxygen, there are two moles of hydrogen, and therefore, the empirical formula of Ribose becomes the same as that of the empirical formula of glucose, i.e., CH2O.
Besides these examples, Acetic acid (molecular formula: C2H4O2), and formaldehyde (molecular formula: CH2O) all have varying molecular formulas but the same empirical formula,i.e., CH2O.
Example 1: For instance, a chemical analysis of the methyl acetate sample gives us the following elemental data:
Carbon (C): 48.64% carbon
Hydrogen (H): 8.16% hydrogen (H), and
Oxygen (O): 43.20%
Solution: For the purpose to calculate the empirical formula, assume that we have 100 grams of the compound. Also, the percentages equal the mass of each element in grams.
Now, we will follow the step-by-step procedure to obtain the desired empirical formula.
Step 1: Transform each percentage corresponding to an expression of the mass of each element in grams.
So, 48.64% C becomes 48.64 g of Carbon, 8.16% H becomes 8.16 g of Hydrogen, and 43.20% O becomes 43.20 g of Oxygen.
Step 2: Now, we will convert the amount of each element in grams to its amount in moles in the following manner:
12.00 g C = 1 mol of C
48.64 g C = ? mol
? = 1/12.00 * 48.64 = 4.0533
So, 48.64 g C = 4.0533 mol of C
1 g H = 1 mol
8.16 g H = ? mol
? = 1/1.00 * 8.16 = 8.16 mol
Therefore, 8.16 g H = 8.16 mol of H
16 g O = 1 mol
43.20 g O = ? mol
? mol = 1/16.00 * 43.20
Therefore, 43.20 g O = 2.7 mol of O
Step 3: Here, the smallest of all values of mol is 2.7, so we will divide these three values in mol by 2.7 in order to get the empirical formula.
4.0533/2.7 = 1.5
8.16/2.7 = 3.0322 ≈ 3
2.7/2.7 = 1
Final Step of the Empirical Formula
Since the first value, i.e., 1.5 is not a whole number because the concept of empirical formula says that it is a positive integer ratio of atoms in a compound. So, we multiply these values with an integer to get a whole number:
1.5 x 2 = 3
3 x 2 = 6
1 x 2 = 2
So, the desired empirical formula obtained out of our calculation is C3H6O2. This is also the Methyl Acetate formula, which means our calculation meets the criteria, i..e., matches with the exact molecular formula.
Example 2: Let’s say that you heated 0.297 g of Mg and obtained 0.493 g of the oxide. So, the empirical formula of magnesium oxide is:
Solution: Magnesium oxide mass = mass of Mg + mass of O
0.493 g = 0.297 g + mass of O
Mass of O = 0.493 g – 0.297 g
Therefore, mass of O = 0.196 g
Moles of Mg = 0.297 *124.305 = 0.01222 mol of Mg
Moles of O = 0.493 * 116.000 = 0.0125 mol of O
Now, dividing both the values in mol by 0.01222
0.012220.0122 = 1, and
0.012250.0122 = 1.022 1
Finally, we get the empirical formula of magnesium oxide as MgO.
Empirical Formula Statistics
The Empirical formula statistics say that almost 95% of the observations in a normal distribution have a place within three Standard Deviations from the Mean. Therefore, the empirical formula rule is a very important rule and it helps in forecasting.
The formula displays the predicted percentage of observations that may lie within each Standard Deviation from the Mean. So, the rules go like this:
68% of the observations may lie within 土 1 Standard Deviation (SD) from the mean
95% of the observations may lie within 土 2 SD from the mean
However, 7% of observations may lie within 土 3 SDs from the mean.
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Now, how to use an empirical formula in statistics, let’s understand this part.
Empirical Rule Statistics Example
Example: Mr. Xavier is trying to find the average number of years his patient will survive after retirement, assuming the retirement age to be 60. If her mean survival is 50 years and random observations are 20 years and SD is 3, then find out the probability that she will draw her pension for more than 23 years.
As per the empirical rule formula, 68% of the observations will probably be within 1 Standard Deviation from the Mean. Here the Mean of all the observations is 20.
Therefore, 68% of the observations will lie within 20 土 1 SD, which is 20 土 3. So the range is from 17 to 23.
This means that there is a 68% chance that the minimum years a lady survives after retirement ranges from 17 to 23.
Now the percentage that is lying outside the given range is (100 – 68), i.e., 32%. Here, 32 is distributed equally on both sides of the graph, which means there is a 16% chance that the minimum years will be below 17, while a16% chance that minimum years will be over 23.
So, the probability that a lady will draw more than 23 years of pension will be 16%.
The Empirical formula helps us in calculating the formula of a chemical compound with a given percentage of the amount of each atom in that compound. Also, it is a statistical concept that helps display the probability of observations, which is very useful for finding an approximation of a large population of any place. However, we must note that these are just approximations. Besides this, there are always chances of outlying values, i.e., the values that don’t fall in the distribution. So our findings are not accurate, and therefore, some precautionary measures should be taken when making the forecast.