For any chemical compound an empirical formula is the simplest positive integer ratio of atoms present in that compound (Reger et al, 135). As oppose to the molecular formula, Empirical formula takes the lowest number ration in the elements in a compound and not the actual number. For example carbon dioxide is written as CO2.
Significance of Empirical formula
The process of finding empirical formula involves computing the composition percentage represented by each compound’s mass. Below are some steps followed when determining empirical formula:
The first step is stating the number of grams in every element.
The second step involves computation of molar mass through conversion of each element’s mass to moles.
The next step involves division of every mole by using the smallest count of moles that was found in the second step.
The values are then rounded off to whole numbers in case there are decimal points.
The percentage composition; this is the percent of the mass due to the specific component divided by the molar mass of that compound times 100 (Zumdahl et al, 70)
For example, below is an example of calculating the percentage composition of CO2
Molar mass of compound:
Mass due to carbon: 12.01 g/mol
Percent composition of carbon:
Mass ratio is the proportion of the mass of one element say A to the proportion of another element says B. Mass ratio is important since it helps in chemical properties and chemical reactions of an elements. Mass ratio also helps in calculating the empirical formula. The most important thing in calculating the mass ratio is writing down the chemical formula. Below is an example of getting mass ratio which in turn helps to write the empirical formula.
Given an organic compound of 15.25g was combusted in oxygen produced 34.71 g of CO2 and 14.2 g of water. It was then found out that the compound contained 27.59% oxygen. Determine the empirical compound’s formula by using the given information.
Solution
34.71g CO244gmolCO2= 0.7889 mol of carbon dioxide
14.2 g H2O18gmolH2O= 2*0.7889 mol of water=1.5778
27.59/100 * 15.25g sample= 4.207/16g/mol = 0.2630 mol O
0.7889mol C0.2630 mol O= the ration 3 C:1 O
1.5778 mol H0.2630 mol O= the ratio 6 H: 1O
= C3H6O
Differences between empirical formula and molecular formula
Empirical formula represents the simplest positive integer possessed by an atom contained in a compound while molecular formula is given by the actual number of atoms of each element in a molecule of a compound (Zumdahl et al, 34). Empirical formula and molecular formula are mathematically related. That is
Molecular formula = n * empirical formula
Provide some real life examples of applying the % composition of a compound and empirical formula includes.
The amount of fertilizer to apply in your garden
The relative amount of nutrients
The percentage of nitrogen and potassium
For example knowing the percentage composition of every element in this compound Ca(NO3)2.
Step 1
Finding the molar mass of Ca(NO3)2
Ca 40.1 grams/mole = 40.1 g/mol
N 2 x (14.0) grams/mole = 28.0g/mol
O 6 x (16.0) grams/mole = 96.0 g/mol
= 164.1 g/mole
Step 2
Dividing the mass of each element with molar mass
% Ca = 40.1g x 100% = 24.4 % Calcium
164.1g
% N = 28.0g x 100% = 17.1% Nitrogen
164.1 g
% O = 96.0g x 100% = 58.5 % Oxygen
164.1g
Combustion analysis is the most common method for determining empirical formula in the real – world determination. The weighed compound is normally burned in oxygen and its products are then analyzed using the gas chromatogram.
Work cited
Reger, Daniel L, Scott R. Goode, and David W. Ball. Chemistry: Principles and Practice. Belmont, CA: Brooks/Cole, Cengage Learning, 2010. Print.
Zumdahl, Steven S, Donald J. DeCoste, Iris Stovall, Steven S. Zumdahl, Steven S. Zumdahl, and Steven S. Zumdahl. Study Guide for Introductory Chemistry: A Foundation ; Introductory Chemistry ; Basic Chemistry. Boston: Houghton Mifflin Co, 2004. Print.
