Q. Find the percent composition of each element in water. Molar mass of NaOH=23+16+1=40 urea= NH2-CO-NH2
Solution. We know that the chemical formula for water is H2O. Let us now calculate the molar mass of water. The molar mass of Oxygen = 16.00 × 1 = 16 g/mole and of Hydrogen = 1.01 × 2 = 2.02 g/mole.
Now, using the molar mass of each of the given elements, we find out the percentage composition of each element in H2O. It is given as the ratio of the grams of the element to the grams of the total element in the compound, multiplied by 100.
Calculating the percentage composition of Hydrogen:
% H = 2.02/18.02 × 100
Therefore, % H= 11.21 %
Calculating the percentage composition of Oxygen:
Therefore, % O = 16/18.02 × 100 = 88.79 %
➢ Try Yourself!
% composition of oxygen= 16/40×100 = 40%
mass of nitrogen = 14,
mass of 2 molecules hence
N%= mass of nitrogen/mass of urea X 100
= 28/60 X 100
Molar mass of NaOH=23+16+1=40
The empirical formula of a compound can be determined by the following steps:
(i) Write the name of detected elements in the column 1 present in the compound. Write the corresponding atomic mass in column 2.
(ii) Write the experimentally determined percentage composition by weight of each element present in the compound in column 3.
(iii) Divide the percentage of each element by its atomic weight to get the relative number of atoms of each element in column 4.
(iv) Divide each number obtained for the respective elements in step (iii) by the smallest number among those numbers so as to get the simplest ratio in column 5.
(v) If any number obtained in step (iv) is not a whole number then multiply all the numbers by a suitable integer to get the whole-number ratio. This ratio will be the simplest ratio of the atoms of different elements present in the compound. The empirical formula of the compound can be written with the help of this ratio in column-6.
Example 1. A compound contains C = 71.23%, H = 12.95% and O = 15.81%. What is the empirical formula of the compound?
Example 2. The simplest formula of a compound containing 50% of element X (Atomic mass = 10) and 50% of the element Y (Atomic mass = 20) is:
Hence (B) is correct.
Example 3. A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?
Empirical formula = CH2Cl
Molecular formula = n * Empirical formula
n = 98.96 / 48.5 = 2.04 = 2 (approx)
Molecular formula = 2 * CH2Cl = C2H4Cl2
Example 4. A compound of carbon, hydrogen and nitrogen contains these elements in the ratio 9:1:3.5. Calculate the empirical formula. If its molecular mass is 108, what is the molecular formula?
Empirical formula = C3H4N
Empirical formula mass = (3 × 12) + (4 × 1) + 14 = 54
n = mol. mass/Emp. mass
= 108/54 = 2
Thus, molecular formula of the compound = 2 × Empirical formula
= 2 × C3H4N
Example 5. 2.38 gm of uranium was heated strongly in a current of air. The resulting oxide weighed 2.806 g. Determine the empirical formula of the oxide. (At. mass U = 238; O = 16).
Step 1: To calculate the percentage of uranium and oxygen in the oxide.
2.806 g of the oxide contain uranium = 2.38 g
∴ Percentage of uranium = 2.38/2.806 × 100 = 84.82
Hence, the percentage of oxygen in the oxide
= 100.00 – 84.82 = 15.18
Step 2: To calculate the empirical formula
Hence, the empirical formula of the oxide is U3O8.
Example 6. Chemical analysis of a carbon compound gave the following percentage composition by weight of the elements present. Carbon 10.06%, hydrogen 0.84%, chlorine 89.10%. Calculate the empirical formula of the compound.
Step 1: Percentage of the elements present.
Step 2: Dividing the percentage compositions by the respective atomic weights of the elements.
Step 3: Dividing each value in step 2 by the smallest number among them to get simple atomic ratio.
Step 4: Ratio of the atoms present in the molecule.
∴ The empirical formula of the compound C1H1Cl3 or CHCl3.
Example 7. A carbon compound on analysis gave the following percentage composition. Carbon 14.5%, hydrogen 1.8%, chlorine 64.46%, oxygen 19.24%. Calculate the empirical formula of the compound.
Step 1: Percentage composition of the elements present in the compound.
Step 2: Dividing by the respective atomic weights.
Step 3: Dividing the values in step 2 among them by the smallest number.
Step 4: Multiplication by a suitable integer to get the whole-number ratio.
∴ The simplest ratio of the atoms of different elements in the compound.
C : H : Cl : O = 2 : 3 : 3 : 2
∴ The empirical formula of the compound C2H3Cl3O2.
Example 8. The empirical formula of a compound is CH2O. Its molecular weight is 90. Calculate the molecular formula of the compound. (Atomic weights C = 12, H = 1, O = 16)
Solution. Empirical formula = CH2O
Empirical formula weight = (12 + 2 + 16) = 30
n = molecular weight/empirical formula weight
∴ n = 90/30 = 3
The molecular formula (CH2O)3 = C3H6O3
Relation between the two:
Molecular formula= Empirical formula × n
Check out the importance of each step involved in calculations of the empirical formula.
(a) Absolute density
(b) Relative density
Absolute density = Mass/volume
Relative density = density of substance/density of a standard substance
Specific gravity = density of substance/density of H2O at 4oC
Vapour density: It is defined only for gas.
It is a density of gas with respect to H2 gas at the same temp & pressure.
V.D = Mgas / MH2
V.D = M/2
V.D = Molecular wt of gas/Molecular wt of H2 gas
The density of Cl2 gas with respect to O2 gas = Molecular wt. of Cl2 gas/Molecular wt. of O2 gas
Q.1. A crystalline hydrated salt, on being rendered anhydrous, loses 45.6% of its mass. The percentage composition of the anhydrous salt is: Al = 10.5%, K = 15.1%, S =24.8% and oxygen = 49.6%. Find the empirical formula of the anhydrous and crystalline hydrated salt. [K = 39; Al = 27; S = 32; O = 16; H = 1]
Ans. The empirical formula of anhydrous salt = KAlS2O8
Hydrated salt composition: % anhydrous part = 54.4% and % H2O = 45.6%
The empirical formula of hydrated salt = KAIS2O8.12H2O
Q.2. A colorless crystalline compound has the following percentage composition: Sulphur 24.24%, nitrogen 21.21%, hydrogen 6.06% and the rest is oxygen. Determine the empirical formula of the compound. If the molecular mass is 132, what is the molecular formula of the compound? Name the compound if it is found to be sulphite.
Ans. Empirical formula SN2H8O4
Molecular formula = SN2H8O4
Name- Ammonium sulphite: (NH4)2SO3
Q.3. A gaseous hydrocarbon contains 85.7% carbon and 14.3% hydrogen. 1 litre of the hydrocarbon weighs 1.26 g at NTP. Determine the molecular formula of the hydrocarbon.