Composition of a Compound

A compound contains two or more elements combined in a certain fixed ratio.

The composition of a compound can be described by the mass percentage of each element present in it.

This may be determined by any of the methods described below.

 

Determining the composition of a compound

Composition of a compound is commonly expressed as the mass percentage of each element present in it.

To determine the mass percentage of an element in any compound, we should know the mass of that element present in a certain known mass of the compound.

Let us suppose, W g of a compound contain w g of an element A.

Then,

To do this, a known mass of the compound is chemically analysed, and the mass of each element present in it is determined by a suitable chemical method.

EXAMPLE 1.

5.85 g of a compound contains 3.55 g of chlorine, and 2.30 g of sodium. Determine the mass percentage of sodium and chlorine in the compound.

Solution: Mass of the compound = 5.85 g

       Mass of sodium = 2.30 g

       Mass of chlorine = 3.55 g

Then, 

  

Thus, mass percentage of sodium and chlorine in the given compound are 39.3 and 60.7, respectively.

Percentage composition of a compound determined from its formula

The percentage composition of a compound means the mass percentage of each element present in the compound. The mass percentage of each element can be determined from the molecular or empirical formula of the compound as follows.

Step 1. Write the molecular or empirical formula of the compound.

Step 2. Write the name, number of atoms, and the total mass of each element present in one molecule (or formula) of the compound.

Step 3. Obtain the molecular mass or formula mass of the compound by adding the masses of all the atoms present in it.

Step 4. Calculate the mass percentage of each element by using the relationship:

This method of calculating the percentage composition of a compound from its molecular formula is illustrated through the following examples:

EXAMPLE 5.2

Calculate the mass percentage of each element present in water (H₂O).

Solution: The molecular formula of water is H₂O.

A molecule of water is made up of two hydrogen atoms and one oxygen atom. Thus, one can write,

Molecular formula of water: H₂O = 2 atoms of H + 1 atom of O

   Atomic mass of H = 1 u      and   Atomic mass of O = 16 u

Therefore, Mass of two H atoms = 2 × 1 = 2 u

       Mass of one O atom = 1 × 16 = 16 u

and     Molecular mass of water = 2 u + 16 u = 18 u

So,

 

Therefore, water (H₂O) contains 11.11% of hydrogen (H), and 88.89% of oxygen (O).

EXAMPLE 5.3

Calculate the mass percentage of oxygen in (a) HgO  (b) K₂Cr₂O₇ (c) Al₂(SO₄)₃.
The atomic masses (in u units) are: Hg=200.6, O=16, K=39, Cr=52, Al=27, S=32

Solution: (a) HgO = 1 atom of Hg + 1 atom of O

So, Molecular mass of HgO = (200.6 u + 16 u) = 216.6 u

Therefore,

Thus, HgO contains 7.39% oxygen.

 (b) K₂Cr₂O₇ = 2 atoms of K + 2 atoms of Cr + 7 atoms of O

        2 × 39 u       2 × 52 u           7 × 16 u

                          78 u            104 u               112 u

So, Molecular mass of K₂Cr₂O₇ = 78 u + 104 u + 112 u = 294 u

Thus, 294 units of K₂Cr₂O₇ contain 112 units of oxygen. Therefore,

Thus, K₂Cr₂O₇ contains 38.1% oxygen.

 (c) Al₂(SO₄)₃ = Al₂S₃O₁₂ = 2 atoms of Al + 3 atoms of S + 12 atoms of O

                                        2 × 27 u         3 × 32 u         12 × 16 u

                                          54 u                96 u             192 u

So, Molecular mass of Al₂(SO₄)₃ = (54 + 96 + 192) u = 342 u

Thus, 342 u of Al₂(SO₄)₃ contain 192 u of oxygen (O).

Therefore,

Thus, Al₂(SO₄)₃ contains 56.1% of oxygen.

EXAMPLE 5.4

How many grams of Cr are there in 85 g of Cr₂S₃?       Atomic masses are: Cr = 52 u and S = 32 u.

Solution: Cr₂S₃ = 2 atoms of Cr + 3 atoms of S

                         2 × 52 u            3 × 32 u

                           104 u                96 u

        Molecular mass of Cr₂S₃ = (104 + 96) u = 200 u

So,      200 u of Cr₂S₃ contain = 104 u of Cr

or       200 g of Cr₂S₃ contain = 104 g of Cr

Thus,

Do It Yourself

1.   Calculate the percentage of sulfur in sodium sulfate, Na₂SO₄.

2.  Calculate the percentage of iron in the common ore, haematite, Fe₂O₃.

3.   Calculate the percentage of barium in barium hydroxide, Ba(OH)₂.

4.   Calculate the percentage of each element in:

                  a. silver nitrate, AgNO₃                b. ammonium phosphate, (NH₄)₃PO₄

5.   Calculate the percentage of nitrogen in:

           a.   ammonium sulfate, (NH₄)₂SO₄   b. urea, CH₄ON₂

6. A popular explosive has the molecular formula, C₇H₅N₃O₆. Calculate the percentage of nitrogen in the compound.

[Atomic masses: H = 1.0 u, C = 12.0 u, N = 14.0 u, O = 16.0 u, P = 31.0 u,
S = 32.0 u, Na = 23.0 u, Fe = 56.0 u, Ba = 137.0 u and Ag = 108.0 u]

Calculations based on Chemical Equations

A balanced chemical equation provides information regarding the relative amount, mass or volume of the reactants and products.

The fundamental behind all these relationships is the mole concept. The number of moles of any substance is directly related to its mass and its number of molecules or atoms. Therefore, different relationships can be written between the reactants and products. Some typical relationships are given alongside and below.

Some useful relationships are,

How to solve numerical problems based on chemical equations

The calculations based on chemical equations can be done as follows.

§  Write down the given amount, mass or volume of the substance, and identify the substance whose amount, mass or volume is to be found out.

§  Write down the balanced chemical equation.

§  Write down the amount, mass or volume of the concerned reactant and product, as indicated by the balanced chemical equation.

§  Calculate the unknown quantity by applying the unitary method.

Calculations based on mole-mole relationship

In a balanced chemical equation, the coefficients placed before any symbol or formula gives the number of moles of that substance involved in the reaction.

These coefficients therefore give the relative amounts (no. of moles) of the reactants and products. This relationship is illustrated through the following examples:

EXAMPLE 1.

A reaction proceeds in accordance with the chemical equation.

3CaCl₂        +      2K₃PO₄           ®     Ca₃(PO₄)₂    +      6KCl

calcium                  potassium               calcium                  potassium

chloride                  phosphate              phosphate              chloride

How many moles of potassium phosphate are needed to produce 0.076 moles of potassium chloride?

Solution: The balanced chemical equation is

                        3CaCl₂ + 2K₃PO₄ ® Ca₃(PO₄)₂ + 6KCl

                                     2 mol                      6 mol

This gives the relationship,

6 mol of KCl are obtained from 2 mol of K3PO4

1 mol of KCl is obtained from 2/6 mol of K3PO4

        0.076 mol of KCl is obtained from (2 × 0.076) / 6 mol of K3PO4

= 0.025 mol of K3PO4

Thus, to produce 0.076 mol of potassium chloride, 0.025 mol of potassium phosphate (K₃PO₄) is required.

Calculations based on mole-mass and mass-mass relationships

For making these calculations, the number of moles of the reactants and products are converted into mass by using their molar masses. Some typical calculations are illustrated below.

EXAMPLE 2.

Hydrogen reacts to form water as follows.

2H₂ + O₂ ® 2H₂O

How many grams of hydrogen are needed to react completely with 6.4 g of oxygen gas?                             (Atomic masses are : H = 1 u, O = 16 u.)

Solution: The chemical equation for the reaction, and the mass-mole relationship is,

2H₂            +      O₂            ®     2H₂O

                2 mol                 1 mol                 2 mol

        2 × (2 × 1) g       1 × (2 × 16) g              2 × (2 × 1 + 16) g

                4 g                    32 g                  36 g

Thus,

32 g of oxygen reacts completely with 4 g of hydrogen

1 g of oxygen reacts completely with (4 / 32) g of hydrogen

\ 6.4 g of oxygen reacts completely with (6.4 × 4 / 32) g = 0.8 g of hydrogen

Thus, 6.4 g of oxygen gas reacts completely with 0.8 g of hydrogen.

EXAMPLE 3.

Hydrazine (N₂H₄) and hydrogen peroxide (H₂O₂) are used together as a rocket fuel. The products of the reaction are nitrogen and water. How many grams of H₂O₂ are needed per kg (1000 g) of hydrazine carried by the rocket?

Solution: The reaction is,

Hydrazine + Hydrogen peroxide ® Nitrogen + Water

The balanced chemical equation is,

N₂H₄          +      2H₂O₂        ®     N₂     +      4H₂O

                1 mol                 2 mol

From the chemical equation of the reaction,

                1 mol of N₂H₄ = 2 mol of H₂O₂

or (2 × 14 + 4 × 1.0) g of N₂H₄ = 2 × (2 × 1.0 + 2 × 16.0) g of H₂O₂

                 (28 + 4) g of N₂H₄ = 2 × (34) g of H₂O₂

                        32 g of N₂H₄ = 68 g of H₂O₂

Thus, 32 g of N₂H₄ requires 68 g of H₂O₂

 1 g of N₂H₄ requires (68 / 32) g of H₂O₂

 1000 g of N₂H₄ requires (1000 × 68 / 32) g = 2125 g of H₂O₂

Therefore, mass of H₂O₂ required in the reaction is 2125 g, or 2.125 kg.

Calculations based on mass-volume and

volume-volume relationships

The following relationships are useful for solving problems based on mass-volume, and volume-volume relationships:

Molar volume of a gas at NTP = 22.4 L mol^–1                   (= 22400 mL)

 i.e., 1 mol of each gas under NTP conditions occupies a volume of 22.4 L.

These calculations are illustrated through the following examples:

EXAMPLE 4.

2.3 g of metallic sodium reacts with excess of water. Calculate the mass of sodium hydroxide formed. What is the volume of hydrogen evolved under NTP conditions?

Solution: Sodium reacts with water according to the equation,

        2Na(s)        +      2H₂O         ®     2NaOH(aq)         +      H₂(g)

        2 mol                 2 mol                                                 1 mol

        2 × 23 g                                     2 (23 + 16 + 1)            2 g

        46 g                                          80 g                          22.4 L (at NTP)

Thus, 2.3 g of sodium gives (2.3 × 80 / 46) = 4.0 g of NaOH

and 2.3 g of sodium gives (22.4 × 2.3 / 46) = 1.12 L of H₂ gas

Therefore,    Mass of NaOH formed = 4.0 g

        Volume of H₂ evolved at NTP = 1.12 L

EXAMPLE 5.

Calculate the volume of oxygen at NTP obtained by decomposing 12.26 g of KClO₃.

Solution: KClO₃ decomposes as follows,

                     2 mol                                                 3 mol

        2 × (39 + 35.5 + 48) g                                      3 × 22.4 L

        (2 × 122.5) g = 245g                                         67.2 L

Thus,

 12.26 g of KClO₃ gives (12.26 × 67.2 / 245) L = 3.36 L of oxygen

So, 12.26 g of KClO₃ on decomposing gives 3.36 L of oxygen gas at NTP.