Stoichiometry

Ans: The chemical formula of barium nitride is Ba₃N₂.

Ans: Data:
Molar mass = 180 g/mol
Empirical formula = CH₂O
Empirical formula mass = C + 2H + O
= 12 + 2(1) + 16
= 12 + 2 + 16
= 30g

Solution:
To find molecules formula, we will find 'n' (n=0,1,2,3....) which will be multiplied with empirical formula to get molecules formula

n = molar mass / Empirical formula mass
= 180/30
= 6

So,
Molecular formula = (Empirical formula)
= (CH₂O)₆
= C₆H₁₂O₆

Ans. Data:
Mass of H₂O = 1.5g
Molar mass of H₂O = 2H+O
=2 (1) +16
= 2 + 16
= 18 g/mol

To Find:
No of molecules = ?

Solution:
Firstly no. of moles will be calculated which will be multiplied with N_A to get no. of molecules

Moles = mass / Molar mass
= 1.5/18
= 0.083

no. of molecules = moles × N_A
= 0.083 × 6.02 × 10²³
= 5.02 × 10²²

Result:
5.02 × 10²³ molecules of H₂O will be present in 1.5g of water.

Ans:

Copper + Sulphuric acid → Copper sulphate + Sulphur dioxide+ water

Ans: The chemical equation for the reaction between copper and sulphuric acid to form copper sulphate, sulphur dioxide, and water can be written as follows:

Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

This equation shows the reactants on the left side and the products on the right side.

Ans.

Ans. Molecular and empirical formula of water = H₂O
Molecular and empirical formula of Methane = CH₄

Ans.

1. Molecular formula of Glucose= C₆H₁₂O₆
Empirical formula of Glucose = CH₂O

2. Molecular formula of Acetic Acid= CH₃COOH
Empirical formula of Acetic Acid = CH₂O

3. Molecular formula of benzene = C₆H₆
Empirical formula of benzene = CH

Ans. Empirical formula of compound = CH₂O
Empirical formula mass=30gmol⁻¹
Its molar mass = 180g mol⁻¹
Its molecular formula will then be:

n = 180/30
n = 6

Molecular formula = (CH₂O)₆
= C₆H₁₂O₆

Ans. Empirical formula of compound = CH₂O
Empirical formula mass=30gmol⁻¹
Its molar mass = 60g mol⁻¹
Its molecular formula will then be:

n = 60/30
n= 2

Molecular formula = (CH₂O)₂
= C₂H₄O₂

Ans: It has 3H, 1P and 4O atoms
Its molecular formula will be H₃PO₄.

Ans. It has 3C, 1O and 8H atoms.
Its molecular formula will be C₃H₈O

Ans. It has 1Ca, 1C and 3O atoms.
Its formula will be CaCO₃.

Ans. Molar mass of H₃PO₄
Atomic mass of H=1g
Atomic mass of P=31g
Atomic mass of O=16g
Molar Mass = 1 (3) + 31 (1) + 16 (4)
= 3 + 31 + 64
= 98g/mol.

Molar mass of SiO₂
Atomic mass of Si=28g
Atomic mass of O=16g
Molar Mass = 28 (1) + 16 (2)
= 28 + 32 = 60g/mol

Molar mass of C₁₂H₂₂O₁₁
Atomic mass of C=12g
Atomic mass of H=1g
Atomic mass of O = 16g
Molar Mass = 12 (12) + 1 (22) + 16 (11)
= 144 + 22 + 176
= 342 g/mol

Molar mass of N₂O₄
Atomic mass of N=14g
Atomic mass of O=16g
Molar Mass = 14 (2) + 16 (4)
= 28 + 64 => 92 g/mol

Molar mass of MgCO₂
Atomic mass of Mg=24g
Atomic mass of C=12g
Atomic mass of O = 16g
Molar mass = 24 (1) + 12 (1) + 16 (2)
= 24 + 12 + 32
= 68 g/mol

Ans. A chemical formula is a symbolic representation of a chemical compound. It tell us about the type and no. of atoms present in a compound. For example chemical formula of water is H₂O. It is composed of two atom of hydrogen and one atom of oxygen.

Ans. The type of formula which shows the simplest whole number ratio of atoms present in a compound is called empirical formula, e.g. glucose (C₆H₂O₆) has simplest ratio 1:2:1 of carbon, hydrogen and oxygen respectively. Hence its empirical formula is CH₂O.

Ans. The composition of all the chemical products we use in our lives, such as shampoos, perfumes, soaps and fertilizers are formed using stoichiometric calculations. Without stoichiometry the chemical industry does not exist.

Ans. In order to write down the formula of an ionic compound, first identify the cations and anions and the number of charges present on them. Finally combine the two ions together to form an electrically neutral compound.

Ans. Empirical Formula
It is the formula which shows the simplest whole number ratio of atoms present in a compound.

e.g. Glucose has simplest ratio 1:2:1 of carbon, hydrogen and oxygen respectively. Hence its empirical formula is CH₂O.

Molecular Formula
The formula which shows actual number of atoms of each element present in a molecule of that compound is called molecular formula, e.g. Molecular formula of benzene is C₆H₆.

Ans. Molecular formula of a compound can be found out if we know its empirical formula. To calculate the empirical formula of a compound, you need to determine the simplest whole-number ratio of atoms in the compound. This can be done by using experimental data on the mass percent composition of the compound. Molecular formula is then calculated by the following relationship.

Molecular formula = n (Empirical Formula)

where, n = Molar Mass / Empirical Formula mass

For example, the empirical formula of hydrogen peroxide is HO. Its molar mass is 34. Its molecular formula will then be

n = 34/17 = 2

Molecular formula = (HO)₂ = H₂O₂

If for a compound the value of n is one, then its molecular formula is the same as its empirical formula.

Ans. Solution: Empirical Formula = CH
Empirical formula mass=13gmol⁻¹
Molar mass = 78 g mol⁻¹
Molecular formula =n(Empirical formula)

n = molar mass / empirical formula mass
= 78/13 = 6

Molecular Formula = (CH)₆ = C₆H₆

Ans. In order to deduce the molecular formula from the structural formula the following steps are taken.

i. Write down the structural formula of the compound.
ii. Count the number of atoms of each type in the structural formula.
iii. Write the symbols of all the elements.
iv. Write the total number of atoms of each kind as a subscript.
v. Remove the subscript 1.

Ans. It has 2 H, 1 S and 4O atoms.
Its molecular formula will be H₂SO₄

Ans. It has 2C, 4H, 2O atoms
Its molecular formula will be C₂H₄O₂

Ans. Avogadro's number is a collection of 6.02 × 10²³ particles. It is represented by symbol 'N_A'. Hence, the 6.02 × 10²³ number of atoms, molecules or formula units are called Avogadro's number that is equivalent to one 'mole' of respective substance.

e.g: 6.02 × 10²³ atoms of carbon = 1 mole of carbon.

Ans. Mole is important because atoms and molecules are so small. The mole concept allows us to count atoms and molecules by weighing macroscopically small amounts of matter.

Ans. The mass of one mole of a substance is called as molar mass.

e.g. Molar mass of O-atom = 16 g.

Ans. A mole is defined as the amount (mass) of a substance that contains 6.02 × 10²³ number of particles (atoms, molecules or formula units). It is abbreviated as 'mol'.

e.g. 1 mole of carbon = 12 g = 6.02 × 10²³ atoms of carbon.

(a) H₂SO₄ (Sulphuric acid)
(b) C₆H₁₂O₆ (Glucose)

Ans. (a) H₂SO₄
Atomic mass of H=1
Atomic mass of S= 32
Atomic mass of O = 16
Molar mass
= 2(1) + 1(32) + 4(16) = 98 g mol⁻¹

(b) C₆H₁₂O₆
Atomic mass of C = 12
Atomic mass of H=1
Atomic mass of O = 16
Molar mass
= 6(12) + 12(1) + 6(16) = 180 g mol⁻¹

Ans. The chemists have developed a very suitable way of representing a chemical change in terms of symbols of elements and formulas of compound. Representing a chemical change in this way is called a chemical equation.

Ans. Molecular formula of Benzene = C₆H₆
Molecular formula of Acetylene= C₂H₂
But Both have same empirical formula = CH

Ans.

e.g. 2H₂ +O₂ → 2H₂O
(Reactants)    (Product)

Ans. Sometimes a chemical reaction moves both ways. In other words, the reactants react to give the products and the products, in turn, react to give the reactants back. Such reactions are called as reversible reactions and are indicated by (⇌) e.g.

N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

Ans. A complete and balanced chemical equation tells us the mole ratio or molar mass ratio between the reactants and the products. With the help of this ratio, we can find out the molar masses of the products provided we know the molar masses of the reactants. Similarly the molar masses of the reactants can also be found out if we know the molar masses of the products.

For example, the following equation tells us that one mole (100 g) of calcium carbonate reacts with two moles (73 g) of hydrochloric acid to produce one mole (111 g) of CaCl₂, one mole (18 g) of water and one mole (44 g) of carbon dioxide.

CaCO_3(s) + 2HCl_(aq) ⟶ CaCl_2(aq) + H₂O_(l) + CO₂
1 mole   2 moles   1 mole   1 mole   1 mole
100 g   73 g   111 g   18 g   44 g

The total masses of the reactants are equal to the total masses of the products.

Ans: Different compounds can indeed have the same empirical formula but different molecular formulae. The empirical formula represents the simplest whole number ratio of atoms of each element in a compound.

For example, the empirical formula for both glucose and formaldehyde is CH₂O. This means that both compounds contain the same ratio of carbon, hydrogen, and oxygen atoms, but they differ in the actual number of those atoms.

On the other hand, the molecular formula provides the actual number of each type of atom in a molecule. In the case of glucose, the molecular formula is C₆H₁₂O₆, while for formaldehyde, it is CH₂O.

Ans: Here are the chemical formulas for the compounds:

1. Calcium phosphate: Ca₃(PO₄)₂
2. Aluminium nitride: AlN
3. Sodium acetate: CH₃COONa
4. Ammonium carbonate: (NH₄)₂CO₃
5. Bismuth sulphate: Bi₂(SO₄)₃

Ans: Avogadro's number, which is approximately 6.022 x 10²³no. of particles, is immensely important in chemistry for several reasons.

Firstly, it provides a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. When chemists work with substances, they often deal with grams and liters but the reactions and interactions occur at the atomic or molecular level. Avogadro's number allows chemists to convert between the number of particles and the amount of substance in grams, making it easier to calculate and understand chemical reactions.

Secondly, it helps in determining the molar mass of substances. The molar mass is the mass of one mole of a substance, and knowing Avogadro's number enables chemists to relate the mass of a sample to the number of moles it contains. This is crucial for stoichiometric calculations in chemical reactions, where the ratios of reactants and products need to be precisely measured.

Ans: To calculate the percentage of each element in the compound, the following formula can be used:

Percentage of an element = (mass of the element / total mass of the compound) × 100

• Carbon (C):
- Mass of carbon = 5.217 g
- Total mass of the compound = 8.657 g
- Percentage of carbon = (5.217 g / 8.657 g) × 100
- Percentage of carbon = 60.3%

• Hydrogen (H):
- Mass of hydrogen = 0.962 g
- Percentage of hydrogen = (0.962 g / 8.657 g) × 100
- Percentage of hydrogen = 11.1%

• Oxygen (O):
- Mass of oxygen = 2.478 g
- Percentage of oxygen = (2.478 g / 8.657 g) × 100
- Percentage of oxygen = 28.6%

Ans: The following example to illustrate how to calculate the masses of products formed in a reversible reaction.

Consider the reversible reaction of nitrogen gas (N₂) with hydrogen gas (H₂) to form ammonia (NH₃):

N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

• Chemical Equation: First of all, write the balanced chemical equation.

• Identify Known Quantities: Suppose we start with 28 g of nitrogen (N₂) and 6 g of hydrogen (H₂).

• Calculate Moles of Reactants:
- Molar mass of N₂ = 28 g/mol
- Molar mass of H₂ = 2 g/mol
Moles of N₂ = 28 g / 28 g/mol = 1 mol
Moles of H₂ = 6 g / 2 g/mol = 3 mol

• Use Stoichiometry: According to the balanced equation, 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. We have exactly 1 mol of N₂ and 3 mol of H₂, which means we have enough hydrogen to react completely with nitrogen.

• Calculate Moles of Products:
From the balanced equation:
1 mol of N₂ produces 2 mol of NH₃.
Therefore, 1 mol of N₂ will produce 2 mol of NH₃.

• Calculate Mass of Products:
Molar mass of NH₃ = 14 g/mol (N) + 3 g/mol (H) = 17 g/mol
Moles of NH₃ produced = 2 mol
Mass of NH₃ = 2 mol × 17 g/mol = 34 g

In this example, starting with 28 g of nitrogen and 6 g of hydrogen, we can produce 34 g of ammonia.