9th Class Chemistry Chapter 4 Stoichiometry Long Questions

Question 1

Which conditions must be fulfilled before writing a chemical equation for a reaction?

Accepted Answer

What a chemical equation is Representing a chemical change in terms of symbols of elements and formulas of compounds is called a chemical equation. It tells us the reactants (written on the left) and the products (written on the right); an arrow from reactants to products separates the two. 1. Law of conservation of mass A chemical equation must obey the law of conservation of mass. No atom should be destroyed or produced during a chemical change. The total number and the type of atoms must remain the same, so the total number and kind of atoms on both sides of the equation must be equal: the chemical equation must be balanced. 2. Correct formulas The formulas of elements and compounds must be written correctly. A wrong formula cannot be fixed by changing coefficients: water is always H₂O, never HO or H₂O₂. 3. Correct mole ratio A chemical equation must determine the correct mole ratio among the reactants, the products and between the reactants and the products. 4. Direction of change A chemical equation must also point out the direction in which the change is proceeding: → for a reaction that goes one way, ⇌ for a reversible reaction. 5. Physical states It is a usual practice to show the normal physical states of reactants and products as a subscript. Solid, liquid and gas are symbolized as s, l and g; aqueous (aq) represents the solvated ion. Experimental verification A chemical equation can only be written if all the above points are experimentally verified. The nature of the products and their correct formulas must first be ascertained. Example Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g) Zinc reacts with sulphuric acid to give zinc sulphate and hydrogen gas. One mole of zinc reacts with one mole of sulphuric acid to produce one mole of zinc sulphate and one mole of hydrogen gas. The equation is balanced (1 Zn, 2 H, 1 S, 4 O on each side), has correct formulas, a 1 : 1 : 1 : 1 mole ratio, a forward arrow and state symbols. The book writes ZnSO₄(s) here; the zinc sulphate stays dissolved, so it is ZnSO₄(aq).

Question 2

Explain the concepts of Avogadro's number and mole.

Accepted Answer

The problem In a chemical reaction a large number of atoms or molecules of reactants react to give the products. We would like to know the mass ratio in which the reactants react, and to express these masses in grams. It is not possible to weigh individual atoms or molecules because they are very small. Building up to Avogadro's number Consider 2C + O₂ → 2CO. Increasing the number of reacting species (2 × 100 atoms with 100 molecules, 2 × 10 000 atoms with 10 000 molecules) does not change the ratio in which they react, but the numbers are still far too small to weigh. We increase them to 6.022 × 10²³, because 1.00 g = 6.022 × 10²³ amu: particles, 2 × 6.022 × 10²³ atoms, 6.022 × 10²³ molecules, 2 × 6.022 × 10²³ molecules; mass, 2 × 6.022 × 10²³ × 12 amu = 24 g, 6.022 × 10²³ × 32 amu = 32 g, 2 × 6.022 × 10²³ × 28 amu = 56 g Starting from a simple equation we now have a mass ratio, 24 g : 32 g : 56 g, which can conveniently be used in the laboratory. Avogadro's number: The number 6.022 × 10²³, named after the Italian chemist Amedeo Avogadro; it is the number of units in one mole of a substance, represented as Nᴀ. The mole Mole: The quantity of a substance containing Avogadro's number of particles. Mole is a number like a dozen or a gross. A dozen of oranges means 12 oranges; similarly a mole of a substance means its 6.022 × 10²³ particles, which can be atoms, molecules or ions. We must say which particles we mean. A mole of carbon atoms contains 6.022 × 10²³ atoms and weighs 12 g.; A mole of oxygen molecules contains 6.022 × 10²³ molecules and weighs 32 g.; A mole of sodium chloride consists of 6.022 × 10²³ formula units and its mass is 58.5 g. Molar mass The mass of one mole of a substance is called molar mass. Hydrogen atoms 1.008 g; hydrogen molecules 2.016 g. The equation read in moles two atoms, one molecule, two molecules; two moles of C, one mole of O₂, two moles of CO; 24 g, 32 g, 56 g n = m ÷ M;   N = n × Nᴀ

Question 3

(Investigative) The chemical formula for sand is SiO₂ but sand does not exist in the form of discrete molecules like H₂O. How has its formula been determined keeping in view its structure?

Accepted Answer

Structure of sand Sand is mostly silicon dioxide (silica). It is a giant covalent (network) structure, not a collection of small molecules. Each silicon atom is bonded to four oxygen atoms (at the corners of a tetrahedron), and each oxygen atom is shared between two silicon atoms, so the network runs on through the whole crystal. Finding the ratio Each Si "owns" all 4 of its bonds but only half of each O (because each O is shared by two Si).; So each Si has 4 × ½ = 2 oxygen atoms.; Si : O = 1 : 2. The formula is an empirical formula Like the formula units of ionic compounds, SiO₂ is an empirical formula: it shows the simplest whole number ratio of atoms in the network, not a molecule. It can also be found by analysis: silica contains 46.7 % Si and 53.3 % O by mass; 46.7 ÷ 28 = 1.67 and 53.3 ÷ 16 = 3.33, a 1 : 2 ratio. Diamond (C), graphite (C) and sand (SiO₂) are all networks: their formulas give the ratio of atoms only.

Question 4

Define empirical formula and molecular formula. How is the molecular formula found from the empirical formula? Give examples.

Accepted Answer

Empirical formula Empirical formula of a compound shows the simplest whole number ratio of atoms present in that compound. All the ionic compounds are represented by their empirical formulas, which show the simplest ratio between their ions: the empirical formula of calcium fluoride is CaF₂. Molecular formula A molecular formula shows all the types of atoms bonded together in one molecule of the compound, with their actual numbers: H₂O₂, C₆H₆, C₆H₁₂O₆. Relation between the two Molecular formula = n × (Empirical formula);   n = Molar mass ÷ Empirical formula mass If for a compound the value of n is one, its molecular formula is the same as its empirical formula (H₂O, CO₂, CH₄). Example 1: hydrogen peroxide Empirical formula mass = 1 + 16 = 17. n = 34 ÷ 17 = 2. Molecular formula = (HO)₂. H₂O₂ Example 2: benzene Empirical formula mass = 12 + 1 = 13. n = 78 ÷ 13 = 6. Molecular formula = (CH)₆. C₆H₆ Example 3: glucose Empirical formula mass = 12 + 2 + 16 = 30. n = 180 ÷ 30 = 6. Molecular formula = (CH₂O)₆. C₆H₁₂O₆ Same empirical formula, different compounds Acetylene, C₂H₂, CH; Benzene, C₆H₆, CH; Formaldehyde, CH₂O, CH₂O; Acetic acid, C₂H₄O₂, CH₂O; Glucose, C₆H₁₂O₆, CH₂O

Question 5

How are the formulas of binary ionic compounds written? Explain the criss-cross method with examples.

Accepted Answer

Principle 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. Steps Write the symbol of the cation with its charge.; Write the symbol of the anion with its charge.; Cross the numerical value of each charge over to become the subscript of the other ion.; Drop the signs; reduce to the simplest ratio; omit the subscript 1. Example 1: lithium oxide Cation Li⁺, anion O²⁻. Crossing over gives Li₂O₁ → Li₂O (2 × +1 = +2; −2). Example 2: aluminium oxide Cation Al³⁺, anion O²⁻. Crossing over gives Al₂O₃ (+6, −6). Example 3: magnesium nitride Cation Mg²⁺, anion N³⁻. Crossing over gives Mg₃N₂ (+6, −6). More examples Calcium chloride, Ca²⁺, Cl⁻, CaCl₂; Barium nitride, Ba²⁺, N³⁻, Ba₃N₂; Calcium phosphide, Ca²⁺, P³⁻, Ca₃P₂; Magnesium oxide, Mg²⁺, O²⁻, MgO; Iron(III) oxide, Fe³⁺, O²⁻, Fe₂O₃ Equal charges cancel out to a 1 : 1 formula (MgO, AlN); never write Mg₂O₂.

Question 6

What is a chemical equation? What information does it give? Explain reversible and ionic equations.

Accepted Answer

Definition Representing a chemical change in terms of symbols of elements and formulas of compounds is called a chemical equation. It tells us the elements or compounds which are reacting and those which are being produced. The reacting substances are called reactants and those produced are called products. Reactants are written on the left, products on the right, separated by an arrow. Information from a balanced equation The names and correct formulas of reactants and products.; The mole ratio between them, and through molar masses the mass ratio.; The direction of change.; The physical states (s, l, g, aq).; That mass is conserved: atoms of each kind are equal on both sides. Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g) Reversible reactions Sometimes a reaction moves both ways: the reactants react to give the products and the products, in turn, react to give the reactants back. Such reactions are called reversible reactions and are indicated by the sign ⇌. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) Ionic equations Reactions involving ions may also be shown as chemical equations. Both AgNO₃ and NaCl are ionic compounds and are soluble in water. When they are mixed in water they react; AgCl, being insoluble, comes out of the aqueous solution as a precipitate. AgNO₃(aq) + NaCl(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq) Ag⁺(aq) + Cl⁻(aq) → AgCl(s) Ions that appear unchanged on both sides (Na⁺, NO₃⁻) are spectator ions and are left out of the net ionic equation.

Question 7

Explain how calculations are made from a balanced chemical equation, with mass-mass, mole-mole and mass-particle examples.

Accepted Answer

Basis 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 the masses of the products if we know the masses of the reactants, and vice versa. moles, 1, 2, 1, 1, 1; mass, 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 (173 g). Mass-mass: Example 1 100 g CaCO₃ → 111 g CaCl₂. 1 g → 111 ÷ 100 g. 25 g → 111 ÷ 100 × 25 = 27.75 g. 27.75 g CaCl₂ Mole-mole: Example 2 O₂ needed = 3 × 1.8 = 5.4 mol. CO₂ produced = 2 × 1.8 = 3.6 mol. 5.4 mol O₂; 3.6 mol CO₂ Mole-mass: Example 3 4 mol Al need 3 mol O₂. 0.3 mol Al need ¾ × 0.3 = 0.225 mol. Mass = 0.225 × 32 = 7.2 g. 7.2 g O₂ Mass-particles: Example 4 4 g H₂ → 36 g H₂O, so 5 g → 45 g. 18 g (1 mol) H₂O = 6.022 × 10²³ molecules. 45 g = 45 ÷ 18 = 2.5 mol = 2.5 × 6.022 × 10²³. 1.505 × 10²⁴ molecules Every problem has the same spine: balanced equation → convert to moles → use the coefficient ratio → convert to what is asked.

Question 8

How is the molecular formula deduced from a structural formula? Give examples.

Accepted Answer

Steps Write down the structural formula of the compound.; Count the number of atoms of each type in the structural formula.; Write the symbols of all the elements.; Write the total number of atoms of each kind as a subscript.; Remove the subscript 1. Example 1: sulphuric acid H–O–S(=O)(=O)–O–H has 2 H, 1 S and 4 O atoms. Molecular formula H₂SO₄. Example 2: phosphoric acid Phosphorus bonded to three OH groups and one =O: 3 H, 1 P, 4 O. Molecular formula H₃PO₄. Example 3: n-propyl alcohol CH₃–CH₂–CH₂–OH: 3 C, 8 H, 1 O. Molecular formula C₃H₈O (C₃H₇OH). Example 4: calcium carbonate One Ca²⁺ and one carbonate ion (C bonded to three O): CaCO₃. Being ionic, this is a formula unit.

	2C	O₂	2CO
particles	2 × 6.022 × 10²³ atoms	6.022 × 10²³ molecules	2 × 6.022 × 10²³ molecules
mass	2 × 6.022 × 10²³ × 12 amu = 24 g	6.022 × 10²³ × 32 amu = 32 g	2 × 6.022 × 10²³ × 28 amu = 56 g

2C	+ O₂	→ 2CO
two atoms	one molecule	two molecules
two moles of C	one mole of O₂	two moles of CO
24 g	32 g	56 g

Compound	Molecular formula	Empirical formula
Acetylene	C₂H₂	CH
Benzene	C₆H₆	CH
Formaldehyde	CH₂O	CH₂O
Acetic acid	C₂H₄O₂	CH₂O
Glucose	C₆H₁₂O₆	CH₂O

Name	Ions	Formula
Calcium chloride	Ca²⁺, Cl⁻	CaCl₂
Barium nitride	Ba²⁺, N³⁻	Ba₃N₂
Calcium phosphide	Ca²⁺, P³⁻	Ca₃P₂
Magnesium oxide	Mg²⁺, O²⁻	MgO
Iron(III) oxide	Fe³⁺, O²⁻	Fe₂O₃

Long Questions with Answers

Set A: Textbook comprehensive questions 3 questions

Which conditions must be fulfilled before writing a chemical equation for a reaction?

What a chemical equation is

1. Law of conservation of mass

2. Correct formulas

3. Correct mole ratio

4. Direction of change

5. Physical states

Experimental verification

Example

Explain the concepts of Avogadro's number and mole.

The problem

Building up to Avogadro's number

The mole

Molar mass

The equation read in moles

(Investigative) The chemical formula for sand is SiO₂ but sand does not exist in the form of discrete molecules like H₂O. How has its formula been determined keeping in view its structure?

Structure of sand

Finding the ratio

The formula is an empirical formula

Set B: More important long questions 5 questions

Define empirical formula and molecular formula. How is the molecular formula found from the empirical formula? Give examples.

Empirical formula

Molecular formula

Relation between the two

Example 1: hydrogen peroxide

Example 2: benzene

Example 3: glucose

Same empirical formula, different compounds

How are the formulas of binary ionic compounds written? Explain the criss-cross method with examples.

Principle

Steps

Example 1: lithium oxide

Example 2: aluminium oxide

Example 3: magnesium nitride

More examples

What is a chemical equation? What information does it give? Explain reversible and ionic equations.

Definition

Information from a balanced equation

Reversible reactions

Ionic equations

Explain how calculations are made from a balanced chemical equation, with mass-mass, mole-mole and mass-particle examples.

Basis

Mass-mass: Example 1

Mole-mole: Example 2

Mole-mass: Example 3

Mass-particles: Example 4

How is the molecular formula deduced from a structural formula? Give examples.

Steps

Example 1: sulphuric acid

Example 2: phosphoric acid

Example 3: n-propyl alcohol

Example 4: calcium carbonate

The rest of Chapter 4

Long Questions from other Chemistry chapters