50 short questions of 9th Class Chemistry Chapter 4, Stoichiometry, with exam-ready answers in the book's words: the textbook exercise questions plus SLO-based questions asked in board papers. Punjab Board new syllabus.
Exam-ready answers. Learn the meaning, then write them in your own words.
The short questions at the end of the chapter in your book.
Barium forms the cation Ba²⁺ and nitrogen forms the nitride anion N³⁻. Applying the criss-cross method, the numerical value of each ion's charge becomes the subscript of the other ion and the signs are dropped:
Check: 3 × (+2) = +6 and 2 × (−3) = −6, so the compound is electrically neutral.
Molecular formula = n (empirical formula), where n = molar mass ÷ empirical formula mass.
| Mole | Avogadro's number |
|---|---|
| The quantity of a substance containing Avogadro's number of particles. | The number 6.022 × 10²³: the number of units in one mole of a substance. |
| It is an amount of substance, with a mass (molar mass): 1 mol of C weighs 12 g. | It is a pure number, like a dozen (12); it has no mass of its own. |
| Unit: mol. | Written Nᴀ = 6.022 × 10²³ (per mole). |
So a mole is the amount, and Avogadro's number is how many particles that amount contains.
Check: Cu 1 = 1; S 2 = 1 + 1; H 4 = 4; O 8 = 4 + 2 + 2. The acid must be hot and concentrated: copper does not react with dilute sulphuric acid.
The book's in-text exercises first, then short questions on every topic of the chapter (SLO-based), as asked in board papers.
The formula of an element contains only one kind of atom: either the symbol alone for elements that exist as an aggregate of atoms (Na, Fe, C) or a symbol with a subscript for molecular elements (O₂, N₂, O₃). The formula of a compound contains two or more kinds of atoms bonded together (H₂O, NaCl, NH₃, CH₄).
| Ionic | Covalent |
|---|---|
| NaCl sodium chloride | H₂O water; NH₃ ammonia; CH₄ methane |
| CaCl₂ calcium chloride | HCl hydrogen chloride; HF hydrogen fluoride |
| KBr potassium bromide | H₂S hydrogen sulphide; PH₃ phosphine |
| BaCl₂ barium chloride | H₂O₂ hydrogen peroxide; H₂SO₄ sulphuric acid |
| CO₂ carbon dioxide; CO carbon monoxide; C₆H₆ benzene |
Water (H₂O) and carbon dioxide (CO₂): the ratios 2 : 1 and 1 : 2 cannot be simplified, so n = 1. Methane (CH₄) and ammonia (NH₃) are other examples.
| Compound | Molecular | Empirical |
|---|---|---|
| Hydrogen peroxide | H₂O₂ | HO |
| Benzene | C₆H₆ | CH |
| Glucose | C₆H₁₂O₆ | CH₂O |
Empirical formula mass = 30; n = 60 ÷ 30 = 2; molecular formula = (CH₂O)₂ = C₂H₄O₂ (acetic acid, CH₃COOH).
Count the atoms: 3 H, 1 P and 4 O. Molecular formula H₃PO₄.
C: 1 + 1 + 1 = 3; H: 3 + 2 + 2 + 1 = 8; O: 1. Molecular formula C₃H₈O, usually written C₃H₇OH to show the OH group.
The structure has one Ca²⁺ ion and one carbonate ion CO₃²⁻ (one C bonded to three O atoms). The charges balance 1 : 1, so the formula is CaCO₃.
| Compound | Working | Molar mass |
|---|---|---|
| H₃PO₄ | 3(1) + 31 + 4(16) | 98 g mol⁻¹ |
| SiO₂ | 28 + 2(16) | 60 g mol⁻¹ |
| C₁₂H₂₂O₁₁ | 12(12) + 22(1) + 11(16) | 342 g mol⁻¹ |
| N₂O₄ | 2(14) + 4(16) | 92 g mol⁻¹ |
| MgCO₃ | 24 + 12 + 3(16) | 84 g mol⁻¹ |
A chemical formula represents an element or a compound by symbols, with subscripts showing the number of atoms of each kind. H₂O shows that in one molecule of water two atoms of hydrogen are bonded to one atom of oxygen.
A formula unit is the formula of an ionic compound showing the minimum ratio between its ions. Sodium chloride exists as ions bonded together in a crystal, not as molecules; since the ratio between its ions is 1 : 1, it is represented by the formula unit NaCl.
A molecular formula shows all the types of atoms bonded together in one molecule of a compound and their actual numbers. Examples: H₂O, NH₃, CH₄, H₂O₂, C₆H₆.
In ozone three atoms of oxygen are bonded to each other in one molecule, so its formula is O₃. Sodium exists as an aggregate of atoms, not molecules, so it is represented by its symbol alone.
Empirical formula of a compound shows the simplest whole number ratio of atoms present in that compound. All ionic compounds are represented by their empirical formulas.
An empirical formula shows only the simplest ratio, not the actual numbers. C₆H₆ and C₂H₂ both have C : H = 1 : 1, so both have the empirical formula CH.
For H₂O₂: n = 34 ÷ 17 = 2, so (HO)₂ = H₂O₂. If n = 1 the two formulas are the same.
Empirical formula mass = 13; n = 78 ÷ 13 = 6; molecular formula = (CH)₆ = C₆H₆.
Experimental data on the mass percent composition of the compound: the percentage of each element is divided by its atomic mass to get the mole ratio, which is then reduced to the simplest whole numbers.
In this method, the numerical value of each of the ion charges is crossed over to become the subscript of the other ion. Signs of the charges are then dropped. Li⁺ and O²⁻ give Li₂O.
| Name | Ions | Formula |
|---|---|---|
| Lithium oxide | Li⁺, O²⁻ | Li₂O |
| Aluminium oxide | Al³⁺, O²⁻ | Al₂O₃ |
| Magnesium nitride | Mg²⁺, N³⁻ | Mg₃N₂ |
| Name | Ions | Formula |
|---|---|---|
| Calcium oxide | Ca²⁺, O²⁻ | CaO (Ca₂O₂ reduced) |
| Sodium sulphide | Na⁺, S²⁻ | Na₂S |
| Aluminium chloride | Al³⁺, Cl⁻ | AlCl₃ |
Some metals (Cu, Fe, Sn) can lose different numbers of electrons and so show variable valency. Their compounds are named with Roman numerals: copper(I) oxide Cu₂O, copper(II) oxide CuO; iron(II) chloride FeCl₂, iron(III) chloride FeCl₃.
The number 6.022 × 10²³ is called Avogadro's number after the name of the Italian chemist Amedeo Avogadro. Avogadro's number is the number of units in one mole of a substance. It is represented as Nᴀ.
The quantity of a substance containing Avogadro's number of particles (Nᴀ) is called a mole. A mole of carbon atoms contains 6.022 × 10²³ atoms and weighs 12 g; a mole of O₂ contains 6.022 × 10²³ molecules and weighs 32 g; a mole of NaCl has 6.022 × 10²³ formula units and weighs 58.5 g.
The mass of one mole of a substance is called its molar mass. The molar mass of hydrogen atoms is 1.008 g; that of hydrogen molecules (H₂) is 2.016 g.
Because 1.00 g = 6.022 × 10²³ amu. Taking this many particles turns a mass in amu into the same number in grams: one C atom is 12 amu, so 6.022 × 10²³ C atoms weigh 12 g.
A mole of hydrogen atoms (1.008 g) and a mole of hydrogen molecules (2.016 g) are different amounts. When we use the term mole of a substance, we must also refer to what type of particles (atoms, molecules or ions) are present.
M(CO₂) = 12 + 32 = 44 g mol⁻¹. n = 88 ÷ 44 = 2 mol. Molecules = 2 × 6.022 × 10²³ = 1.204 × 10²⁴.
m = n × M = 0.5 × 58.5 = 29.25 g.
Molecules = 0.25 × 6.022 × 10²³ = 1.506 × 10²³. Each CH₄ has 5 atoms: atoms = 5 × 1.506 × 10²³ = 7.53 × 10²³.
Atoms and molecules are so small that they cannot be counted. The mole concept allows us to count atoms and molecules by weighing macroscopic amounts of matter, and it turns the ratio in an equation into a mass ratio we can weigh out in the laboratory.
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 (reactants, on the left) and those being produced (products, on the right).
One mole of zinc reacts with one mole of sulphuric acid to produce one mole of zinc sulphate and one mole of hydrogen gas: 1 : 1 : 1 : 1.
Sometimes a chemical 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 shown by ⇌.
Symbols written after formulas to show physical states: (s) solid, (l) liquid, (g) gas and (aq) aqueous (dissolved in water). Example: Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g).
AgCl being insoluble comes out of the aqueous solution in the form of a precipitate; Na⁺ and NO₃⁻ are spectator ions.
Al: 4 = 2 × 2; O: 3 × 2 = 6 = 2 × 3.
A chemical equation can only be written if all the points are experimentally verified. The nature of the products and their correct formulas must first be ascertained. An equation cannot be invented on paper: copper with dilute H₂SO₄ gives no reaction at all, though one could balance an equation for it.
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. 100 + 73 = 173 g = 111 + 18 + 44: the total mass of the reactants equals that of the products.
1 mol of alcohol gives 2 mol CO₂, so 1.8 mol gives 2 × 1.8 = 3.6 mol CO₂ (and uses 5.4 mol O₂).
5 g of H₂ gives 36 ÷ 4 × 5 = 45 g of water.
More of that reactant is present than the equation needs, so it is left over at the end. The other reactant is used up completely and decides how much product forms; that is why calculations start from the reactant that is not in excess.
The percentage by mass of each element in a compound: mass of the element in one mole ÷ molar mass × 100. For CO₂: 12 ÷ 44 × 100 = 27.3 % carbon (and 72.7 % oxygen).
