These questions appear inside the chapter itself, not at the end. The book page number is shown against each one so you can find it while reading.
A student burns 10 g of ethanol in an open beaker. After the reaction, no residue is left in the beaker. Does this mean the Law of Conservation of Mass is violated? Explain.
No, the Law of Conservation of Mass is not violated.
When ethanol burns, it reacts with oxygen from the air. The products of this combustion are carbon dioxide and water vapour — both are gases at the temperature of the flame:
Because the beaker is open, these gaseous products simply escape into the air, so nothing is left behind. The mass has not disappeared — it has moved out of the beaker in the form of gases.
If the same reaction were carried out in a closed container (as in Activity 9.2), we would find that
so mass is conserved. The ‘missing’ mass is only the mass of the escaping gases, exactly like the escaping carbon dioxide in experimental set-up 1 of the chapter.
When 20 g of hydrogen reacts completely with 160 g of oxygen, how much water is formed according to the Law of Conservation of Mass?
By the Law of Conservation of Mass, the total mass of the products equals the total mass of the reactants. Water is the only product, and both reactants are used up completely, so
So 180 g of water is formed. (Check: hydrogen and oxygen are in the ratio 20 : 160 = 1 : 8 by mass, which is exactly the ratio in which they combine to form water.)
A compound consists of 40% sulfur and 60% oxygen by mass. In a sample of the same compound containing 20 g of sulfur, what mass of oxygen must be present to satisfy the Law of Constant Proportions?
By the Law of Constant Proportions, every sample of this compound has sulfur and oxygen in the same ratio by mass:
So the mass of oxygen is always 3 parts for every 2 parts of sulfur. With 20 g of sulfur:
So 30 g of oxygen must be present. Check: the sample has 20 g + 30 g = 50 g in all, and sulfur is 20 out of 50, i.e. 40% — the same composition as stated.
Carbon monoxide (CO) contains carbon and oxygen in the mass ratio of 3:4. How much oxygen will combine with 9 g of carbon to form carbon monoxide?
The ratio carbon : oxygen = 3 : 4 tells us that every 3 g of carbon combines with 4 g of oxygen. For 9 g of carbon:
So 12 g of oxygen combines with 9 g of carbon, forming 9 g + 12 g = 21 g of carbon monoxide.
The Law of Definite Proportions holds true for compounds but not for mixtures. Give reason.
In a compound, the elements are chemically combined. Their atoms join in a fixed whole-number ratio (for example, always 2 hydrogen atoms for every 1 oxygen atom in water), and each kind of atom has a fixed mass. So the ratio of the elements by mass is also fixed — 1 : 8 for hydrogen and oxygen in water, whatever the source of the water.
In a mixture, the components are only physically mixed; no new substance is formed and no atoms are bonded together. The components can therefore be mixed in any proportion — a little sugar or a lot of sugar can be dissolved in the same amount of water, and air can have more or less water vapour in it.
Because a mixture has no fixed composition, the Law of Definite Proportions cannot apply to it; it applies only to compounds, which always have a definite composition by mass.
Students X and Y, both prepared an oxide of copper by combining copper and oxygen in the ratios of 4:1 and 8:2, respectively. Do their results justify the Law of Constant Proportions? Explain.
Yes, their results justify the Law of Constant Proportions. Reduce each ratio to its simplest form:
Both students obtained copper and oxygen in the same ratio, 4 : 1, by mass. Student Y simply used larger amounts (twice as much of each). The law says that a given compound always contains its elements in the same fixed ratio by mass, whatever the amount taken or the method of preparation — and that is exactly what both students found. So both have made the same oxide of copper.
Assertion (A): 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water.
Reason (R): According to Dalton’s Atomic Theory, atoms combine in a simple whole number ratio by mass to form compounds.
Choose the correct option:
Correct option: (iii) A is true, but R is false.
Assertion is true. Hydrogen and oxygen combine in the fixed mass ratio 1 : 8, and 2 : 16 = 1 : 8. The mass is also conserved:
Reason is false. Dalton’s postulate says that atoms combine in a simple whole-number ratio of numbers of atoms — for water, 2 atoms of hydrogen to 1 atom of oxygen. It does not say that the masses combine in a simple whole-number ratio. The mass ratio depends on how heavy the atoms are, and it need not be a simple whole-number ratio at all; for example, sodium and chlorine combine in the mass ratio 23 : 35.5 (Example 9.3). The words ‘by mass’ make the Reason incorrect.
Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule (\(\mathrm{N_2}\)).
A nitrogen atom (atomic number 7, configuration 2, 5) has 5 valence electrons. It needs 3 more electrons to complete its octet.
So each of the two nitrogen atoms shares three of its electrons with the other. This gives three shared pairs (6 shared electrons) between the two atoms, i.e. a triple bond. Each nitrogen atom keeps its remaining 2 electrons as one lone pair.
Counting around each nitrogen: 6 shared electrons + 2 lone electrons = 8, so both atoms now have an octet. The structure is written as N≡N.
The atomic number of fluorine is 9. Explain the formation of the fluorine molecule (\(\mathrm{F_2}\)).
Fluorine has atomic number 9, so its electronic configuration is 2, 7: it has 7 valence electrons and needs just 1 more to complete its octet.
Two fluorine atoms each share one electron with the other. The two shared electrons form one shared pair, which is a single covalent bond. Each atom keeps its other 6 valence electrons as three lone pairs.
Now each fluorine atom has 2 shared + 6 lone = 8 electrons around it (an octet), so the molecule is stable. It is represented as F—F, just like Cl—Cl in Fig. 9.7.
Show the formation of the following molecules:
(i) Carbon dioxide, \(\mathrm{CO_2}\). Carbon (2, 4) has 4 valence electrons and needs 4 more; oxygen (2, 6) has 6 and needs 2 more. Carbon shares two of its electrons with each oxygen atom, and each oxygen shares two of its electrons with carbon. So there are two shared pairs (a double bond) between carbon and each oxygen.
Carbon now has 4 + 4 = 8 electrons around it, and each oxygen has 4 shared + 4 lone = 8. The structure is O=C=O.
(ii) Hydrogen sulfide, \(\mathrm{H_2S}\). Sulfur (2, 8, 6) has 6 valence electrons and needs 2 more; each hydrogen (1) has 1 electron and needs 1 more to complete its duplet. Sulfur shares one electron with each of two hydrogen atoms, forming two single bonds.
Sulfur now has 4 shared + 4 lone = 8 electrons, and each hydrogen has 2. The structure is H—S—H.
(iii) Ammonia, \(\mathrm{NH_3}\). Nitrogen (2, 5) has 5 valence electrons and needs 3 more; each hydrogen needs 1 more. Nitrogen shares one electron with each of three hydrogen atoms, forming three single bonds, and keeps its remaining two electrons as a lone pair.
Nitrogen now has 6 shared + 2 lone = 8 electrons, and every hydrogen has a duplet of 2.
Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.
Neon has atomic number 10, so its electronic configuration is 2, 8. Its outermost (L) shell already holds 8 electrons — a complete octet.
Atoms share, gain or lose electrons only to reach a stable octet (or duplet). Neon is already in that stable state, so it has no tendency to give away, accept or share electrons. Doing so would only make it less stable. That is why neon does not form bonds and exists as single, separate atoms (like helium, which the chapter mentions exists only as atoms because its atoms are stable).
What kind of ion will oxygen (O) form?
Oxygen has atomic number 8, so its configuration is 2, 6: it has 6 valence electrons. It is much easier to gain 2 electrons than to lose 6, so an oxygen atom gains two electrons to complete its octet (2, 8).
It then has 8 protons but 10 electrons, i.e. two extra negative charges. So oxygen forms a negative ion (anion) with a charge of 2−, the oxide ion, \(\mathrm{O^{2-}}\) — as listed in Table 9.1(a) with valency 2.
Fill in the blanks.
Among magnesium and chlorine, magnesium atom can give two electrons to become \(\mathrm{Mg^{2+}}\). However, chlorine can take only one electron to become ________. Now, ________ ion of magnesium and ________ ions of chlorine combine to give magnesium chloride.
Among magnesium and chlorine, magnesium atom can give two electrons to become \(\mathrm{Mg^{2+}}\). However, chlorine can take only one electron to become \(\mathrm{Cl^-}\) (chloride ion). Now, one ion of magnesium and two ions of chlorine combine to give magnesium chloride.
Why one and two? One \(\mathrm{Mg^{2+}}\) ion carries a charge of 2+, and each \(\mathrm{Cl^-}\) carries 1−. Two chloride ions are needed to balance the charge:
So the formula of magnesium chloride is \(\mathrm{MgCl_2}\).
Show the formation of cations of potassium (K) and calcium (Ca) atoms, and the formation of their corresponding chlorides using diagrams.
Electronic configurations: potassium (atomic number 19) is 2, 8, 8, 1; calcium (atomic number 20) is 2, 8, 8, 2; chlorine (atomic number 17) is 2, 8, 7.
Potassium chloride: one \(\mathrm{K^+}\) (charge 1+) balances one \(\mathrm{Cl^-}\) (charge 1−), so the formula is \(\mathrm{KCl}\).
Calcium chloride: one \(\mathrm{Ca^{2+}}\) (charge 2+) needs two \(\mathrm{Cl^-}\) ions to balance it, so the formula is \(\mathrm{CaCl_2}\). In both compounds the oppositely charged ions are held together by the electrostatic force of attraction — the ionic bond.
Illustrate how sodium sulfide (\(\mathrm{Na_2S}\)) is formed.
Sodium (atomic number 11) has the configuration 2, 8, 1: it loses its 1 valence electron to become \(\mathrm{Na^+}\) (2, 8). Sulfur (atomic number 16) has the configuration 2, 8, 6: it needs 2 electrons to complete its octet and becomes the sulfide ion \(\mathrm{S^{2-}}\) (2, 8, 8).
One sodium atom can give only one electron, so two sodium atoms are needed to supply the two electrons that one sulfur atom takes:
The two \(\mathrm{Na^+}\) ions and the one \(\mathrm{S^{2-}}\) ion attract each other strongly (ionic bond). The charges balance:
so the formula is \(\mathrm{Na_2S}\).
Name the following:
Use the prefix system for covalent compounds: the first element keeps its name (no ‘mono-’), the second ends in ‘-ide’, and a prefix gives the number of atoms of the second element.
Write the formula for the following:
Write the formulae for the compounds formed from the following pairs of ions:
What type of chemical bond is present in a solid compound that does not conduct electricity in the solid state but conducts electricity when dissolved in water?
An ionic bond.
Such a compound is made of ions. In the solid, the ions are held in fixed positions in the crystal by strong electrostatic forces, so they cannot move and no current flows. When it is dissolved in water, the ions become free to move through the solution and carry the current. This is exactly the behaviour of ionic compounds such as sodium chloride and copper sulfate (Section 9.6). A covalent compound like sugar may dissolve in water but gives no ions, so it does not conduct even in solution.
Metal M, with two electrons in its valence shell (M shell), reacts with oxygen to form a compound that is slightly soluble in water. Predict its:
The valence shell is the M shell (the third shell), so the configuration of M is 2, 8, 2 — atomic number 12, i.e. M is magnesium. It gives away its two valence electrons to form \(\mathrm{M^{2+}}\); oxygen (2, 6) takes two electrons to form \(\mathrm{O^{2-}}\).
Find the molecular mass of nitric acid (\(\mathrm{HNO_3}\)). Atomic mass — H = 1 u; N = 14 u; O = 16 u.
One molecule of \(\mathrm{HNO_3}\) contains 1 H atom, 1 N atom and 3 O atoms. Add the masses of all the atoms:
The molecular mass of nitric acid is 63 u.
Find the molecular mass of methane (\(\mathrm{CH_4}\)). Atomic mass — C = 12 u; H = 1 u.
One molecule of \(\mathrm{CH_4}\) contains 1 C atom and 4 H atoms:
The molecular mass of methane is 16 u.
Find the formula unit mass of potassium chloride (\(\mathrm{KCl}\)). Atomic mass — K = 39 u; Cl = 35.5 u.
One formula unit of \(\mathrm{KCl}\) contains one \(\mathrm{K^+}\) ion and one \(\mathrm{Cl^-}\) ion:
The formula unit mass of potassium chloride is 74.5 u.
Find the formula unit mass of magnesium hydroxide, \(\mathrm{Mg(OH)_2}\). Atomic mass — Mg = 24 u; O = 16 u; H = 1 u.
One formula unit of \(\mathrm{Mg(OH)_2}\) contains one Mg and two OH groups, i.e. 2 O atoms and 2 H atoms. Work out the bracket first, then multiply by 2 (as in Example 9.7):
The formula unit mass of magnesium hydroxide is 58 u.
A particular element (A) has one electron in its third shell. There is another element (B) with six electrons in its second shell.
A has one electron in its third shell, so its configuration is 2, 8, 1 (atomic number 11, sodium). B has six electrons in its second shell, which is its outermost shell, so its configuration is 2, 6 (atomic number 8, oxygen).
An element X has six electrons in its outer shell and forms a diatomic molecule.
You want to design a new ionic compound, where the total positive charge is 6+ and the total negative charge is 6−. Which of the following combinations gives the correct number of ions?
Work out the total positive and total negative charge for each option:
Correct option: (iii) 2 \(\mathrm{Fe^{3+}}\) and 3 \(\mathrm{O^{2-}}\), giving the neutral compound \(\mathrm{Fe_2O_3}\) (ferric oxide).
Choose the correct statement(s) and correct the false statement(s).
Write the chemical formulae for the following compounds.
Write the formulae of the compounds formed from the following pairs of ions.
Which of the following, in Fig. 9.18, correctly represents \(\mathrm{Cl^-}\) ion (Atomic number of chlorine = 17)?
Correct option: (ii).
A chlorine atom has 17 electrons, arranged 2, 8, 7. The chloride ion \(\mathrm{Cl^-}\) is formed when the atom gains one electron, so it has
Checking each diagram shell by shell (inner to outer):
Note that the nucleus (17 protons) does not change when the ion is formed; only the number of electrons changes.
Determine the formula unit mass of the following substances.
The question does not list atomic masses, so we use the values used in this chapter — H = 1 u, C = 12 u, N = 14 u, O = 16 u, Na = 23 u — and, for phosphorus (not given in the chapter), the standard value P = 31 u.
(i) \(\mathrm{NH_4NO_3}\) contains 2 N, 4 H and 3 O:
(ii) \(\mathrm{H_3PO_4}\) contains 3 H, 1 P and 4 O:
(iii) \(\mathrm{NaHCO_3}\) contains 1 Na, 1 H, 1 C and 3 O:
So the formula unit masses are 80 u, 98 u and 84 u respectively.
Write the formulae for the compounds formed by the reaction of:
Metals lose their valence electrons to form cations; the non-metals gain electrons to complete their octet. Nitrogen (2, 5) gains 3 electrons to form the nitride ion \(\mathrm{N^{3-}}\); sulfur gains 2 to form \(\mathrm{S^{2-}}\); oxygen gains 2 to form \(\mathrm{O^{2-}}\).
Complete the Table 9.3 by writing the formulae of the compounds formed by the cations on the left and the anions at the top. \(\mathrm{LiNO_3}\) is given as an example.
Table 9.3 (cations on the left; anions at the top: \(\mathrm{NO_3^-}\), \(\mathrm{SO_4^{2-}}\), \(\mathrm{PO_4^{3-}}\)):
| Cation | With \(\mathrm{NO_3^-}\) / \(\mathrm{SO_4^{2-}}\) / \(\mathrm{PO_4^{3-}}\) |
|---|---|
| \(\mathrm{NH_4^+}\) | ___ / ___ / ___ |
| \(\mathrm{Li^+}\) | \(\mathrm{LiNO_3}\) / ___ / ___ |
| \(\mathrm{Al^{3+}}\) | ___ / ___ / ___ |
| \(\mathrm{Cu^{2+}}\) | ___ / ___ / ___ |
Criss-cross the charges (only the numbers) and simplify. The anions have charges \(\mathrm{NO_3^-}\) (1−), \(\mathrm{SO_4^{2-}}\) (2−) and \(\mathrm{PO_4^{3-}}\) (3−). The completed table is written below one pair per row so it fits a phone screen:
| Ion pair | Formula |
|---|---|
| \(\mathrm{NH_4^+}\) + \(\mathrm{NO_3^-}\) | \(\mathrm{NH_4NO_3}\) |
| \(\mathrm{NH_4^+}\) + \(\mathrm{SO_4^{2-}}\) | \(\mathrm{(NH_4)_2SO_4}\) |
| \(\mathrm{NH_4^+}\) + \(\mathrm{PO_4^{3-}}\) | \(\mathrm{(NH_4)_3PO_4}\) |
| \(\mathrm{Li^+}\) + \(\mathrm{NO_3^-}\) | \(\mathrm{LiNO_3}\) (given) |
| \(\mathrm{Li^+}\) + \(\mathrm{SO_4^{2-}}\) | \(\mathrm{Li_2SO_4}\) |
| \(\mathrm{Li^+}\) + \(\mathrm{PO_4^{3-}}\) | \(\mathrm{Li_3PO_4}\) |
| \(\mathrm{Al^{3+}}\) + \(\mathrm{NO_3^-}\) | \(\mathrm{Al(NO_3)_3}\) |
| \(\mathrm{Al^{3+}}\) + \(\mathrm{SO_4^{2-}}\) | \(\mathrm{Al_2(SO_4)_3}\) |
| \(\mathrm{Al^{3+}}\) + \(\mathrm{PO_4^{3-}}\) | \(\mathrm{AlPO_4}\) |
| \(\mathrm{Cu^{2+}}\) + \(\mathrm{NO_3^-}\) | \(\mathrm{Cu(NO_3)_2}\) |
| \(\mathrm{Cu^{2+}}\) + \(\mathrm{SO_4^{2-}}\) | \(\mathrm{CuSO_4}\) |
| \(\mathrm{Cu^{2+}}\) + \(\mathrm{PO_4^{3-}}\) | \(\mathrm{Cu_3(PO_4)_2}\) |
Points to notice: when the two charges are equal they cancel to 1 : 1 (\(\mathrm{CuSO_4}\), \(\mathrm{AlPO_4}\)); brackets appear only when there are two or more of the same polyatomic ion (\(\mathrm{(NH_4)_2SO_4}\), \(\mathrm{Al_2(SO_4)_3}\), \(\mathrm{Cu_3(PO_4)_2}\)).
5.3 g of sodium carbonate and 6.0 g of acetic acid react to produce 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium acetate. Verify whether the law of conservation of mass is valid.
Total mass of reactants:
Total mass of products:
Mass of reactants = mass of products = 11.3 g. Hence the Law of Conservation of Mass is valid for this reaction (as in Example 9.1).
If a species has 11 protons, 12 neutrons and 10 electrons then
Two elements, A and B, have the following configurations — A: 2, 8, 5 B: 2, 8, 7
A (2, 8, 5) has atomic number 15 — phosphorus; B (2, 8, 7) has atomic number 17 — chlorine. Both are non-metals.
Assertion (A): Copper sulfate conducts electricity in the molten state but not in the solid state.
Reason (R): Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely.
Choose the correct option:
Correct option: (iii) A is true, but R is false.
Assertion is true. Copper sulfate is an ionic compound. Like other ionic compounds it does not conduct as a solid but conducts when its ions are free to move, i.e. when molten (or dissolved in water).
Reason is false — it states the situation the wrong way round. In the solid state the \(\mathrm{Cu^{2+}}\) and \(\mathrm{SO_4^{2-}}\) ions are held in fixed positions in the crystal lattice and cannot move; in the molten state the lattice breaks down and the ions move freely, which is why it conducts.
The species \(\mathrm{^{27}Al}\), \(\mathrm{^{80}Br^-}\) and \(\mathrm{^{201}Hg^{2+}}\) have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?
Neutrons = mass number − number of protons. Electrons = protons for a neutral atom; one extra for each negative charge; one fewer for each positive charge.
\(\mathrm{^{27}Al}\) (neutral atom):
\(\mathrm{^{80}Br^-}\) (one extra electron):
\(\mathrm{^{201}Hg^{2+}}\) (two electrons lost):
| Species | Electrons | Neutrons |
|---|---|---|
| \(\mathrm{^{27}Al}\) | 13 | 14 |
| \(\mathrm{^{80}Br^-}\) | 36 | 45 |
| \(\mathrm{^{201}Hg^{2+}}\) | 78 | 121 |
Your orientation request has been received.
Saraswati Vidyamandir will contact you soon.
