Chapter 9: Atomic Foundations of Matter

Given data

• Mass of calcium carbonate added: $$m_{\mathrm{CaCO_3}} = 4.0\,\mathrm{g}$$
• Mass of hydrochloric acid added: $$m_{\mathrm{HCl}} = 2.92\,\mathrm{g}$$
• Mass of products obtained:
  
  • Carbon dioxide: $$m_{\mathrm{CO_2}} = 1.76\,\mathrm{g}$$
  • Water: $$m_{\mathrm{H_2O}} = 0.72\,\mathrm{g}$$
  • Calcium chloride: $$m_{\mathrm{CaCl_2}} = 4.44\,\mathrm{g}$$

Step 1: Calculate the total mass of reactants

The reaction is carried out in a closed container, so every gram of each reactant is present inside the system.

\[\text{Total mass of reactants} = m_{\mathrm{CaCO_3}} + m_{\mathrm{HCl}}\]

Substituting the given values,

$$\text{Total mass of reactants} = 4.0\,\mathrm{g} + 2.92\,\mathrm{g}$$
$$\phantom{\text{Total mass of reactants}} = 6.92\,\mathrm{g}$$

Step 2: Calculate the total mass of products

All the products are weighed after the reaction is complete.

\[\text{Total mass of products} = m_{\mathrm{CO_2}} + m_{\mathrm{H_2O}} + m_{\mathrm{CaCl_2}}\]

Inserting the measured masses,

$$\text{Total mass of products} = 1.76\,\mathrm{g} + 0.72\,\mathrm{g} + 4.44\,\mathrm{g}$$

First add the gases and water:
$$1.76\,\mathrm{g} + 0.72\,\mathrm{g} = 2.48\,\mathrm{g}$$

Then add the solid product:
$$2.48\,\mathrm{g} + 4.44\,\mathrm{g} = 6.92\,\mathrm{g}$$

\[\text{Total mass of products} = 6.92\,\mathrm{g}\]

Step 3: Compare the two totals

$$\text{Total mass before reaction} = 6.92\,\mathrm{g}$$
$$\text{Total mass after reaction}\; = 6.92\,\mathrm{g}$$

Since the two masses are identical, no loss or gain of matter is detected inside the closed container.

Conclusion

The total mass of the reactants equals the total mass of the products. Therefore, the experiment satisfies the Law of Conservation of Mass.

Given chemical equation
Carbon + Oxygen → Carbon dioxide

Step 1 · Write the mass data for the complete reaction

For every $$12 \, \mathrm{g}$$ of carbon the reaction uses $$32 \, \mathrm{g}$$ of oxygen and forms $$44 \, \mathrm{g}$$ of $$\mathrm{CO_2}$$.

Step 2 · Find the fixed mass ratio between carbon and carbon dioxide

The ratio of the mass of carbon dioxide produced to the mass of carbon that reacts is

Step 3 · Apply the ratio to $$2.4 \, \mathrm{g}$$ of carbon

If the reacting carbon is only $$2.4 \, \mathrm{g}$$, then the mass of carbon dioxide formed is

Compute the product:

Check by calculating oxygen mass

The mass ratio of oxygen to carbon from the given data is

Hence for $$2.4 \, \mathrm{g}$$ of carbon the oxygen required is

Total mass of products:

which agrees with the earlier result and satisfies the law of conservation of mass.

Therefore, the mass of carbon dioxide produced is $$8.8 \, \mathrm{g}$$.

The compound sodium chloride contains sodium and chlorine in the fixed mass ratio $$23 : 35.5$$.

That is,

Let the mass of chlorine required to combine with $$46\,\mathrm{g}$$ of sodium be $$m\,\mathrm{g}$$.

Cross-multiplying gives

Calculate the product on the left:

$$46 \times 35.5 = 1633$$

Now isolate $$m$$:

Hence, $$71\,\mathrm{g}$$ of chlorine are needed to react completely with $$46\,\mathrm{g}$$ of sodium to form $$\mathrm{NaCl}$$.

Given data

• Atomic mass of hydrogen, $$\mathrm{H} = 1\,\text{u}$$
• Atomic mass of oxygen, $$\mathrm{O} = 16\,\text{u}$$

Step 1 – Identify the number of each type of atom in water

The chemical formula of water is $$\mathrm{H_2O}$$. This shows:

• 2 atoms of hydrogen (the subscript 2 next to H)
• 1 atom of oxygen (no subscript is written, so it is understood to be 1)

Step 2 – Write the expression for molecular mass

The molecular mass is the sum of the atomic masses of all atoms present in one molecule:

Step 3 – Substitute the known values

Step 4 – Perform the arithmetic

Therefore, the molecular mass of water is

Given data

• Chemical formula of carbon dioxide : $$\mathrm{CO_2}$$
• Atomic mass of carbon (C) = $$12\,\text{u}$$
• Atomic mass of oxygen (O) = $$16\,\text{u}$$

Step 1 — Count the atoms present in one molecule

Step 2 — Multiply the atomic mass by the number of atoms

• Mass contributed by carbon: $$1 \times 12\,\text{u} = 12\,\text{u}$$
• Mass contributed by oxygen: $$2 \times 16\,\text{u} = 32\,\text{u}$$

Step 3 — Add the individual masses to obtain the molecular mass

Total molecular mass = $$12\,\text{u} + 32\,\text{u}$$

Therefore,

\[44\,\text{u}\]

The molecular mass of carbon dioxide is $$44\,\text{u}$$.

Step 1 : Note the atomic masses given
For sodium, $$A_{\mathrm{Na}} = 23 \;\text{u}$$
For oxygen, $$A_{\mathrm{O}} = 16 \;\text{u}$$

Step 2 : Identify the number of each kind of atom in one formula unit of $$\mathrm{Na_2O}$$
The formula $$\mathrm{Na_2O}$$ tells us that one formula unit contains:

• 2 atoms of sodium (Na)
• 1 atom of oxygen (O)

Step 3 : Write the expression for the formula unit mass
The formula unit mass is the sum of the individual atomic masses multiplied by their respective number of atoms:
$$m_{\text{f.u.}} = \bigl(2 \times A_{\mathrm{Na}}\bigr) + \bigl(1 \times A_{\mathrm{O}}\bigr)$$

Step 4 : Substitute the numerical values
$$m_{\text{f.u.}} = \bigl(2 \times 23\bigr) + \bigl(1 \times 16\bigr)$$

Step 5 : Do the arithmetic carefully
$$m_{\text{f.u.}} = 46 + 16 = 62$$

Step 6 : State the unit
The unit for atomic and formula masses is the unified atomic mass unit, written as “u”.

Therefore, the formula unit mass of sodium oxide is 62 u.

Given data

• Atomic mass of calcium, $$\mathrm{Ca}$$  = 40 u
• Atomic mass of nitrogen, $$\mathrm{N}$$  = 14 u
• Atomic mass of oxygen, $$\mathrm{O}$$  = 16 u

Step 1 – Count the number of each atom in one formula unit of $$\mathrm{Ca(NO_3)_2}$$

The subscript ‘2’ outside the bracket means there are two nitrate groups, $$\mathrm{(NO_3)^-}$$, in one formula unit.

Step 2 – Multiply each atomic mass by the number of atoms present

• Mass from calcium: $$1 \times 40 \text{ u} = 40 \text{ u}$$
• Mass from nitrogen: $$2 \times 14 \text{ u} = 28 \text{ u}$$
• Mass from oxygen: $$6 \times 16 \text{ u} = 96 \text{ u}$$

Step 3 – Add all the contributions

Total (formula unit) mass = $$40 \text{ u} + 28 \text{ u} + 96 \text{ u} = 164 \text{ u}$$

The formula unit mass of $$\mathrm{Ca(NO_3)_2}$$ is therefore

\[ 164\,\text{u} \]

Step 1 – Recall the formula of water
Pure water is composed of molecules whose formula is $$\mathrm{H_2O}$$. Each molecule therefore contains 2 atoms of hydrogen and 1 atom of oxygen.

Step 2 – Use atomic masses to find the fixed mass ratio
The relative atomic masses are
$$\text{H} : 1\;\text{u}, \qquad \text{O} : 16\;\text{u}$$
Hence the mass of one molecule is
$$\text{mass of }\mathrm{H_2O}=2(1\,\text{u})+16\,\text{u}=18\,\text{u}$$
The mass ratio of hydrogen to oxygen in any sample of water is therefore \[ \frac{\text{mass of H}}{\text{mass of O}} = \frac{2\,\text{u}}{16\,\text{u}} = \frac{1}{8} \]

Step 3 – State the Law of Definite Proportions
The law states that a chemical compound always contains the same elements combined together in the same fixed proportion by mass, irrespective of the method of preparation or the source.

Step 4 – Apply the law to water from different sources
Whether water is collected from rain, a river, a well, or distilled in the laboratory, after removing suspended matter and dissolved impurities every sample shows the same 1 : 8 mass ratio of hydrogen to oxygen. Thus the chemical composition of water is identical in all cases.

Step 5 – Clarify the role of impurities
Naturally occurring water often carries dissolved salts, gases and microscopic particles. These impurities can give samples different tastes, colours or odours, but they do not change the identity of the $$\mathrm{H_2O}$$ molecules themselves. Once impurities are removed (e.g. by distillation or de-ionisation), only pure water remains, exactly the same for every source.

Conclusion
Therefore, after accounting for and removing any impurities, all samples of water obtained from different natural or laboratory sources are chemically identical.

When chemists write the symbol $$\mathrm{O}$$ and the formula $$\mathrm{O_2}$$ they are talking about two different scales of the same element ‑- the atom and the molecule.

• The symbol $$\mathrm{O}$$ (just one O)
  • represents one atom of oxygen.
  • has an atomic mass of $$16\,\text{u}$$ (u = unified atomic mass unit).
  • 1 mol of oxygen atoms is $$6.022\times10^{23}$$ atoms and has a mass of $$16\,\text{g}$$: $$\text{mass of 1 mol O atoms}=16\,\text{u}\times6.022\times10^{23}=16\,\text{g}$$.

• The formula $$\mathrm{O_2}$$
  • represents one molecule of oxygen gas in which two oxygen atoms are chemically bonded.
  • has a molecular mass of $$16\,\text{u}+16\,\text{u}=32\,\text{u}$$.
  • 1 mol of oxygen molecules is $$6.022\times10^{23}$$ molecules and has a mass of $$32\,\text{g}$$. Each of those molecules contains 2 atoms, so the same 1 mol of molecules actually contains $$2\times6.022\times10^{23}$$ oxygen atoms.

Thus:

$$\mathrm{O}$$ = 1 oxygen atom (or 1 mol of atoms = 16 g)
$$\mathrm{O_2}$$ = 1 oxygen molecule made of 2 atoms (or 1 mol of molecules = 32 g)

That is the essential difference between the two symbols.

Step 1 — Identify the nature of the two solids

• Common salt is sodium chloride, $$\mathrm{NaCl}$$. It is an ionic compound, i.e. it is built from positive sodium ions $$\mathrm{Na^+}$$ and negative chloride ions $$\mathrm{Cl^-}$$.
• Ordinary table sugar is sucrose, $$\mathrm{C_{12}H_{22}O_{11}}$$. It is a covalent (molecular) compound in which all atoms are joined by shared-electron (covalent) bonds and the molecule as a whole is electrically neutral.

Step 2 — What happens on dissolving in water?

• Salt: Water molecules surround the $$\mathrm{Na^+}$$ and $$\mathrm{Cl^-}$$ ions and pull them out of the crystal lattice. This process is called dissociation.
  \[ \mathrm{NaCl_{(s)}} \xrightarrow[\text{in water}]{} Na^+_{(aq)} + Cl^-_{(aq)} \]
• Sugar: Water molecules separate the sugar molecules from one another, but the molecules themselves stay intact. No ions are produced.
  \[ \mathrm{C_{12}H_{22}O_{11 ext{(s)}}} \xrightarrow[\text{in water}]{} C_{12}H_{22}O_{11 ext{(aq)}} \]

Step 3 — Electric current in liquids needs mobile charge carriers

When electrodes from a circuit are dipped into a solution and a potential difference is applied, current will flow only if the liquid contains species that can move and carry charge:

• In a salt solution the free ions $$\mathrm{Na^+}$$ and $$\mathrm{Cl^-}$$ migrate toward the oppositely charged electrodes. Their movement constitutes an electric current.
• A sugar solution has only neutral molecules, which cannot migrate under the electric field to give a net charge flow. Hence almost no current passes.

Step 4 — Conclusion using the idea of electrolytes

• Salt solution is an electrolyte: a substance that furnishes ions in solution and therefore conducts electricity.
• Sugar solution is a nonelectrolyte: a substance that dissolves without forming ions and therefore remains practically non-conducting.

Final answer

Dissolved salt conducts electricity because it dissociates into free ions, whereas dissolved sugar does not produce ions and therefore cannot carry current.

Objective

To verify the Law of Conservation of Mass for the reaction

$$\mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl_2} + \mathrm{H_2}$$

Principle

The total mass of the closed system (reactants + apparatus + air trapped inside) must remain unchanged after the chemical change if mass is conserved.

Apparatus and materials

• 250 mL conical flask with airtight rubber stopper
• Small test tube (≈8 cm length)
• Thread or cotton yarn
• Electronic balance (least-count 0.01 g)
• Measuring cylinder (10 mL)
• ≈2 g zinc granules (washed, dried)
• ≈5 mL dilute hydrochloric acid (≈2 M)
• Dropper, spatula, tissue paper

Experimental set-up (describe to the student artist)

• Draw a conical flask. A small test tube is hanging inside by a thread from the stopper. The test tube contains HCl. Zinc granules lie at the bottom of the flask. The stopper fits tightly; no delivery tube is provided (gas will remain in the flask).

Procedure

1. Dry the flask, stopper and small test tube thoroughly. (Moisture alters mass.)
2. Weigh the empty, dry flask with its stopper and record mass $$m_0$$.
3. Place about $$m_\text{Zn}=2.00\;\text{g}$$ zinc granules in the flask. Record combined mass $$m_1$$.
4. Using a dropper, transfer $$V_{\mathrm{HCl}} = 5.0\;\text{mL}$$ dilute HCl into the small test tube. Do not spill any on the outside.
5. Gently suspend the acid-filled test tube inside the flask with thread, ensuring its mouth stays above the zinc. Replace the stopper airtight. Wipe any fingerprints and weigh the entire closed set-up; record $$m_2$$ (this is the initial mass). Typical value: $$m_2 = 203.14\;\text{g}$$.
6. Hold the flask firmly and tilt it so that the acid spills onto the zinc. Swirl lightly to ensure complete mixing, then keep the flask upright. Effervescence (hydrogen gas) begins.
7. When fizzing ceases (≈5 min) place the flask back on the balance without opening it. Record the mass as $$m_3$$ (the final mass).

Observations

Difference $$\Delta m = m_3 - m_2 = -0.03\;\text{g}$$ (within balance error ±0.05 g).

Calculations for theoretical check

Molar masses: $$M_{\mathrm{Zn}} = 65.4\;\text{g mol}^{-1}$$, $$M_{\mathrm{H_2}} = 2.0\;\text{g mol}^{-1}$$.

Moles of zinc taken: $$n_{\mathrm{Zn}} = \dfrac{2.00}{65.4} \approx 0.0306\;\text{mol}$$.

Stoichiometry predicts equal moles of $$\mathrm{H_2}$$ formed, so

Mass of hydrogen produced inside the flask:

$$m_{\mathrm{H_2}} = n_{\mathrm{H_2}} M_{\mathrm{H_2}} = 0.0306 \times 2.0 \approx 0.061\;\text{g}$$.

This hydrogen remains trapped, so total mass theoretically unchanged. The observed 0.03 g loss is smaller than the balance uncertainty; hence experiment agrees with theory.

Result & Conclusion

• Initial mass $$m_2$$ ≈ Final mass $$m_3$$ within experimental error.
• Therefore, during the reaction of zinc with dilute hydrochloric acid, mass is conserved, validating the Law of Conservation of Mass.

Precautions

• Flask must be perfectly dry before weighing.
• Stopper must fit tightly; apply petroleum jelly if necessary.
• Weigh on the same balance; allow bubbling to finish before final weighing.
• Use dilute acid to avoid vigorous reaction that could force gas past the stopper.

Thus the hypothesis is experimentally supported.

Given data

The student places $$10\,\text{g}$$ of liquid ethanol in an open beaker and ignites it. After the flame goes out, absolutely nothing (no ash, no liquid) is found in the beaker.

This observation might tempt one to think that the initial mass of $$10\,\text{g}$$ has "disappeared", apparently contradicting the Law of Conservation of Mass. Let us test this logically and numerically.

1. What exactly does the Law of Conservation of Mass state?

In any closed system the total mass of all the reactants equals the total mass of all the products: no mass is created and none is destroyed during a chemical change.

Key phrase — closed system: all matter must stay inside the vessel so that we can weigh it before and after the reaction.

2. Chemical equation for the combustion of ethanol

Ethanol: $$\mathrm{C_2H_5OH}$$ burns in oxygen to form carbon dioxide and water vapour.

Unbalanced skeleton: $$\mathrm{C_2H_5OH + O_2 \rightarrow CO_2 + H_2O}$$

Balancing C, H, O step by step:

• Carbon: 2 atoms on LHS, so put 2 in front of $$\mathrm{CO_2}$$.
• Hydrogen: 6 atoms on LHS, so put 3 in front of $$\mathrm{H_2O}$$.
• Oxygen count after these insertions: RHS has $$2\times2 + 3\times1 = 7$$ atoms. LHS already has 1 atom inside ethanol, leaving $$7-1 = 6$$ atoms to be supplied by $$\mathrm{O_2}$$, i.e. $$3\,\mathrm{O_2}$$.

Balanced equation:

\[\mathrm{C_2H_5OH + 3\,O_2 \;\longrightarrow\; 2\,CO_2 + 3\,H_2O}\]

3. Check mass equality for one mole of ethanol

The reactant side and product side each weigh exactly $$142\,\text{g}$$. Hence, at the molecular level, mass is perfectly conserved.

4. Why does the beaker look "empty" then?

1. The oxygen needed (part of the reactants) came from the surrounding air, not from the beaker, so it was never part of the "weighed" system.
2. The products $$\mathrm{CO_2}$$ and hot $$\mathrm{H_2O}$$ vapour are gases at room temperature. Being in an open beaker, they simply diffused out into the air. Therefore nothing remained in the vessel to be seen or weighed.

5. Proper application of the law

If the same combustion were carried out in a sealed but otherwise identical container (so that no gas could enter or leave), weighing the container before ignition and after cooling would give identical readings, thereby verifying the Law of Conservation of Mass experimentally.

Conclusion

The disappearance of the visible liquid does not violate the Law of Conservation of Mass. The apparent loss occurs only because the experiment was performed in an open system; mass was merely transferred to the surrounding air.

Given data

• Mass of hydrogen $$m_{\mathrm H}=20\,\mathrm g$$
• Mass of oxygen $$m_{\mathrm O}=160\,\mathrm g$$

The reaction that forms water is

$$2\;\mathrm{H_2}+\;\mathrm{O_2}\;\longrightarrow\;2\;\mathrm{H_2O}$$

Step 1 – Find the molar masses

• $$M_{\mathrm{H_2}}=2\times1=2\,\mathrm{g\,mol^{-1}}$$
• $$M_{\mathrm{O_2}}=2\times16=32\,\mathrm{g\,mol^{-1}}$$
• $$M_{\mathrm{H_2O}}=2\times1+16=18\,\mathrm{g\,mol^{-1}}$$

Step 2 – Use the stoichiometric mass ratio

From the balanced equation, 2 moles of $$\mathrm{H_2}$$ (mass $$2\times2=4\,\mathrm g$$) react with 1 mole of $$\mathrm{O_2}$$ (mass $$32\,\mathrm g$$) to give 2 moles of $$\mathrm{H_2O}$$ (mass $$2\times18=36\,\mathrm g$$).

Thus for every

$$4\,\mathrm g : 32\,\mathrm g : 36\,\mathrm g$$ (of $$\mathrm{H_2} : \mathrm{O_2} : \mathrm{H_2O}$$)

Step 3 – Scale up to the given masses

The factor by which hydrogen mass is increased is

$$k=\frac{m_{\mathrm H}}{4}=\frac{20}{4}=5$$

Multiply all the stoichiometric masses by this same factor $$k$$:

• Required oxygen $$=32\times5=160\,\mathrm g$$ (matches the given 160 g, so O₂ is exactly sufficient).
• Water formed $$=36\times5=180\,\mathrm g$$.

Step 4 – Check with the Law of Conservation of Mass

Total mass of reactants $$=20\,\mathrm g+160\,\mathrm g=180\,\mathrm g$$.

Total mass of product (calculated) $$=180\,\mathrm g$$.

The two totals are equal, so the calculation obeys the law.

Therefore, the mass of water formed is

\[m_{\mathrm{H_2O}}=180\,\mathrm g\]

Step 1 ⟶ Recall the existing rule

Under ordinary conditions atoms combine according to fixed whole-number ratios. This is stated in the Law of Definite Proportions:

"A given compound always contains the same elements combined together in the same fixed mass ratio."

For example, water is always $$2\,\mathrm{H} : 1\,\mathrm{O}$$ (2 atoms of hydrogen to 1 atom of oxygen) so its molecular formula is $$\mathrm{H_2O}$$ and its constant mass ratio is $$\dfrac{2\times 1\;\text{u}}{16\;\text{u}} = 1:8$$.

Step 2 ⟶ Imagine the rule is removed

Suppose atoms could join in any numerical ratio, e.g.

• $$\mathrm{H_3O}$$, $$\mathrm{H_7O_4}$$, $$\mathrm{H_{11}O_2}$$ instead of just $$\mathrm{H_2O}$$
• $$\mathrm{C_5O_{17}}$$, $$\mathrm{C_{28} O}$$ instead of the fixed $$\mathrm{CO_2}$$ or $$\mathrm{CO}$$

No single molecular formula would be privileged; infinitely many variants would be allowed.

Step 3 ⟶ Consequences for substances

1. No well-defined composition
   Each sample of what we now call “water” could have a different H : O ratio. There could be millions of non-identical “waters”. The concept of a pure substance would fail.
2. Physical properties would drift
   Boiling point, melting point, density, colour, taste, etc. depend on the exact atomic ratio. If the ratio can vary continuously, these properties will also vary continuously. A liquid you boiled at $$100^{\circ}\text{C}$$ yesterday might boil at $$73^{\circ}\text{C}$$ today because its composition changed.
3. Chemical equations could not be balanced
   Balancing relies on set numbers of atoms on both sides, e.g.
   $$\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$$
   If $$\mathrm{H_2O}$$ might just as well be $$\mathrm{H_5O_2}$$, there is no single correct coefficient set, so the law of conservation of mass would become impossible to verify experimentally.
4. Predictability and technology would collapse
   Food, medicines, fuels, building materials, plastics, alloys, etc. owe their reliable behaviour to definite composition. Without it, identical performance could never be guaranteed.
5. The periodic table would lose its organising power
   Because compounds would not display regular valencies, trends in groups and periods would blur; chemistry would become an unmanageable catalogue of arbitrary mixtures.

Step 4 ⟶ Final inference

Fixed combining ratios give rise to distinct substances with reproducible properties. If atoms combined in any ratio, the clear distinction between compounds and mixtures would vanish, and the material world would be chaotic and unpredictable.

Given data

• The compound contains 40 % sulfur (S) by mass.
• The compound contains 60 % oxygen (O) by mass.
• In the sample provided, the mass of sulfur is $$m_\mathrm{S}=20\,\mathrm{g}$$.

Objective

Find the mass of oxygen, $$m_\mathrm{O}$$, that must accompany 20 g of sulfur so that the sample still obeys the Law of Constant (Definite) Proportions.

Step 1 · Set up the fixed mass ratio

According to the Law of Constant Proportions, every pure sample of the compound has the same mass ratio

$$\text{mass ratio} = \frac{\text{mass of S}}{\text{mass of O}}$$

From the percentage composition:

$$\frac{40\,\%}{60\,\%}=\frac{40}{60}$$

Simplify the fraction:

$$\frac{40}{60}=\frac{4}{6}=\frac{2}{3}$$

Therefore, every 2 g of sulfur are combined with 3 g of oxygen.

Step 2 · Apply the ratio to the given mass of sulfur

The proportion is maintained, so

$$\frac{m_\mathrm{S}}{m_\mathrm{O}}=\frac{2}{3}$$

Insert the known value $$m_\mathrm{S}=20\,\mathrm{g}$$ and solve for $$m_\mathrm{O}$$:

$$\frac{20}{m_\mathrm{O}}=\frac{2}{3}$$

Cross-multiply:

$$2\,m_\mathrm{O}=20\times 3$$

$$2\,m_\mathrm{O}=60$$

Divide both sides by 2:

$$m_\mathrm{O}=\frac{60}{2}=30\,\mathrm{g}$$

Step 3 · Verify

The calculated masses are 20 g S and 30 g O. Their percentage in the sample is

$$\%\,\mathrm{S}=\frac{20}{20+30}\times100=40\,\%$$

$$\%\,\mathrm{O}=\frac{30}{20+30}\times100=60\,\%$$

The percentages match the original composition, so the answer is consistent.

Conclusion

To satisfy the Law of Constant Proportions, a 20 g portion of sulfur in this compound must be accompanied by 30 g of oxygen.

Given information

• In carbon monoxide $$\mathrm{CO}$$ the masses of carbon and oxygen are always in the fixed ratio $$3 : 4$$ (Law of Constant Proportions).
• Mass of carbon actually taken = $$9\,\mathrm g$$.

Let the mass of oxygen that will combine with this carbon be $$x\,\mathrm g$$.

Set up the proportion

The ratio of carbon to oxygen in any sample of $$\mathrm{CO}$$ must remain $$3:4$$. Therefore

$$ \dfrac{\text{mass of carbon}}{\text{mass of oxygen}} = \dfrac{3}{4} $$

Substituting the known value (carbon = $$9\,\mathrm g$$):

$$ \dfrac{9}{x} = \dfrac{3}{4} $$

Solve for $$x$$

Cross-multiply:

$$ 9 \times 4 = 3 \times x $$

$$ 36 = 3x $$

Divide both sides by 3:

$$ x = \dfrac{36}{3} = 12 $$

Result

The mass of oxygen required is

Step 1 : Recall the statement of the Law of Definite Proportions

The law says that a given chemical compound always contains exactly the same elements combined together in the same fixed mass ratio, no matter how or where the compound is prepared.

Mathematically, if a compound is made of elements A and B, then the ratio

$$\frac{\text{mass of A in the compound}}{\text{mass of B in the compound}} = \text{a constant value}$$

is the same for every sample of that compound.

Step 2 : Why this law must hold for a compound

• A compound is formed by a chemical combination. Atoms of the constituent elements unite in an integer, stoichiometric ratio, fixed by their valencies or combining capacities.
• Thus every molecule of water is exactly $$\mathrm{H_2O}$$ — 2 atoms of hydrogen combined with 1 atom of oxygen.
• Because atomic masses are fixed (H = 1 u, O = 16 u), the mass ratio will always be \[\frac{2\,(1\,\text{u})}{1\,(16\,\text{u})} = \frac{2}{16} = 1:8\]
• Whether we electrolyse water, burn hydrogen in oxygen, or condense steam from the atmosphere, every pure sample of water is forced by its formula to contain hydrogen and oxygen in the 1 : 8 mass ratio. Hence the law is obeyed.

Step 3 : Contrast with a mixture

• A mixture is produced by a physical combination of two or more substances. No chemical bonds have to form, so the components can be taken in any proportion.
• Example 1: Air is a mixture of $$\mathrm{N_2}$$, $$\mathrm{O_2}$$, $$\mathrm{CO_2}$$, water vapour, etc. The composition changes from place to place and day to day (e.g. humid vs. dry air).
• Example 2: Common salt in water can be 5 g per 100 mL, 20 g per 100 mL or any other strength we choose. Each sample is still a ‘salt solution’, yet the mass ratio of $$\mathrm{NaCl}$$ to $$\mathrm{H_2O}$$ is different every time.
• Because there is no fixed chemical formula governing the entire mixture, the relative masses of its constituents are free to vary, so the mass ratio is not constant.

Step 4 : Final reasoning

The Law of Definite Proportions depends on the existence of a definite chemical formula, which only compounds possess. Mixtures lack such a formula; therefore the law cannot and does not apply to them.

Law of Constant Proportions (Proust’s Law)
Elements in a pure chemical compound are always present in a fixed proportion by mass, irrespective of the method of preparation or the amount of reactants taken.

Data collected

• Student X: copper = $$4\,\text{g}$$, oxygen = $$1\,\text{g}$$
• Student Y: copper = $$8\,\text{g}$$, oxygen = $$2\,\text{g}$$

Step 1 – Calculate the mass ratio \(\mathrm{Cu:O}\) for each student

Step 2 – Compare the ratios

Both students obtained exactly the same simplest ratio of masses:

$$\mathrm{Cu:O}=4:1$$

Conclusion

Since the copper-to-oxygen mass ratio in the oxide is constant, regardless of whether the starting amounts were $$4:1$$ or $$8:2$$, the experimental results satisfy the Law of Constant Proportions.

Step 1 : Verify the Assertion (A)
For water

Step 2 : Examine the Reason (R)
Dalton’s postulate actually states: “Atoms of different elements combine in simple whole-number ratios of atoms to form compounds.”
It refers to the number of atoms, not to their masses. Therefore the Reason, which claims a simple whole-number ratio by mass, is incorrect. Hence Reason (R) is false.

Step 3 : Pick the correct option
A is true and R is false ⇒ Option (iii).

Step 1 – Write the electronic configuration of one nitrogen atom

Nitrogen has atomic number 7, so

$$Z = 7 \;\Rightarrow\; \text{electronic configuration} = 2,\,5$$

This means each nitrogen atom possesses five valence electrons.

Step 2 – State the octet requirement

For stability every nitrogen atom tries to attain the nearest noble-gas configuration, i.e. eight electrons (an octet) in its valence shell.

Step 3 – Decide how many electrons must be shared

• Each N atom already has 5 valence electrons.
• To reach an octet it needs $$8 - 5 = 3$$ more electrons.
• Therefore two nitrogen atoms must share three electron pairs (6 electrons in total).

Sharing three pairs creates a triple covalent bond.

Step 4 – Show the distribution of shared and lone-pair electrons

• Out of the five valence electrons on each N, three are contributed to the bond region (one from each N per bond pair).
• The remaining two electrons on each N stay together as a lone pair.

Step 5 – Draw / describe the structural formula

Write the two N symbols side by side, connect them by three lines (≡) representing the triple bond, and place one lone pair (two dots) on the left of the left N and one lone pair on the right of the right N.

Description for the notebook diagram:

• Put N on the left, N on the right.
• Between them draw three parallel lines to show a triple bond: N ≡ N.
• Add two dots (a lone pair) to the outer side of each N.

The completed structure satisfies the octet rule for both atoms: each N counts 2 (its lone pair) + 6 (three shared pairs) = 8 valence electrons.

Final structural formula

$$\mathrm{N{:}\!\!\,\equiv\,\!\!{:}N}$$  (each “:” stands for a lone pair of electrons)

Step 1 – Determine the electronic configuration of a single fluorine atom
Fluorine has atomic number $$Z = 9$$, so each neutral atom contains 9 electrons. Filling the electrons in the order of increasing energy levels we get
$$1s^2 2s^2 2p^5$$
Thus the outer (second) shell contains
$$2s^2 2p^5 \Rightarrow 2 + 5 = 7$$ valence electrons.

Step 2 – Locate the deficiency to complete the octet
An octet requires 8 electrons in the valence shell. Each F atom already has $$7$$, so it lacks
$$8 - 7 = 1$$ electron.
Therefore each fluorine atom needs one additional electron to acquire the stable noble-gas configuration of neon.

Step 3 – Explain how two F atoms help each other
If two fluorine atoms come close, each can share one electron with the other. The shared pair belongs simultaneously to both atoms and is represented as a single covalent bond.
Verbal Lewis description to draw:
• Write two F symbols side by side.
• Place six dots around each symbol to show the three lone pairs on every atom.
• Place one dot from the left-hand atom and one dot from the right-hand atom between the two symbols; circle or connect this pair to emphasise that it is shared.
This produces the structure
$$\mathrm{F\,–\,F}$$

Step 4 – Check the octet for each atom
After sharing, each fluorine “sees”

• its own 6 non-bonding electrons (3 lone pairs)
• the 2 electrons in the shared pair

Step 5 – State the final molecule
Because exactly one covalent bond is formed between the two identical atoms, the resulting molecule is diatomic fluorine:
\[ \boxed{\mathrm{F_2}} \]

Step 1 — Write the electronic configuration of neon

The atomic number of neon is $$Z = 10$$, which means a neutral neon atom possesses $$10$$ protons and (in the neutral state) $$10$$ electrons.

Electrons are filled in energy levels (shells) according to the rule $$2n^{2}$$, where $$n$$ is the shell number.

• K-shell (first shell, $$n = 1$$) can hold up to $$2 \times 1^{2} = 2$$ electrons.
• L-shell (second shell, $$n = 2$$) can hold up to $$2 \times 2^{2} = 8$$ electrons.

Thus the $$10$$ electrons of neon are arranged as

$$2\,\text{(K-shell)} + 8\,\text{(L-shell)} = 10$$.

Therefore the electronic configuration of neon is often written as

$$\mathrm{Ne\ (2,\,8)}$$.

Step 2 — Identify the valence shell and its occupancy

The valence shell of an atom is its outermost occupied shell. For neon, the L-shell is the valence shell and already contains $$8$$ electrons:

$$\text{Valence electrons of }\mathrm{Ne} = 8$$.

Step 3 — Recall the octet rule

According to the octet rule, atoms tend to achieve a stable configuration of $$8$$ electrons in their valence shell (an octet) by losing, gaining or sharing electrons. When this shell is completely filled, the potential energy of the electrons is at a minimum and the atom is chemically stable.

Step 4 — Explain why transfer or sharing is unnecessary for neon

• Because neon already possesses an octet (its L-shell holds the maximum possible $$8$$ electrons), it has no energetic incentive to gain electrons: adding any extra electron would force it into the higher-energy M-shell.
• Similarly, neon has no tendency to lose electrons; removing even one electron would break the stable octet and require energy input.
• Finally, since both gaining/losing are unfavourable, neon also does not need to share electrons via covalent bonding to reach stability; it is already stable.

Step 5 — State the conclusion

Neon does not take part in chemical reactions that involve transfer (ionic bonding) or sharing (covalent bonding) of its valence electrons. In other words, neon is chemically inert.

Step 1 – Recall why atoms are normally “invisible”

• The diameter of an atom is of the order of $$10^{-10}\,\text{m}$$ (one ångström).
• Visible light has wavelengths $$\lambda \approx 4\times10^{-7}\,\text{m}$$ to $$7\times10^{-7}\,\text{m}$$, about 1 000 times larger than an atom. According to the wave-nature of light we cannot resolve objects that are much smaller than the wavelength used.
• Hence ordinary microscopes cannot show atoms; scientists infer their presence from indirect evidence (Brownian motion, X-ray diffraction, etc.).

Step 2 – Imagine a device that really “photographs” single atoms

Suppose we invent an instrument whose resolving power is better than $$10^{-10}\,\text{m}$$, perhaps using a beam of electrons or some completely new technology, and we can display the image directly on a screen like an everyday camera.

Step 3 – How it would help scientists

1. Direct confirmation of atomic theories
   Seeing the exact arrangement of $$\mathrm{H}$$ and $$\mathrm{O}$$ atoms in a single molecule of water would give visual proof of the atomic-molecular model that we now accept mainly on indirect grounds.
2. Faster materials research
   Engineers could watch in real time how atoms migrate in a metal on heating, or how dislocations move when the metal is bent, and therefore design stronger alloys much more quickly.
3. Observation of chemical reactions frame by frame
   At present reaction mechanisms are deduced from products, rate data and spectroscopy. A true atomic-camera would let chemists “film” the breaking and making of bonds, verifying transition-state theories directly.
4. Biology at atomic resolution
   Proteins, viruses and cell membranes could be inspected atom by atom without the need for complex crystallisation or computer reconstruction used in X-ray crystallography and cryo-EM.
5. Quality control at the ultimate scale
   Chip manufacturers could count the exact number and position of dopant atoms in a transistor channel, pushing Moore’s law further.

Step 4 – Challenges and new problems it would create

1. Disturbance of the very atoms observed
   To see an object you must interact with it. A particle beam energetic enough to resolve $$10^{-10}\,\text{m}$$ carries enormous momentum $$p=\dfrac{h}{\lambda}$$. It could knock the atoms out of their natural positions, altering the system you wanted to study.
2. Radiation damage
   High-energy photons/electrons would break chemical bonds and create ions and free radicals, destroying delicate biological samples.
3. Unmanageable data volume
   A cubic centimetre of solid contains roughly $$N=\dfrac{\rho N_A}{M}\approx10^{23}$$ atoms. Storing even one grayscale pixel per atom every microsecond would flood global data centres within minutes.
4. Loss of privacy at the nanoscale
   If such imaging became portable, industrial espionage could copy patented nanodevices simply by “looking” at them; new legal frameworks would be needed.
5. Ethical and safety limits
   Observing living tissue atom-by-atom might require radiation doses fatal to the organism. Society would have to balance knowledge gained versus harm done.

Step 5 – Balanced conclusion

Therefore, being able to see atoms directly would be a revolutionary tool confirming theories, speeding innovation and deepening our grasp of matter. Yet the very act of imaging at that scale introduces technical, ethical and safety challenges that scientists would have to solve before routine use becomes possible.

Step 1 — Write the electronic configuration of oxygen

The atomic number of oxygen is 8, so its electrons are arranged as

$$K\text{-shell}:\;2 \quad L\text{-shell}:\;6$$

or, in the long form, $$1s^2\,2s^2\,2p^4$$.

Step 2 — Locate the valence electrons

Only the outermost (L) shell decides chemical behaviour. Oxygen has

$$6$$ valence electrons.

Step 3 — Reach the octet

To attain the stable octet (8 electrons), oxygen can

• lose its 6 valence electrons, or
• gain 2 additional electrons.

Losing 6 electrons would require far more energy than gaining 2, so oxygen gains electrons.

Step 4 — Determine charge on the ion

Each electron gained adds one unit of negative charge. Gaining $$2$$ electrons gives

$$\text{charge} = -2$$.

Step 5 — Name and symbol of the ion

An ion carrying a negative charge is called an anion. Thus oxygen forms a divalent anion known as the oxide ion:

\[\mathrm{O^{2-}}\]

First, recall how atoms form ions in order to obtain the stable electronic configuration of the nearest noble gas.

• Magnesium: atomic number = 12. Its electronic configuration is $$2,\;8,\;2$$. By losing its two outer-shell electrons it achieves the neon configuration $$2,\;8$$ and becomes the divalent cation

\[ \mathrm{Mg \;\longrightarrow\; Mg^{2+} + 2e^-} \]

• Chlorine: atomic number = 17. Its electronic configuration is $$2,\;8,\;7$$. By gaining one electron it attains the argon configuration $$2,\;8,\;8$$ and becomes the monovalent anion

\[ \mathrm{Cl + e^- \;\longrightarrow\; Cl^-} \]

Thus chlorine forms the chloride ion, $$\mathrm{Cl^-}$$.

To combine into an electrically neutral compound, the total positive charge must balance the total negative charge. One $$\mathrm{Mg^{2+}}$$ ion carries $$+2$$ charge, while one $$\mathrm{Cl^-}$$ ion carries $$-1$$ charge. Therefore we need two chloride ions to balance one magnesium ion:

\[ \mathrm{Mg^{2+} + 2\,Cl^- \;\longrightarrow\; MgCl_2} \]

Magnesium chloride thus consists of
one $$\mathrm{Mg^{2+}}$$ ion and two $$\mathrm{Cl^-}$$ ions.

Hence the completed statement is:

“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 $$\mathrm{Mg^{2+}}$$ ion of magnesium and two $$\mathrm{Cl^-}$$ ions of chlorine combine to give magnesium chloride.”

1  Potassium atom → Potassium cation

• Atomic number of potassium = 19.
  Electron distribution: $$2,\,8,\,8,\,1$$ (the fourth shell has one valence electron).
• To achieve the stable octet of argon, the atom loses this single valence electron:
  $$\mathrm{K \;\longrightarrow\; K^+ + e^-}$$
• Electron distribution after loss: $$2,\,8,\,8$$ (same as argon).
• Diagram to draw: one large circle for the K nucleus, four concentric shells with electrons 2, 8, 8, 1; an arrow showing the outermost electron leaving; a second picture of the ion with three shells (2, 8, 8) and the symbol K⁺ inside.

2  Calcium atom → Calcium cation

• Atomic number of calcium = 20.
  Electron distribution: $$2,\,8,\,8,\,2$$.
• It attains the noble-gas configuration by losing two valence electrons:
  $$\mathrm{Ca \;\longrightarrow\; Ca^{2+} + 2e^-}$$
• Electron distribution after loss: $$2,\,8,\,8$$.
• Diagram to draw: four-shell picture for Ca with two outer electrons; two arrows showing both electrons leaving; second picture of Ca²⁺ with three shells (2, 8, 8).

3  Formation of potassium chloride (KCl)

• Chlorine atom: atomic number 17, electron distribution $$2,\,8,\,7$$. It needs one electron to complete its octet.
• Ionic bond formation:
    $$\mathrm{K^+ + Cl^- \;\longrightarrow\; KCl}$$
• Dot-and-cross diagram to draw:
  1. Write K with one cross (its valence electron).
  2. Write Cl with seven dots (its own electrons).
  3. Arrow from the cross to the Cl symbol.
  4. Show the resulting bracketed ions: [K]⁺ and [··Cl··]⁻ with eight dots around Cl.
  5. Place the two ions side by side and label the compound KCl.

4  Formation of calcium chloride (CaCl₂)

• Each Ca atom provides two electrons; each Cl atom needs one. Hence one Ca atom combines with two Cl atoms.
• Ionic equation:
  $$\mathrm{Ca^{2+} + 2Cl^- \;\longrightarrow\; CaCl_2}$$
• Dot-and-cross diagram to draw:
  1. Write Ca with two crosses.
  2. Draw two separate Cl atoms, each with seven dots.
  3. Arrow from each Ca cross to a different Cl atom.
  4. Resulting ions: [Ca]²⁺ (no dots) and two [··Cl··]⁻ (each with eight dots).
  5. Arrange them as [Ca]²⁺ between the two chloride ions to represent the formula CaCl₂.

Key points to remember

• Metals form cations by losing electrons, non-metals form anions by gaining electrons.
• Ionic compounds are electrically neutral: total positive charge = total negative charge.

Step 1 – Write the atomic numbers and electronic configurations

For sodium, $$Z_{\mathrm{Na}} = 11$$, so the electronic configuration is $$2,\;8,\;1$$ (one valence electron).

For sulphur, $$Z_{\mathrm{S}} = 16$$, so the electronic configuration is $$2,\;8,\;6$$ (six valence electrons).

Step 2 – Recognise the tendency of each atom

• Each sodium atom can obtain a stable octet by losing its single valence electron, becoming a monovalent cation $$\mathrm{Na^{+}}$$.
• The sulphur atom attains a stable octet by gaining two electrons, becoming a divalent anion $$\mathrm{S^{2-}}$$.

Step 3 – Show the electron-transfer equations

Loss of an electron by each sodium atom:

$$\mathrm{Na} \;\longrightarrow\; \mathrm{Na^{+}} + e^{-}$$

Since two electrons are required in total, the process happens twice:

$$2\bigl( \mathrm{Na} \longrightarrow \mathrm{Na^{+}} + e^{-}\bigr)$$

Gain of two electrons by sulphur:

$$\mathrm{S} + 2e^{-} \;\longrightarrow\; \mathrm{S^{2-}}$$

Step 4 – Combine the half-equations

Adding the two sodium losses to the sulphur gain gives the overall ionic reaction

Step 5 – Write the formula from the ion charges

The compound must be electrically neutral. Two monovalent $$\mathrm{Na^{+}}$$ ions (total charge $$+2$$) balance one divalent $$\mathrm{S^{2-}}$$ ion (charge $$-2$$), so the simplest ratio is $$2{:}1$$, giving the empirical formula $$\mathrm{Na_2S}$$.

Step 6 – Electron-dot (Lewis) representation

• Draw the sulphur symbol with six dots (its valence electrons).
• Place one crossed electron from each of the two sodium atoms next to the sulphur, showing their transfer.
• After transfer, show [S] surrounded by eight electrons (an octet) and square brackets with charge $$2-$$. Each sodium symbol is shown with no valence electrons inside [ ]⁺.

This diagram makes it clear that an ionic bond is produced—electrostatic attraction between the opposite charges—resulting in the crystalline ionic compound $$\mathrm{Na_2S}$$.

Hence, sodium sulfide is formed by the transfer of one electron from each of two sodium atoms to a single sulphur atom.

Step 1 – Recall the electrical behaviour of different bond types

• Covalent (molecular) solids consist of neutral molecules. They lack mobile charge-carriers both in the solid state and after dissolving in water (most do not even dissolve), so they do not conduct electricity in either situation.
• Metallic solids contain a ‘sea’ of delocalised electrons. These electrons are mobile inside the crystal, so metallic solids do conduct in the solid state.
• Ionic (electrovalent) solids are built from oppositely charged ions held in a rigid lattice. In the solid state the ions are fixed in position, therefore no conduction occurs. When the lattice melts or when the substance dissolves in water, the ions become free to move; the resulting melt or aqueous solution does conduct electricity.

Step 2 – Match the given behaviour with the descriptions

The compound in the question

• “does not conduct electricity in the solid state”  ➔  rules out metallic bonding;
• “conducts electricity when dissolved in water”  ➔  requires mobile ions in solution, a feature of ionic compounds but not of covalent ones.

Step 3 – State the bond type

The only bond type that fits both observations is the ionic (electrovalent) bond.

Step 1 ▶ Determine the valency of the metal M
Valence-shell data: the atom has two electrons in its outermost (M) shell.
This is the same electronic situation as the alkaline–earth metals (Group 2).
Hence the metal atom needs to lose those two electrons to acquire the stable noble-gas configuration.
Therefore its valency is
$$\text{valency of }\mathrm M = 2$$
and the ion produced is the divalent cation $$\mathrm{M^{2+}}$$.

Step 2 ▶ Work out the formula with oxygen
Oxygen (group 16) has six valence electrons and attains the octet by gaining two electrons, giving the oxide ion $$\mathrm{O^{2-}}$$; its valency is 2.
To obtain an electrically neutral compound the charges must cancel:

The simplest whole-number ratio is therefore 1 : 1, giving the empirical formula
\[\mathrm{MO}\]

Step 3 ▶ Nature of bonding
Because electrons are transferred from the metal atom to the oxygen atom, the bond formed is an ionic (electrovalent) bond.

Step 4 ▶ Electrical behaviour of its aqueous solution
The solid $$\mathrm{MO}$$ is only slightly soluble in water, but the small amount that dissolves reacts with water to form the hydroxide:
$$\mathrm{MO + H_2O \longrightarrow M(OH)_2}$$
which dissociates as
$$\mathrm{M(OH)_2 \rightarrow M^{2+} + 2\,OH^-}$$
The presence of free $$\mathrm{M^{2+}}$$ and $$\mathrm{OH^-}$$ ions makes the solution capable of conducting an electric current. (Because the solubility is low, the conductivity will be modest, yet it is definitely a conductor.)

Final predictions

1. Formula of the oxide: $$\mathrm{MO}$$
2. Type of bond: ionic (electrovalent)
3. Conductivity of aqueous solution: conducts electricity (ionic conductor)

Step 1 | Write the molecular formula
For nitric acid the molecular formula is $$\mathrm{HNO_3}$$.

Step 2 | Count the number of each type of atom

Step 3 | Write the given atomic masses
Hydrogen: $$1\,\text{u}$$    Nitrogen: $$14\,\text{u}$$    Oxygen: $$16\,\text{u}$$

Step 4 | Multiply atomic mass by the number of atoms

• Hydrogen contribution: $$1\times1=1\,\text{u}$$
• Nitrogen contribution: $$14\times1=14\,\text{u}$$
• Oxygen contribution: $$16\times3=48\,\text{u}$$

Step 5 | Add all contributions to get the molecular mass
$$1\,\text{u}+14\,\text{u}+48\,\text{u}=63\,\text{u}$$

The molecular mass of nitric acid, therefore, is

\[63\,\text{u}\]

For any molecule, its molecular mass is obtained by adding up the atomic masses of all the atoms present in one molecule of the substance.

Given the molecular formula of methane is $$\mathrm{CH_4}$$, each molecule contains

• 1 atom of carbon (C)
• 4 atoms of hydrogen (H)

Step 1  Write the atomic masses provided: $$M_{\text{C}} = 12\,\text{u}, \; M_{\text{H}} = 1\,\text{u}$$

Step 2  Multiply each atomic mass by the corresponding number of atoms in the molecule:

• Contribution of carbon: $$1 \times 12\,\text{u} = 12\,\text{u}$$
• Contribution of hydrogen: $$4 \times 1\,\text{u} = 4\,\text{u}$$

Step 3  Add the contributions to obtain the molecular mass:

$$\text{Molecular mass of } \mathrm{CH_4} = 12\,\text{u} + 4\,\text{u} = 16\,\text{u}$$

The final result is therefore

\[\boxed{16\,\text{u}}\]

Concept  recall: For an ionic compound such as $$\mathrm{KCl}$$ we talk about formula-unit mass, not molecular mass, because the crystal does not contain discrete molecules. The formula-unit mass is obtained by adding the atomic masses of all atoms present in the empirical formula.

1. Write the empirical (chemical) formula of potassium chloride:
   $$\mathrm{KCl}$$ consists of one potassium atom (K) and one chlorine atom (Cl).

2. Note the given atomic masses:
   Potassium, $$\mathrm{K} : 39\,\text{u}$$
   Chlorine, $$\mathrm{Cl} : 35.5\,\text{u}$$

3. Add the atomic masses because one atom of each element is present in the formula unit:
   $$\text{Formula-unit mass} = 39\,\text{u} + 35.5\,\text{u}$$

4. Carry out the addition step by step:
   $$39 + 35.5 = 74.5$$

The formula-unit mass of potassium chloride is therefore

\[74.5\,\text{u}\]

We have to add the atomic masses of every atom present in one formula unit of $$\mathrm{Mg(OH)_2}$$.

1. Count the atoms.
   In one formula unit of $$\mathrm{Mg(OH)_2}$$ there are:
   
   • Mg : 1 atom
   • O : 2 atoms (the group $$\mathrm{OH}$$ occurs twice)
   • H : 2 atoms (for the same reason)
2. Write the given atomic masses.
   $$\text{Mg} = 24\,\text{u}$$,  $$\text{O} = 16\,\text{u}$$,  $$\text{H} = 1\,\text{u}$$
3. Multiply by the number of atoms.
   

   Element	No. of atoms	Mass per atom (u)	Total mass (u)
   Mg	1	24	$$1\times24=24$$
   O	2	16	$$2\times16=32$$
   H	2	1	$$2\times1=2$$

4. Add all the contributions.
   $$24\,\text{u}+32\,\text{u}+2\,\text{u}=58\,\text{u}$$

Hence,

\[\text{Formula unit mass of } \mathrm{Mg(OH)_2}=58\,\text{u}\]

Given goal: the ionic compound must be electrically neutral, i.e. the algebraic sum of all positive and negative charges must be zero.
Therefore we need:

\[\text{Total positive charge}= +6, \qquad \text{Total negative charge}= -6\]

Analyse every combination one by one.

1. Option (i): $$2\,\mathrm{Al^{3+}}$$ and $$3\,\mathrm{Cl^-}$$

   • Positive charge: $$2\times(+3)=+6$$
   • Negative charge: $$3\times(-1)=-3$$

   The algebraic sum is $$+6+(-3)=+3\neq0$$.
   Not neutral  ⇒  reject.

2. Option (ii): $$3\,\mathrm{Mg^{2+}}$$ and $$1\,\mathrm{PO_4^{3-}}$$

   • Positive charge: $$3\times(+2)=+6$$
   • Negative charge: $$1\times(-3)=-3$$

   Sum: $$+6+(-3)=+3\neq0$$.
   Not neutral  ⇒  reject.

3. Option (iii): $$2\,\mathrm{Fe^{3+}}$$ and $$3\,\mathrm{O^{2-}}$$

   • Positive charge: $$2\times(+3)=+6$$
   • Negative charge: $$3\times(-2)=-6$$

   Sum: $$+6+(-6)=0$$.
   Charges cancel exactly  ⇒  neutral compound obtained.

4. Option (iv): $$3\,\mathrm{Ca^{2+}}$$ and $$2\,\mathrm{SO_4^{2-}}$$

   • Positive charge: $$3\times(+2)=+6$$
   • Negative charge: $$2\times(-2)=-4$$

   Sum: $$+6+(-4)=+2\neq0$$.
   Not neutral  ⇒  reject.

Only option (iii) satisfies the requirement of total $$+6$$ and total $$-6$$, hence gives a neutral ionic compound $$\mathrm{Fe_2O_3}$$.

Step 1 — Recall the data for chlorine
For a neutral chlorine atom: atomic number = 17, therefore it possesses $$17$$ protons and (neutral state) $$17$$ electrons.
Electronic configuration (maximum 2 electrons in K-shell, 8 in L-shell, etc.):
$$\text{Cl (neutral)}: \; 2,\;8,\;7$$

Step 2 — Convert the atom into the chloride ion
A chloride ion $$\mathrm{Cl^{-}}$$ is formed when the atom gains one extra electron:
$$17\;e^- + 1\;e^- \;\longrightarrow\; 18\;e^-$$
Thus, for $$\mathrm{Cl^{-}}$$

• protons (positive charges) = 17 (unchanged, they reside in the nucleus);
• electrons = 18.

Step 3 — Distribute the 18 electrons into shells
Applying the 2 $$n^{2}$$ rule (maximum electrons a shell n may hold):

• K-shell (n = 1): maximum 2 ⇒ fill 2.
• L-shell (n = 2): maximum 8 ⇒ fill 8.
• M-shell (n = 3): electrons left = 18 − (2+8)=8 ⇒ place 8.

Step 4 — Match with the given diagrams
You must choose the diagram in which the nucleus is surrounded by three concentric shells containing 2, 8 and 8 electrons respectively (total 18).
Examining Fig. 9.18, only option (iv) shows exactly this 2–8–8 arrangement, while the other three options either end with 7 electrons in the outermost shell or have the wrong total count.

Conclusion
Diagram (iv) is the correct representation of the $$\mathrm{Cl^{-}}$$ ion.

Key idea: An ionic compound is electrically neutral, so the total positive charge supplied by the cation(s) must exactly cancel the total negative charge supplied by the anion(s). To obtain the simplest (empirical) formula, we look for the lowest common multiple (LCM) of the magnitudes of the two charges and then decide how many ions of each kind are needed to reach that LCM.

In symbolic language, if the cation has charge $$+m$$ and the anion has charge $$-n$$, the LCM of $$m$$ and $$n$$ is $$\text{LCM}(m,n)$$. The number of cations required is $$\dfrac{\text{LCM}(m,n)}{m}$$ and the number of anions required is $$\dfrac{\text{LCM}(m,n)}{n}$$. These become the subscripts in the formula. Any subscript equal to 1 is omitted, and poly-atomic ions are enclosed in round brackets if the subscript is greater than 1.

We carry this out for every pair in Table 9.3.

The required entries are now placed in the original grid.

All cells have been filled with the correct neutral chemical formulae.

Law of conservation of mass: In a chemical reaction, the total mass of the reactants equals the total mass of the products.

Given data

• Mass of sodium carbonate $$\mathrm{(Na_2CO_3)} = 5.3 \, \mathrm{g}$$
• Mass of acetic acid $$\mathrm{(CH_3COOH)} = 6.0 \, \mathrm{g}$$
• Mass of carbon dioxide $$\mathrm{(CO_2)} = 2.2 \, \mathrm{g}$$
• Mass of water $$\mathrm{(H_2O)} = 0.9 \, \mathrm{g}$$
• Mass of sodium acetate $$\mathrm{(CH_3COONa)} = 8.2 \, \mathrm{g}$$

Step 1 – Total mass of reactants

Reactants are sodium carbonate and acetic acid.

$$\text{Total mass of reactants} = 5.3 \, \mathrm{g} + 6.0 \, \mathrm{g}$$

$$ = 11.3 \, \mathrm{g}$$

Step 2 – Total mass of products

Products are carbon dioxide, water and sodium acetate.

$$\text{Total mass of products} = 2.2 \, \mathrm{g} + 0.9 \, \mathrm{g} + 8.2 \, \mathrm{g}$$

$$ = 11.3 \, \mathrm{g}$$

Step 3 – Comparison

\[11.3 \, \mathrm{g \; (reactants)} = 11.3 \, \mathrm{g \; (products)}\]

Since the total mass before and after the reaction is the same, the law of conservation of mass is verified.

Step 1 – Check the Assertion (A)
In solid $$\mathrm{CuSO_4}$$ each $$\mathrm{Cu^{2+}}$$ and $$\mathrm{SO_4^{2-}}$$ ion is locked in a rigid ionic lattice. Because the ions cannot leave their fixed positions, no charge can flow, so solid copper sulfate is a non-conductor.
On melting, the lattice breaks down and the ions become mobile in the liquid. Free ions can carry charge from one electrode to the other, therefore molten $$\mathrm{CuSO_4}$$ does conduct electricity.
Hence the Assertion is true.

Step 2 – Check the Reason (R)
R says: “Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely.”
This is exactly the reverse of what really happens (ions are fixed in the solid and free in the molten state). Therefore the Reason is false.

Step 3 – Match with the options
• A is true.
• R is false.
The only choice that fits is option (iii).

Key ideas to be used

• For any nuclide the mass number is $$A = Z + N$$, where $$Z$$ is the number of protons and $$N$$ is the number of neutrons.
• The symbol always writes $$A$$ at the top-left of the element symbol, e.g. $$\mathrm{^{27}Al}$$ means $$A = 27$$.
• The electronic charge:
• Neutral atom → electrons $$= Z$$.
• Negative ion with charge $$n^-$$ → electrons $$= Z + n$$ (extra electrons).
• Positive ion with charge $$n^+$$ → electrons $$= Z - n$$ (electrons lost).

We now apply these facts one species at a time.

1. $$\mathrm{^{27}Al}$$

Given: $$Z = 13$$, $$A = 27$$. It is neutral.

Neutrons: $$N = A - Z = 27 - 13 = 14$$.

Electrons (neutral): $$13$$.

2. $$\mathrm{^{80}Br^-}$$

Given: $$Z = 35$$, $$A = 80$$, ion charge $$= -1$$.

Neutrons: $$N = A - Z = 80 - 35 = 45$$.

Electrons: neutral atom would have $$35$$, the extra $$1$$ electron gives
$$\text{electrons} = 35 + 1 = 36$$.

3. $$\mathrm{^{201}Hg^{2+}}$$

Given: $$Z = 80$$, $$A = 201$$, ion charge $$= +2$$.

Neutrons: $$N = A - Z = 201 - 80 = 121$$.

Electrons: the neutral atom has $$80$$, losing $$2$$ electrons gives
$$\text{electrons} = 80 - 2 = 78$$.

Summary table