Textbook - ATOMS AND MOLECULES Questions in English

(A) The Law of Conservation of Mass states that mass can neither be created nor destroyed during a chemical reaction.
This means that the total mass of the reactants must be equal to the total mass of the products.
Step $1$: Calculate the total mass of the reactants $(LHS)$:
$\text{Mass of reactants} = \text{Mass of sodium carbonate} + \text{Mass of acetic acid} = 5.3 \, g + 6 \, g = 11.3 \, g$.
Step $2$: Calculate the total mass of the products $(RHS)$:
$\text{Mass of products} = \text{Mass of sodium acetate} + \text{Mass of carbon dioxide} + \text{Mass of water} = 8.2 \, g + 2.2 \, g + 0.9 \, g = 11.3 \, g$.
Step $3$: Compare $LHS$ and $RHS$:
Since $LHS = RHS = 11.3 \, g$,the total mass remains conserved.
Therefore,these observations are in agreement with the law of conservation of mass.

(B) The 'Law of Constant Proportions' states that the elements in a chemical compound are always present in a definite proportion by mass.
According to the given ratio,$1 \,g$ of hydrogen gas combines with $8 \,g$ of oxygen gas to form water.
Therefore,to react completely with $3 \,g$ of hydrogen gas,the required mass of oxygen gas is:
$3 \times 8 \,g = 24 \,g$.
Thus,$24 \,g$ of oxygen gas is required.

(A) The postulate of Dalton's atomic theory that is the result of the law of conservation of mass is: 'Atoms are indivisible particles,which cannot be created or destroyed in a chemical reaction.'
This law states that mass can neither be created nor destroyed in a chemical reaction,which directly aligns with the idea that the total number of atoms remains constant before and after the reaction.

(B) The postulate of Dalton's atomic theory that explains the 'law of definite proportions' is:
'The relative number and kinds of atoms are constant in a given compound.'
This means that in any chemical compound,the elements are always present in a fixed ratio by mass,which is a direct consequence of the fixed ratio of the number of atoms of different elements forming the compound.

(N/A) One atomic mass unit $(amu)$ is a mass unit equal to exactly one-twelfth $(1/12^{th})$ the mass of one atom of carbon-$12$.
The relative atomic masses of all the elements have been determined with respect to an atom of carbon-$12$.

(B) An atom is extremely small in size,making it impossible to see with the naked eye.
Generally,the radius of an atom is on the order of nanometers $(nm)$. For example,the atomic radius of a hydrogen atom is approximately $10^{-10} \, m$ (or $0.1 \, nm$).

(N/A) To write the chemical formulae,we use the valencies of the constituent ions:
$(i)$ Sodium $(Na^+)$ has a valency of $1$,and Oxide $(O^{2-})$ has a valency of $2$. By criss-crossing the valencies,we get $Na_2O$.
$(ii)$ Aluminium $(Al^{3+})$ has a valency of $3$,and Chloride $(Cl^-)$ has a valency of $1$. By criss-crossing the valencies,we get $AlCl_3$.
$(iii)$ Sodium $(Na^+)$ has a valency of $1$,and Sulphide $(S^{2-})$ has a valency of $2$. By criss-crossing the valencies,we get $Na_2S$.
$(iv)$ Magnesium $(Mg^{2+})$ has a valency of $2$,and Hydroxide $(OH^-)$ has a valency of $1$. By criss-crossing the valencies,we get $Mg(OH)_2$.

(N/A) $(i) Al_{2}(SO_{4})_{3}$ is Aluminium sulphate.
$(ii) CaCl_{2}$ is Calcium chloride.
$(iii) K_{2}SO_{4}$ is Potassium sulphate.
$(iv) KNO_{3}$ is Potassium nitrate.
$(v) CaCO_{3}$ is Calcium carbonate.

(N/A) chemical formula of a compound (or element) is the symbolic representation of its composition. It represents:
$(i)$ The number and kind of atoms present per molecule of the compound.
$(ii)$ One mole of the compound.
$(iii)$ The molar mass of the compound.

(A) $(i)$ In a $H_2S$ molecule,there are $2$ atoms of hydrogen and $1$ atom of sulphur.
Total atoms = $2 + 1 = 3$ atoms.
$(ii)$ In a $PO_4^{3-}$ ion,there is $1$ atom of phosphorus and $4$ atoms of oxygen.
Total atoms = $1 + 4 = 5$ atoms.

(C) $(i)$ Molecular mass of $H_2$ (hydrogen) $= \text{Atomic mass of hydrogen} \times 2 = 1 \times 2 = 2\,u$.
$(ii)$ Molecular mass of $O_2$ (oxygen) $= \text{Atomic mass of oxygen} \times 2 = 16 \times 2 = 32\,u$.
$(iii)$ Molecular mass of $Cl_2$ (chlorine) $= \text{Atomic mass of chlorine} \times 2 = 35.5 \times 2 = 71\,u$.

(D) $(i)$ Molecular mass of $CO_2$ (carbon dioxide) = (Atomic mass of carbon $\times 1$) + (Atomic mass of oxygen $\times 2$) = $12 + (16 \times 2) = 12 + 32 = 44\,u$.
$(ii)$ Molecular mass of $CH_4$ (methane) = (Atomic mass of carbon $\times 1$) + (Atomic mass of hydrogen $\times 4$) = $12 + (1 \times 4) = 12 + 4 = 16\,u$.
$(iii)$ Molecular mass of $C_2H_6$ (ethane) = (Atomic mass of carbon $\times 2$) + (Atomic mass of hydrogen $\times 6$) = $(12 \times 2) + (1 \times 6) = 24 + 6 = 30\,u$.

(A) $(i)$ Molecular mass of $C_2H_4$ (ethene):
$= (\text{Atomic mass of C} \times 2) + (\text{Atomic mass of H} \times 4)$
$= (12 \times 2) + (1 \times 4) = 24 + 4 = 28\,u$.
$(ii)$ Molecular mass of $NH_3$ (ammonia):
$= (\text{Atomic mass of N} \times 1) + (\text{Atomic mass of H} \times 3)$
$= (14 \times 1) + (1 \times 3) = 14 + 3 = 17\,u$.
$(iii)$ Molecular mass of $CH_3OH$ (methanol):
$= (\text{Atomic mass of C} \times 1) + (\text{Atomic mass of H} \times 4) + (\text{Atomic mass of O} \times 1)$
$= (12 \times 1) + (1 \times 4) + (16 \times 1) = 12 + 4 + 16 = 32\,u$.

(B) $(i)$ Formula unit mass of $ZnO = 65 + 16 = 81\,u$.
$(ii)$ Formula unit mass of $Na_2O = (23 \times 2) + 16 = 46 + 16 = 62\,u$.
$(iii)$ Formula unit mass of $K_2CO_3 = (39 \times 2) + 12 + (16 \times 3) = 78 + 12 + 48 = 138\,u$.

(C) One mole of carbon atoms contains $6.022 \times 10^{23}$ atoms (Avogadro's number).
The molar mass of carbon is $12 \, g/mol$.
Therefore,the mass of $6.022 \times 10^{23}$ carbon atoms is $12 \, g$.
The mass of $1$ carbon atom is calculated as:
$\text{Mass of } 1 \text{ atom} = \frac{\text{Molar mass}}{\text{Avogadro's number}}$
$\text{Mass of } 1 \text{ atom} = \frac{12}{6.022 \times 10^{23}} \, g$
$\text{Mass of } 1 \text{ atom} \approx 1.99 \times 10^{-23} \, g$.

(D) The number of atoms in a given mass of an element is calculated using the formula: $\text{Number of atoms} = \frac{\text{Given mass}}{\text{Atomic mass}} \times N_A$,where $N_A$ is Avogadro's number $(6.022 \times 10^{23} \text{ mol}^{-1})$.
For $100\,g$ of Sodium $(Na)$:
$\text{Number of atoms} = \frac{100}{23} \times 6.022 \times 10^{23} \approx 2.618 \times 10^{24}$ atoms.
For $100\,g$ of Iron $(Fe)$:
$\text{Number of atoms} = \frac{100}{56} \times 6.022 \times 10^{23} \approx 1.075 \times 10^{24}$ atoms.
Comparing the two values,$2.618 \times 10^{24} > 1.075 \times 10^{24}$.
Therefore,$100\,g$ of sodium contains a greater number of atoms than $100\,g$ of iron.

(A) Mass of the compound $= 0.24 \, g$
Mass of boron $= 0.096 \, g$
Mass of oxygen $= 0.144 \, g$
Percentage of boron $= \frac{\text{Mass of boron}}{\text{Mass of compound}} \times 100$
$= \frac{0.096 \, g}{0.24 \, g} \times 100 = 40\%$
Percentage of oxygen $= \frac{\text{Mass of oxygen}}{\text{Mass of compound}} \times 100$
$= \frac{0.144 \, g}{0.24 \, g} \times 100 = 60\%$
Alternative method:
Percentage of oxygen $= 100\% - \text{Percentage of boron}$
$= 100\% - 40\% = 60\%$

(A) First,we determine the mass ratio of carbon and oxygen in carbon dioxide $(CO_2)$.
The atomic mass of carbon is $12 \,u$ and oxygen is $16 \,u$. In $CO_2$,the ratio of carbon to oxygen is $12 : (2 \times 16) = 12 : 32$,which simplifies to $3 : 8$.
This means that $3.00 \,g$ of carbon requires exactly $8.00 \,g$ of oxygen to react completely to form $11.00 \,g$ of $CO_2$.
When $3.00 \,g$ of carbon is provided with $50.00 \,g$ of oxygen,only $8.00 \,g$ of oxygen will react with the carbon,and the remaining $42.00 \,g$ of oxygen will remain unreacted.
Therefore,the mass of $CO_2$ formed remains $11.00 \,g$.
This result is governed by the 'Law of Constant Proportions',which states that in a chemical substance,the elements are always present in definite proportions by mass.

(N/A) polyatomic ion is a group of atoms carrying a fixed net charge (either positive or negative) that behaves as a single unit during chemical reactions.
Examples:
$(i)$ Carbonate ion: $CO_3^{2-}$
$(ii)$ Sulphate ion: $SO_4^{2-}$
$(iii)$ Ammonium ion: $NH_4^+$
$(iv)$ Phosphate ion: $PO_4^{3-}$

(N/A) $(i)$ Magnesium has a valency of $+2$ and chloride has $-1$. By criss-cross method,the formula is $MgCl_2$.
$(ii)$ Calcium has a valency of $+2$ and oxide has $-2$. The ratio is $1:1$,so the formula is $CaO$.
$(iii)$ Copper $(II)$ has a valency of $+2$ and nitrate has $-1$. By criss-cross method,the formula is $Cu(NO_3)_2$.
$(iv)$ Aluminium has a valency of $+3$ and chloride has $-1$. By criss-cross method,the formula is $AlCl_3$.
$(v)$ Calcium has a valency of $+2$ and carbonate has $-2$. The ratio is $1:1$,so the formula is $CaCO_3$.

(N/A) Quick lime is Calcium oxide $(CaO)$. The elements present are Calcium $(Ca)$ and Oxygen $(O)$.
$(b)$ Hydrogen bromide is $(HBr)$. The elements present are Hydrogen $(H)$ and Bromine $(Br)$.
$(c)$ Baking powder is Sodium hydrogen carbonate $(NaHCO_3)$. The elements present are Sodium $(Na)$,Hydrogen $(H)$,Carbon $(C)$,and Oxygen $(O)$.
$(d)$ Potassium sulphate is $(K_2SO_4)$. The elements present are Potassium $(K)$,Sulphur $(S)$,and Oxygen $(O)$.

(B) Molar mass of $C_2H_2 = (2 \times \text{Atomic mass of } C) + (2 \times \text{Atomic mass of } H)$
$= (2 \times 12) + (2 \times 1) = 24 + 2 = 26\,u$.
$(b)$ Molar mass of $S_8 = 8 \times \text{Atomic mass of } S$
$= 8 \times 32 = 256\,u$.

(C) Molar mass of $P_4 = 4 \times$ Atomic mass of $P = 4 \times 31 = 124\,u$.
$(b)$ Molar mass of $HCl = $ Atomic mass of $H + $ Atomic mass of $Cl = 1 + 35.5 = 36.5\,u$.
$(c)$ Molar mass of $HNO_3 = $ Atomic mass of $H + $ Atomic mass of $N + (3 \times$ Atomic mass of $O) = 1 + 14 + (3 \times 16) = 15 + 48 = 63\,u$.

(D) Molar mass of $N$ atom $=$ Atomic mass of $N = 14\, g/mol$.
Mass of $1$ mole of $N$ atoms $= 1 \times 14 = 14\, g$.
$(b)$ Molar mass of $Al$ atom $= 27\, g/mol$.
Mass of $4$ moles of $Al$ atoms $= 4 \times 27 = 108\, g$.
$(c)$ Molar mass of $Na_2SO_3 = (23 \times 2) + 32 + (16 \times 3) = 46 + 32 + 48 = 126\, g/mol$.
Mass of $10$ moles of $Na_2SO_3 = 10 \times 126 = 1260\, g$.

(A) For oxygen gas $(O_2)$:
Molar mass of $O_2 = 16 \times 2 = 32\, \text{g/mol}$.
Number of moles $= \frac{\text{Given mass}}{\text{Molar mass}} = \frac{12\, \text{g}}{32\, \text{g/mol}} = 0.375\, \text{mol}$.
$(b)$ For water $(H_2O)$:
Molar mass of $H_2O = (2 \times 1) + 16 = 18\, \text{g/mol}$.
Number of moles $= \frac{20\, \text{g}}{18\, \text{g/mol}} \approx 1.11\, \text{mol}$.
$(c)$ For carbon dioxide $(CO_2)$:
Molar mass of $CO_2 = 12 + (2 \times 16) = 44\, \text{g/mol}$.
Number of moles $= \frac{22\, \text{g}}{44\, \text{g/mol}} = 0.5\, \text{mol}$.

(B) The molar mass of oxygen atoms $(O)$ is $16 \, g/mol$.
Mass of $0.2$ mole of oxygen atoms $= \text{Number of moles} \times \text{Molar mass} = 0.2 \times 16 = 3.2 \, g$.
$(b)$ The molar mass of water molecules $(H_2O)$ is $(2 \times 1) + 16 = 18 \, g/mol$.
Mass of $0.5$ mole of water molecules $= \text{Number of moles} \times \text{Molar mass} = 0.5 \times 18 = 9.0 \, g$.

(B) Molar mass of sulphur $(S_8) = 32 \times 8 = 256 \, g/mol$.
Number of moles of $S_8$ in $16 \, g = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{16}{256} = 0.0625 \, mol$.
Number of molecules = $\text{Number of moles} \times \text{Avogadro constant} (N_A)$.
Number of molecules = $0.0625 \times 6.022 \times 10^{23} = 3.76375 \times 10^{22} \approx 3.76 \times 10^{22}$ molecules.
Wait,re-calculating: $0.0625 \times 6.022 \times 10^{23} = 0.376375 \times 10^{23} = 3.76 \times 10^{22}$.
Correction: The provided option $3.76 \times 10^{22}$ is not listed,but $3.76 \times 10^{22}$ is $0.376 \times 10^{23}$. Let us re-verify: $16/256 = 1/16 = 0.0625$. $0.0625 \times 6.022 \times 10^{23} = 0.376375 \times 10^{23}$. Thus,the correct value is $0.376 \times 10^{23}$.

(D) $1$. Calculate the molar mass of aluminium oxide $(Al_{2}O_{3})$:
Molar mass = $(2 \times 27) + (3 \times 16) = 54 + 48 = 102 \, g/mol$.
$2$. Determine the number of moles in $0.051 \, g$ of $Al_{2}O_{3}$:
Moles = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{0.051}{102} = 0.0005 \, mol = 5 \times 10^{-4} \, mol$.
$3$. Relate moles of $Al_{2}O_{3}$ to moles of $Al^{3+}$ ions:
One mole of $Al_{2}O_{3}$ contains $2$ moles of $Al^{3+}$ ions.
Therefore,$5 \times 10^{-4} \, mol$ of $Al_{2}O_{3}$ contains $2 \times 5 \times 10^{-4} = 10^{-3} \, mol$ of $Al^{3+}$ ions.
$4$. Calculate the number of ions:
Number of ions = $\text{Moles} \times \text{Avogadro's number} = 10^{-3} \times 6.022 \times 10^{23} = 6.022 \times 10^{20}$ ions.