Q: Question 2When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Given,C 3 g+O2 8 g⟶CO211 g\underset{\text{ 3 g}}{\text{C}} + \underset{\text{ 8 g}}{\text{O}_2} \longrightarrow \underset{\text{11 g}}{\text{CO}_2} 3 gC+ 8 gO2⟶11 gCO2When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.Now,C 3 g+O2 50 g⟶CO2?\underset{\text{ 3 g}}{\text{C}} + \underset{\text{ 50 g}}{\text{O}_2} \longrightarrow \underset{\text{?}}{\text{CO}_2} 3 gC+ 50 gO2⟶?CO2Carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8. Hence, 3 g of carbon will react with 50 g of oxygen but only 8 g of it will be used in producing 11 g of CO2 and 42 g [i.e., 50 g - 8 g] of oxygen will be left unused.The above answer is governed by the law of constant proportions.

Question 1

Question 1A 0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Accepted Answer

Given,Mass of the sample compound = 0.24 g,mass of boron = 0.096 g,mass of oxygen = 0.144 gTo calculate the percentage composition of the compound,Percentage of boron := mass of boronmass of the compound\dfrac{	ext{mass of boron}}{	ext{mass of the compound}}mass of the compoundmass of boron​ x 100= 0.0960.24\dfrac{0.096}{0.24}0.240.096​ x 100= 40%Percentage of oxygen= mass of oxygenmass of the compound\dfrac{	ext{mass of oxygen}}{	ext{mass of the compound}}mass of the compoundmass of oxygen​ x 100= 0.1440.24\dfrac{0.144}{0.24}0.240.144​ x 100= 60%Hence, percentage of boron = 40% and percentage of oxygen = 60% in the compound.

Question 2

Question 2When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Accepted Answer

Given,C 3 g+O2 8 g⟶CO211 g\underset{	ext{ 3 g}}{	ext{C}} + \underset{	ext{ 8 g}}{	ext{O}_2} \longrightarrow \underset{	ext{11 g}}{	ext{CO}_2} 3 gC​+ 8 gO2​​⟶11 gCO2​​When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.Now,C 3 g+O2 50 g⟶CO2?\underset{	ext{ 3 g}}{	ext{C}} + \underset{	ext{ 50 g}}{	ext{O}_2} \longrightarrow \underset{	ext{?}}{	ext{CO}_2} 3 gC​+ 50 gO2​​⟶?CO2​​Carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8. Hence, 3 g of carbon will react with 50 g of oxygen but only 8 g of it will be used in producing 11 g of CO2 and 42 g [i.e., 50 g - 8 g] of oxygen will be left unused.The above answer is governed by the law of constant proportions.

Question 3

Question 3What are polyatomic ions? Give examples.

Accepted Answer

Cluster of atoms that act as an ion are called polyatomic ions. They carry a fixed charge on them.Example: Hydroxide [OH-], Cyanide [CN-]

Question 4

Question 4Write the chemical formula of the following.(a) Magnesium chloride(b) Calcium oxide(c) Copper nitrate(d) Aluminium chloride(e) Calcium carbonate

Accepted Answer

(a) Magnesium chlorideStep 1 — Write each symbol with its valencyMg2+↗Cl1−\text{Mg}^{2+} \phantom{\nearrow} \text{Cl}^{1-}Mg2+↗Cl1−Step 2 — Interchange the valenciesMg22 ↘↙ Cl1⇒Mg11 ↘↙ Cl2\overset{\phantom{2}{2}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{Cl}} \[0.5em]Mg22​ ↘↙ Cl1⇒11Mg​ ↘↙ 2Cl​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Magnesium chloride : MgCl2\bold{MgCl_2}MgCl2​(b) Calcium oxideStep 1 — Write each symbol with its valencyCa2+↗O2−\text{Ca}^{2+} \phantom{\nearrow} \text{O}^{2-}Ca2+↗O2−Step 2 — Interchange the valenciesCa22 ↘↙ O2⇒Ca12 ↘↙ O2\overset{\phantom{2}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{1}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{O}} \[0.5em]Ca22 ↘↙ O2⇒12Ca​ ↘↙ 2O​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Calcium oxide : CaO\bold{CaO}CaO(c) Copper nitrateStep 1 — Write each symbol with its valencyCu2+↗NO31−\text{Cu}^{2+} \phantom{\nearrow} \text{NO}_3^{1-}Cu2+↗NO31−​Step 2 — Interchange the valenciesCu22 ↘↙ NO31⇒Cu11 ↘↙ NO32\overset{\phantom{2}{2}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{NO}_3} \Rightarrow \underset{\phantom{1}{1}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{NO}_3} \[0.5em]Cu22 ↘↙ NO3​1​⇒11Cu​ ↘↙ 2NO3​​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of copper nitrate : Cu(NO3)2\bold{Cu(NO_3)_2}Cu(NO3​)2​(d) Aluminium chlorideStep 1 — Write each symbol with its valencyAl3+↗Cl1−\text{Al}^{3+} \phantom{\nearrow} \text{Cl}^{1-}Al3+↗Cl1−Step 2 — Interchange the valenciesAl23 ↘↙ Cl1⇒Al11 ↘↙ Cl3\overset{\phantom{2}{3}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{Cl}} \[0.5em]Al23 ↘↙ Cl1⇒11Al​ ↘↙ 3Cl​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Aluminium chloride : AlCl3\bold{AlCl_3}AlCl3​(e) Calcium carbonate –Step 1 — Write each symbol with its valencyCa2+↗CO31−\text{Ca}^{2+} \phantom{\nearrow} \text{CO}_3^{1-}Ca2+↗CO31−​Step 2 — Interchange the valenciesCa22 ↘↙ CO31⇒Ca11 ↘↙ CO32\overset{\phantom{2}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{CO}_3} \Rightarrow \underset{\phantom{1}{1}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{CO}_3} \[0.5em]Ca22 ↘↙ CO3​1​⇒11Ca​ ↘↙ 2CO3​​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of copper nitrate : Ca(CO3)2\bold{Ca(CO_3)_2}Ca(CO3​)2​

Question 5

Question 5Give the names of the elements present in the following compounds.(a) Quick lime(b) Hydrogen bromide(c) Baking powder(d) Potassium sulphate

Accepted Answer

(a) Quick lime (CaO) — Calcium and oxygen(b) Hydrogen bromide (HBr) — Hydrogen and bromine(c) Baking powder (NaHCO3) — Sodium, hydrogen, carbon and oxygen(d) Potassium sulphate (K2SO4) — Potassium, sulphur and oxygen,

Question 6

Question 6Calculate the molar mass of the following substances.(a) Ethyne, C2H2(b) Sulphur molecule, S8(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)(d) Hydrochloric acid, HCl(e) Nitric acid, HNO3

Accepted Answer

(a) Molar mass of Ethyne C2H2 = (2 x Mass of C) + (2 x Mass of H) = (2 × 12) + (2 × 1) = 24 + 2 = 26 g(b) Molar mass of Sulphur molecule S8 = (8 x Mass of S) = 8 x 32 = 256 g(c) Molar mass of Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124 g(d) Molar mass of Hydrochloric acid, HCl = Mass of H + Mass of Cl = 1 + 35.5 = 36.5 g(e) Molar mass of Nitric acid, HNO3 = Mass of H + Mass of N + (3 x Mass of O) = 1 + 14 + (3 × 16) = 15 + 48 = 63 g

Question 7

Question 1In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.Sodium carbonate + acetic acid ⟶ Sodium acetate + carbon dioxide + water

Accepted Answer

As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of products.Sodium carbonate + acetic acid ⟶ Sodium acetate + carbon dioxide + water5.3 g + 6 g ⟶ 8.2 g + 2.2 g + 0.9 g⇒ 11.3 g ⟶ 11.3 gAs per the above reaction, L.H.S. = R.H.S. = 11.3 gHence, the observations are in agreement with the law of conservation of mass.

Question 8

Question 2Hydrogen and oxygen combine in a ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Accepted Answer

Given,hydrogen : water = 1 : 8So, for every 1 g of hydrogen, 8 g of oxygen is required∴ For 3 g of hydrogen, oxygen required is 81\dfrac{8}{1}18​ x 3 = 24 gHence, 24 g of oxygen would be required for the complete reaction with 3 g of hydrogen gas.

Question 9

Question 3Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?

Accepted Answer

One of the postulates of Dalton's atomic theory states that : 'Atoms cannot be created nor be destroyed in a chemical reaction'. This law is the result of law of conservation of mass.

Question 10

Question 4Which postulate of Dalton's atomic theory can explain the law of definite proportions?

Accepted Answer

The postulate of Dalton's atomic theory that can explain the law of definite proportions is : 'Relative number and kinds of atoms are equal in given compounds.'

Question 11

Question 1Define the atomic mass unit.

Accepted Answer

Atomic mass unit is defined as 1⁄12 th the mass of an carbon atom C-12.

Question 12

Question 2Why is it not possible to see an atom with naked eyes?

Accepted Answer

It is not possible to see an atom with naked eyes because:Atoms are very small in size, measured in nanometres.Except for atoms of noble gases, they do not exist independently.

Question 13

Question 1Write down the formulae of(i) sodium oxide(ii) aluminium chloride(iii) sodium sulphide(iv) magnesium hydroxide

Accepted Answer

(i)	Sodium oxide:Step 1 — Write each symbol with its valencyNa1+↗O2−\text{Na}^{1+} \phantom{\nearrow} \text{O}^{2-}Na1+↗O2−Step 2 — Interchange the valenciesNa21 ↘↙ O2⇒Na12 ↘↙ O1\overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{1}{2}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{O}} \[0.5em]Na21 ↘↙ O2⇒12Na​ ↘↙ 1O​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Sodium oxide : Na2O\bold{Na_2O}Na2​O(ii) Aluminium chlorideStep 1 — Write each symbol with its valencyAl3+↗Cl1−\text{Al}^{3+} \phantom{\nearrow} \text{Cl}^{1-}Al3+↗Cl1−Step 2 — Interchange the valenciesAl23 ↘↙ Cl1⇒Al11 ↘↙ Cl3\overset{\phantom{2}{3}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{Cl}} \[0.5em]Al23 ↘↙ Cl1⇒11Al​ ↘↙ 3Cl​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Aluminium chloride : AlCl3\bold{AlCl_3}AlCl3​(iii) sodium sulphideStep 1 — Write each symbol with its valencyNa1+↗S2−\text{Na}^{1+} \phantom{\nearrow} \text{S}^{2-}Na1+↗S2−Step 2 — Interchange the valenciesNa21 ↘↙ S2⇒Na12 ↘↙ S1\overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}} \Rightarrow \underset{\phantom{1}{2}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{S}} \[0.5em]Na21 ↘↙ S2⇒12Na​ ↘↙ 1S​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Sodium sulphide : Na2S\bold{Na_2S}Na2​S(iv) Magnesium hydroxideStep 1 — Write each symbol with its valencyMg2+↗OH1−\text{Mg}^{2+} \phantom{\nearrow} \text{OH}^{1-}Mg2+↗OH1−Step 2 — Interchange the valenciesMg22 ↘↙ OH1⇒Mg11 ↘↙ OH2\overset{\phantom{2}{2}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{OH}} \Rightarrow \underset{\phantom{1}{1}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{OH}} \[0.5em]Mg22​ ↘↙ OH1⇒11Mg​ ↘↙ 2OH​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Magnesium hydroxide : Mg(OH)2\bold{Mg(OH)_2}Mg(OH)2​

Question 14

Question 2Write down the names of compounds represented by the following formulae:(i) Al2(SO4)3(ii) CaCl2(iii) K2SO4(iv) KNO3(v) CaCO3

Accepted Answer

(i) Al2(SO4)3 — Aluminium sulphate(ii) CaCl2 — Calcium chloride(iii) K2SO4 — Potassium sulphate(iv) KNO3 — Potassium nitrate(v) CaCO3 — Calcium carbonate

Question 15

Question 3What is meant by the term chemical formula?

Accepted Answer

The chemical formula of a compound is a symbolic representation of its composition. It denotes in a compound, the number of atoms of each element present.

Question 16

Question 4How many atoms are present in a(i) H2S molecule and(ii) PO43- ion ?

Accepted Answer

(i) In one molecule of H2S, 2 atoms of hydrogen and 1 atom of sulphur are present. Hence, 3 atoms in total are present.(ii) In one ion [PO43-], 1 atom of phosphorus and 4 atoms of oxygen are present. Hence, 5 atoms in total are present.

Question 17

Question 1Calculate the molecular masses ofH2O2Cl2CO2CH4C2H6C2H4NH3CH3OH.

Accepted Answer

The molecular mass of H2 = 2 x atoms atomic mass of H = 2 x 1u = 2uThe molecular mass of O2 = 2 x atoms atomic mass of O = 2 x 16u = 32uThe molecular mass of Cl2 = 2 x atoms atomic mass of Cl = 2 x 35.5u = 71uThe molecular mass of CO2 = atomic mass of C + (2 x atomic mass of O) = 12 + (2 × 16)u = 12 + 32 = 44uThe molecular mass of CH4 = atomic mass of C + (4 x atomic mass of H) = 12 + (4 x 1) = 16uThe molecular mass of C2H6 = (2 x atomic mass of C) + (6 x atomic mass of H) = (2 x 12) + (6 x 1) = 24 + 6 = 30uThe molecular mass of C2H4 = (2 x atomic mass of C) + (4 x atomic mass of H) = (2 x 12) + (4 x 1) = 24 + 4 = 28uThe molecular mass of NH3 = atomic mass of N + (3 x atomic mass of H) = 14 + (3 x 1) = 17uThe molecular mass of CH3OH = atomic mass of C + (3 x atomic mass of H) + atomic mass of O + atomic mass of H = [12 + (3 × 1) + 16 + 1] = (12 + 3 + 17) = 32u

Question 18

Question 2Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and O = 16u.

Accepted Answer

Given:Atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and O = 16uThe formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O = 65u + 16u = 81uThe formula unit mass of Na2O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u + 16u = 46u + 16u = 62uThe formula unit mass of K2CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u

Atoms and Molecules

Exercises

Intext Questions 1

Intext Questions 2

Intext Questions 3

Intext Questions 4