5. Calculate the atomicity of a gas 'X', if 1g. of 'X' occupies 11,200 cc at s.t.p. [at wt. of 'K' =lJ

Solution: [1 mole of any substance = 1 gm molecular weightof it and occupies 22.4 litres at s.t.p.]

a] 1 weighs ? will occupy 22,400 cc. at s.t.p.[MOLE] [G. MOL. WT.]

b] 1 g. occupies 11,200 cc. at s.t.p. = 22400 x 1 = 2 [g. mol. wt.]..WT. 11200

Atomicity = g. mol. wt. = 2 [calculated] =2 :. Atomicity of 'X' = 2 - Ans.At. wt .. 1 [given]

6. 0.48g of a gas forms 100cm3 of vapours at s.t.p. Calculate the gram molecular wt of the gas.

Solution: [22.4 lit. of gas at s.t.p. == 1 g. mol. wt. of the gas]

100 cm3 of the gas weighs 0.48 grams at s.t.p. a] 1 ? 22,400 cc.- will occupy at s.t.p.

c ' [MOLE] [G. MOL. WT.] [VOLUME]:. (22.4 x 1000)cm3 of the gas weighs 22400 x 0.48100 0.48 g. 100 c.c.

Gram mol. weight of the g~s'= 107.52 g - Ans. b] .. occupies at s.t.p.WEIGHT [VOLUME]

7. Calculate the weight of a substance X which in gaseous fonn occupies 10 litres at 270C and 700mmpressure. The molecular weight of X is 60.

Solution: Convert the volume to s.t.p. using the gas equation

Initial conditions Final Conditions (s.t.p.)PI = 700 mm ofHg Pz = 760 mm ofHgVI = 10 litres Vz = X litres [volume at s.t.p.]TI = 27+273 = 300 K Tz = 273K

Using the gas equation PlVI PzVz=TI Tz

Substituting the values: 700 x 10 760 x X :. V2 = 700 x 10 x 273 = 8.38 litres300 = 273 300 x 760 [vol. at s.t.p.]

19. mol. wt. of the gas occupies 22.4lits. at s.t.p. a]1 60 g.

:. 60 g of gas X occupies 22.4 litres at s.t.p. - occupies 22.4 lits. at s.t.p.? g of gas X occupied by 8.38 litres at s.t.p. [MOLE] [G. MOL. WT.]

ie. 8.38 x 60 = 22.45 g b]? g. occupied by 8.38 lits. at s.t.p...

22.4 WEIGHT [calculated above]

:. The weight of the substance X is 22.45 g. - Ans.

8. Calculate the gram atoms present in 8g of oxygen [0=16J.Gram atom is the relative atomic mass of an element expressed in grams.

G t Mass in grams [ofoxygen] 8 = 0.5 g. atoms - Ans.ram a oms = =Relative atomic mass [At. Wt.] 16

9. Calculate the gram molecules present in 45 g of water. [H=l, 0=16JGram molecule is the relative molecular mass of a substance expressed in grams.

Mass in grams [ofwater] 45 = 2.5 g. molecules - Ans.Gram molecules = = -Relative molecular mass [Mol. Wt.] 18

81

1 Mole of gas' A' weighs 'x' g. and contains.[REL. MOL. MASS]

10 g. of gas' A'

6.023 X 1023 molecules.

I

ii] 64 g. of 502 occupies 24lits. at room temperature

:.80 g. of 502 occupies? lits. at room temperature =

iii] 64 g. of 502 contains N molecules of 502

:.0.64 g. of 502 contains? molecules of 502 =

24x8064

= 30 dm3 of S02 Ans.

N x 0.64 = ~molecules of S02 Ans.~ 100

iv] Gram molecules

:.0.5 gm. molecules

= Mass in grams of 502Mol. wt. of 502

?64

64 x 0.5 = 32 g. of S02 Ans.= =

11. Calculate i] the number of moles ii] the total number of molecules iii] the total number of atomsiv] the number of hydrogen atoms - in 294g. of sulphuric acid. [H = 1, S = 32, 0 = 16]Solution: Relative molecular mass i.e. $. mol. wt. of H2504 = 2 + 32 + 16 x 4 = 98

i] _1_ of H2504 weighs 98 contains 6.023x 1023

molecules[MOLE] [G. MOL. WT.] [AVOG. NO.]

:. ? moles of H2SO4 will weigh 294 g. 1 x 294=98

= 3 moles of H2S04 - Ans.

ii] 1 of H2504 weighs 98g. contains 6;023 x 1023 molecules[MOLE]

294 g. contains ? molecules = 6.023 x 1023

x 29498

= 3 x 6.023 X 1023molecules - Ans,

iii] 1 molecule of H2504 contains 7 atoms. [2(H) + 1(5) + 4(0) = 7]:. 3 x 6.023 x 1023molecules of H2504 contains 7 x 3 x 6.023 X 1023 = 21 x 6.023 x 1023atoms - Ans.

iv] 1 molecule of H2504 contains 2 atoms of hydrogen ..'. 3 x 6.023x 1023molecules of H2504 contains 2 x 3 x 6.023X 1023= 6 x 6.023x 1023atoms of hydrogen - Ans.

12. If gases 'A', 'B', 'C' are arranged in increasing in order of their relative molecular mass & the mass ofeach gas is 10g. at s.t.p. State which gas will contain the least number of molecules &which the most.

contains 6.023 X 1023x 10 molecules.'X'

5ince gases' A', 'B' and 'C' are in increasing order of relative molecular mass we assume the relativemolecular mass of gas 'B' as '2X' and gas'C' as '3X'. Hence

10 g. of gas 'B contains 6.023 x 1023x 10 molecules & 10 g. of gas -c contains 6023 x 1023x 10 molecules.2X 3X

Gas'C' contains least number and gas' A' the maximum no. of molecules. - Ans.

82

13. Calculate the number of moles of zinc [Zn2+J ions and chloride [CIl-] ions which will be obtained from272 g. of ZnCI2. [Zn = 65, Cl = 35.5JSolution: Relative molecular mass of ZnCl2 = 65 + 35.5 x 2 = 136 g.

1 of ZnCl weighs 136g. of ZnCl2[MOLE] 2 [G. MOL. WT.]

. . ? moles of ZnCl2 which weigh 272 g.

.. ZnCl2 -7 Zn2+ + 2Cll-1 mole 1 mole 2 mole

:. 2 mole 2 mole 4 mole

:.372 g. of ZnCl2 furnishes 2 moles of Zn2+ ions & 4 moles of Cl1- ions - Ans.

i.e. 1 x 272136

= 2 moles

MOLE CONCEPT AND AVOGADRO'S LAW - PROBLEMS14. If 100 cc of a gas A contains Y molecules. How many molecules of gas B will be present in 50 cc of B

and of gas C in 25 cc. of C. The gases A, B, & C are under the same conditions of temperature &pressure.According to Avogadro's Law - Under the same conditions of temperature and pressure equal volumes ofall gases contain the same number of molecules.

:. If 100 cc of gas A contains Y molecules then 100 cc of gas B and gas C also contains Y molecules.

:. 50 cc of gas B contains Y/z. molecules and 25 cc of gas C contains Y/4 molecules - Ans.

15. Under the same conditions of temperature and pressure 021 CI21S021 CO2 contain the same numberof molecules represented by 'Y'. The molecules of oxygen gas occupy V litres and have a mass of 16g.Under the same conditions, state the volume occupied by - i] Ymolecules of chlorine ii] 3Ymoleculesof S02 iii] State the mass of CO2 it}grams.According to Avogadro's Law - under the same conditions of temperature and pressure equal volumes ofall gases contain the same number of molecules.

., If gases under the same conditions have same no. of molecules, then they must have the same volume.Hence if : 'Y' molecules of 02 occupy V litres, 'Y' molecules of all gases must occupy the same vol. i.e. 'V' litres.

i) Y' molecules of Cl2 also occupies V litres - Ans. ii] 3'Y' molecules of S02 occupies 3V litres - Ans.

iii] 1 mole of 02 weighs . [G. :~r.·WT.] ,occupies 22.4lits. at s.t.p.

:. 16g. of 02 occupies ? i.e 2;24 x 16 = 11.2 lits.

If 16 g. of oxygen occupies 11.2lits at s.t.p. :. the volume of Cl2' S02 and CO2 is also 11.2lits at s.t.p.44g.

[G. MOL. WT.]

? g. of CO2

1 mole of CO2 weighs occupies 22.4lits. at s.t.p.

11 2 lit t t 44 x 11.2 22 f CO Aoccupy . 1 s. a s.. p. = 22.4 = g. 0 2- ns.I l

:)'" . /' :.

~VAPOUR DENSITY & MOLECULAR WEIGHT.- PROBLEMS16. A gas culinder filled with hydrogen holds 50 g of the gas. The same cylinder holds 200 g of a gas A

and 500 g of gas B. Considering the same conditions of temperature and pressure in the cylinder,calculate the relative molecular masses [molecular uieights] of gas A and B.VAPOUR DENSITY = wt. of a certain volume of a gas or Weight of 1 litre of gas Same conditionsof a gas Wt. of an equal volume of H2 Weight 1 litre of H2 of temp. & press.

For Gas A: Vapour Density = 250~= 4 :. Molecular Weight = 2 x Vapour Density = 2 x 4 = 8 g.

For Gas B: Vapour Density = 5~00= 10 :. Molecular Weight = 2 x Vapour Density = 2 x io = 20 g.

The relative molecular mass of gas A is 8 g and gas B is 20 g. - Ans.

83

330= 533.37 ml.[v6.f.,.at s.t.p.]

17. A gas cylinder can hold 1kg of hydrogen at room temperature and pressure. Calculatea} The weight of carbon dioxide it can hold under similar conditions of temperature and pressure,b} If the number of molecules of hydrogen in the cylinder is X. State the number of molecules of

carbon dioxide in the culinder. [C=12, 0=16, H=l}Solution: Molecular weight of carbon dioxide = 12 + 32 = 44

Molecular weight = 2 x Vapour Density

44 = 2 x VD.' 44:. V.D. of CO2 =2

= 22

Wt. of a certain volume of gas [i.e. C02]Vapour Density =Wt. of the same volume of H2 [same conditions]

:. 22 = ? [wt: of carbon dioxide] = 221

a] Weight of carbon dioxide is 22 kg. - Ans.

b] The number of molecules of CO2 in the cylinder is X (Avogadro's Law - under the same conditionsof temperature & pressure equal volumes of all gases contain the same number of molecules) - Ans.

18. A gas occupies .700ml at a pressure of 700mm of Hg and a temperature of 5'JOC.If at s.t.p. the massof the gas is 1.5g find the vapour density and the molecular weight of the gas.(Given 1litre of hydrogen weighs 0.09 g at s.t.p.)Solution: A} Convert the volume to s.t.p. using the gas equationInitial conditions . . Final Conditions (s.t.p.)PI = 700mmofHg P2 = 760mmofHgVI = 700 ml V2 = X ml [volume at s.t.p.}TI = 57+273 = 330 K Ti = 273 K

Using the gas equation PIVI = P2V2TI T2

700 x 700Substituting the values : =

760 x X273

700 x 700 x 273. V -.. 2 - 760 x 330

Bl Calculate the vapour density and the molecular weight of the gas.533.37 ml. [vol. at s.t.p. calculated above] of the gas at s.t.p. weighs 1.5 g.

:. 1000 ml of the gas weighs 1.5 x 1000 = 2.8123 g.533.37

[given in the problem]

VAPOUR DENSITY = Wt. of 1000 ml [1 litre] of the gas at s.t.p.Wt. of 1000 ml [1 litre] of H2 at s.t.p.

= 2.8123

0.09= 31.25

:. Molecular weight = 2 x VD. = 2 x 31.25 = 62.5 g.Vapour density of the gas is 31.25 and molecular weight is 62.5 g. - Ans.

19. KMn04decomposesonheatingaccordingtotheequation- 2KMn04 ~ KzMn04 + Mn02 + 02[K2Mn04 &;Mn02 are the solid residues}. On heating KMn04' 1litre of oxygen was collected at roomtemp. & it was found the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under thesame conditions of temp. &press. has a mass of 0.0825g. Calculate the relative molecular mass of oxygen.Solution: .'

Vapour Density of oxygen

Weight of 1 litre of gas [02] under the same conditions ofWeight of 1 litre of H2 temperature & pressure

1.32 g. [since on collecting 1 litre of 02 the loss in mass is 1.32 g.]

1.32 = 160.0825

VAPOUR DENSITY - of oxygen =

Weight of 1 litre of oxygen ==

Relative molecular mass of oxygen = 2 x Vapour Density = 32 g. - Ans.

84

ADDITIONAL PROBLEMS

GAYJ~USSAC'SLAW - AVOGADRO'S LAW - MOL:ECONCEPT/ \ LAWS & TERMS• Gay Lussac's Law - When gases react they do so in volumes which bear a -

simple whole number ratio to one another and to the volumes of the products, if gaseous -provided the temperature and pressure of the reacting gases and their products remain constant.

• Avogadro's Law - Under the same conditions of temperature and pressure -equal volumes of all gases contain the same number of molecules.

• Relative atomic mass [Atomic weight] - Relative atomic mass of an element is the-number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon [C12].

• . Gram atomic mass [gram atom] - The relative atomic mass of an element expressed in-grams is known as - gram atomic mass or gram atom of that element.

• . Relative molecular mass [Molecular weight] - Relative molecular mass of an element or compound is the -number of times one molecule of the substance is heavier than 1/12 the mass of an atom of carbon [C12].

• Gram molecular mass - [gram molecule] - The relative molecular mass of a substance expressed in -grams is known as - gram molecular mass or gram molecule of that element.

• Avogadro's Number - The number of atoms - present in 12 g [gram atomic wt.] of carbon 6C12.

• Vapour density - Is the ratio of the mass of a certain volume of gas or vapour -to the mass of the same volume of hydrogen [volumes measured under same condition of temp. & press.].

• Mole - Is the amount of substance which contains -the same number of units as the number of atoms in 12.000 g of carbon - 12 [6C12].

• Atomicity - It is the no. of atoms present in - one molecule of that element. e.g. He [mono atomic]

• Molar volume - It is the volume occupied by - 1 gm. molecular weight of a gas at s.t.p.

Q.l LUSSAC'S LAW

1. Nitrogen reacts with hydrogen to give ammonia. Calculate the volume of the ammonia gas formedwhen nitrogen reacts with 6 litres of hydrogen. All volumes' measured at s.t.p. [4lts.]

2. 2500 cc of oxygen was burnt with 600 cc of ethane [C2H6]. Calculate the volume of unused oxygenand the volume of carbon dioxide formed. [400 cc, 1200 cc]

3. 20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide are exploded in an enclosure.What will be the volume and composition of the mixture of the gases when they are cooled to roomtemperature. [Oz 5 ml, COz 10 ml]

4. 224 cm3 of ammonia undergoes catalytic oxidation in presenc;:e of Pt to give nitric oxide and watervapour. Calculate the volume of oxygen required for the reaction. All volumes measured at roomtemperature and pressure. [Ans. 280 cc. of 0z]

5. Acetylene [C2H2]burns in air forming carbon dioxide and water vapour. Calculate the volume of airrequired to completely burn 50cm3 of acetylene. [Assume air contains 20% oxygen]. [Ans. 625cm3]

6. On igniting a mixture of acetylene [C2H2] & oxygen, 200 cm3 of CO2 is collected at s.t.p. Calculate thevol. of acetylene & 02 at s.t.p. in the original mixture. [Ans. Acetylene = 100 cm3; Oxygen = 250 cm3]

7. Ammonia is formed from the reactants nitrogen and hydrogen in presence of a catalyst under suitableconditions+Assuming all volumes are measured in litres at s.t.p. Calculate the volume of ammonia

. formed if only 10% conversion has taken place. [Ans. 0.2 litres or 20% or 1/5th of vol. of Nz & Hz]

8. 100 cc. each of water gas and oxygen are ignited and the resultant mixture of gases cooled to roomtemp. Calculate the composition of the resultant mixture. [Water gas contains. CO & H2 in equal ratio]

. [Ans. :. 50 cc. of 0z + 50 cc. of COz]

85

State which of the following:23. Has higher no. of moles: 5 g. of N20 or 5 g. of NO [N = 14,° = 16]24. Has a higher mass: 1 mole of CO2 or 1 mole of CO [C = 12,° = 16]25. Has a higher no. of atoms: 1 g. of 02 or 1 g. of Cl2 [0 = 16, Cl = 35.5]

Q.3 VAPOUR DENSITY AND MOLECULAR WEIGHT1. 500 rnl. of a gas 'X' at s.t.p. weighs 0.50 g. Calculate the vapour density and molecular weight of the

gas. [1 lit. of H2 at s.t.p. weighs 0.09 g]. [Ans. 11.1, 22.2g.]2. A gas cylinder holds 85 g of a gas 'X'. The same cylinder when filled with hydrogen holds 8.5 g of

hydrogen under the same conditions of temperature and pressure.Calculate the molecular weight of 'X'. [Ans. 20]

3. Calculate the relative molecular mass [molecular weight] of 290 ml. of a gas' A' at 17°C and 1520mmpressure which weighs 2.73 gat s.t.p. [1 litre of hydrogen at s.t.p. weighs 0.09 g.] [Ans. 111.11g.]

4. State the volume occupied by 40 g. of a hydrocarbon - CH4 at s.t.p. if its V.D. is 8. [Ans.56lits.]5. Calculate the atomicity of a gas X [at. no. 35.5] whose vapour density is equal to its

relative atomic mass. [Ans.2]6. Calculate the relative molecular mass and vapour density of methyl alcohol [CH30H] if 160 g. of the

alcohol on vaporization has a volume of 112 litres at s.t.p. . [Ans. 32 g., i6]

[Ans. = NO][Ans. = CO~

[Ans. = O~

Q.2 MOLE CONCEPT - AVOGADRO'S LAW - AVOGADRO'S NUMBERCalculate the following: [all measurements at s.t.p. or as stated in the problem]1. The mass of 2.8 litres of CO2, [C = 12,° = 16] [Ans. = 5.5 g.]2. The volume occupied by 53.5 g. of C12.[Cl = 35.5] [Ans. = 16.87 lit.]3. The number of molecules in 109.5 g. of HCl. [H = 1, Cl = 35.5] [Ans. = 3 x 6.023 x 1023]

4. The number of i] molecules [5 = 32] [Ans. = 0.75 x 6.023 x 1023molecules]il] atoms in 1929. of sulphur. [Ss] [6 x 6.023 x 1023atoms]

5. The mass of Na which will contain 6.023 x 1023atoms. [Na = 23] [Ans. = 23 g. of sodium]6. The no. of atoms of potassium present in 117g. of K. [K = 39] [Ans. = 3 x 6.023 X 1023atom]7. The no. of moles & molecules in 19.86 g. of Pb(N03h [Ans. = 0:06 moles &

[Pb = 207, N = 14, 0=16] 0.06 x 6:023 x 1023molecules]

8. The mass of an atom of lead [Pb = 202] [Ans. = 33.53 x 10-23g.]

9. The no. of molecules in 1Yz.litres of water. [Ans. = 83.33 x 6.023 x 1023molecules][density of water 1.0 g./ cc. - :. mass of water = vol. x density]

10. The gram-atoms in 88.75 g. of chlorine [Cl = 35.5] [Ans. = 2.5 g. atoms]11. The no. of hydrogen atoms in 0.25 mole of H2S04. [Ans. = 0.5 x 6.023 x 1023 particles]12. The gram molecules in 21 g. of nitrogen [N = 14] [Ans. = 0.75 g. molecules]13. The number of atoms in 10 litres of ammonia [N = 14, H = 1] [Ans. = 1.786x 6.023x 1023 atoms]14. The number of atoms in 60 g. of neon [Ne = 20] [Ans. = 3 x 6.023 X 1023atoms]15. The number of moles of 'X' atoms in 93 g. of 'X' [X is phosphorus = 31] [Ans. = 3 moles]16. The volume occupied by 3.5 g. of 02 gas at 27°C & 740 mm. press. [0 = 16] [Ans. = 2.76lits.]17. The moles of sodium hydroxide contained in 160 g of it. [Na=23, 0=16, He I], [Ans. 4 moles]18. The weight in g. of 2.5 moles of ethane [C2H6]. [C=12, He l] [Ans. 75 g.]19. The molecular weight of 2.6 g of a gas which occupies 2.24lits. at OOC& 760 mm press.[Ans. 26 g.]20. The gram atoms in 46 g of sodium [Na=23] [Ans. 2 g. atoms]21. The no. of moles of KCl03 that will be required to give 6 moles of oxygen. [Ans. 4 moles]22. The weight of the substance if it's molecular weight is 70 and in the gaseous form occupies 10 lits.

at 27°C and 700 mm pressure. [Ans. 26.18 g.]

86

/' Part B - STOICHIOMETRYP/ENTAGE COMPOSITION - EMPIRICAL & MOLECULAR FORMULA

CHEMICAL EQUATION CALCULATIONSV

1. PERCENTAGE COMPOSITION-v• Percentage composition - is the percentage by weight of each element in the compound .

Percentage composition =Wt. of the element in one molecule of the compound x 100

Gram molecular weight of the compound[% composition is also the % by mass of atoms of an element present in one mole of the compound.]

PEl}CENTAGE COMPOSITION - PROBLEMSVCalculate the percentage by weight of the following : a] Potassium in potassium dichromate

[K=39, Cr=52, 0=16] b) Phosphorus in calcium phosphate [CalPO,J2] [Ca=40, P=31, 0=16]a] Molecular weight of potassium dichromate [K2Cr207] = 78 + 104 + 112 = 294

294 g of potassium dichromate contains 78 g of potassium.

:. 100 g of K2Cr207 contains 78 x 100 = 26.53% of potassium - Ans.294

b] Molecular weight of calcium phosphate [Ca3(P04}z] = 3 x 40 + 62 + 8 x 16 = 310

310 g of calciuo:; phosphate contains 62 g of phosphorus

:. 100 g of Ca3(P04}z contains 62 x 100 = 20% of phosphorus - Ans.310

2. Calculate the mass of nitrogen supplied to the soil by 5 kg of urea. [CO(NHZ}JJ[N=14,C=12,0=16, H=1]Molecular weight of urea [CO(NH2}z] = 12 + 16 + 28 + 4'= 60 g

60 g of urea contains 28 g of nitrogen

5 x 1000 g of urea contains ? 28 x 5 x 1000 = 2333.3 g. of nitrogen - Ans ... 60

3. Calculate the percentage of water of crystallisation in washing soda Na2C03.10H2O.[ Na=23, C=12, 0=16, H=1].Molecular weight of Na2C03.10H20 = 23 x 2 + 12 + 16 x 3 (t) 10 (18) = 286

286 g of Na2C03.10H20 contains 180 g of water of crystallisation

100 g of Na2C03.10H20 contains ? 180 x 100 = 62.94% ofwaterofcrystallisation-Ans... 286

4. Calculate the percentage of pure iron in 10 kg. of iron [Ill] oxide [F.e20~ of 80% purity. [Fe=56, 0=16].Molecular weight of Fe20~ = 56 x 2 + 16 x 3 = 160

160 g. of pure Fe203 contains 112 g of pure iron.

8000 g. of pure Fe203 contains ? 112 x 8000:::! 5600g. or 5.6 kg ... 160

.. % of pure Fe in 10 kg. of Fe203 = ~'gx 100 = 56% of pure iron - Ans .I

5. Calculate the number of molecules of water of crystallisation in copper sulphate crystals, if 10g. ofhydrous copper sulphate crystals gives 6.4 g. of anhydrous Cu504 on heating. [Cu=64, 5=32, 0=16]Mass of CuS04.XH20 = 10 g.; Mass of anhydrous CuS04 = 6.4 g.; .'. Mass of XH20 = 3.6 g.Mol. wt. of Cu504 = 64 + 32 + 16 x 4 = 160; Mol. wt. of H20 = 18

Mass of water = 18x = 3.6 :. X = 3.6 x 160 = 5 molecules - Ans... Mass of anhydrous CuS04 160 6.4 6.4 x 18

R7

~2a] EMPIRICAL FORMULA• Empirical formula - is the formula of a compound which shows the-.simplest.uihol: p .. ratio between the alom~ of the elements in the compound.

MOLECULAR WHOLE NUMBERS SIMPLEST RATIO bFCOMPOUND FORMULA EMPIRICAL FORMULA

Glucose' C6H1206 1 2 1 CHzOBenzene C6H6 1 1 CH

/DETERMINATION OFEMPIRICAL FORMULA OFA COMPOUNDTo determine the empirical formula of a compound of sodium, sulphur and oxygen havingthe percentage composition Na=29.11 %, 5=40.51 %, 0=30.38% [Na=23, 5=32, 0=16]

Step I Write down the percentage composition [wt.] and the-atomic weight of each element present in the given compound

Element % Composition At. Wt.Sodium 29.11 23Sulphur 40.51 32Oxygen 30.38 "16

StepIl Divide the % composition [wt.] of each element by its atomic weight -The ratio gives the number of atoms of each element or relative number of atomsin the compound [egofor element sodium % comp. (29.11) + At. wt. (23)]

Element % Composition At. Wt. Relative No. of atomsSodium 29.11 23 29.11 = 1.266

23Sulphur 40.51 32 40.51 = 1.266 ;

32Oxygen 30.38 16 30.38 = 1.898

, . 16~Select the smallest ratio - amongst the relative no. of atoms & divide the.Step III aJremaining ratios by it to give the simplest ratio of atoms present in thecompound,[eg. the smallest ratio amongst 1.266, 1.266 and 1.898 is 1.266,

. hence divide each ratio by 1.266] .

bJ If the simplest ratio is not a whole number - multiply each ratio by thesmallest suitable integer so that a whole number ratio is obtained.

c] Write the empirical formula - showing the atoms [elements J in the propersimple ratio of whole numbers.

Element % Composition At. Wt. Relative No. of atoms Simplest RatioSodium 29.11 23 29.11 = 1.266 1.266 = lx2 = 2

23 1.266Sulphur 40.51 32 40.51 = 1.266 1.266 = 1x2 = 2

32 1.266Oxygen 30.38 16 30.38 = 1.898 1.898 = 1.5x2=3--16 1.266

. [Simplest ratio of whole numbers is 2:2:3.J Hence empirical formula is Na2S203•

88

2b] MOLECULAR FORMULA• Molecular formula - is the chemical formula which represents -the actual number of atoms of each element present in a molecule of the compound.

COMPOUND MOLECULARFORMULA

ACTUAL NUMBER OF ATOMS -PRESENT IN A COMPOUND

Glucose

Sulphuric acid

DETERMINATION OFMOLECULAR FORMULA OFA COMPOUND -FROM ITS EMPIRICAL FORMULA

Example : To determine the molecular formula of a compound having the percentagecomposition C=26.59%, H=2.22%, 0=71.19%. Vapour density of thecompound = 45. [C=12, H=l, 0=16J

The empirical formula of the compound is determined as explained already.

Step I Calculate the empirical formula weight from the empirical formula.

Empirical formula of above compound was calculated to be CH02.

:. empirical formula weight = 12 + 1 + 16 x 2 = 45[C] [H] [2(0)]

Step Il

Step III

StepN

6 atoms of carbon, 12 atoms of hydrogen, 6 atoms of oxygen

2 atoms of hydrogen, 1 atom of sulphur, 4 atoms of oxygen

Record the molecular weight [or calculate it from the V.D. of the compound]

. Molecular weight from vapour density.

V. D. is given = 45 .'. molecular weight = 2 x V. D. = 2 x 45 = 90

Determine the value of n an integer by applying the formula.Molecular weight = n x Empirical formula weight

Molecular weightor , n =

Empirical formula weight

90:. n = 45 = 2

Calculate the molecular formula by applying the formula

Molecular formula = [Empirical formulaj.,

ie. [CH02hHence molecular formula = C2H204

89

EMPIRICAL & MOLECULAR FORMULA - PROBLEMS

1. A compound of carbon, hydrogen and oxygen is found to contain 40% of carbon, 6.7% of hydrogen and53.3% of oxygen. Calculate its empirical formula. If its vapours density is 30, calculate the molecularformula. [C=12,u-t, 0=16J

Element % Composition At. Wt. Relative No. of Atoms Simplest Ratio of[At Ratio] whole numbers

40 = 3.33 3.33 = 1- -12 3.33

6.70 = 6.70 6.70 = 2- --I 3.33

53.3 = 3.33 3.33 = 116 3.33

Carbon 40% 12

Hydrogen 6.70% 1

Oxygen 53.3% 16

Hence the Empirical formula of the compound = CH20 - Ans ..-----------------~--------~IMolecular weight = 2 x VD. = 2 x 30 = 60 Molecular formula = Empirical formula x nEmpirical formula weight = 12 + 2 + 16 = 30 [n is an integer]

n = Molecular Weight = 60 = 2 n = Molecular Weight [or 2 x V.D.]Empirical Formula Weight 30 Empirical Formula Weight

Molecular formula = Empirical formula x n [n is a integer]CH20x2C2H402

=

=

Molecular formula of the compound = CzH402- Ans.

2. A chemical reaction showed that 10.47g. of the compound contained 6.21 g. of metal 'X' and the restof a non-metal 'Y'. Calculate the empirical formula of the compound formed between 'X' and 'Y'.[At. wt. of X = 207, Y = 35.5JSolution:

Calculate the % composition of each element

a] 10.47 g. of the compound contains 6.21 g. of metal 'X'

:. 100 g. of the compound contains 6.21 x 100 = 59.31 % of metal 'X'10.47

b] 10.47 g. of the compound contains 4.26 g. of non-metal 'Y' [10.47 - 6.21 = 4.26 g.]

:. 100 g. of the compound contains 4.26 x 100 = 40.69% of non-metal 'Y'10.47

Element % Composition At. wt. Atomic Ratio Simplest ratio of whole numbers

.X 59.31% 207 ~~#1 = 0.286

0.286 10.286 =

Y ,40.69% 35.5 ~~:~9 = 1.1461.146

40.286 =

:. The empirical formula of the compound in XY4- Ans.

90

Element % Composition At. wt. Relative No. of Atoms [At Ratio] Simplest Ratio of whole numbers

Sodium 18.60% 23 18.60 = 0.80 0.80 = 123 0.80

Sulphur 25.80% 32 25.80 = 0.80 0.80 = 132 0.80

Hydrogen 4.03% 1 ·4.03 = 4.03 4.03 = 51 0.80

Oxygen 51.58% 16 51.58 = 3.22 3.22 = 416 0.80

3. A compound has the following percentage composition: Na = 18.60%, S = 25.80%, H = 4.03% ando = 51.58%. Calculate the molecular formula of the crystalline salt assuming that all the hydrogenin the compound is in combination with the oxygen as water of crystallisation. Molecular weight ofthe compound is 248. [Na = 23, S = 32, H = 1, 0 = 16J

Empirical formula of the compound = NaSHsO 4Molecular weight = 248 [given]

.. Empirical formula weight = 23 + 32 + [1 x 5] + [16 x 4] = 124

n = Molecular Weight = 248 = 2Empirical Formula Weight 124

. . Molecular formula = Empirical formula x n [n is an integer]

= NaSHs04 x 2 - Na2S2HIOOgSince all the hydrogen in the compound is in combination with the oxygen as water of crystallisation.

.. 10 atoms of Hand 5 atoms of 0= 5H20 and hence 3 atoms of oxygen remain.

.. The molecular formula of the compound is NaZSZ03.5HzO - Ans.

4. Empirical formula of a compound is XY2• If its empirical formula weight is equal to its vapourdensity, calculate the molecular formula of the compound.Molecular formula = Empirical formula x n ie. [XY2x n].. n = Molecular Weight = 2xV.D.

Empirical Formula Weight Empirical Formula Weight

but Vapour Density = Empirical Formula Weight [given in problem]

.. n=2

:. Molecular Formula = XY2x 2 = X2Y4 :. Molecular formula of the compound = X2Y4- Ans.

5. State the empirical formula of each compound whose molecular formula is - a] C.sHI0 b]H2C02•a] Molecular formula = CSHlO b] Molecular formula = H2C02

:. Ratio of C & H is 5 : 10 :. Simplest ratio is 1 : 2 :. Ratio of H, C and ° is 2 : 1 : 2

:. Empirical Formula = CH2- Ans. :. Empirical Formula = H2C02 - Ans.

6. Calculate the empirical formula of a compound whose molecular formula is C8H60 4 and empiricalformula weight is 83. [C=12,H=l, 0=16]Molecular weight of CgH604 is = 96 + 6 + 64 = 166

.. n = Molecular Weight = 166 = 2Empirical Formula Weight 83

But molecular formula = Empirical formula x n

ie. CgH604 = empirical formula x n [n=2] :. Empirical formula of the compound = C4H302- Ans.

91

2H202 [1 x 2 + 16]2 [18] = 36 g

-7 2H2 + 02

[2 x 16]32g

3 'CALCULATIONS BASED ON - CHEMICAL EQUATIONS• . Chemical equation - is the balanced chemical transition reaction.

In a chemical reaction a rearrangement of particles results in -changes in the properties of the substances thereby the -properties of the product formed are entirely different from those of the reactants.

All reactants and products in a chemical equation are -represented by a set of formula thus giving the -identity of each element taking part in the reaction either as the reactant or as the product.

• Information from chemical equation

A chemical equation gives the information regarding:

a] The molecular proportion b] The relative mass c]The relative volumes [if gaseous]of the reactant and the products taking part in the chemical reaction.

Example

PROCEDURE INVOL VED FOR SOLVING PROBLEMS ON CHEMICAL EQUATIONS

Step I

Step II

Step III

SteplV

Calculate the weight and volume of oxygen at s. t.p. which will be evolved onelectrolysis of 18g of water. [Hel, 0=16J

Write fully the balanced equation of the reaction.

2H20 ~ 2H2 + 02

a] Find the molecular weight of each substance which is obtained byadding the weight of all the atoms in the molecules [neglect the molecularweights not asked] ,

b] The product of the number of molecules and the molecular weight of eachsubstance is written below each formula.

a] 36 g. of water liberates 32 g. of oxygen.

b] :. 18 g of water liberates 32 x 18 = 16 g.36

The volume of the gaseous product is calculated by multiplying the numberof molecules of the gaseous substance with 22.4 litres [1 mole of a gas occupies22.4 litres at s.t.p.].

2H20 . ~ 2H2 + 0236g 32 g = 1 mole of 02= 22.4 litres at s.t.p.

a] 36 g. of water liberates 22.4 litres of 02 at s.t.p.

b] :. 18 g of water liberates 22.4 x 18 = 11.2 litres of 02 at s.t.p.36

92

CHEMICAL EQUATIONS - PROBLEMSProblems based on a] Weight-Weight relationship, b] Weight-Volume relationship.1. Calculate the weight of potassium nitrite formed by thermal decomposition of 15.15 g of potassiumnitrate. [K=39, N=14, 0=16].Solution:The chemical equation for the reaction is

/!,.2KN03 ) 2KN02

2[39 + 14 :t 3 x 16] 2[39 + 14 + 2 x 16]2 x 101 2[85]

:. 2 x 101 gms of KN03 yield 2 x 85 gms of KN02.

:. 15.15 gms. of KN03 yields 15.15 x 85 x 2 = 12.75 g of KN02.101 x2

: .. Weight of potassium nitrite formed is 12.75 g. - Ans.

2. .Copper on reacting with cone. Hz504 produces copper sulphate. If 1.28 gm of copper is to beconverted to copper sulphate. Find i] the weight of the copper sulphate formed and ii] the weight ofthe acid required. [Cu=64, 5=32, 0=16].Solution:The chemical equation for the reaction isa] Cu + 2H2S04 ---7 CuS04 + 2H20 + S02 a]

64 g [64 + 32 + 4 x 16 = 160 g]64 g of Cu yields 160 g of CuS04

. . 1.28 g of Cu will yield 160 x 1.2864

Wt. of CuS04 formed =.3.2 g of CuS04 - Ans.

b] Cu + 2H2S04 ----7 CuS04 + 2H20 + S0264 g 2[2x1 + 1x32 + 4x16]

2[98] = 196. . 196 g of H2S04 are required to react with 64 g of CuHence ? g of H2S04 are required to react with 1.28 g ?f CuThe weight of acid required = 1.28 x 196 = 3.92 g. of H2~04 - Ans.

64

+ °2 [g]

! .. ", .2KN03 ----=-7 2KN02 "+ 02

a] 2x101 g. 2x85 g.[mol. wt.] [mol. wt.]

b] 15.15 g.[wt.]

:

? g.[wt.]

Cu + 2H2504 ~ CuS04 + 2H20+ 502

b] 1.28 g.

[wt.]

64g.[wt.]

2 [98] = 1969.[mol. wt.]

160g.[mol. wt.]

I

? g.

[wt.]

? g.

[wt.]

3. From the equation CaC03 + 2HCI-) CaClz +HzO +COz' Calculate the weight ofCaClz obtained from10g. of CaC03 and the volume at s.t.p. of COz obtained at a same time. [Ca=40, C=12,0=16, CI=35.5]Solution:CaC03 + 2HCI -) CaCl2 + H20 + CO2[40+12+48] [40+71] 1 mole 100g.

. a]100 g 111 g 22.4ltrs. at s.t.p. [mol. wt.]

a] lOO g of CaC03 gives 111 g of CaCl2

b] :. 10 g of CaC03 gives 111 x 10 = 11.1 g.100 of CaCl2 -Ans .

CaC03 + 2HCl ~ CaCI2 + 1I20 + CO2

b] ·10 g.[wt.]

.Similarly

100 g of CaC03 liberate 22.4 litres of CO2 at s.t.p.

:. 10 g of CaC03 liberate 22.4 x 10 = 2.24 litres of CO2 - Ans.100

111g.[mol. wt.]

22.4 Iits. [ ]-;::;-;-;--:-- s.t.p.[Vol.]

? g.[wt.]

? Iits,

[Iits.]

b] ? g.

[wt.] i

233 g.[mol. wt.]

1.74 g.

[wt.]

4. Combustion of butane takes place as follows: 2C4H10 + 130z· 7 8COz + 10HzO. Calculate a] thenumber of moles of oxygen needed for complete combustion of 58g of butane, bJ the volume of carbondioxide formed at s.t.p. at the same time. [H=l, C=,.1_2-=--J. ---,I.

2C4H1O + 13°2 ~ 8C02 + 10H2OSolution:

a] 2 x 58 = 116 g. 13 moles 8 x 22.4 lits. [s.t.p.]2C4H1O + 13°2 7 8C02 + iOH2O [mol. wt.] [moles] [Vol.]

2[48+10] 13 moles 8 moles 58 g. ? moles ? lits.116 g 22.4 x 8 litres at s.t.p. b] -- -- --

[wt.] [moles] [lits.]

a] 116 g of C4H10 needs for combustion 13 moles of 02b) . . 58 g of C4H10 needs 13 x 58 = 6.5 moles of 02 - Ans.

116Similarly116 g of C4H10 liberates 8 x 22.4 litres of CO2 at s.t.p.:. 58 g of C4H10 liberates 22.4 x 8 x 58 = 89.6 litres at s.t.p. of CO2 - Ans.

116

5. Thermal decomposition of calcium nitrate takes place as follows: 2Ca(NO:;)z 7 2CaO +4NOz + O2,If the relative molecular mass of calcium nitrate is 164. aJ Calculate the volume of nitrogen dioxideobtained at s.t.p. and bJ the weight of calcium oxide obtained when 16.4 g of calcium nitrate is heatedto constant weight. [Ca=40, 0=16, N=14J.

2Ca(~03)2 7 2CaO + 4~02 + °2Solution:

a] 2 x 164 = 328 g. 2x56 = 112g. 4 x 22.4 lits. [s.t.p.]2Ca(N03h 7 2CaO + 4N02 + °2 [mol. wt.] [mol. wt.] [Vol.]

2[164] 2[40+16] 4 moles 16.4 g. ? g. ? lUs.328 g 112g 4 x 22.4 litres at s.t.p. b] -- -- --

[wt.] [wt] [lits.]

a] 328 g of Ca(N03h liberates 4 x 22.4lits. of N02 at s.t.p.b) : ..16.4 g of Ca(N03h liberates 4 x 22.4 x 16.4 = 4.48 litres of N02 at s.t.p. - Ans.

328Similarly328 g of Ca(N03h gives 112 g of CaO:.16.4 g of Ca(N03h gives 112 x 16.4 = 5.6 g of CaO - Ans.

328

6. 2.12 g. of an impure mixture containing anhydrous sodium sulphate is dissolved in water. An excessof barium chloride solution is added when 1.74 g. of barium sulphate is obtained as a dry precipitate.Calculate the percentage purity of the impure sample. [Na = 23, S = 32, °= 16, Ba = 137].

Na2S04 + BaCl27 BaS04

[2 x 23 + 32 + 4 x 16] [137 + 32 + 64]

142 g. 233g.

+ 2NaCI a]142g.

[mol. wt.]

Solution:

a] 233 g. of BaS04 is obtained from 142 g. of Na2S04b] :. 1.74 g. of BaSO 4 is obtained from 142x 1.74 = 1.06 g. - mass of pure Na2S04 in2.12g. of impure mixture.

233:. % purity of the impure sample = 1.06 x 100 = 50%.

2.12

:. % purity of the impure sample = 50% - Ans.

94

LAWS & TERMS

PERCENTAGE COMPOSITION EMPIRICAL & MOLECULAR FORMULACHEMICAL EQUATION CALCULATIONS

Percentage composition - Is the percentage by weight of - each element present in the compound.

Empirical formula - Is the formula of a compound which shows the -simplest whole number ratio between the atoms of the elements in the compound.Molecular formula -1s the chemical formula which represent the -actual number of atoms of each element present in a molecule of the compound.

ADDITIONAL PROBLEMSQ.l PERCENTAGE COMPOSITION

1. Calculate the percentage by weight of: a] C in carbon dioxide, b] Na in sodium carbonate, c] Al inaluminium nitride. [ C=12, 0=16, n-i. Na=23, AI=27~ N=14 ][Ans.27.3%of C, 43.4%ofNa, 65.85%of Al]

2. Calculate the percentage of iron in K3Fe(CN)6' [K=39, Fe=56; C=12, N=14] [Ans. 17.02% of Fe]

3. Calculate which of the following - calcium nitrate or ammonium sulphate has a higher % of nitrogen.[Ca=40,0= 16,S = 32,N = 14] [Ans.(NH4)2S04 % ofN =21.21% more than Ca(N03)Z %ofN =17.07%]

4. Calculate the percentage of pure aluminium in.If) kg. of aluminium oxide [AI203] of 90% purity.[AI = 27, ° = 16] [Ans.47.64%]

5. State which of the following are better fertilizers- i] Potassium phosphate [K3P04] or potassiumnitrate [KN03] ii] Urea [NH2CONH2] or ammonium phosphate [(NH4)3P04] .[K=39,P~31,0= 16,N =14,H= 1] [Ans.K3POC % ofK=55.18% morethanKN03% ofK=38.61%]

[Ans. Urea % of N = 46.67% more than (NH4hP04 % of N = 28.19%],6. Calculate the percentage of carbon in a 55% pure sample of carbon carbonate. [Ca = 40, C = 12, o = 16]

[Ans. Carbon content = 6.6%]

7. Calculate the percentage of water of crystallisation in hydrated copper sulphate [CuS04.5H20].[Cu = 63.5, S = 32, ° = 16, H = 1] . [Ans. 36.07%]

8. Hydrated calcium sulphate [CaS04.xH20] contains 21 % of water of crystallisation.Calculate the number of molecules of water of crystallisation i.e. 'X' in the hydrated compound.[Ca = 40, S = 32, °= 16, H = 1] [Ans. CaS04.2H20]

Q.2 EMPIRICAL & MOLECULAR FORMULA

1. A compound gave the following data: C=57.82%, 0=38.58% and the rest hydrogen. Its vapourdensity is 83. Find its empirical and molecular formula. [C=12, 0=16, H=l ][Ans. C4H30Z & C8H60~

2. Four g of a metallic chloride contains 1.89 g of the metal 'X' . Calculate the empirical formula of themetallic chloride. [At. wt. of 'X' = 64, Cl = 35.5 ] [Ans. XCI2]

3. Calculate the molecular formula of a compound whose empirical formula is CH20 and vapourdensity is 30. [Ans. C2H40zl

4. A compound has the following percentage composition. Al = 0.2675 g.; P = 0.3505 g.; ° = 0.682 g. Ifthe molecular weight of the compound is 122 and its original weight which on analysis gave the aboveresults 1.30 g. Calculate the molecular formula of the compound. [AI=27, P=31, 0=16] [Ans. AIP04]

5. Two organic compounds 'X' and 'Y' containing carbon and hydrogen only have vapour densities 13and 39 respectively. State the molecular formula of 'X' and 'Y'. [C = 12, H = 1] [Ans. C2HZ; C6H6]

6. A compound has the following % composition. Zn = 22.65%; S = 11.15%; °= 61.32% and H = 4.88%.Its relative molecular mass is 287 g. Calculate its molecular formula assuming that all the hydrogenin the compound is present in combination with oxygen as water of crystallization.[Zn = 65, S = 32, ° = 16, H = 1] [Ans. ZnS04.7H20]

Q.3 CHEMICAL EQUATIONS

1. What mass of silver chloride will be obtained by adding an excess of hydrochloric acid to a solutionof 0.34 g of silver nitrate. [Cl=35.5, Ag=108, N=14,. 0=16, H=l ] [Ans. 0.287 g]

2. What volume of. oxygen at s.t.p. will be obtained by the action of heat on 20 g. of -KCl03.[K=39, CI=35.5, 0=16] [Ans. 5.486 Its.]

3.

7. A hydrocarbon contains 82.8% of carbon. Find its molecular formula if its vapour density is 29.[H = 1, C = 12] [Ans. C4H10]

8. An organic compound on analysis gave H = 6.48% and 0= 51.42%. Determine its empirical formulaif the compound contains 12 atoms of carbon. [C = 12, H = 1, ° = 16] [Ans. C12H240d

9. A hydrated salt contains Cu =25.50%, S = 12.90%, ° = 25.60% and the remaining % is water of 'crystallization. Calculate the empirical formula of the salt. [Cu = 64, S = 32, ° = 16, H = 1]

.. [Ans. CuS04.5H20]

10. A gaseous hydrocarbon weighs 0.70 g. and contains 0.60 g. of carbon. Find the molecular formula ofthe compound if its molecular weight is 70. [C = 12, H = 1] [Ans. CSH10]

11. A salt has the following % composition:- Al = 10.50%, K= 15.1%,S = 24.8% and the remaining oxygen.Calculate the empirical formula of the salt. [AI = 27, K = 39, S = 32, °= 16] [Ans. AIK(S04h]

From the equation: 3Cu + 8HN03 -7 3Cu(N03h + 4HzO + 2NO. Calculate(i) the mass of copper needed to react with 63 g of nitric acid [Ans. 24 g,](ii) the volume of nitric oxide collected at the same time. [Cu=64, H=l, 0=16, N=14][Ans. 5.61ts.]

Zinc blende [ZnS] is roasted in air. Calculate:a] the number of moles of sulphur dioxide liberated by 776 g of ZnS and [Ans. 8 moles]b] The weight of ZnS required to produce 22.4lits of SOz at s.t.p. [S=32, Zn=65, 0= 16] [Ans. 97 g.]

Ammonia reacts with sulphuric acid to give the fertilizer ammonium sulphate. Calculate the volumeof ammonia [at s.t.p.] used to form 59 g of ammonium sulphate.· [Ans. 20.021ts.][N=14, H=l, S=32, 0=16 ].

Heat on lead nitrate gives yellow lead [Il] oxide, nitrogen dioxide & oxygen. Calculate the total volumeof NOz' & 0z produced on heating 8.5 of lead nitrate. [Pb = 207, N = 14, ° = 16].

[Ans. 1.15 of N02 & 0.287 of 02 (1.4371ts.)]

A· A .2KCl03 ) 2KCI + 30z; C + 0z .) COz' Calculate the amount of KCl03 which on thermaldecomposition gives 'X' vol. of 0z, which is the volume required for combustion of 24 g. of carbon.[K = 39, Cl = 35.5, ° = 16, C = 12]. [Ans. 163.33 g.]

Calculate the weight of ammonia gas.a] Required for reacting with sulphuric acid to give 78 g. of fertilizer ammonium sulphate.b] Obtained when 32.6 g. of ammonium chloride reacts with calcium hydroxide during the

laboratory preparation of ammonia. [2NH4Cl + Ca(OHh ) CaClz + 2HzO + 2NH3][N = 14, H = 1, ° = 16, S = 32, Cl = 35.5]. [Ans. a) 20.09 g. b) 10.36 g.]

Sodium carbonate reacts with dil. HZS04 to give the respective salt, water and carbon dioxide.Calculate the mass of pure salt formed when 300 g. of NaZC03 of 80% purity reacts with dil. HZS04.[Na = 23, C = 12, °= 16, H = 1, S = 32]. [Ans. 321.51 g.]

10. Sulphur burns in oxygen to give sulphur dioxide. If 16 g. of sulphur burns in 'x' cc. of oxygen,calculate the amount of potassium nitrate which must be heated to produce 'x' cc. of oxygen.[S = 32, K = 39, N =14, ° = 16]. [Ans. 101 g.]

4.

5.

6.

7.

8.

9.

11: Sample of impure magnesium is reacted with dilute sulphuric acid to give the respective salt andhydrogen. If 1 g. of the impure sample gave 298.6 cc. of hydrogen at s.t.p. Calculate the % purity ofthe sample. [Mg = 24, H = 1]. [Ans.31.99%]

96

SUMMARY - Types of ProblemsA. PROBLEMS BASED ON - LUSSAC'S LAW IPROBLEM:4000cc. of 0z was burnt with 300 cc.of ethane. Calculate the vol. of unused 0z and COzfarmed.

• WRITE TIlE BALANCED EQUA TION- As per instructions in the calculations, with volumes below-I 300 cc. 4000 cc.

2 C2H6 + 7°2 ) 4C02 + 6H2O2 vols. 7 vols. 4 vols. + 6 vols.

. • REPRESENT THE EQUATION AS PER LUSSAC'S LAW - When gases react, they do so in vols.which bear a simple whole no. ratio to one another and to the products [temp., press., constant].

300 cc. 4000 cc.

2C2H6 + 7°2 ) 4C02 + 6H2O2 vols. 7 vols.' 4 vols. + 6 vols.

As per Lussac's Law 2 : 7 : 4 : 62 X 150 = 300 cc.Since ratio is 2 : 7 :4 : 6 multiply each ratio by 150

i.e. 2X150 : 7X150 : 4X150 : 6X150[1050 c.c.] [600 c.c.]

Ans. : 600 cc. of CO2 is formed: 2950 cc. of unused oxygen remains [4000 cc. (original) -1050 cc. (used)]

B. PROBLEMS BASED ON - MOLE CONCEPT & AVOGADRO'S NUMBERPROBLEM:Calculations based on -

iul The number of molecules:

iJ The number of moles iiJ The mass iiil The volumevJ The gram molecular weight.

a] 1 MOLE weighs ? GRAMS occupies22.4lits.or 22,400 cc., contains 6.023 X 1023MOLECULES-- [atoms/ ions][MOLE] [G. MOL. WT.] [VOLUME AT S.T.P.] [AVOG. NO.]

b] ? ? ? ? [atoms/NO. OF MOLES WEIGHT [mass] VOLUME NO. OF MOLECULES. ions]

C. PROBLEMS BASED ON - MOLE CONCEPT & 1\VOGADRO'S LAWPROBLEM:If 30 lits. of 0z contains iX' no. of molecules, state the no. of molecules in 10 lits. of H2I60 lits. of Clz and 5 lits. of NH3• All gases collected under the same conditions of temp. & press.As per Ayogadro's Law - Under the same conditions of temperature and pressure equal volume of all gasescontain the same number of molecules.

@30 tits. 'X' mOlecules@lO tits. X/3 molecules ®60 tits. 2X molecules 65 tits. X/6 molecules

, D. PROBLEMS BASED ON - VAPOUR DENSITY & MOLECULAR WEIGHTa] VAPOUR DENSITY = Weight of a certain vol. of gas or 1000 ml. [1 litre] of gas [Same conditions

OFAGAS Weight of an equal vol. of H2 i.e. 1000 ml. [1 litre] of H2 of temp. & press.]

b] MOLECULAR WEIGHT = 2 X vapour density

97

SUMMARY - Types of Problems [Contd.]E. PROBLEMS BASED ON - PERCENTAGE COMPOSITIONPROBLEM:Calculate the percentage of boron [BJ in borax - Na2B407·10H20 [Hel, B::=11,0=16, Na=23j

• PERCENTAGE = Wt. of element [or substance] in 1 molecule of the compound x 100COMPOSITION Gram molecular weight of the compound

• Weight of element [boron] in molecule of the compound [borax] = 11 X 4 = 44• Gram mol. wt. of compound [borax] = 23 X 2 + 11 X 4 + 16 X 7 + 10 [2 + 16] = 382

• PERCENTAGE COMPOSITION = 44 X 100 =<' 11.5% Ans.382

F. PROBLEMS BASED ON;... EMPIRICAL FORMULA & MOLECULAR FORMULAPROBLEM:A compound has the following % Composition: C = 40%, H = 6.7%, 0= 53.3%, the vapourdensity of the compound is 30, calculate its molecular formula [C = 12, H = 1, °= 16J.Element - Write the names of the elements Element % Comp. At. Wt At. ratio Simplest ratic

% Comp. - Write their % composition as given of whole nos.

40 = 3.33 3.33At. Wt. - Write the at. wts. of the elements Carbon 40 12 =1

12 3.33

At. ratio - Atomic ratio . = % Composition Hydrogen 6.7 1 6.7 = 6.70 6.70 =2[rel. no. of atoms] At. weight 1 3.33Simplest _ Simplest ratio = Each at. ratio Oxygen 53.3 16 53.3 = 3.33 3.33 =1ratio of whole numbers Smallest at. ratio 16 3.33

• Empirical formula = Cl H2 01 i.e. CH20• Molecular formula = Empirical formula X n [integar] I

n = Molecular wt. or 2 X VD.Empirical formula weight2XVD. 2X30 60 2n = = = - =CH20 12+2+16 30

• Molecular formula = CH20 X 2 = C2H402

G. PROBLEMS BASED ON - CHEMICAL EQUATioNSPROBLEM:Copper reacts with dilute nitric acid to give copper nitrate, water and nitric oxide. Calculateil the mass of copper needed to react with 126g. of HN03ii] vol. of nitric oxide obtained at the same time [Cu = 64, H = 1,0 = 16, N = 14J.

• WRITE THE BALANCED EQUATION - Complete the corresponding columns a] and b]3Cu + 8HN03 ) 3Cu (N03h+ 4H20 + 2NO

a] 3 X 64 = 192 8 [1 + 14 + 48] = 504 2 X 22.4 = 44.81it [1 1-g. g. S. mo e-22.4lits]

b] ? 126 ? lits.g. g.

i] 504 g. of HN03 reacts with 192 g. of Cu ii] 504 g. of HN03liberates 44.8lits. of NO. 192

126 g. of HN03 liberates 44.8 X 126 = 11.2 lits126 g. of HN03 reacts with -- X 126 = 48 e'504 of u 504 of NO

98

QUESTIONS"",.

For OB]ECTNE and ICSE BOARD TYPE QUESTIONS - Refer Dr. Viraf J. Dalal's'OBJECTIVE WORKBOOK FOR SIMPLIFIED I C S E CHEMISTRY' FOR STD. X &

'SIMPLIFIED ICSE CHEMISTRY - SOLVABLE QUESTION BANK & 25 TEST PAPERS' FOR STD. X

A. PROBLEMS BASED ON - LUSSAC'S LAW1999 1. ·4NH3 + S02 -7 4NO + 6H20 - If 27 litres of reactants are consumed, what volume of nitrogen

monoxide is produced at the same temperature and pressure. [12 lits.]2001 1. 4N20 + CH4 -7 C0...J, + 2H,0 + 4N2 If all volumes are measured at the same temp. & press. Calculate

the volume of N2U required to give ISO cm3 of steam. [300 cc.]2. What volume of oxygen would be required for the complete combustion of 100 litres of ethane

according to the following equation. 2C2H6 + 702 -7 4C02 + 6H20 [3S0 lits.]2003 1. What vol?-me of 02 is required to burn completely a mixture of 22.4 dm ' of CH4 & 11.2 dm3 of ~2;

The reactions are: CH4 + 202 -7 CO2 + 2H20; 2H2 + 02 -7 2H20 [All vols. at s.t.p.] [SO.4lits.]2006 1. S60 ml. of carbon monoxide is mixed with SOOml of oxygen and ignited. Calculate the volume of

oxygen used and carbon dioxide formed in the above reaction. [280 ml., S60 ml.]2009 1. 200 cm3 of acetylene is formed from a certain mass of calcium carbide. Find the volume of oxygen

required and carbon dioxide formed during its complete combustion. The combustion reaction canbe represented as: 2C2H2[g] + S02[g] -;. 4C02[g] + 2H20[g] [SOOcm-' of °2,400 cm3 of CO2]

2010 1. 10 litres of a mixture of propane [C3HS] [60%] and butane [C4HlO][40%] is burnt. Calculate thetotal volume of carbon dioxide formed. Combustion reactions of the mixture are represented as -C3HS(g)+ S02(g) -7 3C02(g) + 4H20(g) ; 2C4H10(g) + 1302(g) -7 8C02(g) + 10H20(g) [34 lits.]

. 2011 1. State Cay-Lussac's Law.B.PROBLEMS BASED ON MOLE CONCEPT -AVOGADRO'S NUMBER2004 1. A flask contains 3.2 g. of sulphur dioxide. Calculate the following:-

i] The moles of sulphur dioxide present in the flask. !O.OSmoles]ii The number of molecules of S02 present in the flask. [O.OSx 6.023 x 102 molecules]iii] The volume occupied by 3.2 g. of sulphur dioxide at s.t.p. [S = 32, ° = 16] [1.12 litres]

2. 2KMn04 -7 K2Mn04 + Mn02 + 02 Given that the molecular mass of KMn04 is lS8, what volumeof oxygen [measured at room temp.] would be obtained by the complete decomposition oflS.8 g. of potassium permanganate. [Molar volume at room temperature is 24 litres.] [1.2 litres]

2005 1. The volumes of gases A, B, C & D are in the ratio, 1:2:2:4 under the same conditions of temp. & press.i] Which sample of gas contains the maximum number of molecules. [D]ii] If the temperature and the pressure of gas A are kept constant, then what will happen to the

volume of A when the number ofmolecules is doubled. [Doubles]iii] If this ratio of gas vols. refers to reactants & products of reaction - gas law observed is_. [pg. 71Jiv] If the volume of' A' is actually S.6 dm3 at s.t.p., calculate the number of molecules in the actual

volume of 'D' at s.t.p. [Avogadro's number is 6 x 1023).Using your answer, state the mass of'D' if the gas is dinitrogen oxide [N20]. [N = 14, °= 16] [6 x 1023;44 g.]

2006 1. Calculate the number of moles and the number of molecules present in 1.4 g. of ethylene gas [C2H4].What is the vol. occupied by the same amount of ethylene. [O.OSmoles, 3x1022molecules, 1.12 lit.]What is the vapour density of ethylene. [Avogadro's Number = 6 x 1023;Atomic weight of C = 12, B = 1; Molar volume = 22.4 litres at s.t.p.] [14]

2008 1. The equation for the burning of octane is: 2 CsHlS + 2S 02 -7 16 CO2 + 18 H20i] How many moles of carbon dioxide are produced when one mole of octane burns. [8 moles]ii] What volume, at s.t.p. is occupied by the number of moles determined in 1. i]. [179.2lits.]iii] If the relative molecular mass of carbon dioxide is 44, what is the mass of [704 g.]

carbon dioxide produced by burning two moles of octane.2009 1. Define the term - Mole. A gas cylinder contains 24 x 1024molecules of nitrogen gas. If Avogadro's

number is 6 x 1023 and the relative atomic mass of nitrogen is 14, calculate:[i] Mass of nitrogen gas in the cylinder. [ii] Volume of nitrogen at STP in dm3 [1120 g., 896 dm-']

2. Gas 'X' occupies a volume of 100 cm3 at S. T. P. and weighs 0.5 g. find its relative molecular mass. [112g.]2010 1. Dilute hydrochloric acid [HCl] is reacted with 4.5 moles of calcium carbonate. Give the equation for the

said reaction. Calculate i] The mass of 4.5 moles of CaC03• ii] The volume of CO?, liberated at stp.

iii] The mass of CaClz formed. iv] The number of moles of the acid HCl used in the reaction[relative molecular mass of CaC02 is 100 and of CaC~ is 111]. [4S0 g., 100.8lits., 499.5 g., 9 moles]

2011 1. Calculate the mass of - i] 102 atoms of sulphur. ii] 0.1 mole of carbon dioxide.[S=32, C=12 and 0=16 & Avogadro's number = 6 x 1023] [0.533 g., 4.4 g.]

2. Calculate the volume of 320 g of S02 at stp. [S = 32 and 0= 16]. [112 lits.]

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C. PROBLEMS BASED ON - MOLE CONCEPT -AVOGADRO'S LAW1996 1.

D. PROBLEMS BASED ON - VAPOUR DENSITY AND MOLECULAR WEIGHT1996 1. Find the relative molecular mass of a gas, 0.546 g of which occupies 360 cm3 at 87°C and 380 mm

Hg pressure. [1 litre of hydrogen at s.t.p. weighs 0.09g] [88.89 g]2001 1. Mention the term defined by the following :- The mass of a given volume of gas compared to the

mass of an equal volume of hydrogen. [pg.78]2004 1. 2KMn04 -7 K2Mn04 + Mn02 + 02 [K2.Mn04. + Mn02 is the solid residue]

Some potassium permanganate was heated in a test tube. Atter collecting one litre of oxygen at roomtemperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre ofhydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g.Calculate the relative molecular mass of oxygen. . [32g.]

2009 1. A gas cylinder of capacity of 20 dm3 is filled with gas X the mas.s of which is 10 g. When the samecylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogenis 2 g., hence the relative molecular mass of the gas is : A] 5 B] 10 C] 15 D] 20 [10]

E. PROBLEMS BASED ON - PERCENTAGE COMPOSITION1996 1. Find the total percentage of oxygen in magnesium nitrate crystals: Mg(N03h. 6H20.

[0 = 16, N = 14, H = 1, Mg = 24] [75 %]What is the mass of nitrogen in 1000 kg of urea [CO(NH2}z]. [C=12] [Answer to nearest kg.][467 kg.]Calculate the % of boron [B] in borax Na2B407.10H20. [H = I, B = 11, ° = 16, Na = 23]. [11.5%]If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass in kg. of thefertilizer calcium nitrate would be required to replace the nitrogen in a 10 hectare field.[N=14; 0=16; Ca=40]. [1171 kg.]

2001 1. Calculate the percentage of phosphorus in the fertilizer superphosphate Ca(H2P04}z. [correctto Idp][H = 1; ° = 16; P = 31; Ca = 40] [26.5%]

2002 1. Calculate the percentage of platinum in ammonium chloroplatinate (NH4)2.PtCl6.. .-.[Give your answer correct to the nearest whole number]. [N = 14, H = I, Cl = ::>5.5,l't = 195] [44%]

2005 1. Calculate the percentage of nitrogen in aluminium nitride. [AI = 27, N = 14] [34.15%]2006 1. Calculate the percentage of sodium in sodium aluminium fluoride [Na3AIF6] correct to the nearest

whole number. [F = 19; Na = 23; Al = 27] [33%]2007 1. Determine the percentage 0'£ oxygen in ammonium nitrate [0 = 16] [60%]2010 1. If the relative molecular mass of ammonium nitrate is 80, calculate the percentage of nitrogen

and oxygen in ammonium nitrate. [N = 14, H = I, ° = 16] . . [35%, 60%]

F.PROBLEMSBASEDON-EMPIRICALFORMULAANDMOLECULARFORMULA2000 1.' Determine the empirical formula of the compound whose composition by. mass is: 42% nitrogen,

48% oxygen and 9% hydrogen. [H = 1; N = 14; °= 16]. [NOH3]2001 1. A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative

to hydrogen [i.e.YD.] is 162.5, find the molecular formula of the chloride. [M=56; C1=35.5][M2Cl6]2002 1. The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium,

18.9% phosphorus and 39% oxygen. Find the empirical formula of the compound [work to twodecimal places]. [Na = 23, P = 31, 0= 16] . [Na3P04]

2004 1. An experiment showed that in a lead chloride solution, 6.21g. of lead combined with 4.26 g. ofchlorine. What is the empirical formula of this chloride. [Pb = 207; Cl = 35.5] [PbCI4]

1999 1.

2001 1.

2002 1.

2005 1.2008 1.

2009 1.

1997 1.1998 1.1999 1.

Under the same conditions of temp. and press. you collect 2 litres of CO2, 3 litres of Cl2' 5 litres ofH2' 4 litres of N2 and 1 litre of S02' In which gas sample will there be the greatest number of moleculesand the least number of molecules. Justify your answer. [pg. 74, H2' SO~A vessel contains N molecules of oxygen at a certain temperature & pressure. How many moleculesof sulphur dioxide can the vessel accommodate at the same temperature & pressure. [N molecules]The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditionsof temperature and pressure. If 20 litres of nitrogen contain 'X' no. of molecules state the no. ofmolecules in 10 litres of chlorine, 20 litres of ammonia & 5 litres of sulphur dioxide. [x/2, x, x/4]Samples of 02' N2, CO and CO~nder the same conditions of temp. &press. contain the same numberof molecules represented by X.The molecules of oxygen occupy V litres and have a mass of 8g. Underthe same conditions of temp. & press., what is the volume occupied by: i] X molecules of N2;ii] 3X molecules of CO. iii] What is the mass of CO2 in grams. [V litres]; [3V litres][l1 g.Jiv] In answering the above questions, whose law has been used. [C = 12, N = 14, °= 16] [pg.74]Define the term' atomic weight'. ~ [pg.75]The gas law which relates the volume of a gas to the number.of molecules of the gas is [A]A: Avogadro's Law B: Gay-Lussac's Law C: Boyle's Law D: Charles'LawCorrect the following statement - Equal masses of all gases under identical conditions contain thesame number of molecules. [pg.74]

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2006 1. Determine the empirical formula of a compound containing 47.9% potassium, 5.5% beryllium and46.6% fluorine by mass. [At. weight of Be = 9; F = 19; K = 39] Work to one decimal place. [K2BeF4]

2007 1. A compound X consists of 4.8% carbon and 95.2% bromine by mass. i] Determine the empiricalformula of this compound working correct to one decimal place. ii] If the vapour density of thecompound is 252, what is the molecular formula of the compound. [C = 12; Br = 80][CBr3, C2Br6]

2008 1. What is the empirical formula of octane. [CSHlS] [C4H9]2. A compound contains - Carbon 14.4%, hydrogen 1.2% and chlorine 84.5 %. Determine the empirical

formula of this compound. Work correct to 1 decimal place. The relative molecular mass of thiscompound is 168, so what is its molecular formula. [C = 12; H = 1; Cl = 35.5] [CHCl2, C2H2C14]

2009 1. A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find themolecular formula of the compound if its relative molecular mass is 37. [N = 14, H = 1] [N2H4]

2011 1. An organic compound has vapour density 94. It contains C = 12.67%, H = 2.13%, and Br = 85.11 %.Find the molecular formula of the organic compound. [C = 12, H = I, Br = 80] [C2H4Br2]

G.PROBLEMS BASED ON CHEMICAL EQUATIONS1999 1. P + 5HN03 ~ H3P04 + H20 + 5N02

i] What mass of phosphoric acid can be prepared from 6.2 g of phosphorus. [19.6 g]ii] What mass of nitric acid will be consumed at the same time. [63 g]iii] What would be the volume of steam produced at the same time if measured at s.t.p.

[H = 1; N = l4; °= 16; P = 31] [4.48lits.]2000 1. Washing soda has the formula Na2C03.lOH2~' What mass of anhydrous sodium carbonate is left

when all the water of crystallization is expelled by heating 57.2 g of washing soda. [21.2 g]2. Na2S04 + Pb(N03}z ~ PbS04 + 2NaN03. When excess lead nitrate solution was added to a solution

of sodium sulphate, 15.15 g of lead sulphate were precipitated. What mass of sodium sulphate waspresent in the original solution. [H = 1; C = 12; ° = 16; Na = 23; 5 = 32; Pb = 207] [7.1 g]

2001 1. From the equation :- (l'{H4~ Cr207 ~ Cr203 + 4H20 + N2 Calculate:i] the vol. of nitrogen at STP, evolved when 63g. of ammonium dichromate is heated. [5.6lits.]ii] the mass of Cr203 formed at the same time. [N = 14, H = I, Cr = 52, °= 16]. [38 g.]

2003 1. 10 g. of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. Anexcess of barium chloride solution is added and 6.99 g. of barium sulphate is precipitated accordingto the equation'- Na2S04 + BaCl2 ~ BaS04 + 2NaCl. Calculate the percentage of sodium sulphatein the original mixture. [0 = 16; Na = 23; S = 32; Ba = 137] [42.6%]

2004 1. The reaction of potassium permanganate with acidified iron [Il] sulphate is given below:-2KMn04 + 10FeS04 + 8H2S04 -7 K2S04 + 2MnS04 + 5 Fe2(S04h + 8H.,,0.If 15.8 g. of potassium permanganate was used in the reaction, calculate the mass of iron [I1]sulphateused in the above reaction. [K = 39, Mn = 55, Fe = 56, 5 = 32, 0= 16] [76 g.]

2005 1. The equations given below relate to the manufacture of sodium carbonate [Mol. wt. of Na2C03 = 106]i] NaCl + NH3 + CO~ + H20 ~ NaHC03 + NH4Cl ii] 2NaHC03 ~ NaZC03 + H20 + CO2

Questions (a) and (b) are based on the production of 21.2 g. of sodium carbonate .. (a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g. of sodium

carbonate [Molecular weight of NaHC03 = 84]. [33.6 g.](b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of

carbon dioxide, measured at s.t.p., would be required. [8.96 lit.]2006 1. Given that the relative molecular mass [molecular weight] of copper oxide is 80, what volume of

ammonia [measured at s.t.p.] is required to completely reduce 120 g. of copper oxide.The equation for the reaction is: 3CuO + 2NH3 ~ 3Cu + 3H20 + N2. [22.4 lit.]

2007 1. A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at stp).NH4N03 -- N20 + 2H20 i] What volume of dinitrogen oxide is produced at the same timeas 8.96 litres of steam. ii] What mass of ammonium nitrate should be heated to produce8.96 litres of steam [Relative molecular mass of NH4N03 is 80] [4.48 litres., 16 g.]

2008 1. From the equation: C + 2H250 4 ~ CO2 + 2H20 + 2502 . Calculate:i] The mass of carbon oxidized by 49 g. of sulphuric acid [C = 12; rel. mol. mass of H2504 = 98].ii] The volume of sulphur dioxide measured at s.t.p., liberated at the same time.

[Volume occupied by 1 mole of a gas at s.t.p. is 22.4 dm3]. [3 g., 11.2 dm ']2009 1. Commercial sodium hydroxide weighing 30 g. has some sodium chloride in it. The mixture on

dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitateweighing 14.3 g. What is the percentage of sodium chloride in the commercial sample of sodiumhydroxide. The equation for the reaction isNaCl + AgN03 -- AgCl + NaN03. [Relative molecular mass of NaCl = 58; AgCl = 143] [19.33%]

2011 Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane [C3HS]'[C = 12, ° = 16, H = I, Molar Volume = 22.4 dm-' at stp] [22.4lits.]

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