Atoms and Molecules

Problems Based On Moles Of Molecules

We know that:
1 mole of molecules =Gram molecular mass =$6.022x1023$ molecules of a substance of the substance
This gives us three elations :
(i) The first relation is :
1 mole of molecules = Gram molecular mass
This relation can be used to convert the number of moles of molecules into mass in grams,and to convert the mass in grams into moles of molecules.
(ii) The second relation is :
1 mole of molecules = $6.022x1023$ molecules
By using this relation,we can convert the moles of molecules into number of molecules,and the number of molecules can be converted into moles of molecules.
(iii) The third relation is :
Gram molecular mass =$6.022x1023$ molecules
This relation is used to find out the number of molecules in a given mass of the substance and also to calculate the mass of a given number of molecules.
In many of the problems which we are going to solve now,we will need the mass of one mole of molecules of a substance. And to know the mass of one mole of molecules of a substance, we should know its molecular mass. So, first of all we will have to calculate the molecular mass of the substance from its molecular formula by using the atomic masses of the various atoms present in it (This is because the molecular mass of a substance is equal to the atomic masses of all the atoms present in one molecule of the substance).The molecular mass of a substance expressed in grams becomes the "gram molecular mass" of the substance and it gives the mass of 1 mole of the substance (or mass of $6.022x1023$ molecules of the substance).
The atomic masses of the elements which are required for the calculation of molecular mass are usuallygiven with the problem.But we should also remember the atomic masses of the common elements ourselves because sometimes these are not given in the problem.Keeping these points in mind,we will now solve some problems based on the moles of molecules.
Sample Problem 1.What is the mass of each one of the following ?
(a) 1 mole of water ($H$₂O)(b) 1 mole of ethanol ($C$₂H₆O)
(c) 1 mole of glucose ($C$₆H₁₂O₆) (d) 1 mole of cane sugar ($C$₁₂H₂₂O₁₁)
(Atomic masses :H = lu;C = 12u;0 = 16u)
Solution. The mass of 1 mole of each one of the given substances will be equal to their respective molecular masses expressed in 'grams Now :
(a)1 mole of $H$₂O = Molecular mass of $H$₂O in grams
= Mass of Hx2+Mass of O
= 1x2+16
= 2 +16
= 18 grams
So,the mass of 1 mole of water($H$₂O) is 18 g.
(b)1 mole of $C$₂H₆O= Molecular mass of$C$₂H₆Oin grams
= Mass of Cx2 +Mass of Hx6 +Mass of O
= 12x2 + 1x6 + 16
= 24 + 6 + 16
= 46 grams
So,the mass of 1 mole of ethanol ($C$₂H₆O)is 46 g.
(c)1 mole of $C$₆H₁₂O₆= Molecular mass of $C$₆H₁₂O₆in grams
= Mass of Cx6 + Mass of Hx12 + Mass of Ox6
= 12x6 + 1x12 + 16x6
= 72 +12+96
= 180 grams
So,the mass of 1 mole of glucose ($C$₆H₁₂O₆)is 180 g
(d) 1 mole of $C$₁₂H₂₂O₁₁ = Molecular mass of $C$₁₂H₂₂O₁₁ in grams
=Mass of Cx12 + Mass of Hx22 + Mass of Ox11
= 12x12 +1x22 +16x11
= 144+22+176
= 342 grams
So, the mass of 1 mole of cane sugar ($C$₁₂H₂₂O₁₁)is 342 g.
Sample Problem 2.Convert 22 g of carbon dioxide ($CO$₂) into moles,
Solution. Here we have been given that the atomic mass of carbon (C) is 12 u and the atomic mass of oxygen (O) is 16 u. So, first of all we will calculate the mass of 1 mole of carbon dioxide by using these values of atomic masses. The mass of 1 mole of carbon dioxide ($CO$₂) will be equal to its molecular mass expressed in grams.That is :
1 mole of $CO$₂ = Molecular mass of$CO$₂ in grams
=Mass of C+Mass of Ox2
= 12 + 16x2
= 12 + 32
= 44 grams
So,the mass of 1 mole of carbon dioxide is 44 grams.
Now,44 g of carbon dioxide = 1 mole
So,22 g of carbon dioxide=$\frac{1}{44}$x22 mole
=$\frac{1}{2}$
= 0.5 mole (or 0.5 mol)
Thus,22 grams of carbon dioxide are equal to 0.5 mole.
Note.The above problem can also be solved directly by using the formula
Number of moles
of molecules = $\frac{\mathrm{Mass\; of\; substance\; in\; grams}}{\mathrm{Gram\; molecular\; mass\; of\; substance}}$
= $\frac{22}{44}$
= $\frac{1}{2}$
=0.5 mole
Thus, 22 grams of carbon dioxide constitute 0.5 mole of carbon dioxide.
A yet another way of writing the above formula is by using the term 'molar mass' in place of gram molecular mass'.That is :Number of moles
of molecules =$\frac{\mathrm{Mass\; of\; substance}}{\mathrm{Molar\; mass\; of\; substance}}$
In the above case, mass of substance is 22 grams and molar mass of the substance is 44 g/mol.
Sample Problem 3.What is the mass of 0.5 mole of water ($H$₂O).
(Atomic masses :H=1u;0 = 16u)
Solution. In order to solve this problem, we should know the mass of 1 mole of water.This can be obtained by using the given values of the atomic masses of hydrogen and oxygen as follows :
1 mole of water ($H$₂O)=Molecular mass of$H$₂O in grams
= Mass of 2H atoms+Mass of O atom
=2x1+16
= 2+16
= 18 grams
Thus, the mass of 1 mole of water is 18 grams.
Now,Mass of 1 mole of water = 18 g
S0 Mass of 0.5 mole of water=18x0.5 g
= 9g
Thus, the mass of 0.5 mole of water ($H$₂O) is 9 grams.
Sample Problem 4. What is the number of molecules in 0.25 moles of oxygen ?
Solution. We know that:
1 mole of oxygen contains = $6.022x1023$ molecules
So,0.25 mole of oxygen contains = $6.022x1023$x0.25
= $1.505x1023$ molecules
Thus,0.25 mole of oxygen contains $1.505x1023$ molecules.
Sample Problem 5.Convert $12.044x1022$ molecules of sulphur dioxide into moles.
Solution.We know that:
$6.022x1023$ molecules of sulphur dioxide = 1 mole
So, $12.044x1022$ molecules of sulphur dioxide = $\frac{1}{\mathrm{6.022x10²³}}$x$12.044x1022$
=$\frac{2}{10}$
= 0.2 mole
Thus, $12.044x1022$ molecules of sulphur dioxide are 0.2 mole.
Sample Problem 6.What is the number of water molecules contained in a drop of water weighing 0.06g ?
Solution.We know that 1 mole of water contains $6.022x1023$ water molecules. So, in order to calculate the number of molecules in 0.06 gram of water, we should first calculate the mass of 1 mole of water in grams.
Now,Molecular mass of water,$H$₂O =2+16
= 18 u
SO Gram molecular mass of water = 18 grams(or 1 mole of water)
Thus, 1 mole of water has a mass of 18 grams and it contains$6.022x1023$molecules of water.
Now 18g water contains = $6.022x1023$ molecules
So 0.06 gwater contains=$\frac{\mathrm{6.022x10²³}}{18}$x0.06
= $2.007x1021$ molecules
Thus,a drop of water weighing 0.06 gram contains$2.007x1021$ moleculesof water in it.
Sample problem 7.Calculate the mass of $3.011x1024$
Solution. First of all we should find out the mass of 1 mole of nitrogen molecules ($N$₂).Now,
1 mole of math>N₂ = Molecular mass of $N$₂ in grams
= Mass of 2 N atoms in grams
= 2x14 grams
= 28 grams
Thus, the mass of 1 mole of nitrogen molecules is 28 grams. Now, 1 mole of nitrogen molecules contains $6.022x1023$ molecules. This means that the mass of $6.022x1023$ molecules of nitrogen is 28 grams.
Now,Mass of $6.022x1023$ molecules of $N$₂ = 28 g So, Mass of $3.011x1024$ molecules of $N$₂=$\frac{28}{\mathrm{6.022x10²³}}$x$3.011x1024$
$\frac{\mathrm{28x10}}{2}$
= 140 g
Thus, the mass of $3.011x1024$ molecules of nitrogen is 140 grams.
Sample Problem 8. The absolute mass of one molecule of a substance is $5.32x10-23$ What is its molecular mass ? What could this substance be ?
Solution.The molecular mass of a substance is numerically equal to the mass of 1 mole of its molecules. Since 1 mole of molecules is equal to $6.022x1023$ molecules,it means that the molecular mass of a substance is numerically equal to the mass of its $6.022x1023$6.022x10²³ Now,Mass of 1 molecule of bstance =$5.32x10-23$ g So, Mass of $6.022x1023$ molecules of substance =$5.32x10-23$ x6.022 x$6.022x1023$
= 32 g
Thus, the molecular mass of the given substance is 32 u. The substance having a molecular mass of 32 u is oxygen, having the formula $O$₂.
Sample Problem 9.In which one of the following cases the number of hydrogen atoms is more ?Two moles of HCl or One mole of $NH$₃
Solution (i) Two moles of HCl can be written as 2HCl.We can see that the two moles of HCl contain 2 moles of H atoms (or hydrogen atoms).
(ii) One mole of $NH$₃ contains 3 moles of H atoms (or hydrogen atoms).Now,two moles of HCl contain 2 moles of hydrogen atoms whereas one mole of $NH$₃ contains 3 moles of hydrogen atoms. It is obvious that 1 mole of $NH$₃ contains more hydrogen atoms.
Sample Problem 10.Calculate the mass of 1 mole of each one of the following :
(a) NaCl (b) $CaCO$₃ (c)$FeSO$₄.$7H$₂O (d)$Na$₂O₂
Solution. The mass of 1 mole of each one of the given ionic substances will be equal to their respective molecular masses (or formula masses) expressed in grams. In this problem,NaCl is sodium chloride,$CaCO$₃is calcium carbonate,$FeSO$₄.$7H$₂O is ferrous sulphate heptahydrate or iron (II)sulphate heptahydrate and $Na$₂O₂ is sodium peroxide.Now :
(a) 1 mole of NaCl = Formula mass of NaCl in grams
= Mass of Na + Mass of Cl
= 23+35.5
=58.5 g
Thus,the mass of 1 mole of NaCl = 58.5 g
(b) 1 mole of $CaCO$₃ =Formula mass of $CaCO$₃in grams
= Mass of Ca+Mass of C+Mass of Ox3
= 40+12+16x3
= 40 + 12+48
= 1oo g
Thus,the mass of 1 mole of
= 56 +32+16x11+1xl4
= 56 + 32 + 176 + 14
= 278g
Thus,the mass of 1 mole of$FeSO$₄.$7H$₂O is 278 g.
(d) 1 mole of Na202 = Formula mass of$Na$₂O₂ in grams
= Mass of Nax2 + Mass of Ox2
= 23x2 +16 x2
= 46 + 32
= 78 g
Thus,the mass of 1 mole of $Na$₂O₂ is 78 g.
Sample Problem 11.Which contains more molecules,4g of methane ($CH$₄) or 4 g of oxygen ($O$₂) ?
(Atomic masses :C = 12u,H=1 u, O = 16 u)
Solution.In this problem, we should convert 4 g of methane into moles and 4 g of oxygen should also be converted into moles by using the respective gram molecular masses.The substance having more moles will contain more molecules in it.
(i)Here,Mass of methane ($CH$₄) =4 grams
Gram molecular mass of$CH$₄ = Mass of C + Mass of 4 H
= 12 + 4x1
= 16 grams Now, Number of moles of $CH$₄ = $\frac{\mathrm{Mass\; of\; CH₄in\; grams}}{\mathrm{Gram\; molecular\; mass\; of\; CH₄}}$
=$\frac{4}{16}$
= $\frac{1}{4}$
= 0.250 mole
(ii) Here,mass of oxygen($O$₂) = 4 grams
Gram molecular mass of $O$₂=Mass of 2'O'atoms
= 2 x 16
= 32 grams
Now,number of moles of$O$₂ =$\frac{\mathrm{Mass\; of\; O₂in\; grams}}{\mathrm{Gram\; molecular\; mass\; of\; O₂}}$
= $\frac{4}{32}$
= $\frac{1}{8}$
= 0.125 mole
We find that 4 g of CH4 contains more moles of molecules (0.250 moles), whereas 4 g of$O$₂ contains less moles of molecules (0.125 moles). Since 4 g of methane has more moles,it contains more molecules (than 4 g of oxygen).
Sample Problem 12.If 1 g of sulphur dioxide contains x molecules,what will be the number of molecules in 1 g of methane ? (S = 32u,O=16 u,C=12 u, H = 1 u)
Solution.The ratio of molecules in sulphur dioxide and methane will be the same as the ratio of their moles.So,first of all we should find out the number of moles of sulphur dioxide in 1 gram of sulphur dioxide,and the number of moles of methane in 1 gram of methane.This can be done as follows :
(i)The molecular formula of sulphur dioxide is $SO$₂.
So,1 mole of$SO$₂ = Mass of S+Mass of 2'O'
= 32 + 2 x 16
=64 grams
Now,64 g of sulphur dioxide = 1 mole
So,1 g of sulphur dioxide = $\frac{1}{64}$ mole
Thus,we have $\frac{1}{64}$ mole of sulphur dioxide and it contains x molecules in it.Now,since equal moles of all the substances contain equal number of molecules,therefore,$\frac{1}{64}$ mole of methane will also contain xmolecules of methane.
We will now calculate the number of moles in 1 gram of methane.
(ii) Molecular formula of methane is $CH$₄.
So,1 mole of CH4 = Mass of C+Mass of 4H
= 12 + 4x1
= 16 grams
Now,16g of methane =1 mole
So,1 g of methane =$\frac{1}{16}$ mole
We know that:
$\frac{1}{64}$ mole of methane contains = x molecules
So,$\frac{1}{16}$ mole of methane will contain=$\frac{\mathrm{xX64}}{16}$ molecules
= 4x molecules
Thus,if 1 g of sulphur dioxide contains x molecules,then 1 g of methane contains 4x molecules.
Sample Problem 13.How many grams of oxygen gas contain the same number of molecules as 16 grams of sulphur dioxide gas ?(O= 16 u,S= 32 u)
Solution.This problem will be solved by using the fact that equal moles of all the gases contain equal number of molecules.Let us first convert 16 grams of sulphur dioxide into moles.1 mole of sulphur dioxide,$SO$₂= Mass of S+Mass of 2'O'
=32+2x16
= 64 grams
Now,64 g of sulphur dioxide = 1 mole
So,16 g of sulphur dioxide =$\frac{1}{64}$x 16 mole
= $\frac{1}{4}$
Now,$\frac{1}{4}$ mole of sulphur dioxide will have the same number of molecules as $\frac{1}{4}$ mole of oxygen.So,we should now convert mole of oxygen into mass in grams.
1 mole of oxygen,$O$₂= Mass of 2'O'atoms
= 2 x 16
= 32 grams
Now,1 mole of oxygen = 32 grams
So,$\frac{1}{4}$ mole of oxygen = 32x$\frac{1}{4}$ grams
= 8 grams
Thus,8 grams of oxygen will contain the same number of molecules as 16 grams of sulphur dioxide.
Sample Problem 14.Calculate the number of aluminium ions present in 0.051 g of aluminium oxide ($Al$₂O₃).(Atomic masses: A1=27 u;O = 16 u)
Solution.This problem involves aluminium ions.Please note that the mass of an aluminium ion is the same as that of an aluminium atom.In order to solve this problem,first of all we have to find out the mass of aluminium atoms in 0.051 g of aluminium oxide (which will give us the mass of aluminium ions)This can be done as follows:
1 mole of $Al$₂O₃=Formula mass of $Al$₂O₃ in grams
=Mass of A1x2+Mass of Ox3
= 27x2+16x3
=54 + 48
= 102 grams
Now,1 mole of$Al$₂O₃ contains 2 moles of Al.
So,Mass of Al in 1 mole of A1203 = Mass of Al x 2
= 27x2
= 54grams
Now,102 g aluminium oxide contains = 54 g Al
So,0.051 g aluminium oxide contains = $\frac{54}{102}$x0.051g Al
=0.027g Al
The atomic mass of aluminium is eiven to be 27 U.This means that 1 mole of aluminium atoms (or aluminium ions) has a mass of 27 grams,and it contains$6.022x1023$ aluminium ions.
Now,27g of aluminium has ions =$6.022x1023$So,0.027 g of aluminium has ions =$\frac{\mathrm{6.022X10²³}}{27}$x0.027
=$6.022x1020$
Thus,the number of aluminium ions($Al3+$)in 0.051 gram of aluminium oxide is$6.022x1020$
We are now in Position toanSWer the foIlowing questions:

Atoms and Molecules - Notes

1. Drawbacks of Dalton’s Atomic Theory

Show Notes

2. Monovalent Cations (Cations Having a Valency of 1+)

Show Notes

3. Divalent Cations (Cations Having a Valency of 2+)

Show Notes

4. Dalton’s Symbols of Elements

Show Notes

5. Trivalent Cations (Cations Having a Valency of 3+)

Show Notes

6. Formulae of Some Molecular Compounds

Show Notes

7. An Important Discussion

Show Notes

8. an example, let us give the significance of symbol C

Show Notes

9. Molecular Formulae of Some Common Elements

Show Notes

10. Chemical Formulae

Show Notes

11. 2. A Negatively Charged Ion is Known as Anion

Show Notes

12. As an example, let us give the significance of the formula H₂O

Show Notes

13. Molecules

Show Notes

14. Molecules of Compounds

Show Notes

15. Molecular Mass

Show Notes

16. Explanation of the Law of Conservation of Mass

Show Notes

17. Some Ionic Compounds

Show Notes

18. Atoms, Molecules and Ions

Show Notes

19. Atoms

Show Notes

20. Formula Mass

Show Notes

21. Formulae of Elements

Show Notes

22. Mole of Molecules

Show Notes

23. Simple Ions and Compound Ions (Polyatomic Ions)

Show Notes

24. 1. A Positively Charged Ion is Known as Cation

Show Notes

25. Symbols of Elements

Show Notes

26. 1. Molecules of Elements

Show Notes

27. Formulae of Compounds

Show Notes

28. Gram Atomic Mass And Gram Molecular Mass

Show Notes

29. Mole of Atoms

Show Notes

30. John Dalton

Show Notes

31. Ions

Show Notes

32. Explanation of the Law of Constant Proportions

Show Notes

33. Dalton's Atomic Theory

Show Notes

34. Significance of the Formula of a Substance

Show Notes

35. Some Common Ions

Show Notes

36. Molecular Masses of Some Common Elements

Show Notes

37. Laws Of Chemical Combination

Show Notes

38. Ionic Compounds

Show Notes

39. Calculation of Molecular Mass

Show Notes

40. Divalent Anions (Anions Having a Valency of 2-)

Show Notes

41. Formula Unit of Ionic Compounds

Show Notes

42. Atoms And Molecules

Show Notes

43. How Do Atoms Exist

Show Notes

44. Writing Of Formulae Of Ionic Compounds

Show Notes

45. Problems Based On Moles Of Molecules

Show Notes

46. Gram Molecular Mass

Show Notes

47. Experiment to Verify Law of Conservation of Mass

Show Notes

48. Gram Atomic Mass

Show Notes

49. Laws Of Conservation of Mass

Show Notes

50. Writing Of Formulae Of Molecular Compounds

Show Notes