Download 1ST Chapter text-book-questions

1st year chemistry notes
Chapter No. 1
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BASIC CONCEPTS
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TEXT BOOK EXERCISE
Select the most suitable answer from the given ones in each question.
The mass of one mole of electrons is
(a)
Properties which depend upon mass
(b)
Arrangement of electrons in orbital
(c)
Chemical properties
(d)
The extent to which they may be affected in electromagnetic field
(ii)
Which of the following statements is not true?
(a)
isotopes with even atomic masses are comparatively abundant
(b)
isotopes with odd atomic masses and even atomic number are
comparatively abundant
(c)
atomic masses are average masses of isotopes.
(d)
Atomic masses are average masses of isotopes proportional to their
relative abundance
(iii)
Many elements have fractional atomic masses, this is because
(a)
The mass of the atom is itself fractional
(b)
Atomic masses are average masses of isobars
(c)
Atomic masses are average masses of isotopes.
(d)
Atomic masses are average masses of isotopes proportional to their
relative abundance
(iv)
The mass of one mole of electrons is
(a)
008mg (b) 0.55mg
(c) 0.184mg
(d) 1.673mg
(v)
27g of Al will react completely with how much mass of O2 to produce
Al2O3
(a) 8g of oxygen
(b)
16g of oxygen
(c) 32g of oxygen
(d)
24g of oxygen
(vi)
The number of moles of CO2 which contain 8.0 g of oxygen.
(a)
0.25 (b)
0.50 (c)
1.0
(d)
1.50
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Q1.
(i)
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1st year chemistry notes
The largest number of molecules are present in
(a)
3.6g of H2 O
(b)
4.8g of C2H5 OH
(c)
2.8 g of CO
(d)
5.4g of N2O5
(viii) One mole of SO2 contains
(a)
6.02x1023 atoms of oxygen
(b)
18.1x1023 Molecules of SO2
(c)
6.02x1023 atoms of sulphur
(d)
4 gram atoms of SO2
(ix)
The volume occupied by 1.4 g of N2at STP is
(a)
2.24 dm3
(b)
22.4dm3
(c)
1.12 dm3
(d)
112 cm3
(x)
A limiting reactant is the one which
(a) is taken in lesser quantity in grams as compared to other reactants
(b) is taken in lesser quantity in volume as compared to the other reactants
(c) give the maximum amount of the product which is required
(d) give the minimum amount of the product under consideration
Ans: (i)a
(ii)d
(iii)d (iv)b (v)d
(vi)a (vii)a (viii)c
(ix)c (x)d
Q2: Fill in the blanks:
(i)
The unit of relative atomic mass is----------(ii)
The exact masses of isotopes can be determined by -----------spectrograph.
(iii)
The phenomenon of isotopes was first discovered by -------------(iv)
Empirical formula can be determined by combustion analysis for those
compounds which have-----------and -----------in them.
(v)
A limiting reagent is that which controls the quantities of ------------(vi)
I mole of glucose has-----------atoms of carbon ---------------of oxygen
and ----------of hydrogen.
(vii) 4g of CH4 at OoC and I atm pressure has ---------molecules of CH4.
(viii) Stoichiometry calculations can by performed only when -------------law is
obeyed.
Ans: (i)
amu (ii)
mass (iii)
Soddy (iv) carbon, hydrogen
(v)
Products
(vi) 6NA,6NA,12NA
23
(vii)
1.505x10
(viii) conservation and multiple
proportion
Q3: Indicate true or false as the case may be:
(i)
Neon has three isotopes and the fourth one with atomic mass 20.18 amu.
(ii)
Empirical formula gives the information about he total number of atoms
present in the molecule
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(vii)
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1st year chemistry notes
During combustion analysis Mg(CIO4)2 is employed to absorb water
vapors.
(iv)
Molecular formula is the integral multiple of empirical formula and the
integral multiple can never be unity.
(v)
The number of atoms in 1.79 g of gold and 0.023g of sodium are equal.
(vi)
The number of electrons in the molecules of CO an dN2 are 14 each, so
1 mg go each gas will have same number of electrons.
(vii) Avogadro’s hypothesis is applicable to all types of gases, i.e., ideal and
non-ideal .
(viii) Actual yield of a chemical reaction may by greater than the theoretical
yield.
Ans. (i)
False (ii)
False (iii)
True (iv)
False
(v)
False (vi)
True (vii) False (viii) False
Q4: What are ions? Under What condition are they produced?
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Ans: Ions can be produced by the following processes:
(i)
By dissolving ionic compounds in water
(ii)
By X-rays
(iii)
In mass spectrometry
(iv)
By removing or adding electron in atom
Q4:
(a)
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What are isotopes? How do you deduce the fractional atomic masses of
elements form the relative isotopes abundance? Give two examples in
support of your answer. ( See detail in Sublime subjective)
(b)
How does a mass spectrograph show the relative aboundace of isotopes of
an element?
. ( See detail in Sublime subjective)
(c)
What is the justification of two strong peaks in the mass spectrum for
bromine; while for iodine only one peak at 127 amu , is indicated?
Ans
The two strong peak in the mass spectrum for bromine represent two
different isotopes of bromine having nearly equal natural abundances. Only
one peak at 127 amu in the mass spectrum for iodine indicates that it has
only one isotope of atomic mass 127 amu.
Remember that!
Height of the peaks ∞ Relative abundance of isotopes
No. of peaks = No. of isotopes
Q5:
Silver has atomic number 47 and has 16 known isotopes but two occur
naturally I,e, Ag _____107 . and Ag _____109 . Given the following mass
spectrometric data, calculated the average atomic mass of silver,
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1st year chemistry notes
Isotopes mass (amu) percentage abundance
107
Ag
106.90509
51.84
109
Ag
108.90476
48.16
Solution:
The mass contribution for silver are:
Isotopes
Fractional abundance isotopic mass mass contribution
Ag
107
0.5184x107=55.4688
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107
107
0.4816x109=52.4944
Fractional atomic mass of silver
=107.9632
Hence the fractional atomic mass of silver is =107.9632
Ans.
Q6:
Boron with atomic number 5 has two naturally occurring isotopes.
Calculate the percentage abundance of 10B and 11B from the following
information.
Average atomic mass of boron
=10.81 amu
10
Isotopic mass of B
=10.0129 amu
11
Isotopic mass of B
=11.0093
10
Solution:
Let, the fractional abundance of B =x
The fractional abundance of 11B
=1-x
Remember that the sum of the fractional abundances of isotopes must be
equal to one, now, The equation to determine the atomic mass of element is
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109Ag
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(fractional abundance) (isotopic mass) (fractional abundance of 10B)(isotopic
mass of 10B )+(fractional abundance of 11B) (isotopic mass of 11B)
=Average atomic mass of Boron
(x)(10.0129)+(1-x)(11.0093) =10.81
10.0129x+11.00093x
=10.81
10.0129x-11.00093x =10.81-11.0093
-0.9964x
=-0.1993
x
=
10
Fractional abundance of B =0.2000
Fractional abundance of 11B =(1-0.2000)=0.8000
By percentage the fractional abundance of isotope is
%of 10B
=0.2000x100 =20% Answer
11
% of B
=0.8000x100 =80%Answer
Q7: Define the following terms and give three examples of each.
(i)
Gram atom
(ii)
Gram molecular mass
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(iii)
Gram molecular mass
(iv)
Gram ion
(v)
molar volume
(vi)
Avogadro’s number
(vii) Stoichiometry
(viii) Percentage yield
Q8: Justify the following statements:
(a)
23 g of sodium and 238g of uranium have equal number of atoms in the (b)
Mg atom is twice heavier than that of carbon
(c)
180g of glucose and 342 g of sucrose have the same number of molecules
but different number of atoms present in them.
(d)
4.9g of H2 SO4 when completely ionized in water , have equal number of
positive and negative charges but the number of positively charged ions are
twice the number of negatively charged ions.
(e)
One mg of K2 Cr O4 has thrice the number of ions than the number of
formula units when ionized in water.
(f)
Two grams of H2 , 16 g of ch4 and 44g of CO2 occupy separately the
volumes of 22.414 dm3 , although the sizes and masses of molecules of three
gases are very different from each other.
Solution:
(a)
23g of Na
=1 mole of Na
=6.02x1023 atoms of Na
238g of U
=1 mole of U
=6.02x1023 atoms of U.
Since equal number of gram atoms(moles) of different elements contain
equal number of atoms. Hence, 1 mole (23g ) of sodium and 1 mole (238)g of
uranium contain equal number of atoms , i , e ,6.02x1023 atoms.
(b)
Since the atomic mass of Mg (24) is twice the atomic mass of carbon (12)
therefore, Mg atom is twice heavier than that of carbon. Or
Mass of 1 atom of Mg=
Mass of 1 atom of C =
Since the mass of one atom of Mg is twice the mass of one atom of C, therefore,
Mg atom is twice heavier than that of carbon.
(c)
180 g of glucose = 1 mole of glucose =6.02x10 23 molecules of glucose 342
g of sucrose=1mole of sucrose =6.02x1023 molecules of sucrose
Since one mole of different compounds has the same number of molecules.
Therefore 1 mole (180g) of glucose and I mole (342g) of sucrose contain the
same number (6.02x1023)of molecules. Because one molecule of glucose, C6H12O6
contains 45 atoms whereas one molecules of glucose, C 12 H22 O11 contains 24
atoms. Therefore, 6.02x1023 molecules of glucose contain different atoms as
compound to6.02x1023 molecules of sucrose. Hence, 180 g of glucose and 342g og
sucrose have the same number of molecules but different number of atoms present
in them.
(d)
H2 SO4
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2H+ + SO
1st year chemistry notes
When one molecules of H2 SO4 completely ionizes in water it produces two
+
H ion and one SO ion,. Hydrogen ion carries a unit positive charge whereas SO
(e)
H2 SO4
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ion carries a double negative charge. To keep the neutrality, the number of
hydrogen are twice than the number of soleplate ions. Similarly the ions produced
by complete ionization of 4.8g of H2 SO4 in water will have equal number of
positive and negative but the number of positively charged ions are twice the
number of negatively charged ions.
2H+ + SO
(i)
Solution:
(a)
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K2 Cr O4 when ionizes in water produces two k+ ions one C O ion. Thus
each formula unit of K2 Cr O4produces three ions in solution .Hence one mg of K2
Cr O4 has thrice the number of ion than the number of formula units ionized in
water.
(f)
2g of H2 =1 mole of H2 =6.02x1023 molecules of H2 at STP =22.414dm3 16g
of CH4 =1mole of CH4 =6.02x1023 molecules of CH4 at STP =22.414dm3 144g of
CO2 =1mole of CO2 =6.02x1023 molecules of CO2 at STP =22.144dm3
Although H2 , CH4 and CO2 have different masses but they have the same
number of moles and molecules . Hence the same number of moles or the same
number of molecules of different gases occupy the same volume at STP. Hence 2 g
of H2 ,16g of CH4 and 44 g of CO2 occupy the same volume 22.414 dm3 at STP.
The masses and the sizes of the molecules do not affect the volumes.
Q10: Calculate each of the following quantities
(a)
Mass in grams of 2.74 moles of KMnO4 .
(b)
Moles of O atoms in 9.0g of Mg (NO3)2 .
(c)
Number of O atoms in 10.037g of Cu SO4 .5H2 O.
(d)
Mass in kilograms of 2.6x 1020 molecules of SO2 .
(e)
Moles of C1 atoms in 0.822g C2H4C12 .
(f)
Mass in grams of 5.136 moles of silver carbonate .
(g)
Mass in grams of 2.78x1021 molecules of CrO2 C12 .
(h)
Number of moles and formula units in 100g of KC1O3 .
Number of K+ ions C1O ions, C1 atoms, and O atoms in (h)
No of moles of KMnO4
=2.74moles
formula mass of KMnO4
=39+55+64=158g mol -1
Mass of KMnO4
=?
Formula used:
Mass of KMnO4
= no of mole of KMnO4 x formula mass of KMnO4
=2.74 mol x 158 g mol-1
=432.92g Answer
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1st year chemistry notes
Mass of Mg (NO3)2 =9g
Formula mass of Mg (NO3)2 =24+28+96=148g mol -1
No of moles of O atoms
=?
Formula used:
No of mole of Mg (NO3)2
=
Now, I mole of Mg (NO3)2 contains = 6moles of O atoms
0..06 moles of Mg (NO3)2contains =6x0.6
=0.36 moles of O atoms
Alternatively,
148g of Mg (NO3)2 contains =6moles of O atoms
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(b)
=
=0.36 mole Answer
4
(c)
Mass of CuSO . 5H2O=10.037g
Formula mass of CuSO4. 5H2O=63.54+32+64+90
=249.546g mol -1
No of moles of CuSO4. 5H2O
=?
4
No of moles of CuSO . 5H2O
=
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g of Mg (NO3)2contains
=
atoms
(d)
Now,
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1 mole of CuSO4 .5H2O contains
9moles of O atoms
0.04 mole of CuSO4 .5H2O contains=9x0.04
=0.36 moles of O atoms
I mole of O atoms contains
=6.02x1023 O atoms
0.36 mole of O atoms contains
=6.02x1023 x0.36 oxygen
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Now,
No of molecules of SO2 .
Molecular mass of SO2 .
Avogadro’s number , NA
Mass of SO2 molecules
=2.17x1023 oxygen atoms
=2.17x1023 atoms Answer
=2.6x1020 molecules
=32+32=64 g mol-1
=6.02x1023 molecules of SO2
=27.64x10-3 g
=
=27.64x10-6 kg
=2.764x10-3 kg Answer
(e)
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Mass of C2 H4C1
= 0.822g
1st year chemistry notes
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Molecular mass of C2 H4C1
=24+4+71=99 g mol-1
No of moles of C2 H4C1
=
Now, 1 mole of C2 H4C1 contains
=2moles of C1 atoms
8.3x10-3mole of C2 H4C1 contains =2x8.3x10-3 mole of atom
=16.6x10-3
=0.0166mole of C1 atom
=0.017 mole Answer
(f)
No of mole of Ag2 CO3
=5.136moles
Formula mass of Ag2 CO3
=215.736+12+48=275.736 g mol-1
Mass of Ag2 CO3=No of moles of Ag2 CO3xformula mass of Ag2 CO3
=5.136molx275.736 g mol-1
=416.18g
=1416.2 g Answer
(g)
Molecular mass of CrO2C12 =52+32+71=155g mol-1
NA
=6.02x1023 molecules mol-1
Molecules of CrO2C12==2.78x1021 molecules
Now,
mass of CrO2C12
=
=
=71.578x10-2 g
=0.71578
=0.716 g Answer
(h)
Mass of KCIO3
=100g
Formula mass of KCIO3
=39x35.5+48=122g mol-1
No of moles of KCIO3
=?
No of moles of KCIO3
No of formula units
(i)
No of K+ ions
=
=
=0.816mole Answer
No of moles x Avogadro,s No
=0.816mole x 6.02x1023 formula units
=4.91x1023 formula units
=4.91x1023 Answer
=4.91x1023 Answer
=4.91x1023 Answer
= 4.91x1023 x3
=14.73x1023 =1.473x1024 Answer
Q 11 Aspartame he artificial sweetener, has a molecular formula of C14 H18
N2O5 .
(a)
What is the mass of one mole of aspartame?
No of CIO ions
No of CIO ions
No of O atoms
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1st year chemistry notes
(b)
No of moles of aspartame
=
=
e.c
=0.1768 mol
=0.177 mol Answer
No moles of aspartame
= 10.122 moles
Molecular mass of aspartame =294g mol-1
Mass of aspartame
=No of moles x Molar mass
=10.122mol x 294g mol-1
=2975.87 g Answer
Mass of aspartame
=243g
Molar mass of aspartame
=294g mol -1
No of molecules of aspartame=?
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(a)
(b)
How many moles are present in 52g of aspartame?
(c)
What is the mass in grams of 10.122 moles of aspartame?
(d)
How many hydrogen atoms are present in 2.34g of aspartame?
Molecular mass of aspartame =168+18+28+80=295g mol-1
Mass of 1 mole of aspartame =294g mol-1 Answer
Mass of aspartame
=52g
Molecular mass of aspartame =294g mol-1
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No of molecules of aspartame=
xNA
=
=
=4.98x1021 molecules.
Now,1 molecule of aspartame contains
=18 H atoms
4.98x 1022 molecules =18x4.98x1021 H atoms
=89.64x1021H atoms
=8.964x1022 H atoms Answer
Q 12: A sample of 0.600 mole of a metal M reacts completely with excess of
fluorine to from 46.8g MF2 .
(a)
How many moles of F are present in the sample of MF2 that forms.
(b)
which elements is represented by the symbol M ?
Solution:
(a)
Formula of compound
=MF2
No of moles of M
=0.6 mol
Mass of MF2 =46.8g
The molar of M:F in the compounds;
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1st year chemistry notes
=0.6x2=1.2mol Answer
=No of moles of Fx At . mass of F
=1.2x19=22.8g
Mass of compound =46.8g
Mass of metal, M
=46.8-22.8
=24
At mass of M
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No of moles of F
Mass of F
=
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=
(b)
The atomic mass of the elements, M
=40
The metal is calcium, Ca Answer
Q 13 : In each pair , choose the larger of the indicated quantity ,or state if the
samples are equal.
(a)
Individual particles: 0.4 mole of oxygen molecules or0.4mole of
oxygen atom.
(b)
Mass: 0.4 mole of ozone molecules or0.4mole of oxygen atoms
(c)
Mass: 0.6 mole of C2 H4 or 0.6mole of 12
(d)
Individual particles: 4.0g N2O4 or 3.3g SO2
(e)
Total ions: 2.3 moles of NaC1O3 or 2.0mole of MgC12
(f)
Molecules: 11.0g of H2Oor 11.0g H2O2
(g)
Na+ ion: 0.500 moles of NaBr or 0.0145kg NaC1
(h)
Mass: 6.02x1023 atoms of 235U or 6.02x1023 atoms of 238U
Ans:
(a)
Number of molecules
=moles x NA
Number of O2 molecules
=0.4x6.02x1023 =2.408x1023 molecules
No of O atoms=0.4x6.02x1023=2.108x1023 atoms
There are equal number of individual particles in 0.4 mole of oxygen
molecules and 0.4 mole of oxygen atom. In general, equal number of moles of
different substances contains equal number of particles.
Both are equal
Answer
(b)
Mass of substance
= moles x molar mass
Mass of oxygen atoms =0.4x16=64g
Mass of ozone, O3 molecules =0.4x48=19.2g
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1st year chemistry notes
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0.4 moles of ozone molecules have larger mass than 0.4mole of oxygen atoms.
Ozone Answer
(c)
Mass of C2H4
=0.6x28=1.68g
Mass of 12
=0.6x127=254g
0.6mole of 12 have larger mass than 0.6 mole of C2H4
12 Answers
(d)
No of molecules
=
e.c
(e)
No of molecules in N2 O4 = x6.02x1023 =2.62 x1023 molecules
No of molecules in SO2 =x6.02x1023
=3.1x1022 molecules
3.3g of SO2 have larger number of individual particles than 4.0 g of N2 O4 .
SO2 Answer
No of formula units
=Moles x NA
3
No of formula units of NaC1O
=2.3x6.02x1023=1.38x1024 formula
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units
No of molecules
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No of ions in 1 formula units of NaC1O3=2
Total no of ions in MgC12
=2x1.38x1023=2.76x1024 ions
No of formula units of MgC12
=2.0x6.02x1023 x3=3.6x1024 ions
No .of ions in one formula unit of MgC12 =3
Total no of ions in MgC12
=1.20x1024 x3=3.6x1024 ions
2.0moles of MgC12 contain lager number of total ions than 2.3 moles of NaC1o3MgC1 Answer
No of molecules in H2 O2
=
=
NA
x6.02x1023=3.68x1023 molecules
No of molecules in H2 O2
x6.02x1023=1.95x1023 molecules
=
11.0g of H2 O2contains larger number of molecules than 11.0g of H2 O2
H2 O2Answer
(g)
No of formula units
=moles xNA
No of formula units NaBr
=0.5x6.02x1023=3.01x1023 formula units
One formula units o NaBr contain Na+ ions =1
3.01 x1023 formula unit of NaBr contains Na +ions =3.01x1023 Na+ ions
No of formula units of NaC1
=
x6.02x1023 =1.49x1023formula
units
One formula unit of NaC1 contains Na+ ions =1
1.49x1023 formula units of NaC1 contains
=1.49x1023 Na+ ions
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1st year chemistry notes
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0.500 moles of NaBr contains lager number of Na+ ions than 0.0145kg ofNaC1.
NaBr Answer
(h)
Mass of atoms of an element =
Mass of 235Uatoms
=x6.02x1023 =235g
Mass of 238U atoms =x6.02x1023=238g
238
U Answer
Q 13:
(a)
Calculate the percentage of nitrogen in the four important fertilizer i.e.,
(i)NH3 (ii)NH2CONH2(Urea) (iii)(NH4)2SO4
(iv)NH4 NO3
(b)
Calculate the percentage of nitrogen and phosphorus in each of the
following:
(i)
NH4H2PO4
(ii)
(NH4)) PO4 (iii)
(NH4)4 PO4
Solution:
(a)
Mol-mass of NH3
=14+4=17g
Mass of N
=14g
% of N
=x100
=82.35% Answer
(b)
Mol-mass of NH2 CONH2
=28+4+12+16=60g
Mass of N
=28g
%of N
=x100
=46.35% Answer
(c)
Mol-mass of (NH2 )2 SO4
=28+8+32+64=132g
Mass of N
=28g
% of N
=x100
=21.21% Answer
(d)
Mol-mass of (NH2 )2 SO4 =28+4+48=80g
Mass of N
=28g
%of N
=x100
=35% Answer
(I)
Mol-mass of (NH2 )2 SO4
=14+6+31+64=115g
Mass of N
=14g
Mass of P
=31g
%of N
=x100=12.17% Answer
%of P
==26.96% Answer
(II)
Mol-mass of ((NH2 )2 SO4
=28+9+31+64=132g
Mass of N
=28g
Mass of P
=
%of N
= =21.21% Answer
%of P
= =23.48% Answer
(III) Mol-mass of (NH2 )2 SO4
=42+12+31+64=149g
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1st year chemistry notes
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Mass of N
=42g
Mass of P
=31g
%of N
=
%of P
=
Q 14: Glucose C6 H12 O6 is the most important nutrient in the cell for generating
chemical potential energy. Calculate the mass% of each element in glucose and
determine the number of C,H and O atoms in 10.5g go the sample.
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Solution:
Mol-mass of glucose C6 H12 O6
=72+12+96=180g
Mass of C
=72
Mass of H
=12
Mass of O
=96
% of C
= =40% Answer
% of H
= =6.66% Answer
% of O
= =53.33% Answer
Mass of C6 H12 O6
=10.5g
Mol-mass of C6 H12 O6
=180g
Mol-mass of
=180g mol-1
No of moles of C6 H12 O6
=
No of molecules of glucose =No of moles x NA
=0.058 molx 6.02x1023 molecules mol-1
=0.35x1023 molecules
=3.5x1022 molecules
Now, 1 molecule of glucose contains
=6C-atoms
22
3.4x10 molecules of glucose contains
=6x3.5x1022 C-atoms
=21x1022 =2.1x1023 C atoms Answer
1 molecules of glucose contains
=12H-atoms
22
3.5x10 molecules glucose contains =12x3.5x1022
=4.2x1023 H- atoms Answer
1 molecule of glucose contains
=6 O –atoms
22
3.5 x 10 molecules of glucose contains
=6x3.5x1022
=2.1x1023 O-atoms Answer
Q 16: Ethylene glycol is used as automobile antifreeze .It has 38.7% carbon,
9.7% hydrogen and 51.6% oxygen. Its molar mass is 62.1 grams mol -1 .Determine
its molecular formula.
Solution:
% of C=38.37 g
% of H =9.7g
% of O=51.6g
-1
-1
At. Mass of C=12g mol
At. Mass of H=1.008g mol
At. Mass of O =16g
-1
mol
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1st year chemistry notes
=31
e.c
1
:3
:1
3
Empirical formula =CH O
Empirical formula mass
om
No of moles of C
=
No of moles of H
=
No of moles of O
=
Atomic ratio is obtained by dividing the moles with 3.23, which is the smallest
ratio.
C
:H
:O
ww
w.a
llon
line
fre
n=
Molecular formula
=2x CH3 O
=C2 H6 O2 Answer
Q 16: Serotonin (Molecular mass= 176g mol-1 ) is a compound that conducts
nerve impulses in brain and muscles. It contains 68.2 % C, 6.86% H, and 9.08%
O. What is its molecular formula?
Solution:
No of moles of C
=
No of moles of H
=
No of moles of N
=
No of moles of O
=
C
:
H
:
N
:
O
Atomic
:
:
:
Ratio
10
:
12
:
2
:
1
Empirical formula
=C10 H12 N2 O
Empirical formula mass
=120+12+28+16=176g mol-1
Molecular mass
=176g mol-1
n=
Q17: An unknown metal M reacts with S to from a compound with a formula
M2S3 .If 3.12 g of M reacts with exactly 2.88 g of sulphur, what are the names of
metal M and the compound M2 S3 .
Solution:
Formula of compound
= M2 S3
Mass of M
=3.12g
Mass of S
=2.88g
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1st year chemistry notes
om
Atomic mass of S
=32g mol-1
No of moles of S
=
No of moles of S
=
The molar ratio of M: S in the compound is :
No of moles of M
=
fre
Now,
=
=0.06 mole
e.c
No of moles of M
llon
line
At. Mass M =
The mass of M used in the formation of M2S3 is 3.12g. The product M2S3 therefore
also contains 3.12g of M, because mass is conserved. The amount of M before and
after reaction must be the same. Since we know both the number of moles of M
and the mass of M , we can cal calculate the atomic mass of M as follows:
=
=52
Atomic number, Z
=52
Q19: The octane present in gasoline burns according to the following equation.
2C8 H18 (i)
+ 2502(g)
16CO 2(g) + 18H2O (i)
(a)
How many moles of O2 are needed to react fully with 4 moles of
actane?
(b)
How many moles of CO2 can be produced from one mole of actane?
(c)
How many moles of water are produced by the combustion of 6 moles
of octane?
(d)
If this reaction is to be used to synthesize 8 moles of CO 2 how many
grams of oxygen are needed? How many grams of octane will be used?
Solution:
ww
w.a
At. Mass of M
4 moles
2C8 H18 (i)
(a)
2 moles
+ 2502(g)
25 moles
2 moles of C8 H18
4 moles of C8 H18
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16CO 2(g) + 18H2O (i)
=25 moles of O2
=
1st year chemistry notes
=50moles of O2 Answer
(b)
1 moles
2C8 H18 (i)
2 moles
+ 2502(g)
Now, 2 moles of C8 H18
1 mole of C8 H18
16CO 2(g) + 18H2O (i)
=16 moles of CO2
=
=8 moles of CO2 Answer
+ 2502(g)
Now, 2 moles of C8 H18
6 moles of C8 H18
(d)
16CO 2(g) + 18H2O (i)
=18 moles of H2 O(i)
=
=54 moles of H2 O
e.c
2C8 H18 (i)
2 moles
om
6 moles
6 moles
2C8 H18 (i)
2 moles
+ 2502(g)
Now, 16 moles of CO2
8 moles of CO2
16CO 2(g) + 18H2O (i)
fre
(c)
1800moles
=25 moles of O2
ww
w.a
llon
line
=
=12.5 moles of CO2
Mol-mass of O2
=32g mol-1
=12.5 molx 32g mol-1
=400g of O2
Now, 16moles of CO2
=2moles of C8 H18
8 moles of CO2
=
=1 mole of C8 H18
Mol-mass of C8 H18 =96+18=114g mol-1
Mass of C8 H18
=No of moles of C8 H18xMol.mass ofC8
H18
=1 molx 114 g mol-1
114g Answer
Q19: Calculate the number of grams of A12 S3 which can be prepared by the
reaction of 20 g of A1 and 30 g of sulphur. How much the non-limiting reaction is
in excess?
Solution:
Mass of A1
=20g
Molar mass of A1
=27g mol-1
No of moles of A1
=
Mass of S
= 30g
Molar mass of S
=32g mol-1
No of moles of S
=
0.74 mole 0.94 mole
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1st year chemistry notes
2A1
+
3S
2 mole
A12 S3
1 mole
3 mole
Now, 2 moles of A1
0.74 moles of A1
Now,
3 moles of S
0.94 moles of S
=1 mole of A12 S3
=
=0.37 mole of A12 S3
=1 moles of A12 S3
=
line
fre
e.c
om
=0.313 mole of A12 S3
Since S give the least number of moles of A12 S3 therefore, it is the limiting
reactant.
No of moles of A12 S3
=0.313 mole
Molar mass of A12 S3
=150g mol-1
Mass of A12 S3=No of moles of A12 S3xMolar mass of A12 S3
=0.313molx 150 g mol-1
=46.95 g of A12 S3 Answer
The non-limiting reactant is A1 which is in excess. Now mass of A1 required
reacting completely with 0.94 moles of S can be calculated as:
0.94 mole
2A1
+
2 mole
3S
A12 S3
3 mole
3 moles of S
=2 moles of A1
0.94 moles of S
=
=
Mass of A1
=No of moles of A1 x molar mass of A1
=0.63x 27
=17g of A1
Mass of A1available =20g
Mass of A1 which reacts completely =17g with available S
Excess of A1
=20-17=3g
Q20: A mixture of two liquids, hydrazine N2H4 and N2 O4 are used as a fuel in
rockets. They produce N2 and water vapors. How many grams of N2 gas will be
formed by reacting 100g of N2 O4 and 200g g of N2 O4.
2N2H4 + N2O2
3N2 +4 H2O
Solution:
Mass of2N2H4 =100g
Mass of N2O2
=200g
Molar mass of 2N2H4
=28+4=32g mol-1
Molar mass of N2O2
=28+64=92g mol-1
No of moles of N2H4
=
No of moles of N2O2
=
ww
w.a
llon
Now,
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1st year chemistry notes
3.125moles 2.174 moles
2N2H4 + N2O2
2 moles
1mole
3N2 +4 H2O
3moles
=3moles of N2
=
=4.69 mole of N2
Now , 1 mole of N2O2
=3moles of N2
2.174 moles of N2O4
=
=6.52 mole of N2O2
Since N2H4gives the least number of moles of N2, hence it is the limiting reactant.
Amount of N2 produced
=4.69 moles
2
Molar mass of N
=28g mol-1
Mass of N2
=4.69g molx 28g mol-1
=131032 g Answer
Q21: Silicon carbide (SiC) is an important ceramic material . It is produced by
allowing sand (SiO2 )to react with carbon at high temperature.
SiO2 +
3C
SiC +
2CO
When 100kg sand isn reacted with excess of carbon, 51.4 kg of Sic is produced.
Solution:
Mass of SiO2
=100 kg=100000g
Mass of SiC produced
=5.14 kg =51400g
100000g
SiO2
+
60g
SiC +
2CO
40g
60g of SiO2
100000g of SiO2
=40g of SiC
=
=66666.67 g
Actual yield of Sic
=51400 g
Theoretical yield of SiC
=66666.67g
% yield =
=
=77.1%
Q22: (a)
What is Stoichiometry? Give its assumptions? Mention two
important law, which help to perform the Stoichiometry calculations.
(b)
What is a limiting reactant? How does it control the quantity of
the product formed? Explain with three examples
Q 23: (a)
Define yield. How do we calculate the percentage yield of a
chemical reaction?
(b)
What are the factors which are mostly responsible for the low yield
of the products in chemical reactions.
ww
w.a
Now,
3C
llon
line
fre
e.c
om
Now , 2moles of N2H4
3.125moles of N2H4
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1st year chemistry notes
1 mole
ww
w.a
llon
line
fre
e.c
om
Q24: Explain the following with reasons.
(j)
Law of conservation of mass has to be obeyed during Stoichiometric
calculations.
(ii)
Many chemical reactions taking place in our surrounding involves the
limit reactants.
(iii)
No individual neon atom in the sample of the element has a mass of
20.18amu.
(iv)
One mole of H2 SO4 should completely react with two moles of NaOH.
How does Avogadro, s number help to explain it.
(v)
One mole H2 O has two moles of bonds , three moles of atoms , ten
moles of electrons and twenty eight moles of the total fundamental particles
present in it.
(vi)
N2 and CO have the same number of electrons, protons and neutrons.
Ans. (i)
According to law of conservation of mass, the amount of each
element is conserved in a chemical reaction. Chemical equations are written
and balanced on the basis of law of conversation of mass. Stoichiometry
calculations are related with the amounts of reactants and products in a
balanced chemical equation. Hence, law of conservation of mass has to be
obeyed during stoichiometric calculations.
(ii)
In our surrounding many chemical reactions are taking place which
involve oxygen. In these reactions oxygen in always in excess quantity while
other reactant are in lesser amount. Thus other reactants act as limiting
reactants.
(iii)
Since the overall atomic mass of neon in the average of the determined
atomic masses of individual isotopes present in the sample of isotopic
mixture .Hence, no individual neon atom in the sample has a mass of
20.18amu.
(iv)
H2 SO4
+2NaOH
Na2 SO4
+ 2H2 O
2moles
2 moles of H+ ions
2x6.02x1023 H+ ions
(v)
2 moles of OH ions
2x6.02x1023 OH ions
Once mole of H2SO4 consists of 2 moles of H+ ions that contains twice the
Avogadro’s number of H+ ions. For complete neutralization it needs 2 moles
of one mole of H2 SO4 should completely react with two moles of NA OH.
Since one molecule of H2O has two covalent bonds between H and O
atoms. Three atoms, ten electrons and twenty eight total fundamental
particles present in it. Hence, one mole of H2 O has two moles of bond, three
moles of atoms, ten moles of electrons and twenty eight moles of total
fundamental particle present in it.
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1st year chemistry notes
ww
w.a
llon
line
fre
e.c
om
In N2 there are 2 N atoms which contain 14 electrons (2x7),14 protons (2x7) and 14
neutrons (2x7) . In CO, there are one carbon and one oxygen atoms. It contains 14
electrons (6carbon e +8 oxygen e), 14 protons (6 C proton +8 O proton ) and 14
neutrons (6 neutrons +8 O neutrons).Hence , N2 and CO have the same number of
electrons, protons and neutrons. Remember that electrons, protons and neutrons of
atoms remain conserved during the formation of molecules in a ch emical reaction.
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