From this chapter, you will understand the role of chemistry in different spheres of life. You will also understand characteristics of three states of matter. You can also classify substances into elements,compounds and mixtures. You can explain various laws of chemical combination along with description of mole and molar mass.You will also be able to determine empirical formula and molecular formula for a compound from the given experimental data.
Molecules are made up of atoms & are quite small, therefore the actual mass of a molecule cannot be detrmined.It is expressed as the relative mass.C-12 isotope is used to express the relative molecular masses of substances.Thus molecular mass of a substance may be defined as: the average relative mass of its molecule as compared to the mass of carbon atom taken as 12 amu.
(i) H2O:
The molecular mass of water, H2O
= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
= [2(1.0084) + 1(16.00 u)]
= 2.016 u + 16.00 u
= 18.016
= 18.02 u
(ii) CO2:
The molecular mass of carbon dioxide, CO2
= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)
= [1(12.011 u) + 2 (16.00 u)]
= 12.011 u + 32.00 u
= 44.01 u
(iii) CH4:
The molecular mass of methane, CH4
= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)
= [1(12.011 u) + 4 (1.008 u)]
= 12.011 u + 4.032 u
= 16.043 u
The molecular formula of sodium sulphate is (Na2SO4).
Molar mass = atomic mass of Na(23) + atomic mass of S(32.066) + atomic mass of O(16)
Molar mass of (Na2SO4) = [(2 × 23.0) + (32.066) + 4 (16.00)] = 142.066
Mass percent of an element 
∴ Mass percent of sodium:
Mass percent of sulphur:
Mass percent of oxygen:
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of the atoms of various elements present in one molecule of the compound.
Form the available data Percentage of iron = 69.9
Percentage of oxygen= 30.1 Total percentage of iron & oxygen= 69.9+30.1= 100%
Step 1 calculation of simplest whole number ratios of the elements
|
Element |
Percentage |
Atomic mass |
Atomic ratio |
Simplest ratio |
Simplest whole no ratio |
|
Fe |
69.9 |
55.84 |
69.9/55.84=1.25 |
1.25= 1 |
2 |
|
O |
30.1 |
16 |
30.1/16 = 1.88 |
1.88=1.5 |
3 |
Step 2 writing the empirical formula of the compound
The empirical formula of the compound = Fe2 O3
The chemical equation for the reaction is:
i) when 1 mole of carbon is burnt in 1 mole(32 g) of air(dioxygen) ,then according to chemical equation (1) 1mole(44 g) of carbon dioxide is produced.
ii) 32g of Oxygen upon heating gives CO2 = 44g
1 g of Oxygen upon heating gives CO2 = 44/32
Therefore 16 g of Oxygen upon heating gives CO2 = 44/32*16 = 22 g
(iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.
0.375 M solution of (CH3COONa) = 0.375 moles(grams) of (CH3COONa) dissolved in 1000 ml of solvent.
But according to question ,we have to make a 500 ml solution of (CH3COONa)
∴Number of moles of sodium acetate in 500 mL
Molar mass of sodium acetate = 82.0245 g mole–1 (Given)
∴ Required mass of sodium acetate = (82.0245 g mol–1) (0.1875 mole)
= 15.38 g
Mass of nitric acid solution = 69 g (because 69% means 69 g dissolved in 100 ml of solvent)
Density of nitric acid solution = 1.41 g/ml
Volume of nitric acid solution = mass/density= 100/1.41 = 70.92 ml
No of moles of nitric acid = mass of nitric acid/molar mass of nitric acid=69/63= 1.09
Volume of the solution = 70.92
Molarity of the solution = moles of nitric acid/volume of solution in litres
= 1.09*1000/70.92= 15.37 moles/liter
∴Concentration of nitric acid = 15.37 mol/L
1 mole of CuSO4 contains 1 mole of copper.
Molar mass of CuSO4 = atomic mass of Cu + atomic mass of S + 4X atomic mass of O
Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00 = 159.5 amu
Gram molecular mass of CuSO4= 159.5 g
159.5 g of CuSO4 contains Cu = 63.5
1 g of CuSO4 contains Cu= 63.5/159.5* 1
Therefore 100g of CuSO4 contains Cu = 63.5/159.5*100 = 39.81 g
From the available data Percentage of iron = 69.9
Percentage of oxygen= 30.1
Total percentage of iron & oxygen= 69.9+30.1= 100%
Step 1 calculation of simplest whole number ratios of the elements
|
Element |
Percentage |
Atomic mass |
Atomic ratio |
Simplest ratio |
Simplest whole no ratio |
|
Fe |
69.9 |
55.84 |
69.9/55.84=1.25 |
1.25= 1 |
2 |
|
O |
30.1 |
16 |
30.1/16 = 1.88 |
1.88=1.5 |
3 |
Step 2 Writing the empirical formula of the compound
The empirical formula of the compound = Fe2 O3
Step 3 determination of molecular formula of the compound
Empirical formula mass = 2 X69.9 + 3 X16=187.8 amu
Molecular mass of oxide= 159.69g/mol(given)
Now we know molecular formula = n x Empirical formula
And n= molecular mass / empirical formula mass= 159.69/187.8 = 0.85 = approx 1
Therefore molecular formula = n x empirical formula
=1 x(Fe2O3) = Fe2O3
The molecular formula of the oxide is Fe2O3
Calculate the atomic mass (average) of chlorine using the following data:
| % Natural Abundance | Molar Mass | |
| 35Cl | 75.77 | 34.9689 |
| 37Cl | 24.23 | 36.9659 |
Atomic mass of first isotope = 34.9689
Natural abundance of first isotope = 75.77% or 0.757
Atomic mass of second isotope= 36.9659
Natural abundance of second isotope= 24.23% or 0.242
Now average atomic mass of chlorine
= [34.9689x 0.757 + 36.9659x 0.242]/(0.757+0.242) =35.4521
So, the average atomic mass of chlorine = 35.4527 u
Mole is a latin term which means a pile or mass of stones placed in sea. In Chemistry a mole represents Avogadro’s number(6.022 X 1023 particles)
(i) 1 mole of C2H6 contains 2 moles of carbon atoms(because 2 carbons are present in the formula of ethane)
∴ Number of moles of carbon atoms in 3 moles of C2H6
= 2 × 3 = 6
(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms(because 6 hydrogen atoms are present in ethane)
∴ Number of moles of carbon atoms in 3 moles of C2H6
= 3 × 6 = 18
(iii) 1 mole of C2H6 contains 6.023 × 1023 molecules of ethane(because 1mole = 6.022 x 1023)
∴ Number of molecules in 3 moles of C2H6
= 3 × 6.023 × 1023 = 18.069 × 1023
Mass of sugar= 20g
Molecular mass of sugar=12x12 +1x22 +16x11=342 g
Mass of water= 2 liter
Molarity (M) = number of moles of solute/volume of solution in liters
= 20/(342x2) =0.02924 mol/liter
Therefore Molar concentration of sugar = 0.03 mol L–1
Let the volume of 0.25 M of ethanol needed = x Molar mass of methanol (CH3OH) = (1 × 12) + (4 × 1) + (1 × 16) = 32 g mol–1 = 0.032 kg mol–1
Molarity of methanol solution 
= 24.78 mol L–1 (Since density is mass per unit volume)
Applying, M1V1 = M2V2
(Given solution) (Solution to be prepared)
(24.78 mol L–1) x = (2.5 L) (0.25 mol L–1)
x = 0.0252 L
x = 25.22 mL
Pressure is defined as force acting per unit area of the surface.
= 1.01332 × 105 kg m–1s–2
We know,
1 N = 1 kg ms–2 Then,
1 Pa = 1 Nm–2 = 1 kg m–2s–2
1 Pa = 1 kg m–1s–2
1 Pressure = 1.01332 × 105 Pa
The system of measurement is the most common system employed all over the world.The 11th General Conference of the Weights & Measures held in France in 1960 introduced this system called Systeme International d’ units, abbreviated as SI system.
The SI unit of mass is kilogram (kg). 1 Kilogram is defined as the mass equal to the mass of the international prototype of kilogram.
|
SNO |
PREFIXES |
SYMBOLS |
MULTIPLES |
|
1 |
Micro |
μ |
10-6 |
|
2 |
Deca(deka) |
da |
101 |
|
3 |
Mega |
M |
106 |
|
4 |
Giga |
G |
109 |
|
5 |
Femto |
f |
10-15 |
Significant figures are those meaningful digits that are known with certainty.
The significant figures in any number are all certain digits plus one doubtful digit. In order to report a scientific data, the data is expressed in significant figures in which all the digits reported are certain except the last one which is uncertain or doubtful.For eg 20.86 has four digits In all.Out of them 2,0 & 8 are certain digits while the last digit 6 is uncertain.Thus the number 20.86 has all the four digits as significant figures.Out of them 2, 0 & 8 are certain while 6 has some uncertainity about it.
Hence, significant figures are defined as the total number of digits in a number including the last digit that represents the uncertainty of the result.
(i) 1 ppm is equivalent to 1 part out of 1 million (106) parts.
∴ Mass percent of 15 ppm chloroform in water
(ii) molality (M) = no of moles of solute/mass of solvent in g *1000
Therefore mass of chloroform= 12 + 1+3(35.5) = 119.5 g/mol
100 g of the sample contains 1.5 × 10–3 g of CHCl3.
⇒ 1000 g of the sample contains 1.5 × 10–2 g of CHCl3.
m = 1.5 x 10-3/119.5 * 1000 = 1.25x 10-4 m
In scientific notation every number is written as N x 10n, where N = a number with single non zero digit to the left of the decimal point n = exponent of 10, which may be positive, negative integer or zero
(i) 0.0048 = 4.8× 10–3
(ii) 234, 000 = 2.34 ×105
(iii) 8008 = 8.008 ×103
(iv) 500.0 = 5.000 × 102
(v) 6.0012 = 6.0012
The significant figures in any number are all certain digits plus one doubtful digit.Certain rules are followed which are as under:
1 all non zero digits in anumber are significant
2 the zeros between two non zero digits are always significant
3 the zeros written to the left of the first non zero digit in a number are non significant.They simply indicate the position of the decimal point.
4 all zeros placed to the right of a decimal point in a number are significant.
(i) 0.0025
There are 2 significant figures.
(ii) 208
There are 3 significant figures.
(iii) 5005
There are 4 significant figures.
(iv) 126,000
There are 3 significant figures.
(v) 500.0
There are 4 significant figures.
(vi) 2.0034
There are 5 significant figures.
Rounding off a number means that the digits which are not significant have to be dropped.The rules are as follows:
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810
Let us fix 14 parts by weight of nitrogen as fixed weight.
Now let us calculate the weights of oxygen which combine with 14 parts by weight of nitrogen
|
Sno |
No of parts by weight of nitrogen |
No of parts by weight of oxygen |
14 parts of nitrogen as fixed weight |
No of parts by weight of oxygen which combine with 14 parts by weight of nitrogen |
|
1 |
14g |
16g |
14g |
16 |
|
2 |
14g |
32g |
14g |
32 |
|
3 |
28g |
32g |
14g |
32 |
|
4 |
28g |
80g |
14g |
80 |
(a) If we fix the mass of dinitrogen at 14 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 16 g, 32 g, 32 g, and 80 g.
The masses of dioxygen bear a whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the law of multiple proportions.
This law was given by Dalton in 1804. The law states that if two elements combine to form 2 or more compound, then the weight of one element which combines a fixed weight of other element in these compounds,bears a simple whole number ratio by weight.
(b) (i) We know 1km=1000m
Or 1m = 1000 mm
Therefore 1km = 1000x 1000mm= 106 mm
1 km = 1 km ×
1 km = 1015 pm
Hence, 1 km = 106 mm = 1015 pm
(ii) We know 1kg = 1000mg
Or 1000mg= 1kg
Or 1mg= 1/1000* 1= 0.01 kg
1 mg = 1 mg × 
⇒ 1 mg = 106 ng
1 mg = 10–6 kg = 106 ng
(iii) We know 1000 ml=l L
Or 1ml=1/1000*1= 0.01L
1 mL = 1 cm3 = 1 cm3
⇒ 1 mL = 10–3 dm3
1 mL = 10–3 L = 10–3 dm3
According to the question:
Time taken to cover the distance = 2.00 ns
= 2.00 × 10–9 s
Speed of light = 3.0 × 108 ms–1
We know speed = distance/time
Or distance = speed x time
Therefore Distance travelled by light in 2.00 ns
= Speed of light × Time taken
= (3.0 × 108 ms –1) (2.00 × 10–9 s)
= 6.00 × 10–1 m
= 0.600 m
while solving the problems relating to chemical equation,the reactant react according to the balanced chemical equation.Quite often, one of the reactant is present in lesser amount while the other may be present in higher amount.The reactant which is present in lesser amount is known as limiting reactant or reagent.
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.Here atom B is in lesser amount(200)
(ii) According to the reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A will not be consumed. Hence, A is the limiting reagent.
(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.
(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent because B is less as compared to A
(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.
(i) Balancing the given chemical equation,
Total mass of Ammonia = 2((14) +3(1))= 34 g
From the chemical equation,we can write
28gm of N2 reacts with 6gm of H2 to produce ammonia= 34g
Or 1 gm of N2 reacts with 1gm of H2 to produce ammonia= 34/28*1
Or when 2.00x103 g of N2 reacts with 1.00x103 gm of H2 to produce ammonia
=34/28 *2.00x103=2428.57g
hence 2.00 × 103 g of dinitrogen will react with 1.00x103 g of dihydrogen to give 2428.57 g of ammonia
Given, Amount of dihydrogen = 1.00 × 103 g
Hence, N2 is the limiting reagent.
(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain unreacted.
(iii) Mass of dihydrogen left unreacted = 1.00 × 103 g – 428.6 g
= 571.4 g
Molar mass of Na2CO3 = (2 × 23) + 12.00 + (3× 16) = 106 g mol–1
1 mole of Na2CO3 = 106 g
0.5 mol of Na2CO3 
= 53 g Na2CO3
⇒ 0.50 M of Na2CO3 = 0.50 mol/L Na2CO3
Hence, 0.50 mol of Na2CO3 is present in 1 L of water or 53 g of Na2CO3 is present in 1 L of water.
Reaction of dihydrogen with dioxygen can be written as:
6H2 + 3O2 = 6H2O
Now, SIX volumes of dihydrogen react with THREE volume of dihydrogen to produce SIX volumes of water vapour.
Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.
(i) 28.7 pm:
1 pm = 10–12 m
28.7 pm = 28.7 × 10–12 m
= 2.87 × 10–11 m
(ii) 15.15 pm:
1 pm = 10–12 m
15.15 pm = 15.15 × 10–12 m
= 1.515 × 10–12 m
(iii) 25365 mg:
1 mg = 0.01 kg
Therefore 25365 mg = 0.01/1 x 25365= 253.65 kg
(i) Gram atomic mass of Au= 197 g
Or
197g of Au contains = 6.022 x 1023
Therefore 1gm of Au contains = 6.022 x 1023/197*1 = 3.06 x 1021 atoms
(ii) Gram atomic mass of Na = 23 g
Or
23 g of Na contains atoms = 6.022x 1023
Or
1gm of Na contains atoms = 6.022x1023 /23 *1 = 26.2 x1021 atoms
(iii) Gram atomic mass of Li = 7
Or
7g of Li contains atoms = 6.022 x 1023
Or
1g of Li contains atoms = 6.022 x 1023/7 *1= 86.0 x 1021 atoms
(iv) Gram atomic mass of Cl = 71 Or 71g of Cl contains atoms = 6.022x1023
Or
1 g of Cl contains atoms = 6.022x1023 /71 * 1= 8.48 x 1021 atoms
Hence, 1 g of Li (s) will have the largest number of atoms.
Mole fraction of C2H5OH=
.................1
Let the moles of C2H5OH= X
Now density of water = 1 (given)
And the weight of 1000ml of water = volume * density (from density = mass/volume)
= 1000 x 1=1000g
Therefore moles of water = 1000/18= 55.55 mol (18g is molecular mass of water)
Also mole fraction of C2H5OH= 0.040 (given)
Putting the values in equation 1,we get
=0.040=X/X+55.55
=0.040X+2.222 = X
OR X= 2.3145 mol
Molarity of solution = 2.314 M
1 mole of carbon= 6.022x 1023 carbon atoms = 12 g carbon
Or 6.022x 1023 atoms of carbon = 12 g of carbon
Therefore 1 atoms of carbon = 12/6.022x 1023 x 1= 1.993 x 10-23 g of carbon
Rules for determining the significant figures
In addition or subtraction of the numbers having different precisions, the final result should be reported to the same number of decimal places as in the term having least number of decimal places
(i)
= 1.648
Least precise number of calculation = 298.15
Number of significant figures in the answer
= Number of significant figures in the least precise number = 2 i.e 1.65
(ii) 5 × 5.364 = 26.820
Least precise number of calculation = 5.364
Number of significant figures in the answer = Number of significant figures in 5.364 = 3 i.e 26.820
(iii) 0.0125 + 0.7864 + 0.0215 = 0.8204
Since the least number of decimal places in each term is four, the number of significant figures in the answer is also 4 i.e 0.8204
Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
|
Isotope |
Isotopic molar mass |
Abundance |
|
36Ar |
35.96755 gmol–1 |
0.337% |
|
38Ar |
37.96272 gmol–1 |
0.063% |
|
40Ar |
39.9624 gmol–1 |
99.600% |
Molar mass of argon =
Atomic mass of 36Ar = 35.96755 & abundance = 0.337
Or
total atomic mass of 36Ar = 35.96755 * 0.337= 0.121g/mol
Atomic mass of 38Ar = 37.96272 & abundance = 0.063
Or
total atomic mass of 38Ar = 37.96272 * 0.063= 0.024g/mol
Atomic mass of 40Ar =39.9624 & abundance = 99.600
Or
total atomic mass of 40Ar = 39.9624 * 99.600= 39.802g/mol
Therefore molar mass of argon = total mass of 36Ar+ total mass of 38Ar + atomic mass of 40Ar
= 0.121 + 0.024 + 39.802= 39.947 g/mol
(i) 1 mole of Ar = 6.022 × 1023 atoms of Ar
52 mol of Ar = 52 × 6.022 × 1023 atoms of Ar
= 3.131 × 1025 atoms of Ar
(ii) Atomic mass of He = 4amu
Or
4amu is the mass of He atoms = 1
Therefore 52 amu is the mass of He atoms= ¼*52 = 13 atoms of He
(iii) Gram atomic mass of He = 4g
Or
4g of He contains = 6.022x 1023 atoms
Therefore 52 g of He contains = 6.022x 1023 /4 * 52 = 7.83 x 1024 atoms
(i) percentage of C can be calculated as follows:
CO2 = C
i.e 44 parts of CO2= 12 parts of C
OR
44g of CO2 = 12 g of C
Therefore according to question
3.38 g of CO2 contains C = 12/44 * 3.38 = 0.921 g
18 g of water contains hydrogen = 2g
Therefore 0.690 g of water contains hydrogen = 2/18 * 0.690 = 0.0767g
0.690 g of water will contain hydrogen
= 0.0767 g
Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:
= 0.9217 g + 0.0767 g = 0.9984 g
Now percentage of carbon = weight of carbon/weight of compound * 100
=0.921 /0.998 * 100= 92.32 %
Also percentage of hydrogen = weight of hydrogen/weight of compound *100
=0.0766/0.998 * 100 = 7.68 %
(ii) Given,
Weight of 10.0L of the gas (at S.T.P) = 11.6 g
Weight of 22.4 L of gas at STP
= 25.984 g
≈ 26 g
Hence, the molar mass of the gas is 26 g.
(iii) empirical formula
| Element | Percentage | Atomic mass | Atomic ratio | Simplest ratio | Simplest whole no ratio |
| C | 92.32 | 12 | 92.32/12 = 7.69 | 7.69/7.65 = 1.00 | 1 |
| H | 7.65 | 1 | 7.65/1 = 7.65 | 7.65/7.65 = 1 | 1 |
Empirical formula of the compound = CH
Now molecular formula calculation
Empirical formula mass = 12 + 1 = 13 amu
Also molecular mass = 26 g (calculated in previous step)
Therefore n = molecular mass/empirical formula mass = 26/13 = 2
Now molecular formula = n x empirical formula = 2 x CH = C2H2
0.75 M of HCl ≡ 0.75 mol of HCl X molecular weight of HCl dissolved in 1000 ml of water
Or
[(0.75 mol) × (36.5 g mol–1)] HCl is present in 1 L of water
≡ 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contains Hcl = 27.375g
Or
1 ml of solutions contains Hcl = 27.375/1000 * 1
And 25 ml of solutions contains Hcl = 27.375/1000 * 25 = 0.6844 g.
From the given chemical equation,
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO3 (100 g).
Amount of CaCO3 that will react with 0.6844 g
= 0.9639 g
Molecular weight of Mno2 = 87 g
Molecular weight of Hcl = 4 * 36.5 = 146 g
According to chemical reaction
1 g of Mno2 reacts with 4g of Hcl togive = 146/87 * 1
Or
5 g of Mno2 reacts with 4g of Hcl to give = 146/87 *5 = 8.4 g of Hcl
Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.