Chapter 1 Some Basic Concepts of Chemistry

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:

 

 

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

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  11^th 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 CHCl_3.

 

⇒ 1000 g of the sample contains 1.5 × 10^–2 g of CHCl_3.

 

 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:

1. if the digit to be dropped is more than 5, then add 1 to the preceding significant figure.For eg in the number 11.06, the digit to be dropped is 6.Therefore the preceding digit 0 is increased by one & the final result is 11.1.
2. if the digit to be dropped is less than 5,then it is deleted as such without bringing any change in the preceding significant figure.For eg 43.123 is to reported upto four significant figures,then the last digit 3 is dropped & the final answer is 43.12.
3. if the digit to be dropped is 5 then the preceding significant digit in the number may be left unchanged if it is even & may be increased by 1 in case if it is odd.Eg 1.6145, if reported upto four significant figures will become 1.614 after rounding off. Similarly 1.6175 will become 1.618.
4. if during rounding off, more than one digit is to be dropped form a particular number, then they are dropped on at a time by following the above rules.

(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

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,

CaCO_3(s) + 2 HCl_(aq) → CaCl_2(aq) + CO_2(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.