2.Introduction to Analytical Chemistry

1. Choose correct option

A. The branch of chemistry which deals

with study of separation, identification,

and quantitaive determination of the

composition of different substances is

called as .............

a. Physical chemistry

b. Inorganic chemistry

c. Organic chemistry

d. Analytical chemistry

B. Which one of the following property of

matter is Not quantitative in nature ?

a. Mass b. Length

c. Colour d. Volume

C. SI unit of mass is ........

a. kg. b. mol

c. pound d. m³

D. The number of significant figures in

1.50 × 104 g is ...........

a. 2. b. 3

c. 4 d. 6

E. In Avogadro’s constant 6.022 × 1023

mol-1, the number of significant figures

is ..........

a. 3 b. 4

c. 5 d. 6

F. By decomposition of 25 g of CaCO3

,the amount of CaO produced will be

............

a. 2.8 g. b. 8.4 g

c. 14.0 g. d. 28.0 g

G. How many grams of water will be

produced by complete combustion of 12

g of methane gas

a. 16 b. 27 c. 36 d. 56

H. Two elements A (At. mass 75) and B (At.mass 16) combine to give a compound having 75.8 % of A. The formula of the compound is

a. AB b.A2B c. AB2 d. A2B3

I. The hydrocarbon contains 79.87 %

carbon and 20.13 % of hydrogen. What

is its empirical formula ?

a. CH. b. CH2

c. CH3. d. C2H5

J. How many grams of oxygen will be

required to react completely with 27 g

of Al? (Atomic mass : Al = 27, O = 16)

a. 8. b. 16

c. 24 d. 32

K. In CuSO4.5H2O the percentage of water is ......

(Cu = 63.5, S = 32, O = 16, H = 1)

a. 10 %. b. 36 %

c. 60 % d. 72 %

L. When two properties of a system are

mathematically related to each other,

the relation can be deduced by

a. Working out mean deviation

b. Plotting a graph

c. Calculating relative error

d. all the above three

Qno.2. Answer the following questions

A. Define : Least count(1mark)

Ans:- The smallest quantity that can be measured by the measuring equipment is called least count.

B. What do you mean by significant

figures? State the rules for deciding

significant figures.(3mark)

Ans:-

i.The significant figures in a measurement or result are the number of digits known with certainty plus one uncertain digit.

ii. Rules for deciding significant figures:

a. All non-zero digits are significant.

e.g. 127.34 g contains five significant figures which are 1, 2, 7, 3 and 4.

b. All zeros between two non-zero digits are significant.

e.g. 120.007 m contains six significant figures.

c. Zeros on the left of the first non-zero digit are not significant. Such a zero indicates the position of the decimal point.

e.g. 0.025 has two significant figures, 0.005 has one significant figure.

d. Zeros at the end of a number are significant if they are on the right side of the decimal point.

e.g. 0.400 g has three significant figures and 400 g has one significant figure.

e. In numbers written is scientific notation, all digits are significant.

e.g. 2.035 × 10²has four significant figures and 3.25 × 10⁵has three significant figures.

C. Distinguish between accuracy and

precision.

Ans:-

No. Accuracy Precision

i. Accuracy refers to nearness of the measured value to the true value. Precision refers to closeness of multiple readings of the same quantity.

ii. Accuracy represents the correctness of the measurement. Precision represents the agreement between two or more measured values.

iii. Accuracy is expressed in terms of absolute error and relative error. Precision is expressed in terms of absolute deviation and relative deviation.

iv. Accuracy takes into account the true or accepted value. Precision does not take into account the true or accepted value.

v. Accuracy can be determined by a single measurement. Several measurements are required to determine precision.

vi. High accuracy implies smaller error. High precision implies reproducibility of the readings.

No.	Accuracy	Precision
i.	Accuracy refers to nearness of the measured value to the true value.	Precision refers to closeness of multiple readings of the same quantity.
ii.	Accuracy represents the correctness of the measurement.	Precision represents the agreement between two or more measured values.
iii.	Accuracy is expressed in terms of absolute error and relative error.	Precision is expressed in terms of absolute deviation and relative deviation.
iv.	Accuracy takes into account the true or accepted value.	Precision does not take into account the true or accepted value.
v.	Accuracy can be determined by a single measurement.	Several measurements are required to determine precision.
vi.	High accuracy implies smaller error.	High precision implies reproducibility of the readings.

D. Explain the terms percentage

composition, empirical formula and

molecular formula.(3 marks)

Ans:-

i. Percentage composition:

a. The percentage composition of a compound is the percentage by weight of each element present in the compound.

b. From the percentage composition, the ratio of the atoms of the constituent elements in the molecule can be calculated.

ii. Empirical formula: The simplest ratio of atoms of the constituent elements in a molecule is called the empirical formula of that compound.

e.g. The empirical formula of benzene is CH.

iii. Molecular formula:

a. Molecular formula of a compound is the formula which indicates the actual number of atoms of the constituent elements in a molecule.

e.g. The molecular formula of benzene is C6H6.

b. Molecular formula can be obtained from the empirical formula if the molar mass is known.

Molecular formula = r × Empirical formula

r= Molecular mass

Empirical formula mass

E. What is a limiting reagent ? Explain.

Ans:-

Limiting reagent:

i. The reactant which gets consumed and limits the amount of product formed is called the limiting reagent.

ii. When a chemist carries out a reaction, the reactants are not usually present in exact stoichiometric amounts, that is, in the proportions indicated by the balanced equation.

iii. This is because the goal of a reaction is to produce the maximum quantity of a useful compound from the starting materials. Frequently, a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted into the desired product.

iv. The reactant which is present in lesser amount gets consumed after some time and subsequently, no further reaction takes place, whatever be the amount left of the other reactant present.

Hence, limiting reagent is the reactant that gets consumed entirely and limits the reaction.

F. What do you mean by SI units ? What is

the SI unit of mass ?(2marks)

Ans:-

i. In 1960, the general conference of weights and measures proposed revised metric system, called International system of Units i.e. SI units, abbreviated from its French name.

ii. The SI unit of mass is kilogram (kg).

G. Explain the following terms(1 mark each)

(a) Mole fraction

Ans:-

i.Mole fraction is the ratio of number of moles of a particular component of a solution to the total number of moles of the solution.

ii. If a substance 'A' dissolves in substance 'B' and their number of moles are nA and nB , respectively, then the mole fraction of A and B are given as:

Mole fraction of A

= Number of moles of A

Number of moles of solution

=. nA.

nA + nB

Number of moles of B

= Mole fraction of B

Number of moles of solution

= nB.

nA +nB

(b) Molarity

Ans:- Molarity is defined as the number of moles of the solute present in I litre of the solution. It is the most widely used unit and is denoted by M.

Molarity is expressed as follows:

Molarity (M) = Number of moles of solute Volume of solution in litres

(c) Molality

Ans:- Molality is the number of moles of solute present in 1 kg of solvent. It is denoted by m.

Molality is expressed as follows:

Molality (m)

= Number of moles of solute Mass of solvent in kilograms

H. Define : Stoichiometry(1mark)

Ans:-

The study of quantitative relations between the amount of reactants and/or products is called stoichiometry.

I. Why there is a need of rounding off

figures during calculation ?(2marks)

Ans:-

i. When performing calculations with measured quantities, the rule is that the accuracy of the final result is limited to the accuracy of the least accurate measurement.

ii. In other words, the final result cannot be more accurate than the least accurate number involved in the calculation.

iii.The final result of a calculation often contains figures that are not significant.

iv. When this occurs, the final result is rounded off.

J. Why does molarity of a solution depend

upon temperature ?(1mark)

Ans:-

i. Molarity is the number of moles of the solute present in 1 litre of the solution. Therefore, molarity depends on the volume of the solution.

ii. Volume of the solution varies with the change in temperature. Hence, molarity of a solution depends upon temperature.

M. Define Analytical chemistry. Why

is accurate measurement crucial in

science?

Ans:-

i.The branch of chemistry which deals with the study of separation, identification, qualitative and quantitative determination of the compositions of different substances, is called analytical chemistry.

ii. The accuracy of measurement is of a great concern in analytical chemistry. This is because faulty equipment, poor data processing or human error can lead to inaccurate measurements. Also, there can be intrinsic errors in the analytical measurement.

iii. When measurements are not accurate, this provides incorrect data that can lead to wrong conclusions. For example, if a laboratory experiment requires a specific amount of a chemical, then measuring the wrong amount may result in an unsafe or unexpected outcome.

iv. Hence, the numerical data obtained experimentally are treated mathematically to reach some quantitative conclusion.

v. Also, an analytical chemist has to know how to report the quantitative analytical data, indicating the extent of the accuracy of measurement, perform the mathematical operation and properly express the quantitative error in the result.

Qno.3. Solve the following questions

A. How many significant figures are in

each of the following quantities ?(1/2mark each)

a. 45.26 ft

b. 0.109 in

c. 0.00025 kg

d. 2.3659 × 10-8 cm

e. 52.0 cm³

f. 0.00020 kg

g. 8.50 × 10⁴mm

h. 300.0 cg

Ans:-

i. 4

ii. 3

iii. 2

iv. 5

v. 3

vi. 2

vii. 3

viii. 4

B. Round off each of the following

quantities to two significant figures :

a. 25.55 mL

b. 0.00254 m

c. 1.491 × 10⁵mg

d. 199 g

Ans:-

i. 26mL

ii. 0.0025m

iii. 1.5 × 10⁵ mg

iv. 2.0 × 10² g

C. Round off each of the following

quantities to three significant figures :

a. 1.43 cm³

b. 458 × 10²cm

c. 643 cm²

d. 0.039 m

e. 6.398 × 10-³km

f. 0.0179 g

g. 79,000 m

h. 42,150

i. 649.85;

j. 23,642,000 mm

k. 0.0041962 kg

Ans:-

i. 1.43 cm³

ii. 4.58 × 10⁴ cm

iii.643 cm² (or 6.43 × 10²cm²)

iv. 0.0390 m (or 3.9 × 10-² m )

v. 6.4 × 10-³km

vi. 0.0179 g (or 1.79 × 10-² g )

vii. 7.9 × 10⁴ m

viii. 4.22 × 10⁴(or 42,200)

ix. 6.5 × 10²

x. 2.36 × 10⁷ mm

xi. 0.00420 kg (or 4.20 × 10-³kg)

D. Express the following sum to appropriate

number of significant figures :

a. 2.3 × 10³mL + 4.22 × 10⁴mL + 9.04 ×10³mL + 8.71 × 10⁵ mL;

b. 319.5 g - 20460 g - 0.0639 g - 45.642 g

- 4.173 g

Solution:-

To perform addition/subtraction operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added/subtracted.

i. (0.23 ×10⁴mL) + (4.22 × 10⁴ mL) (0.904 × 10⁴ mL) + (8.71 × 10⁴ mL)

= (0.23 + 4.22 + 0.904 + 8.71) ×10⁴mL

= 92.454 × 10⁴ mL = 9.2454 × 10⁵

= 9.2 × 10⁵ mL

ii. 319.5g - 20460 g - 0.0639 g - 45.642g - 4.173 g

= -20190.3789 g = -20190 g

Ans:

i. Sum to appropriate number of significant figures 9.2 x 10⁵ mL

ii.Sum to appropriate number of significant figures= -20190 g

Qno.4. Solve the following problems

A. Express the following quantities in

exponential terms.(1/2 mark each)

a. 0.0003498 b. 235.4678

c. 70000.0 d. 1569.00

Ans:-

i. 0.0003498 = 3.498 × 10-⁴

ii. 235.4678 = 2.354678 × 10²

iii.70000 = 7 ×10⁴

iv. 1569 = 1.569 × 10³

B. Give the number of significant figures

in each of the following(1/2 mark each)

a. 1.230 × 10⁴ b. 0.002030

c. 1.23 × 10⁴d. 1.89 × 10-⁴

Ans:-

i. 4

ii. 4

iii. 3

iv. 3

C. Express the quantities in above (B) with

or without exponents as the case may

be.

i. 1.230 × 10⁴ ii.0.002030

iii. 1.23 × 10⁴ iv.1.89 × 10-⁴

Ans:-

i.12300. ii. 2.030 × 10-³

iii.12300. iv. 0.000189

D. Find out the molar masses of the

following compounds :

a. Copper sulphate crystal (CuSO4.5H2O)

b.Sodium carbonate, decahydrate

(Na2CO3.10H2O)

c. Mohr’s salt [FeSO4(NH4)2SO4.6H2O]

(At. mass : Cu = 63.5; S = 32; O = 16;

H = 1; Na = 23; C = 12; Fe = 56; N = 14)(3 marks)

Ans:-

i. Molar mass of CuSO4.5H2O =(1x At. mass Cu) + (1 x At. mass S) + (9x At. mass O) + (10 x At. mass H) (1x 63.5) +(1×32) + (9×16) + (10 × 1) 63.5 + 32 + 144 + 10 = 249.5 g mol-¹

ii. Molar mass of Na2CO3.10H2O =(2x At. mass Na) + (1 x At. mass C) +(13 x At. mass O) + (20 x At. mass H) = (2 x 23) + (1 x 12) + (13×16) + (20×1) =46+12+208+20=286 g mol-¹

iii. Molar mass of FeSO4 (NH4)2SO4.6H2O (1 x At. mass Fe) + (2 × At. mass S) + (2x At. mass N) + (14 x At. mass O) +(20x At. mass H)

= (1x56)+(2×32)+(2×14)+(14×16) +(20×1)

=56+64+28+224 + 20 = 392 g mol-¹

Ans:

i. Molar mass of CuSO4.5H2O=249.5 g mol¹

ii. Molar mass of Na2CO3.10H2O=286 g mol-¹

iii. Molar mass of FeSO4 (NH4)2SO4.6H₂O = 392 g mol-¹

E. Work out the percentage composition of constituents elements in the following

compounds :

a. Lead phosphate [Pb3(PO4)2],

b. Potassium dichromate (K2Cr2O7),

c. Macrocosmic salt - Sodium

ammonium hydrogen phosphate,

tetrahydrate (NaNH4HPO4.4H2O)

(At. mass : Pb = 207; P = 31; O = 16;

K = 39; Cr = 52; Na = 23; N = 14)(2 marks each)

Solution:-

Given: Atomic mass Pb = 207; P= 31; 0 = 16;K = 39 ; Cr = 52; Na=23;N = 14

To find : The percentage composition of constituent elements

Formula: Percentage (by weight)Mass of the elements in

= 1 mole of compound × 100

Molar mass of the compound

Calculation:

i. Lead phosphate [Pb3(PO4)2]

Molar mass of Pb (PO4)

= 3x(207)+2×(11)+8 (16)

= 621+ 62+ 128=811 g mol-¹

Percentage of Pb= 621 × 100 = 76.57%

811

Percentage of P= 621 × 100= 7.64%

811

Percentage of O= 128 x100 =15.78%

811

ii. Potassium dichromate (K₂Cr₂O7)

Molar mass of (K₂Cr₂O7)

=2 x (39)+2 x (52)+7 × (16)

= 78 + 104 + 112 = 294 g mol-¹

Percentage of K.= 78. ×100 = 26.53%

294

Percentage of Cr= 104 × 100 = 35.37%

294

Percentage of O= 112 x100= 38.10%

294

iii. Microcosmic salt - Sodium ammonium

hydrogen phosphate ,tetrahydrate (NaNH4,HPO4, 4H20)

Molar mass of NaNH4HPO4, 4H2O

=1x(23)+1x(14)+1×(31)+13×(1)+8×(16)

=23+14+31+13+128 = 209 g mol-¹

Percentage of Na= 23 × 100= 11.00%

209

Percentage of N= 14 × 100= 6.70%

209

Percentage of P = 31 × 100 = 14.83%

209

Percentage of H = 13 × 100 = 6.22%

209

Percentage of O = 128 × 100 = 61.24%

209

Ans:

i. Mass percentage of Pb,P and O in lead phosphate [Pb3(PO4)2] are 76.57%, 7.64% and 15.78% respectively.

ii. Mass percentage of K, Cr and O in potassium dichromate (K2Cr2O7) are 26.53%, 35.37% and 38.10% respectively.

iii. Mass percentage of Na, N, P , H and O in NaNH4HPO4.4H2O are 11.00%, 6.70%,

14.83%,6.22% and 61.24% respectively.

F. Find the percentage composition

of constituent green vitriol crystals

(FeSO4.7H2O). Also find out the mass

of iron and the water of crystallisation

in 4.54 kg of the crystals. (At. mass : Fe

= 56; S = 32; O = 16)

mass of 7H2O = 2.058 kg)

(3 marks)

Solution:-

Given: i. Atomic mass: e = 56; S = 32; O = 16

ii. Mass of crystal = 4.54 kg

To find:-

i. Mass percentage of Fe, S, H and O

ii. Mass of iron and water of

crystallisation in 4.54 kg of crystal Formula: Percentage (by weight)

Mass of the element in

= 1 mole of compound × 100

Molar mass of the compound

i. Molar mass of FeSO4.7H2O

= 1 x (56) + 1 x (32) + 14 × (1) X + 11 × (16)

= 56 + 32 + 14 + 176

= 278 g mol-¹

Percentage of Fe = 56. × 100

278

= 20.14%

Percentage of S = 32. × 100

277

= 11.51%

Percentage of H = 14 × 100

278

= 5.04%

Percentage of O = 176 ×100

278

= 63.31%

ii. 278 kg green vitriol = 56 kg iron

Therefore, 4.54 kg green vitriol = x

Therefore, x = (56 × 4.34)/278. = 0.915 kg

Mass of 7H2O in 278 kg green vitriol

= 7x18=126 kg

Therefore,4.54 kg green vitriol = y

Therefore , y = 126 x 4.54 = 2.058 kg

278

Ans:

i. Mass percentage of Fe, S, Hand O in FeSO4.7H2O are 20.14%, 11.51%, 5.04% and 63.31% respectively.

ii. Mass of iron in 4.54 kg green vitriol = 0.915 kg

Mass of water of crystallisation in 4.54 kg green vitriol = 2.058 kg

G. The red colour of blood is due to a

compound called “haemoglobin”. It

contains 0.335 % of iron. Four atoms

of iron are present in one molecule of

haemoglobin. What is its molecular

weight ? (At. mass : Fe 55.84)

(Ans.: 66674.6 g/mol)

Solution:

Given: Iron percentage in haemoglobin = 0.335%

To find: Molecular weight of haemoglobin

Calculation: There are four atoms of iron in a molecule of haemoglobin. Four atoms of iron contribute 0.335% mass to a molecule of haemoglobin.

Mass of one Fe atom = 55.84 น

Mass of 4 Fe atoms = 55.84 x 4 = 223.36u = 0.335%

Let molecular weight of haemoglobin be x.

Hence, 223.36/x × 100 = 0.335%

x = 223.36/0.335 × 100 = 6674.6gmol-¹

Ans: Molecular weight of haemoglobin

= 66674.6gmol-¹

H. A substance, on analysis, gave

the following percent composition:

Na = 43.4 %, C = 11.3 % and O = 45.3 %. Calculate the empirical formula. (At. mass Na = 23 u, C = 12u, O = 16 u).

(Ans.: Na2CO3)(3marks)

Solution:-

Given: Atomic mass of Na 23 u, C = 12 u, and O = 16 u

Percentage of Na, C and O = 43.4%, 11.3% and 45.3% respectively.

To find: The empirical formula of the compound

Calculation:

Moles of Na = % of Na.

Atomic mass of Na

= 43.4 = 1.89 mol

Moles of C = % of C.

Atomic mass of C

= 11.3 = 0.94 mol

Moles of O = % of O

Atomic mass of O

=45.3 = 2.83 mol

Hence, the ratio of number of moles of Na:C:O is

1.89 =2,0.94= 1,2.83=3

0.94 0.94. 0.94

Hence, empirical formula is Na2CO3.

Ans: Empirical formula of the compound = Na2CO3

I. Assuming the atomic weight of a metal

M to be 56, find the empirical formula

of its oxide containing 70.0% of M.

(3 marks)

Solution:-

Given: Atomic mass of M = 56

Percentage of M = 70.0%

To find: The empirical formula of the compound

Calculation: % M = 70.0%

Hence, % O 30.0%, Atomic mass of O

=16 u

Moles of M = % of M

Atomic mass of M

= 70.0 = 1.25 mol

Moles of O = % of O

Atomic mass of O

= 30.0

= 1.875 mol

Hence the ratio of number of moles of M:O is

1.25 = 1 and 1.875 = 1.5

1.25. 1.25

Convert the ratio into whole number by multiplying by the suitable coefficient, i.e., 2.

Therefore, the ratio of number of moles of M:O is 2:3.

Hence, the empirical formula is M₂Os.

Ans: Empirical formula of the compound M₂O₃

J. 1.00 g of a hydrated salt contains

0.2014 g of iron, 0.1153 g of

sulfur, 0.2301 g of oxygen and 0.4532 g of water of crystallisation.Find the empirical formula,(At.wL: Fe = 56; S= 32;O =16)(3marks)

Solution:

Given: Atomic mass of Fe =56 ,S=32, and

O=16

Mass of iron, sulphur, oxygen and water 0.2014 g, 0.1153 g. 0.2301 g and 0.4532 respectively.

To find: The empirical formula of the compound

Calculation: Since the mass of crystal is 1 g the iron, sulphur, oxygen and water 20.14%, 11.53%, 23.01% and 45.32% respectively.

Moles of Fe= % of Fe

Atomic mass of Fe

= 20.14 = 0.360%

Moles of S = % of O.

Atomic mass of S

= 11.53 =0.360 mol

Moles of O = % of O.

Atomic mass of O

= 23.01 = 1.438mol

Moles of water= % of water

Molar mass of water Hence, the ratio of number of moles of Fe:S:O:water is

0.360 =1, 0.360 =1, 1.438 =4 and 2.518= 7

0.360 0.360 0.360 0.360

Hence, empirical formula is FeSO4.7H2O.

Ans: Empirical formula of the compound = FeSO4.7H2O

K. An organic compound containing

oxygen, carbon, hydrogen and nitrogen

contains 20 % carbon, 6.7 % hydrogen

and 46.67 % nitrogen. Its molecular

mass was found to be 60 g mol-¹. Find the

molecular formula of the compound.

(3 marks)

Solution:-

Given: Percentage of carbon, hydrogen, nitrogen = 20%, 6.7%, 46.67% respectively. Molar mass of the compound 60 g mol

To find: The molecular formula of the compound

Calculation: % carbon + % hydrogen +% nitrogen = 20 + 6.7 + 46.67 = 73.37%

This is less than 100%.

Hence, compound contains adequate oxygen so that the total percentage of elements is 100%.

Hence, % of oxygen = 100 - 73.37

= 26.63%

Moles of C = % of C.

Atomic mass of C

= 20 = 1.667 mol

Moles of H = % of H.

Atomic mass of H

= 6.7 = 6.700 mol

1.0

Moles of N = % of N.

Atomic mass of N

= 46.67 = 3.334 mol

Moles of O = % of O.

Atomic mass of O

= 26.63 =1.664 mol

Hence, the ratio of number of moles of C:H:N:O is

1.667≈1, 6.700 ≈ 4, 3.334 ≈ 2, and

1.667 1.664 1.664

1.664 ≈1

1.664

Hence, empirical formula is CH4N2O Empirical formula mass

= 12 + 4 + 28 + 16 = 60 g mol-¹

Hence, Molar mass = empirical formula mass

Therefore, Molecular formula mass

= Empirical formula

= CH4N2O

Ans: Molecular formula of the compound= CH4N2O.

L. A compound on analysis gave the

following percentage composition by

mass : H = 9.09; O = 36.36; C = 54.55.

Mol mass of compound is 88. Find its

molecular formula.(3marks)

Solution:-

Given: Percentage of H, O, C = 9.09% ,36.36%, 54.55% respectively.

Molar mass of the compound = 88gmol-¹

To find: The molecular formula of the compound

Calculation:

Moles of C = % of C.

Atomic mass of C

= 54.55 = 4.546 mol

Moles of H = % of H.

Atomic mass of H

=9.09 =9.090 mol

1.0

Moles of O = % of O.

Atomic mass of O

Hence, the ratio of number of moles of C:H:O is

4.546 = 2, 9.090 = 4 and 2.273= 1

2.273. 2.273. 2.273

Hence, empirical formula is C2H4O.

Empirical formula mass= 24 + 4 + 16

= 44 g mol-¹

Hence,

r = molar mass.

Empirical formula mass

= 88 g mol-¹ =2

44 g mol-¹

Therefore, Molecular formula = r × empirical formula

Molecular formula= 2× C4H4O= C4H8O2

Ans: Molecular formula of the compound = C4H8O2

M. Carbohydrates are compounds

containing only carbon, hydrogen and

oxygen. When heated in the absence

of air, these compounds decompose to

form carbon and water. If 310 g of a

carbohydrate leave a residue of 124 g

of carbon on heating in absence of air,

what is the empirical formula of the

carbohydrate ?(3 marks)

Solution:

Given: Mass of carbon residue 124 g, mass of carbohydrate = 310 g

To find: Empirical formula of the carbohydrate

Calculation: Since the 310 g of compound decomposes to carbon and water and the mass of carbon produced is 124 g, the remaining mass would be of water. Mass of water 310-124186 g

Moles of C = Mass of C.

Atomic mass of C

=124 = 10.33 mol

Moles of water = Mass of water

Molar mass of water

= 186 = 10.33 mol

The ratio of number of moles of C:water = C:H2O = 1:1

Hence, empirical formula = CH2O

Ans: Empirical formula of the carbohydrate = CH2O

N. Write each of the following in

exponential notation :

a. 3,672,199 b. 0.000098

c. 0.00461 d. 198.75

(1/2 mark each)

Ans:-

i. 3, 672, 199 = 3.672199 × 10⁶

ii.0.000098 = 9.8 × 10-⁵

iii. 0.00461 = 4.61 × 10-⁵

iv. 198.75 = 1.9875 × 10²

O. Write each of the following numbers in

ordinary decimal form :

a. 3.49 × 10-¹¹ b. 3.75 × 10-¹

c. 5.16 × 10⁴ d. 43.71 × 10-⁴

e. 0.011 × 10-³ f. 14.3 × 10-²

g. 0.00477 × 10⁵ h. 5.00858585

(1/2 mark each)

Solution:-

i. 3.49 × 10-¹¹ = 0.00000000349

ii. 3.75 × 10-¹ = 0.375

iii. 5.16 × 10⁴ = 51, 600

iv. 43.71 × 10-⁴ = 0.004371

v. 0.011 × 10-³ = 0.000011

vi. 14.3 × 10-² = 0.143

vii. 0.00477 × 10⁵ = 477

viii. 5.008585855.00858585

P. Perform each of the following

calculations. Round off your answers to

two digits.

a. 1. ; b. 33. ;

3.40 × 1024 9.00 × 10-4

c. 1.4 × 10⁹. ;

(2.77 × 10³) (3.76 × 10⁵)

d. (4 × 10-³) (9.9 × 10-⁷). ;

(789) (1.002 × 10-¹⁰) (0.3 × 10²)

(1 mark each )

Solution:-

i. 1 = 0.2941× 10-²⁴ =2.941 × 10-²⁵

3.40× 10²⁴ = 2.9 × 10-²⁵

ii. 33. =3.66667 × 10⁴ = 3.7 × 10⁴

9.00 × 10-⁴

iii. 1.4 × 10-³.

(2.77 × 10³) × (3.76 × 10⁵)

= 0.1344 × 10⁹-³-⁵

= 0.1344 × 10¹ = 1.3

iv. (4 × 10-³) × (9.9 × 10-⁷).

(789) × (1.002 × 10-¹⁰) × (0.3 × 10²)

= 0.16697 × 10-³-⁷(-¹⁰)-²

= 0.16697 × 10-² = 1.7 × 10-³

Q. Perform each of the following

calculations. Round off your answers to

three digits.

a. (3.26 104) (1.54 106)

b. (8.39 107) (4.53 109)

c. 8.94 × 10⁶

4.35 × 10⁴

d. (9.28 × 10⁹) (9.9 × 10-⁷)

(511) (2.98× 10-⁶)

(1 mark each )

Solution:-

i. (3.26 × 10⁴)(1.54 × 10⁶)

= 5.0204 × 10⁴+⁶

= 5.02 × 10¹⁰

ii. (8.39 × 10⁷)(4.53 × 10⁹)

= 38.0067 × 10⁷+⁹

= 38.0067 × 10¹⁶

= 3.8 × 10¹⁷

iii. 8.94 × 10⁶ = 2.055 × 10⁶-⁴ = 2.06 × 10²

4.35 × 10⁴

iv. (9.28 × 10⁹) × (9.9 × 10-⁷)

(511) × (2.98 × 10-⁶)

= 0.06033 × 10⁹-⁷-(-⁶)

= 0.06033 × 10⁸ = 6.03 × 10⁶

R. Perform the following operations :

a. 3.971 × 10⁷ + 1.98 × 10⁴;

b. 1.05 × 10-⁴ - 9.7 × 10-⁵;

c. 4.11 × 10-³ + 8.1 10-⁴;

d. 2.12 × 10⁶ - 3.5 × 10⁵.

Solution:-

To perform addition/subtraction operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added/subtracted.

i. 3.971 × 10⁷ + 1.98 × 10⁴

= 3.971 × 10⁷ + 0.00198 × 10⁷

= (3.971 + 0.00198) × 10⁷

= 3.97298 × 10⁷

ii. 1.05 × 10-⁴ - 9.7 × 10-⁵

= 10.5 × 10-⁵ - 9.7 × 10-⁵

= (10.5 - 9.7) × 10-⁵

= 0.8 × 10-⁵

iii.4.11 × 10-³ + 8.1 × 10-⁴

= 41.1 × 10-⁴+ 8.1 × 10-⁴

= (41.1 + 8.1) × 10-⁴

= 49.2 × 10-⁴

= 4.92 × 10-³

iv. 21.2 × 10⁵ - 3.5 × 10⁵

= 2.12 × 10⁶ - 3.5 × 10⁵

= (21.2 - 3.5) × 10⁵

= 1.77 × 10⁶

S. A 1.000 mL sample of acetone, a

common solvent used as a paint

remover, was placed in a small bottle

whose mass was known to be 38.0015

g. The following values were obtained

when the acetone - filled bottle was

weighed : 38.7798 g, 38.7795 g and

38.7801 g. How would you characterise

the precision and accuracy of these

measurements if the actual mass of the

acetone was 0.7791 g ?

Solution:-

i. Precision:

Mean = 0.7783 + 0.7780 +0.7786

= 0.7783 g

Mean absolute deviation

= 0+ 0.0003+0.0003= 0.0002

Therefore, Mean absolute deviation

= ±0.0002 g

Relative deviation

= Mean absolute deviation ×100%

Mean

= 0.0002 × 100% = 0.0257%

0.7791

ii. Accuracy:

Actual mass of acettone = 0.7791g

Observed value(average) = 0.7783 g

a. Absolute error

= observed value - True value

= 0.7783 - 0.7791= -0.0008 g

b. Relative error = Absolute error × 100

True value

= -0.0008 × 100% = -0.1.27%

0.7791

Ans: These observed values are close to each other and are also close to the actual mass.Therefore,the result are precise and as well accurate.

i.Relative deviation= 0.0257%

ii. Relative error = - 0.1027%

T. Your laboratory partner was given

the task of measuring the length of

a box (approx. 5 in) as accurately as

possible, using a metre stick graduated

in milimeters. He supplied you with the

following measurements:

12.65 cm, 12.6 cm, 12.65 cm, 12.655 cm,

126.55 mm, 12 cm.

a. State which of the measurements you

would accept, giving the reason.

b. Give your reason for rejecting each of

the others.(2 marks)

Ans:-

i. The metre stick is graduated in millimetres i.e. 1 mm to 1000 mm, and 1 mm = 0.1 cm. Therefore, if length is measured in centimetres, the least count of metre stick is 0.1 cm. The results 12.6 cm has the least count of 0.1 cm and is acceptable result.

ii. Since, the least count of metre stick is 0.1 cm or 1 mm, the results such as 12.65 cm, 12.655 cm, 126.55 mm cannot be measured using this stick and hence, these results are rejected. The result, 12 cm doesn't include the least count and is rejected.

U. What weight of calcium oxide will be

formed on heating 19.3 g of calcium

carbonate ?(2 marks)

(At. wt. : Ca = 40; C = 12; O = 16)

Solution:

Given: Mass of CaCO3 consumed in reaction = 19.3 g

To find: Mass of CaO formed

Calculation: Calcium carbonate decomposes according to the balanced equation,

CaCO3. ∆→ CaO. + CO2↑

40+12+3×16 40+16 12+2×16

=100 parts =56 parts = 44 parts

So, 100 g of CaCO3 produce 56 g of Cal 19.3 g of CaCO3 will produce

= 568g x 19.3 g = 10.81 g of CaO

100 g

Ans: Mass of CaO formed = 10.81 g

V. The hourly energy requirements of an

astronaut can be satisfied by the energy

released when 34 grams of sucrose are

“burnt” in his body. How many grams

of oxygen would be needed to be carried

in space capsule to meet his requirement

for one day ? (2 marks)

Solution:

34 g of sucrose provides energy for an hour. Hence, for a day, the mass of sucrose needed = 34 x 24 = 816 g

The balanced equation is,

C12H22O11(g) + 12O2(g)

342 g 12x32 = 384 g →12CO2(g) + 11H2O(l)

Thus, 342 g of sucrose require 384 g of oxygen.

816 g of sucrose will require = 816 × 384 342

= 916 g of O2

Ans: Astronaut needs to carry 916 g of O2.

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