## Introduction to Analytical Chemistry

### Class-11-Science-Chemistry-Chapter -2 Maharashtra State Board

### Solution

**Question 1. **

**Choose correct option**

**(A) The branch of chemistry which deals with study of separation, identification, and quantitative determination of the composition of different substances is called as .............**

**a- Physical chemistry**

**b- Inorganic chemistry**

**c- Organic chemistry**

**d- Analytical 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) Colour

**(C) SI unit of mass is ........**

**a- kg b- mol**

**c- pound d- m ^{3}**

(a) kg

**(D) The number of significant figures in 1.50 ****× ****10 ^{4} g is ...........**

**a- 2 b- 3**

**c- 4 d- 6**

(b) 3

**(E) In Avogadro’s constant 6.022 ****× ****10 ^{23} mol-1, the number of significant figures is .........**

**a- 3 b- 4**

**c- 5 d- 6**

(b) 4

**(F) By decomposition of 25 g of CaCO _{3}, the amount of CaO produced will be **

**............**

**a- 2.8 g b- 8.4 g**

**c- 14.0 g d- 28.0 g**

(c) 14.0 g

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**(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**

(b) 27

**(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- A _{2}B **

**c- AB _{2} d- A_{2}B_{3}**

(d) A_{2}B_{3}

**(I) The hydrocarbon contains 79.87 % carbon and 20.13 % of hydrogen. What is its empirical formula ?**

**a- CH b- CH _{2}**

**c- CH _{3} d- C_{2}H_{5}**

(c) CH_{3}

**(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**

(c) 24

**(K) In CuSO _{4}.5H_{2}O the percentage of water is ......**

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

**a- 10 % b- 36 %**

**c- 60 % d- 72 %**

(b) 36 %

**(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**

(b) Plotting a graph

**Question 2. **

**Answer the following questions**

**(A) Define : Least count**

**Least count : **It is the smallest quantity or value that can be measured on the measuring equipment or instrument.

** **

**(B) What do you mean by significant figures? State the rules for deciding significant figures.**

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

**Rules for deciding signiﬁcant figures :**

- All nonzero digits are significant. For example 127.34 g contains ﬁve significant figures which are 1, 2, 7, 3 and 4.
- All zeros become significant, if they are placed between two non zero digits. For example, 120.007 m contains six significant figures which are 1, 2, 0, 0, 0 and 7.
- Zeros on the left of the first non zero digit are not significant. Such a zero indicates the position of the decimal point. For example, 0.025 has two significant figures,
- 2 and 5. 0.005 has only one significant figure, 5.
- All zeros placed at the end of a number are significant if they are on the right side of the decimal point. They represent the accuracy or precision of the measuring scale.
- For example, 243.0 has four significant ﬁgures which are 2, 4, 3 and 0.
- Terminal zeros are not significant if there is no decimal point. For example, the measurement 400 g has one significant figure which is 4.
- All digits in numbers written in scientiﬁc notation are significant. For example, 2.035 x 10
^{2}has four significant figures which are 2, 0, 3 and 5. - The measurement, 3.25 x 10
^{-5}has three signiﬁcant ﬁgures viz. 3, 2 and 5.

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**(C) Distinguish between accuracy and precision.**

Accuracy |
Precision |

1- Accuracy refers to the nearness of the observed value or the measured value to the true value.
2- Accuracy expresses the correctness of the measurement. 3-Accuracy depends upon the sensitivity or the least count of the measuring equipment. 4- Accuracy can never be determined exactly because the true value of a quantity can never be known exactly, instead an accepted value must be used to determine accuracy. 5-Larger the accuracy, smaller is the error. 6-Accuracy is expressed in terms of either absolute or relative error. |
1- Precision is defined as the closeness between two or more measured values to each other.
2- Precision describes the agreement between a result and its true value. 3- High precision implies reproducibility of readings. 4- Precision is determined by simply replicating a measurement. 5-Smaller the difference between the individual values of repeated measurements, greater is the precision. 6- Precision is expressed in terms of deviation. |

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**(D) Explain the terms percentage composition, empirical formula and molecular formula.**

**Percentage composition : **The percentage by mass of each element in a compound is called per cent composition.

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

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

**(E) What is a limiting reagent ? Explain.**

**Limiting reagent** : The reactant in a chemical reaction that limits the amount of product that can be formed is called the limiting agent. The reaction will stop when all of the limiting reactant is consumed.

- A balanced chemical equation gives the ideal stoichiometric relationship among reactants and products. However, when a chemist carries out a reaction, reactants for the experiment are not necessarily present in exact stoichiometric amounts. This is because, the goal of the reaction is to produce the maximum quantity of a useful compound or product from the starting material.
- Many a times, a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted into the desired product. Thus, the reactant which is present in lesser amounts gets consumed after some time and subsequently, no further reaction takes place.

**(F) What do you mean by SI units ? What is the SI unit of mass ?**

In 1960, the General Conference of weight and measures adopted an International System of Units, called SI units (French name System internationale d'Unites) which has seven fundamental units.

The SI unit of mass is kilogram (kg).

**(G) Explain the following terms**

**(a) Mole fraction**

**(b) Molarity**

**(c) Molality**

**Mole fraction (x)** : The mole fraction of any component of a solution is defined as the ratio of number of moles of that component present in the solution to the total number of moles of all the components of the solution.

Mole fraction = \(\frac{\text{Number of moles of component}}{\text{Total no of moles of solute and solvent in solution}}\)

Number of moles of a substance = \(\frac{\text{Mass of substance}}{\text{Molar mass of substance}}=\frac{W}{M}\) mol

If a solution contains n_{1} moles of solvent and n_{2} moles of solute, then the total number of moles present in the solution = n_{1} + n_{2}.

Mole fraction of the solvent x_{1} = \(\frac{n_1}{n_1+n_2}\)

Mole fraction of the solute x_{2} = \(\frac{n_2}{n_1+n_2}\)

Mole fraction is independent of temperature and has no units.

**Molarity (M)** : It is deﬁned as the number of moles of a solute present in 1 litre of a solution.

If n moles of a solute are present in V lit. of a solution then,

Volume of solution in lit.

Molarity(M) = \(\frac{\text{Number of moles of solute}}{\text{Volume of solution in lit}}=\frac{n}{V}\) mol L^{-1}

Molarity = \(\frac{\text{Mass of solute}}{\text{Molar mass of solute × Volume of solution in L.}}\)

= \(\frac{\text{Mass of solute×1000}}{\text{Molar mass of solute × Volume of solution in}cm^3}\)

The unit of molarity is mol litre^{-1} (mol L^{-1})

Molarity changes with temperature.

**Molality (m) **; It is defined as the number moles of a solute dissolved in 1 kg (or 1000 gram) of a solvent.

Molality (m) = \(\frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}\)

= \(\frac{\text{Mass of solute}}{\text{Molar mass of solute × Mass of solvent in kg}}\)

= \(\frac{\text{Mass of solute × 1000}}{\text{Molar mass of solute × Mass of solvent in g}}\)

The unit of molality is mol kg^{-1} and denoted by m.

Molality is independent of temperature, since it involves mass of solute and mass of solvent which are independent of temperature.

**(H) Define : Stoichiometry**

**Stoichiometry : **The study of quantitative relations between the amount of reactants and products is called stoichiometry.

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**(I) Why there is a need of rounding off figures during calculation ?**

The final result of a calculation often contains figures that are/not significant. When this occurs the final result is rounded off.

**(J) Why does molarity of a solution depend upon temperature ?**

Molarity of a solution depends upon temperature since it involves volume of the solution which is temperature dependent.

**Note : K & L question not given in text book

**(M) Define Analytical chemistry. Why is accurate measurement crucial in science?**

**Analytical chemistry : **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.

**Importance of accurate measurements :**

- Accurate measurements are important because exact amounts of reagents or chemicals are required for reactions to take place.
- Measurements that are not accurate, provide incorrect data which can lead to wrong or even dangerous conclusions or results.
- For example, if a lab experiment requires a specific amount of a chemical, measuring the wrong amount may result in an unsafe or unexpected outcome.
- Due to all these reasons, accurate measurement is crucial in science.

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**Question 3. **

**Solve the following questions**

**(A) How many significant figures are in each of the following quantities ?**

**a- 45.26 ft b- 0.109 in**

**c- 0.00025 kg d- 2.3659 ****× ****10 ^{-8} cm**

**e- 52.0 cm ^{3} f- 0.00020 kg**

**g- 8.50 ****× ****10 ^{4} mm h- 300.0 cg**

**(a)** 4, **(b)** 3, **(c)** 2, **(d)** 5, **(e)** 3, **(f)** 2, **(g)** 3, **(h)** 4

**(B) Round off each of the following quantities to two significant figures :**

**a- 25.55 mL, b- 0.00254 m, c- 1.491 ****× ****10 ^{5} mg, d- 199 g**

**(a)** The last two digits being 55, we ﬁrst convert first two digits from 25 to 26. For two significant figures the answer is **26 mL.**

**(b)** Since the third digit 4 is less than 5 and we require two significant ﬁgures, we drop the third digit 4 and write **2.5 x 10**^{-}^{3}** m**

**(c)** 1.5 x 10^{5} mg

**(d)** 2.0 x 10^{2} g

**(C) Round off each of the following quantities to three significant figures :**

**a- 1.43 cm ^{3} b- 458 **

**×**

**10**

^{2}cm**c- 643 cm ^{2} d- 0.039 m**

**e- 6.398 ****× ****10 ^{-3} km f- 0.0179 g **

**g- 79,000 m h- 42,150 **

**i- 649.85 j- 23,642,000 mm**

**k- 0.0041962 kg**

**(a)** 1.43 cm^{3}

**(b)** 458 x 10^{2} cm

**(c)** 643 cm^{2}

**(d)** 3.90 x 10^{-}^{2} m

**(e)** 6.40 x 10^{-}^{3} km : In the figure 6.398 × 10^{-3} the fourth digit is more than 5, hence third digit is increased by 1 which makes second digit 4 and final figure with three significant figures becomes 6.40 x 10^{-}^{3} km.

**(f)** 1.79 x10^{-}^{2} g

**(g)** 790 x 10^{2} or 7.90 x 10^{4} m

**(h)** 4.22 x 10^{4}

**(i)** 650

**(j)** 2.36 x 10^{7} mm

**(k)** 4.20 x 10^{-3} kg

**(D) Express the following sum to appropriate number of significant figures :**

**a- 2.3 ****× ****10 ^{3} mL + 4.22 **

**×**

**10**

^{4}mL + 9.04**×**

**10**

^{3}mL + 8.71**×**

**10**

^{5}mL;**b- 319.5 g ****-**** 20460 g ****-**** 0.0639 g ****-**** 45.642 g ****-**** 4.173 g**

(a) 2.3 × 10^{3} mL +4.22 × 10^{4} mL + 9.04 × 10^{3} mL + 8.71 x 10^{5} mL

= (0.23 + 4.22 + 0.904 + 87.1) × 10^{4}

= 92.454 × 10^{4} mL

= 92.5 × 10^{4} mL

= 9.2 × 10^{5} mL

(b) 319.5 g **-** 20460 g **-** 0.0639 g **-** 45.642 g **-** 4.173 g

= **-** 20190.38 g

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**Question 4. **

**Solve the following problems**

**(A) Express the following quantities in exponential terms.**

**a- 0.0003498 b- 235.4678**

**c- 70000.0 d- 1569.00**

(a) 3.498 × 10^{-}^{4}

(b) 2.354678 × 10^{2}

(c) 7.00000 × 10^{4}

(d) 1.56900 × 10^{3}

**(B) Give the number of significant figures in each of the following**

**a- 1.230 ****× ****10 ^{4} b- 0.002030**

**c- 23 ****× ****10 ^{4} d- 1.89 **

**×**

**10**

^{-4}

(a) 4, (b) 4, (c) 3, (d) 3

**(C) Express the quantities in above question (B) with or without exponents as the case may be.**

(a) 12300

(b) 2.030 × 10^{-3}

(c) 12300

(d) 0.000189

**(D) Find out the molar masses of the following compounds :**

**a- Copper sulphate crystal (CuSO _{4}.5H_{2}O)**

**b- Sodium carbonate, decahydrate (Na _{2}CO_{3}.10H_{2}O)**

**c- Mohr’s salt [FeSO _{4}(NH_{4})2SO_{4}.6H_{2}O]**

**(At. mass : Cu = 63.5; S = 32; O = 16; H = 1; Na = 23; C = 12; Fe = 56; N = 14)**

**(a)** Molar mass of Copper sulphate crystal (CuSO_{4}.5H_{2}O)

= Atomic mass of Cu + Atomic mass of S + 4 × Atomic mass of O + 5(2 × Atomic mass of H + Atomic mass of O)

=63.5+32+4×16+5(2×1+16)

=63.5+32+64+5(18)

=249.5 g/mol

**(b)** Molar mass of Sodium carbonate, decahydrate (Na2CO_{3}.10H_{2}O)

= 2 × Atomic mass of Na + Atomic mass of C + 3 × Atomic mass of O + 10(2 × Atomic mass of H + Atomic mass of O)

=2×23+12+3×16+1O (2×1+16)

=46+12+48+10(18) .

=286g/mol

**(c)** Molar mass of Mohr's salt [FeSO_{4}(NH_{4})_{2}SO_{4}.6H_{2}O]

= Atomic mass of Fe + Atomic mass of S + 4 × Atomic mass of O + 2(Atomic mass of N + 4 × Atomic mass of H) + Atomic mass of S + 4 × Atomic mass of O + 6(2 x Atomic mass of H + Atomic mass of O)

=56+32+4×16+2(14+4×1)+32+4 x 16+6 (2×1+16)

=56+32+64+2(18)+32+64+6 (18)

= 392 g/mol

**(E) Work out the percentage composition of constituents elements in the following compounds :**

**a- Lead phosphate [Pb _{3}(PO_{4})_{2}],**

**b- Potassium dichromate (K _{2}Cr_{2}O_{7}),**

**c- Macrocosmic salt – Sodium ammonium hydrogen phosphate, tetrahydrate (NaNH _{4}HPO_{4}.4H_{2}O)**

**(At. mass : Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14)**

**(a)** Percentage composition of the constituent elements in [Pb_{3}(PO_{4})_{2}]

Molecular mass of Lead phosphate [Pb_{3}(PO_{4})_{2}]

= 3 x Atomic mass of Pb + 2 (Atomic mass of P + 4 x Atomic mass of O)

=3 x 207+2 (31+4x16) = 621+2(31+64)

= 811 g/mol

(1) To determine the percentage composition of Pb

Mass of Pb in Pb_{3}(PO_{4})_{2} = 3 x 207 = 621

= \(\frac{\text{Mass of Pb in}Pb_3 (PO_4)_2}{\text{Molecular mass of}Pb_3 (PO_4)_2}\) x 100%

= \(\frac{621}{811}\) x 100% = 76.57%

(2) To determine the percentage composition of P

= Mass of P in Pb_{3}(PO_{4})_{2} = 2 x 31 = 62

= \(\frac{\text{Mass of P in}Pb_3 (PO_4)_2}{\text{Molecular mass of}Pb_3 (PO_4)_2}\) x 100%

= \(\frac{62}{811}\)x 100% = 7.64%

(3) To determine the percentage composition of O

= Mass of O in Pb3(PO4)2 = 2 (4 x 16) = 128

= \(\frac{\text{Mass of O in}Pb_3 (PO_4)_2}{\text{Molecular mass of}Pb_3 (PO_4)_2}\) x 100

= \(\frac{128}{811}\) x 100% = 15.78%

**(b)** Percentage composition of the constituent elements in [K_{2}Cr_{2}O_{7}]

Molecular mass of K_{2}Cr_{2}O_{7}

= 2 x Atomic mass of K + 2 x Atomic mass of Cr + 7 x Atomic mass of O

= 2x39+2x52+7x16 = 294

(1) To determine the percentage composition of K

= Mass of K in K_{2}Cr_{2}O_{7} = 2 x 39 = 78

= \(\frac{\text{Mass of k in}K_2Cr_2O_7}{\text{Molecular mass of}K_2Cr_2O_7}\) x 100%

= \(\frac{78}{294}\) x 100% = 26.53%

(2) To determine the percentage composition of Cr

= Mass of Cr in K_{2}Cr_{2}O_{7} = 2 x 52 = 104

= \(\frac{\text{Mass of Cr in}K_2Cr_2O_7}{\text{Molecular mass of}K_2Cr_2O_7}\)x 100%

= \(\frac{104}{294}\)x 100% = 35.37%

(3) To determine the percentage composition of O

= Mass of O in K_{2}Cr_{2}O_{7 }= 7 x 16 = 112

= \(\frac{\text{Mass of O in}K_2Cr_2O_7}{\text{Molecular mass of}K_2Cr_2O_7}\) x 100%

= \(\frac{112}{294}\) x 100% = 38.10%

**(c)** Percentage composition of the constituent elements in (NaNH_{4}HPO_{4}.4H_{2}O)

Molecular mass of NaNH_{4}HPO_{4}.4H_{2}O

= Atomic mass of Na + Atomic mass of N + 4 x Atomic mass of H + Atomic mass of H + Atomic mass of P + 4 x Atomic mass of O + 4(2 x Atomic mass of H + Atomic mass of O)

= 23+14+4x1+1+31+4x16+4 (2x1+16)

= 209 g/mol

(1) To determine the percentage composition of Na

= Mass of Na in NaNH_{4}HPO_{4}.4H_{2}O = 23

= \(\frac{\text{Mass of Na in}NaNH_4HPO_4.4H_2O}{\text{Molecular mass of}NaNH_4HPO_4.4H_2O}\) x 100%

= \(\frac{23}{209}\) x 100% = 11.00%

(2) To determine the percentage composition of N

= Mass of N in NaNH_{4}HPO_{4}.4H_{2}O = 14

= \(\frac{\text{Mass of N in}NaNH_4HPO_4.4H_2O}{\text{Molecular mass of}NaNH_4HPO_4.4H_2O}\) x 100%

= \(\frac{14}{209}\) x 100% = 6.70%

(3) To determine the percentage composition of H

= Mass of H in NaNH_{4}HPO_{4}.4H_{2}O = 13x1=13

= \(\frac{\text{Mass of H in}NaNH_4HPO_4.4H_2O}{\text{Molecular mass of}NaNH_4HPO_4.4H_2O}\) x 100%

= \(\frac{13}{209}\)x 100% = 6.22%

(4) To determine the percentage composition of P

= Mass of P in NaNH_{4}HPO_{4}.4H_{2}O = 31

= \(\frac{\text{Mass of P in}NaNH_4HPO_4.4H_2O}{\text{Molecular mass of}NaNH_4HPO_4.4H_2O}\)x 100%

= \(\frac{31}{209}\)x 100% = 14.83%

(5) To determine the percentage composition of O

= Mass of O in NaNH_{4}HPO_{4}.4H_{2}O = 8x16 = 128

= \(\frac{\text{Mass of O in}NaNH_4HPO_4.4H_2O}{\text{Molecular mass of}NaNH_4HPO_4.4H_2O}\)x 100%

= \(\frac{128}{209}\)x 100% = 61.24%

**(F) Find the percentage composition of constituent green vitriol crystals**

**(FeSO _{4}.7H_{2}O). Also find out the mass of iron and the water of crystallization in 4.54 kg of the crystals. (At. mass : Fe = 56; S = 32; O = 16)**

(A) Percentage composition of the constituent elements in FeSO_{4}.7H_{2}O

Molecular mass of FeSO_{4}.7H_{2}O

= Atomic mass of Fe + Atomic mass of S + 4 x Atomic mass of O + 7(2 x Atomic mass of H + Atomic mass of O)

=56+32+4x16+7(2x1+16)

= 278 g/mol

(1) To determine the percentage composition of Fe

= Mass of Fe in FeSO_{4}.7H_{2}O = 56

= \(\frac{\text{Mass of Fe in}FeSO_4.7H_2O}{\text{Molecular mass of}FeSO_4.7H_2O}\) x 100%

= \(\frac{56}{278}\) x 100% = 20.14%

(2) To determine the percentage composition of O

= Mass of O in FeSO_{4}.7H_{2}O = 11x16 = 176

= \(\frac{\text{Mass of O in}FeSO_4.7H_2O}{\text{Molecular mass of}FeSO_4.7H_2O}\)x 100%

= \(\frac{176}{278}\)x 100% = 63.31%

(3) To determine the percentage composition of S

= Mass of S in FeSO_{4}.7H_{2}O = 32

= \(\frac{\text{Mass of S in}FeSO_4.7H_2O}{\text{Molecular mass of}FeSO_4.7H_2O}\)x 100%

= \(\frac{32}{278}\)x 100% = 11.51%

(4) To determine the percentage composition of H

= Mass of H in FeSO_{4}.7H_{2}O = 14x1=14

= \(\frac{\text{Mass of H in}FeSO_4.7H_2O}{\text{Molecular mass of}FeSO_4.7H_2O}\)x 100%

= \(\frac{14}{278}\)x 100% = 5.04%

**To ﬁnd out the mass of iron and the water of crystallisation in 4.54 kg of the crystals.**

Weight of crystals = 4.54 kg

Mass of iron = ?

278 g of FeSO_{4}.7H_{2}O contains 56 g of Fe,

∴ 4.54 kg of FeSO4.7H2O contains

\(\frac{4,54×56}{278}\) = 0.915 kg of Fe

Mass of iron in 4.54 kg = 0.915 kg

Mass of water of crystallisation = ?

278 g of FeSO_{4}.7H_{2}O contains 7H_{2}O molecules

i.e 7 x 18 g of H_{2}O= 126 g of H2O

∴ 4.54 kg of FeSO4.7H2O contains

\(\frac{4,54×126}{278}\) = 2.058 kg of water of crystallisation

Mass of water of crystallisation = 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)**

**Given :**

Atomic mass of Fe = 55.84

Percentage of iron in haemoglobin = 0.335 %

One molecule of haemoglobin contains 4 atoms of iron.

Since, 1 atom of Fe = 55.84 g

4 atoms of Fe = 223.36

0.335 g of Fe is present in 100 g of haemoglobin

223.36 g of Fe is present in

=\(\frac{223.36×100}{0.335}\) = 66674.6 g/mol of haemoglobin

Molecular weight of haemoglobin = 66674.6 g/mol

**(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 = 12 u, O = 16 u).**

**Given : **Na = 43.4 %, C = 11.3 % and O = 45.3 %.

To calculate the moles of the constituent elements

Moles of Na = \(\frac{\text{% of Na}}{\text{Atomic mass of Na}}=\frac{43.4}{23}\) = 1.89

Moles of C = \(\frac{\text{% of C}}{\text{Atomic mass of C}}=\frac{11.3}{12}\) = 0.941

Moles of O = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{45.3}{16}\) = 2.83

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

\(\frac{1.89}{0.941}\) = 2 : \(\frac{0.941}{0.941}\) = 1 : \(\frac{2.83}{0.941}\) = 3

Hence the empirical formula is Na_{2}CO_{3}

**(I) Assuming the atomic weight of a metal M to be 56, find the empirical formula of its oxide containing 70.0% of M.**

**Given :**

Atomic weight of M = 56

Percentage of M = 70.0 %

Since, the oxide contains 70.0% of M, the amount of oxygen will be 30.0 %

The atomic weight of Oxygen = 16

Moles of M = \(\frac{\text{% of Na}}{\text{Atomic mass of M}}=\frac{70}{56}\) = 1.25

Moles of O = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{30}{16}\) = 1.875

Hence, the ratio of the number of moles of the metal with oxygen would be,

\(\frac{1.25}{1.25}\) = 1 and \(\frac{1.875}{1.25}\)= 1.5

i.e. 2 : 3

Hence, the empirical formula of the compound would be M_{2}O_{3}

**(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. wt. : Fe = 56; S = 32; O = 16)**

**Given :**

Iron = 0.2014 g, Sulphur = 0.1153 g, Oxygen = 0.2301 g,

Water of crystallisation = 0.4532 g, Wt. of hydrated salt: 1.00 g

To calculate the percentage of the different elements.

Percentage of Fe in the salt = \(\frac{\text{Wt of Fe}}{\text{Wt. of the hydrated salt}}\)x100

= \(\frac{0.2014}{1.00}\) x 100 = 20.14%

Percentage of S in the salt = \(\frac{\text{Wt of S}}{\text{Wt. of the hydrated salt}}\)x100

= \(\frac{0.1153}{1.00}\) x 100 = 11.53%

Percentage of O in the salt = \(\frac{\text{Wt of O}}{\text{Wt. of the hydrated salt}}\)x100

= \(\frac{0.2301}{1.00}\) x 100 = 23.01%

Percentage of H_{2}O in the salt = \(\frac{\text{Wt of }H_2O}{\text{Wt. of the hydrated salt}}\)x100

= \(\frac{0.4532}{1.00}\) x 100 = 45.32%

To calculate the moles of each element

Moles of Fe =\(\frac{\text{% of Fe}}{\text{Atomic mass of Fe}}=\frac{20.14}{56}\) = 0.360

Moles of S =\(\frac{\text{% of S}}{\text{Atomic mass of S}}=\frac{11.53}{32}\) = 0.360

Moles of O = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{23.01}{16}\) = 1.438

Moles of H_{2}O = \(\frac{\text{% of }H_2O}{\text{Atomic mass of} H_2O}=\frac{45.32}{18}\) = 2.518

Hence, the ratio of number of moles of Fe : S : O : H_{2}O

Is \(\frac{0.360}{0.360}\) = 1 : \(\frac{0.360}{0.360}\) = 1 : \(\frac{1.438}{0.360}\) = 3.99 : \(\frac{2.518}{0.360}\) = 7 :

Hence, the empirical formula is **FeSO _{4}.7H_{2}O**

**(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. Find the molecular formula of the compound.**

**Given :**

Carbon = 20 %, Hydrogen = 6.7 %, Nitrogen = 46.67 % Molecular mass = 60

The total percentage of the constituent elements

= % Carbon + % Hydrogen + % Nitrogen

= 20 + 6.7 + 46.67

= 73.37 %

This is less than 100%.

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

Hence, the percentage of oxygen = 100 — 73.37 = 26.63 %

To calculate the moles of the constituent elements

Moles of C = \(\frac{\text{% of C}}{\text{Atomic mass of C}}=\frac{20}{12}\) = 1.66

Moles of H = \(\frac{\text{% of H}}{\text{Atomic mass of H}}=\frac{6.7}{1}\) = 6.7

Moles of N = \(\frac{\text{% of N}}{\text{Atomic mass of N}}=\frac{46.67}{14}\) = 3.33

Moles of O = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{26.63}{16}\) = 1.66

Hence, the ratio of the number of moles of

C : H : N : O is

\(\frac{1.66}{1.66}\) = 1 : \(\frac{6.7}{1.66}\) = 4.03 : \(\frac{3.33}{1.66}\) = 2.00 : \(\frac{1.66}{1.66}\)= 1 :

Hence, the empirical formula is CH_{4}N_{2}O

Empirical formula mass

= Atomic mass of C + 4 x Atomic mass of H + 2 x Atomic mass of N + Atomic mass of O

=12+4x1+2x14+16 = 60

Molar mass = Empirical mass (Since molar mass = Molecular mass)

Molecular formula = Empirical formula CH_{4}N_{2}O

**(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.**

Given :

Carbon = 54.55, % Hydrogen = 9.09 %, Oxygen = 3636 %, Molecular mass = 88

Moles of C = \(\frac{\text{% of C}}{\text{Atomic mass of C}}=\frac{54.55}{12}\) = 4.54

Moles of H = \(\frac{\text{% of H}}{\text{Atomic mass of H}}=\frac{9.09}{1}\) = 9.09

Moles of O = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{36.36}{16}\) = 2.27

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

\(\frac{4.54}{2.27}\) = 2 : \(\frac{9.09}{2.27}\) = 4 : \(\frac{2.27}{2.27}\) = 1

Hence, the empirical formula is C_{2}H_{4}O

Empirical formula mass = 2 x Atomic mass of C + 4 x Atomic mass of H + Atomic mass of O = 2x12+4x1+16 = 44

r = (\frac{\text{Molar mass}}{\text{Emperical formula mass}}\)=\(\frac{88}{44}\) = 2

Molecular formula = r x Empirical formula

Molecular formula = 2 x C_{2}H_{4}O = C_{4}H_{8}O_{2}

**(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 ?**

Given :

Amount of carbohydrate = 310 g

Amount of Carbon residue obtained on heating = 124 g

Amount of H_{2}O formed = Amount of carbohydrate - Amount of carbon as residue

= 310 − 124 = 186 g

Since 18 g of H_{2}O contains 2 g of H

186 g of H_{2}O will contain \(\frac{186×2}{18}\) = 20.66 g of H

Since 18 g of H_{2}O contains 16 g of O

186 g of H_{2}O will contain \(\frac{186×16}{18}\) = 165.33g of O

To determine the moles of the C, H and O in the compound.

Moles of C = \(\frac{\text{% of C}}{\text{Atomic mass of C}}=\frac{124}{12}\) = 10.33

Moles of H = \(\frac{\text{% of H}}{\text{Atomic mass of H}}=\frac{20.66}{1}\) = 20.66

Moles of O = \(\frac{\text{% of O}}{\text{Atomic mass of O}}=\frac{165.33}{16}\) = 10.33

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

\(\frac{10.33}{10.33}\) = 1 : \(\frac{20.66}{10.33}\) = 2 : \(\frac{10.33}{10.33}\) = 1

Hence, the empirical formula is CH_{2}O

**(N) Write each of the following in exponential notation :**

**a- 3,672,199 b- 0.000098**

**c- 0.00461 d- 198.75**

(a) 3.672199 x 10^{6}

(b) 9.8 x 10^{-}^{5}

(c) 4.61 x 10^{-}^{3}

(d) 1.9875 x 10^{2}

**(O) Write each of the following numbers in ordinary decimal form :**

**a- 3.49 ****× ****10 ^{-11} b- 3.75 **

**×**

**10**

^{-1}**c- 5.16 ****× ****10 ^{4} d- 43.71 **

**×**

**10**

^{-4}**e- 0.011 ****× ****10 ^{-3} f- 14.3 **

**×**

**10**

^{-2}**g- 0.00477 ****× ****1 ^{5} h- 5.00858585**

(a) 0.0000000000349

(b) 0.375

(c) 51600

(d) 0.004371

(e) 0.000011

(f) 0.143

(g) 477

(h) 5.00858585

**(P) Perform each of the following calculations. Round off your answers to two digits.**

**a- **** \(\frac{1}{3.40×10^{24}}\) **

**b-\(\frac{33}{9.00×10^{−4}}\) **

**c- \(\frac{1.4×10^9}{(2.77×10^3)(3.76×10^5}\) **

**d- \(\frac{(4×10^{-3})(9.9×10^{-7})}{(789)(1.002×10^{-10})(0.3×10^2}\) **

(a) \(\frac{1}{3.40×10^{24}}\)= 0.2941 x 10^{-24}** = **2.9 x 10^{-25}** **

(b) \(\frac{33}{9.00×10^{−4}}\)= 3.66 x 10^{4}** = **3.7 x 10^{4}** **

(c) \(\frac{1.4×10^9}{(2.77×10^3)(3.76×10^5}\) = \(\frac{1.4×10^9}{10.4152×10^8}\)

= 0.1344 x 10 = 1.344 = 1.3** **

(d) \(\frac{(4×10^{-3})(9.9×10^{-7})}{(789)(1.002×10^{-10})(0.3×10^2}\) = 0.169 x 10^{-2}** = **1.7 x 10^{-3}** **

**(Q) Perform each of the following calculations. Round off your answers to three digits.**

**a- (3.26 10 ^{4}) (1.54 10^{6})**

**b- (8.39 10 ^{7}) (4.53 10^{9})**

**c- \(\frac{8.94×10^6}{4.35×10^4}\) **

**d- \(\frac{(9.28×10^9)(9.9×10^{-7})}{(511)(2.98×10^{-6})}\) **

(a) (3.26 × 10^{4}) (1.54 × 10^{6}) = 5.0204 × 10^{10} = 5.02 × 10^{10}

(b) (8.39 × 10^{7}) (4.53 × 10^{9}) = 38.0067 × 10^{16} = 3.80 × 10^{17}

(c) \(\frac{8.94×10^6}{4.35×10^4}\) = 2.055 × 10^{2} = 2.06 × 10^{2}

(d) \(\frac{(9.28×10^9)(9.9×10^{-7})}{(511)(2.98×10^{-6})}\) = \(\frac{91.872×10^2}{1522.78×10^{-6}}\)

= 0.06033 × 10^{8} = 6.03 x 10^{6}

**(R) Perform the following operations :**

**a- 3.971 ****× ****10 ^{7} + 1.98 **

**×**

**10**

^{4};**b- 1.05 ****× ****10 ^{−}**

^{4}**−**

**9.7**

**×**

**10**

^{−}

^{5}**;**

**c- 4.11 ****× ****10 ^{−}**

^{3}**+ 8.1**

**×**

**10**

^{−}

^{4}**;**

**d- 2.12 ****× ****10 ^{6} - 3.5 **

**×**

**10**

^{5}.

(a) 3.971 × 10^{7} + 1.93 × 10^{4}

=3971 × 10^{4} + 1.93 × 10^{4}

= (3971 + 1.93) × 10^{4}

= 3972.98 × 104

= 3.97293 × 10^{7}

(b) 1.05 × 10^{−}^{4} − 9.7 × 10^{−}^{5}

= 1.05 ×10−4 − 0.97 × 10−4

= (1.05 − 0.97) × 10^{−}^{4}

= 0.08 × 10^{−}^{4} = 8.0 × 10^{−6}

(c) 4.11 × 10^{−}^{3} + 8.1 × 10^{−}^{4}

= 4.11 × 10^{−}^{3} +0.81 × 10^{−}^{3}

= (4.11 + 0.81) × 10^{−}^{3}

= 4.92 × 10^{−}^{3}

(d) 2.12 × 10^{6} -3.5 × 10^{5}

=21.2 × 10^{5} − 3.5 × 10^{5}

= (21.2 − 3.5) × 10^{5}

= 17.7 × 10^{5} = 1.77 × 10^{6}

**(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 characterisethe precision and accuracy of these measurements if the actual mass of the acetone was 0.7791 g ?**

Given :

Volume of acetone taken = 1.000 mL

Mass of bottle = 38.0015 g

Actual mass of acetone = 0.7791 g

To determine the mean values of the weight of acetone (Observed value)

Weight of acetone +
bottle (a) |
Weight of bottle
(b) |
Weight of acetone
c = a - b |

38.7798 g | 38.0015 g | 0.7783 g |

38.7795 g | 38.0015 g | 0.7780 g |

38.7801 g | 38.0015 g | 0.7786 g |

Mean values of the weight of acetone =\(\frac{0.7783+0.7780+0.7786}{3}\) = 0.7783 g

Absolute error = Observed value — True value

= 0.7783 − 0.7791 = − 0.008

**Relative error = **\(\frac{\text{Absolute error}}{\text{True value}}\)** **x 100%

= \(\frac{0.008}{0.7791}\) = **0.1027%**

Accuracy of the measurements is 0.1027 %.

**To determine the average deviation of the measurements : **

Readings | Weight of acetone | Mean values of the weight of acetone | Deviation |

1 | 0.7783 g |
0.7783 g |
0.0000 |

2 | 0.7780 g | 0.0003 | |

3 | 0.7786 g | 0.0003 |

**Mean absolute Deviation** = \(\frac{0.0000+0.0003+0.0003}{3}\) = ± 0.0002

**Relative deviation = **\(\frac{\text{Absolute error}}{\text{True value}}\)** **** **x 100%

= \(\frac{0.0002}{0.7783}\) x 100% = 0.0257 %

**Relative error = 0.1027%**

**Precision of the measurements** = 0.0257%

**(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.**

The measurement **12.6** is acceptable.

**Reason** :

**(a)** 5 inches is approximately 12.7 cm. The value 12.6 is closer to this value, hence this measurement is acceptable.

**(b)** Since the length of the box has been measured using a metre stick graduated in millimeters.

The least count of the scale is 0.1 cm. The other measurements cannot be obtained using a scale with a least count of 0.1 cm, hence the other measurements can be rejected.

**(U) What weight of calcium oxide will be formed on heating 19.3 g of calcium carbonate? (At. wt. : Ca = 40; C = 12; O = 16)**

Given :

Weight of calcium carbonate = 19.3 g

The balanced chemical equation is,

CaCO_{3(s)} > CaO_{(g)} + CO_{2(g)}

Since 100 g of CaCO_{3} produces 56 g of CaO

19.3 g of CaCO_{3} will produce

\(\frac{19.3×56}{100}\) = 10.8 g of caO

**(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 ? **

The balanced chemical equation for the burning of sucrose is,

C_{12}H_{22}O_{11(s)} + 12O_{2(g)} —> 12CO_{2(g)} + 11H_{2}O_{(l)}

342g 12 x (32)

= 384g

342 g of sucrose requires 384 g of oxygen

34 g of sucrose will require

\(\frac{34×384}{342}\) = 38.1754 g of oxygen

The hourly requirement of oxygen is 38.175 g

Therefore, the requirement of oxygen for one day will be = 38.1754 x 24

= 916.209 g

= 916.21g

The oxygen requirement for one day = 916.21 g.

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