Maharashtra Board Class 11 Chemistry Chapter 2 Introduction to Analytical Solutions

Chapter 2: Introduction to Analytical Chemistry

Choose Correct Option

 

Question 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
Answer: (d) Analytical chemistry
In simple words: Analytical chemistry is the branch of science that helps us separate, identify, and measure the different parts that make up any substance.

🎯 Exam Tip: Remember the three key terms—separation, identification, and quantitative determination—to easily define analytical chemistry in exams.

 

Question B. Which one of the following property of matter is Not quantitative in nature ?
(a) Mass
(b) Length
(c) Colour
(d) Volume
Answer: (c) Colour
In simple words: Quantitative properties are those that can be measured and described with numbers, whereas colour is a qualitative property that we describe with words.

🎯 Exam Tip: Always look for properties that cannot be measured with a numerical value and a unit to identify qualitative characteristics like colour or odour.

 

Question C. SI unit of mass is ........
(a) kg
(b) mol
(c) pound
(d) \( \text{m}^3 \)
Answer: (a) kg
In simple words: The standard international unit used worldwide to measure the amount of matter in an object is the kilogram (kg).

🎯 Exam Tip: Do not confuse mass with weight or volume; the official SI base unit for mass is always the kilogram (kg), not the gram (g).

 

Question D. The number of significant figures in \( 1.50 \times 10^4 \text{ g} \) is ...........
(a) 2
(b) 3
(c) 4
(d) 6
Answer: (b) 3
In simple words: In scientific notation, we only count the digits in the decimal number before the power of 10. Since 1.50 has three digits, it has 3 significant figures.

🎯 Exam Tip: Remember that exponential terms like \( 10^4 \) do not affect the number of significant figures; always focus only on the numerical base.

 

Question E. In Avogadro’s constant \( 6.022 \times 10^{23} \text{ mol}^{-1} \), the number of significant figures is ..........
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (b) 4
In simple words: Just like the previous question, we only look at the digits in 6.022. Since there are four digits, the number of significant figures is 4.

🎯 Exam Tip: All non-zero digits and any zeros placed between them are always counted as significant.

 

Question F. By decomposition of 25 g of \( \text{CaCO}_3 \), the amount of CaO produced will be .................
(a) 2.8 g
(b) 8.4 g
(c) 14.0 g
(d) 28.0 g
Answer: (c) 14.0 g
In simple words: Heating 100 grams of calcium carbonate produces 56 grams of calcium oxide. If we use only a quarter of that amount (25 grams), we will get a quarter of the product, which is 14 grams.

🎯 Exam Tip: Always write down the balanced chemical equation first to find the molar ratio between the reactant and the product.

 

Question G. How many grams of water will be produced by complete combustion of 12g of methane gas?
(a) 16 g
(b) 27 g
(c) 18 g
(d) 36 g
Answer: (b) 27 g
In simple words: Burning 16 grams of methane produces 36 grams of water. Therefore, burning 12 grams of methane will produce a proportional amount, which is 27 grams of water.

🎯 Exam Tip: Convert the mass of the given reactant to moles, use the balanced equation to find the moles of the product, and then convert those moles back to grams.

 

Question 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) \( \text{AB} \)
(b) \( \text{A}_2\text{B} \)
(c) \( \text{AB}_2 \)
(d) \( \text{A}_2\text{B}_3 \)
Answer: (d) \( \text{A}_2\text{B}_3 \)
In simple words: By dividing the percentage of each element by its atomic mass, we find the ratio of A to B is 1 to 1.5, which simplifies to the whole number ratio of 2 to 3.

🎯 Exam Tip: Always convert fractional ratios into the simplest whole numbers by multiplying all numbers by the same factor to get the correct empirical formula.

 

Question I. The hydrocarbon contains 79.87 % carbon and 20.13 % of hydrogen. What is its empirical formula ?
(a) \( \text{CH} \)
(b) \( \text{CH}_2 \)
(c) \( \text{CH}_3 \)
(d) \( \text{C}_2\text{H}_5 \)
Answer: (c) \( \text{CH}_3 \)
In simple words: Dividing the percentage of carbon by 12 and hydrogen by 1 gives a ratio of 6.66 to 20.13, which simplifies to 1 carbon atom for every 3 hydrogen atoms.

🎯 Exam Tip: For empirical formula questions, divide the relative number of moles of each element by the smallest value among them to get the simplest ratio easily.

 

Question 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
Answer: (c) 24
In simple words: Aluminum reacts with oxygen to form \( \text{Al}_2\text{O}_3 \). Since 4 moles of Al (108 g) react with 3 moles of oxygen gas (96 g), 27 g of Al (which is one-fourth of 108 g) will need one-fourth of 96 g, which is 24 g of oxygen.

🎯 Exam Tip: Always write the balanced chemical equation \( 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \) first to establish the correct stoichiometric relationship between reactants.

 

Question K. In \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) the percentage of water is ......
(Cu = 63.5, S = 32, O = 16, H = 1)
(a) 10 %
(b) 36 %
(c) 60 %
(d) 72 %
Answer: (b) 36 %
In simple words: To find the percentage of water, we divide the mass of the 5 water molecules by the total mass of the entire copper sulfate crystal and multiply by 100.

🎯 Exam Tip: Remember the molar masses of common elements like Cu (63.5), S (32), O (16), and H (1) to quickly calculate molecular weights in the exam.

 

Question 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 of the options
Answer: (b) Plotting a graph
In simple words: Drawing a graph helps us visually see how one quantity changes when another quantity changes, making it easy to find their mathematical relationship.

🎯 Exam Tip: Graphs are the most effective visual tools in physics and chemistry to deduce direct or inverse relationships between variables.

2. Answer the Following Questions

 

Question A. Define : Least count
Answer: The smallest quantity that can be measured by the measuring equipment is called least count. Knowing the least count helps in estimating the maximum possible error in a measurement.
In simple words: Least count is the smallest value that any measuring instrument, like a ruler or a stopwatch, can accurately measure.

🎯 Exam Tip: Always mention an example like a standard school ruler having a least count of 1 mm to secure full marks.

 

Question B. What do you mean by significant figures? State the rules for deciding significant figures.
Answer:
(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. These rules ensure that scientific measurements are reported with the correct level of precision.
In simple words: Significant figures are the meaningful digits in a number that tell us how precise a measurement is. Non-zero numbers and zeros caught between them are always counted.

🎯 Exam Tip: Pay close attention to the position of zeros; leading zeros are never significant, whereas trapped zeros are always significant.

 

Question C. Distinguish between accuracy and precision.
Answer:
Accuracy:
1. Accuracy refers to nearness of the measured value to the true value.
2. Accuracy represents the correctness of the measurement.
3. Accuracy is expressed in terms of absolute error and relative error.
4. Accuracy takes into account the true or accepted value.
5. Accuracy can be determined by a single measurement.
6. High accuracy implies smaller error.

Precision:
1. Precision refers to closeness of multiple readings of the same quantity.
2. Precision represents the agreement between two or more measured values.
3. Precision is expressed in terms of absolute deviation and relative deviation.
4. Precision does not take into account the true or accepted value.
5. Several measurements are required to determine precision.
6. High precision implies reproducibility of the readings.
In simple words: Accuracy is how close your measurement is to the actual true value, while precision is how close your different measurements are to each other, even if they are all off-target.

🎯 Exam Tip: To score full marks, list at least three distinct points of comparison, highlighting that accuracy is about correctness and precision is about consistency.

 

Question D. Explain the terms percentage composition, empirical formula and molecular formula.
Answer:
1. Percentage Composition: It is the percentage by mass of each constituent element present in a compound. It is calculated using the formula:
\( \text{Mass \% of an element} = \frac{\text{Mass of that element in 1 mole of the compound}}{\text{Molar mass of the compound}} \times 100 \)

2. Empirical Formula: It is the simplest chemical formula that shows the relative number of atoms of each element present in a compound in the simplest whole-number ratio.

3. Molecular Formula: It is the actual chemical formula that represents the exact number of atoms of each constituent element present in one molecule of the compound. It is a whole-number multiple of the empirical formula: \( \text{Molecular Formula} = n \times \text{Empirical Formula} \).
In simple words: Percentage composition tells us how much of each element is in a compound by weight. The empirical formula shows the simplest ratio of these elements, while the molecular formula shows the actual, exact number of atoms in each molecule.

🎯 Exam Tip: Remember the relation \( \text{Molecular formula} = n \times \text{Empirical formula} \), where \( n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} \), as this is frequently used to solve numerical problems.

 

Question E. What is a limiting reagent? Explain.
Answer: Limiting reagent:
• The reactant which gets consumed and limits the amount of product formed is called the limiting reagent.
• 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. This means one reactant will be completely used up before the others, halting the reaction.
In simple words: A limiting reagent is the ingredient in a chemical reaction that runs out first. Once it is completely used up, the reaction stops, just like how you cannot make more sandwiches once you run out of bread.

🎯 Exam Tip: Always identify the limiting reagent by comparing the mole ratio of the reactants given in the problem with the coefficients in the balanced chemical equation.

 

Question F. What do you mean by SI units ? What is the SI unit of mass ?
Answer:
i. In 1960, the general conference of weights and measures proposed a revised metric system, called the International System of Units i.e. SI units, abbreviated from its French name. These units provide a global standard for scientific measurements.
ii. The SI unit of mass is kilogram (kg).
In simple words: SI units are standard measurements used worldwide so that scientists everywhere can understand each other's data. The standard unit for measuring how heavy something is (mass) is the kilogram.

🎯 Exam Tip: Always remember to write both the full name (kilogram) and its standard abbreviation (kg) to secure full marks.

 

Question G. Explain the following terms
(a) Mole fraction
(b) Molarity
(c) Molality
Answer:
(a) Mole fraction: 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. This ratio is dimensionless because it compares similar physical quantities.
If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are \( n_A \) and \( n_B \), respectively, then the mole fraction of A and B are given as:\[ \text{Mole fraction of A} = \frac{\text{Number of moles of A}}{\text{Number of moles of solution}} = \frac{n_A}{n_A + n_B} \]\[ \text{Mole fraction of B} = \frac{\text{Number of moles of B}}{\text{Number of moles of solution}} = \frac{n_B}{n_A + n_B} \]
(b) Molarity: Molarity is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by the symbol M and depends on the temperature of the system.\[ \text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}} \]
(c) Molality: Molality is defined as the number of moles of solute dissolved in one kilogram (1 kg) of the solvent. It is denoted by the symbol m and is independent of temperature.\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]
In simple words: Mole fraction shows the share of one part in a total mixture. Molarity measures how crowded the solute particles are in a certain volume of liquid, while molality measures how crowded they are in a certain weight of the solvent.

🎯 Exam Tip: Remember that molarity changes with temperature because volume changes, whereas molality remains constant because mass does not change with temperature.

Molarity And Molality

(b) Molarity: Molarity is defined as the number of moles of the solute present in 1 litre of the solution. It is the most widely used unit and is denoted by M.
Molarity is expressed as follows:
Molarity (M) = \( \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}} \)

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

 

Question H. Define : Stoichiometry
Answer: The study of quantitative relations between the amount of reactants and/or products is called stoichiometry. This helps chemists calculate exactly how much raw material is needed to make a specific amount of product.
In simple words: Stoichiometry is like a recipe for chemistry. It tells you exactly how much of each ingredient you need to mix together to get the right amount of final product.

🎯 Exam Tip: Always write a balanced chemical equation before performing any stoichiometric calculations to ensure your mole ratios are correct.

 

Question I. Why there is a need of rounding off figures during calculation ?
Answer: 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. In other words, the final result cannot be more accurate than the least accurate number involved in the calculation. Sometimes, the final result of a calculation often contains figures that are not significant. When this occurs, the final result is rounded off to maintain scientific integrity and avoid false precision.
In simple words: When we measure things, our tools aren't perfect. We round off our final answers so we don't pretend our result is more precise than our actual measurements allow.

🎯 Exam Tip: Remember to round off only at the very final step of your calculation to avoid accumulating rounding errors along the way.

 

Question J. Why does molarity of a solution depend upon temperature ?
Answer: 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. Volume of the solution varies with the change in temperature because liquids expand or contract when heated or cooled.
In simple words: Molarity measures how much stuff is dissolved in a specific volume of liquid. Since liquids expand when they get hot and shrink when they get cold, the volume changes with temperature, which changes the molarity.

🎯 Exam Tip: Clearly state that volume is temperature-dependent, whereas mass is not, which is why molality (unlike molarity) remains constant with temperature changes.

Question M. Define Analytical chemistry. Why is accurate measurement crucial in science?
Answer: 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. This precision forms the foundation of all scientific discoveries.
1. The accuracy of measurement is of 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 analytical measurement.
2. 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.
3. Hence, the numerical data obtained experimentally are treated mathematically to reach some quantitative conclusion.
4. 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.
In simple words: Analytical chemistry is the study of finding out what substances are made of and how much of each ingredient is present. Accurate measurements are crucial because even tiny mistakes can lead to wrong results, failed experiments, or dangerous accidents.

🎯 Exam Tip: When defining analytical chemistry, remember to mention both "qualitative" (what is present) and "quantitative" (how much is present) determination to secure full marks.

 

Solve the Following Questions

 

Question 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) \( 2.3659 \times 10^{-8} \) cm
(e) \( 52.0 \text{ cm}^3 \)
(f) 0.00020 kg
(g) \( 8.50 \times 10^4 \) mm
(h) 300.0 cg
Answer:
(a) 45.26 ft has 4 significant figures. All non-zero digits are significant.
(b) 0.109 in has 3 significant figures. The zero between non-zero digits is significant.
(c) 0.00025 kg has 2 significant figures. Leading zeros are not significant.
(d) \( 2.3659 \times 10^{-8} \) cm has 5 significant figures. All digits in the scientific notation coefficient are significant.
(e) \( 52.0 \text{ cm}^3 \) has 3 significant figures. Trailing zeros after a decimal point are significant.
(f) 0.00020 kg has 2 significant figures. Leading zeros are not significant, but the trailing zero is significant.
(g) \( 8.50 \times 10^4 \) mm has 3 significant figures. All digits in the coefficient of scientific notation are significant.
(h) 300.0 cg has 4 significant figures. Trailing zeros after a decimal point are significant.
In simple words: Significant figures are the digits in a number that carry real meaning about its precision. Non-zero numbers are always counted, zeros between numbers are counted, but leading zeros used only as placeholders are never counted.

🎯 Exam Tip: Remember that exponential terms (like \( 10^{-8} \) or \( 10^4 \)) do not contribute to the number of significant figures; only look at the numerical coefficient.

 

Question B. Round off each of the following quantities to two significant figures:
a. 25.55 mL
b. 0.00254 m
c. \( 1.491 \times 10^5 \text{ mg} \)
d. 199 g
Answer:
a. 26 mL
b. 0.0025 m
c. \( 1.5 \times 10^5 \text{ mg} \)
d. \( 2.0 \times 10^2 \text{ g} \)
In simple words: To round to two significant figures, we keep only the first two important digits. If the third digit is 5 or more, we round the second digit up by one.

🎯 Exam Tip: Remember that leading zeros (like in 0.00254) are not significant, so start counting your significant figures from the first non-zero digit.

 

Question C. Round off each of the following quantities to three significant figures:
a. \( 1.43 \text{ cm}^3 \)
b. \( 458 \times 10^2 \text{ cm} \)
c. \( 643 \text{ cm}^2 \)
d. 0.039 m
e. \( 6.398 \times 10^{-3} \text{ km} \)
f. 0.0179 g
g. 79,000 m
h. 42,150
i. 649.85
j. 23,642,000 mm
k. 0.0041962 kg
Answer:
a. \( 1.43 \text{ cm}^3 \)
b. \( 4.58 \times 10^4 \text{ cm} \)
c. \( 643 \text{ cm}^2 \) (or \( 6.43 \times 10^2 \text{ cm}^2 \))
d. 0.0390 m
e. \( 6.40 \times 10^{-3} \text{ km} \)
f. 0.0179 g
g. \( 7.90 \times 10^4 \text{ m} \)
h. \( 4.22 \times 10^4 \)
i. \( 6.50 \times 10^2 \)
j. \( 2.36 \times 10^7 \text{ mm} \)
k. 0.00420 kg
In simple words: To round to three significant figures, we look at the fourth digit from the left starting with the first non-zero number, and round the third digit up if the fourth is 5 or more.

🎯 Exam Tip: When rounding numbers like 79,000 to three significant figures, write them in scientific notation (\( 7.90 \times 10^4 \)) to clearly show that the trailing zero is significant.

 

Question D. Express the following sum to appropriate number of significant figures:
a. \( 2.3 \times 10^3 \text{ mL} + 4.22 \times 10^4 \text{ mL} + 9.04 \times 10^3 \text{ mL} + 8.71 \times 10^5 \text{ mL} \);
b. \( 319.5 \text{ g} - 20460 \text{ g} - 0.0639 \text{ g} - 45.642 \text{ g} - 4.173 \text{ g} \)
Answer:
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. This ensures that the precision of the final result is consistent with the least precise measurement used.
a. \( (0.23 \times 10^4 \text{ mL}) + (4.22 \times 10^4 \text{ mL}) + (0.904 \times 10^4 \text{ mL}) + (87.1 \times 10^4 \text{ mL}) \)
\( = (0.23 + 4.22 + 0.904 + 87.1) \times 10^4 \text{ mL} \)
\( = 92.454 \times 10^4 \text{ mL} \)
\( = 9.2454 \times 10^5 \text{ mL} \)
\( = 9.2 \times 10^5 \text{ mL} \)
b. \( 319.5 \text{ g} - 20460 \text{ g} - 0.0639 \text{ g} - 45.642 \text{ g} - 4.173 \text{ g} \)
\( = -20190.3789 \text{ g} \)
\( = -20190 \text{ g} \)
Ans:
i. Sum to appropriate number of significant figures = \( 9.2 \times 10^5 \text{ mL} \)
ii. Sum to appropriate number of significant figures = \( -20190 \text{ g} \)
[Note: In addition and subtraction, the final answer is rounded to the minimum number of decimal point of the number taking part in calculation. If there is no decimal point, then the final answer will have no decimal point.]
In simple words: When adding or subtracting measurements, your final answer cannot be more precise than the least precise number you started with. So, we round the final result to match the measurement with the fewest decimal places.

🎯 Exam Tip: Always convert all terms to the same power of 10 before adding or subtracting, and remember to round your final answer based on the term with the least number of decimal places.

4. Solve the Following Problems

 

Question A. Express the following quantities in exponential terms.
a. 0.0003498
Answer:
a. \( 3.498 \times 10^{-4} \)
Moving the decimal point four places to the right gives us a coefficient between 1 and 10 with a negative exponent of 4.
In simple words: To write a very small decimal number in scientific notation, we move the decimal point to the right until there is only one non-zero digit before it, and count the steps to find the negative power of 10.

🎯 Exam Tip: When moving the decimal point to the right for numbers less than 1, the exponent of 10 is always negative.

 

Question A. Express the following numbers in scientific notation:
(a) 0.0003498
(b) 235.4678
(c) 70000.0
(d) 1569.00
Answer:
(a) \( 0.0003498 = 3.498 \times 10^{-4} \)
(b) \( 235.4678 = 2.354678 \times 10^2 \)
(c) \( 70000.0 = 7.00000 \times 10^4 \)
(d) \( 1569.00 = 1.56900 \times 10^3 \)
In simple words: Scientific notation writes very large or very small numbers as a decimal between 1 and 10 multiplied by a power of 10.

🎯 Exam Tip: When moving the decimal point to the right, the exponent is negative; when moving it to the left, the exponent is positive.

 

Question B. Give the number of significant figures in each of the following:
(a) \( 1.230 \times 10^4 \)
(b) 0.002030
(c) \( 1.23 \times 10^4 \)
(d) \( 1.89 \times 10^{-4} \)
Answer:
(a) 4
(b) 4
(c) 3
(d) 3
In simple words: Significant figures are the digits that carry meaning contributing to its measurement resolution. Leading zeros are not significant, but trailing zeros after a decimal point are.

🎯 Exam Tip: Remember that exponential terms (like \( 10^4 \)) do not contribute to the number of significant figures; only look at the numerical coefficient.

 

Question C. Express the quantities in above (B) with or without exponents as the case may be.
Answer:
(a) 12300
(b) \( 2.030 \times 10^{-3} \)
(c) 12300
(d) 0.000189
In simple words: This means converting standard decimal numbers into scientific notation, and scientific notation back into regular decimal numbers.

🎯 Exam Tip: Ensure you maintain the correct number of significant figures when converting between exponential and decimal forms.

 

Question D. Find out the molar masses of the following compounds:
(a) Copper sulphate crystal (\( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \))
(b) Sodium carbonate, decahydrate (\( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \))
(c) Mohr’s salt (\( [\text{FeSO}_4(\text{NH}_4)_2\text{SO}_4 \cdot 6\text{H}_2\text{O}] \))
(At. mass : Cu = 63.5; S = 32; O = 16; H = 1; Na = 23; C = 12; Fe = 56; N = 14)
Answer:
(a) Molar mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \)
\( = (1 \times \text{At. mass Cu}) + (1 \times \text{At. mass S}) + (9 \times \text{At. mass O}) + (10 \times \text{At. mass H}) \)
\( = (1 \times 63.5) + (1 \times 32) + (9 \times 16) + (10 \times 1) \)
\( = 63.5 + 32 + 144 + 10 \)
\( = 249.5 \text{ g/mol} \)

(b) Molar mass of \( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \)
\( = (2 \times \text{At. mass Na}) + (1 \times \text{At. mass C}) + (13 \times \text{At. mass O}) + (20 \times \text{At. mass H}) \)
\( = (2 \times 23) + (1 \times 12) + (13 \times 16) + (20 \times 1) \)
\( = 46 + 12 + 208 + 20 \)
\( = 286 \text{ g/mol} \)

(c) Molar mass of \( [\text{FeSO}_4(\text{NH}_4)_2\text{SO}_4 \cdot 6\text{H}_2\text{O}] \)
\( = (1 \times \text{At. mass Fe}) + (2 \times \text{At. mass S}) + (14 \times \text{At. mass O}) + (2 \times \text{At. mass N}) + (20 \times \text{At. mass H}) \)
\( = (1 \times 56) + (2 \times 32) + (14 \times 16) + (2 \times 14) + (20 \times 1) \)
\( = 56 + 64 + 224 + 28 + 20 \)
\( = 392 \text{ g/mol} \)
In simple words: To find the molar mass, we add up the atomic masses of all the individual atoms present in one formula unit of the compound.

🎯 Exam Tip: Don't forget to multiply the water of crystallization (like \( 5\text{H}_2\text{O} \)) by its coefficient when counting the total number of hydrogen and oxygen atoms.

 

Question. Calculate the molar mass of the following:
(b) \(\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}\)
(c) \([\text{FeSO}_4(\text{NH}_4)_2\text{SO}_4\cdot6\text{H}_2\text{O}]\)
Answer:
(b) Molar mass of \(\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}\)
\( = (2 \times \text{At. mass Na}) + (1 \times \text{At. mass C}) + (13 \times \text{At. mass O}) + (20 \times \text{At. mass H}) \)
\( = (2 \times 23) + (1 \times 12) + (13 \times 16) + (20 \times 1) \)
\( = 46 + 12 + 208 + 20 \)
\( = 286\text{ g mol}^{-1} \)
Molar mass of \(\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O} = 286\text{ g mol}^{-1} \)

(c) Molar mass of \([\text{FeSO}_4(\text{NH}_4)_2\text{SO}_4\cdot6\text{H}_2\text{O}]\)
\( = (1 \times \text{At. mass Fe}) + (2 \times \text{At. mass S}) + (2 \times \text{At. mass N}) + (14 \times \text{At. mass O}) + (20 \times \text{At. mass H}) \)
\( = (1 \times 56) + (2 \times 32) + (2 \times 14) + (14 \times 16) + (20 \times 1) \)
\( = 56 + 64 + 28 + 224 + 20 \)
\( = 392\text{ g mol}^{-1} \)
Molar mass of \([\text{FeSO}_4(\text{NH}_4)_2\text{SO}_4\cdot6\text{H}_2\text{O}] = 392\text{ g mol}^{-1} \)
In simple words: To find the molar mass of a compound, we add up the atomic masses of all the individual atoms present in its chemical formula.

🎯 Exam Tip: Always write down the individual atomic masses first and double-check the number of atoms of each element to avoid simple calculation errors.

 

Question E. Work out the percentage composition of constituents elements in the following compounds:
(a) Lead phosphate \([\text{Pb}_3(\text{PO}_4)_2]\)
(b) Potassium dichromate \((\text{K}_2\text{Cr}_2\text{O}_7)\)
(c) Macrocosmic salt – Sodium ammonium hydrogen phosphate, tetrahydrate \((\text{NaNH}_4\text{HPO}_4\cdot4\text{H}_2\text{O})\)
(At. mass : Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14; H = 1)
Answer:
Given: Atomic mass: Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14; H = 1
To find: The percentage composition of constituent elements
Formula:
\( \text{Percentage (by weight)} = \frac{\text{Mass of the element in 1 mole of compound}}{\text{Molar mass of the compound}} \times 100 \)

Calculation:

(a) Lead phosphate \([\text{Pb}_3(\text{PO}_4)_2]\)
Molar mass of \(\text{Pb}_3(\text{PO}_4)_2 = (3 \times 207) + (2 \times 31) + (8 \times 16) = 621 + 62 + 128 = 811\text{ g mol}^{-1}\)
- % of Pb = \( \frac{621}{811} \times 100 = 76.57\% \)
- % of P = \( \frac{62}{811} \times 100 = 7.64\% \)
- % of O = \( \frac{128}{811} \times 100 = 15.78\% \)

(b) Potassium dichromate \((\text{K}_2\text{Cr}_2\text{O}_7)\)
Molar mass of \(\text{K}_2\text{Cr}_2\text{O}_7 = (2 \times 39) + (2 \times 52) + (7 \times 16) = 78 + 104 + 112 = 294\text{ g mol}^{-1}\)
- % of K = \( \frac{78}{294} \times 100 = 26.53\% \)
- % of Cr = \( \frac{104}{294} \times 100 = 35.37\% \)
- % of O = \( \frac{112}{294} \times 100 = 38.10\% \)

(c) Macrocosmic salt \((\text{NaNH}_4\text{HPO}_4\cdot4\text{H}_2\text{O})\)
Molar mass of \(\text{NaNH}_4\text{HPO}_4\cdot4\text{H}_2\text{O} = (1 \times 23) + (1 \times 14) + (13 \times 1) + (1 \times 31) + (8 \times 16) = 23 + 14 + 13 + 31 + 128 = 209\text{ g mol}^{-1}\)
- % of Na = \( \frac{23}{209} \times 100 = 11.00\% \)
- % of N = \( \frac{14}{209} \times 100 = 6.70\% \)
- % of H = \( \frac{13}{209} \times 100 = 6.22\% \)
- % of P = \( \frac{31}{209} \times 100 = 14.83\% \)
- % of O = \( \frac{128}{209} \times 100 = 61.24\% \)
These calculations are fundamental in analytical chemistry to determine the purity of a sample.
In simple words: To find the percentage of each element in a compound, divide the total mass of that element in one molecule by the total molar mass of the compound, and then multiply by 100.

🎯 Exam Tip: Always ensure that the sum of the percentages of all elements in a compound adds up to approximately 100% to verify your calculations.

Question. Calculate the percentage composition of the following compounds:
(i) Lead phosphate [\(Pb_3(PO_4)_2\)]
(ii) Potassium dichromate [\(K_2Cr_2O_7\)]
(iii) Microcosmic salt [\(NaNH_4HPO_4 \cdot 4H_2O\)]
Answer:
(i) Lead phosphate [\(Pb_3(PO_4)_2\)]:
Molar mass of \(Pb_3(PO_4)_2 = 3 \times (207) + 2 \times (31) + 8 \times (16) = 621 + 62 + 128 = 811\text{ g mol}^{-1}\)
Percentage of Pb = \(\frac{621}{811} \times 100 = 76.57\%\)
Percentage of P = \(\frac{62}{811} \times 100 = 7.64\%\)
Percentage of O = \(\frac{128}{811} \times 100 = 15.78\%\)

(ii) Potassium dichromate [\(K_2Cr_2O_7\)]:
Molar mass of \(K_2Cr_2O_7 = 2 \times (39) + 2 \times (52) + 7 \times (16) = 78 + 104 + 112 = 294\text{ g mol}^{-1}\)
Percentage of K = \(\frac{78}{294} \times 100 = 26.53\%\)
Percentage of Cr = \(\frac{104}{294} \times 100 = 35.37\%\)
Percentage of O = \(\frac{112}{294} \times 100 = 38.10\%\)

(iii) Microcosmic salt [\(NaNH_4HPO_4 \cdot 4H_2O\)]:
Molar mass of \(NaNH_4HPO_4 \cdot 4H_2O = 1 \times (23) + 1 \times (14) + 1 \times (31) + 13 \times (1) + 8 \times (16) = 23 + 14 + 31 + 13 + 128 = 209\text{ g mol}^{-1}\)
Percentage of Na = \(\frac{23}{209} \times 100 = 11.00\%\)
Percentage of N = \(\frac{14}{209} \times 100 = 6.70\%\)
Percentage of P = \(\frac{31}{209} \times 100 = 14.83\%\)
Percentage of H = \(\frac{13}{209} \times 100 = 6.22\%\)
Percentage of O = \(\frac{128}{209} \times 100 = 61.24\%\)

Summary of Answers:
(i) Mass percentage of Pb, P and O in lead phosphate [\(Pb_3(PO_4)_2\)] are \(76.57\%\), \(7.64\%\) and \(15.78\%\) respectively.
(ii) Mass percentage of K, Cr and O in potassium dichromate [\(K_2Cr_2O_7\)] are \(26.53\%\), \(35.37\%\) and \(38.10\%\) respectively.
(iii) Mass percentage of Na, N, P, H and O in \(NaNH_4HPO_4 \cdot 4H_2O\) are \(11.00\%\), \(6.70\%\), \(14.83\%\), \(6.22\%\) and \(61.24\%\) respectively.
In simple words: To find the percentage of each element in a compound, we divide the total mass of that element in one mole of the compound by the total molar mass of the compound, and then multiply by 100.

🎯 Exam Tip: Always double-check your molar mass calculations first, as any mistake there will make all subsequent percentage calculations incorrect.

 

Question F. Find the percentage composition of constituent green vitriol crystals (\(FeSO_4 \cdot 7H_2O\)). 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)
Answer:
Given:
(i) Atomic mass: Fe = 56; S = 32; O = 16; H = 1
(ii) Mass of crystal = 4.54 kg

Step 1: Calculate the molar mass of green vitriol (\(FeSO_4 \cdot 7H_2O\)):
Molar mass of \(FeSO_4 \cdot 7H_2O = 1 \times (\text{At. mass of Fe}) + 1 \times (\text{At. mass of S}) + 11 \times (\text{At. mass of O}) + 14 \times (\text{At. mass of H})\)

\(\implies\) Molar mass = \(56 + 32 + (11 \times 16) + (14 \times 1)\)

\(\implies\) Molar mass = \(56 + 32 + 176 + 14 = 278\text{ g mol}^{-1}\)

Step 2: Calculate the percentage composition of each element:
(i) Percentage of Iron (Fe) = \(\frac{56}{278} \times 100 = 20.14\%\)
(ii) Percentage of Sulphur (S) = \(\frac{32}{278} \times 100 = 11.51\%\)
(iii) Percentage of Hydrogen (H) = \(\frac{14}{278} \times 100 = 5.04\%\)
(iv) Percentage of Oxygen (O) = \(\frac{176}{278} \times 100 = 63.31\%\)
(v) Percentage of water of crystallisation (\(7H_2O\)) = \(\frac{7 \times 18}{278} \times 100 = \frac{126}{278} \times 100 = 45.32\%\)

Step 3: Calculate the mass of iron and water of crystallisation in 4.54 kg of crystals:
(i) Mass of iron (Fe) = \(20.14\%\text{ of } 4.54\text{ kg}\)

\(\implies\) Mass of Fe = \(\frac{20.14}{100} \times 4.54\text{ kg} = 0.914\text{ kg}\)

(ii) Mass of water of crystallisation (\(7H_2O\)) = \(45.32\%\text{ of } 4.54\text{ kg}\)

\(\implies\) Mass of water = \(\frac{45.32}{100} \times 4.54\text{ kg} = 2.058\text{ kg}\)

Thus, the percentage composition of the elements is Fe = 20.14%, S = 11.51%, H = 5.04%, O = 63.31%, and the mass of iron and water of crystallisation in 4.54 kg of crystals are 0.914 kg and 2.058 kg respectively.
In simple words: We first find the total molecular weight of the crystal and the weight of each part. Then, we use these proportions to find the exact weight of iron and water in any given bulk amount of the crystal.

🎯 Exam Tip: Remember to include the water of crystallization (\(7H_2O\)) when calculating the total molar mass of green vitriol, as omitting it is a very common mistake.

 

Question 1. Calculate the mass percentage of Fe, S, H and O in green vitriol (\(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\)) and find the mass of iron and water of crystallisation in 4.54 kg of the crystal.
Answer:
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:
\[ \text{Percentage (by weight)} = \frac{\text{Mass of the element in 1 mole of compound}}{\text{Molar mass of the compound}} \times 100 \]

Calculation:
i. Molar mass of \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O} = 1 \times (56) + 1 \times (32) + 14 \times (1) + 11 \times (16)\)
\(= 56 + 32 + 14 + 176\)
\(= 278\text{ g mol}^{-1}\)
Percentage of \(\text{Fe} = \frac{56}{278} \times 100 = 20.14\%\)
Percentage of \(\text{S} = \frac{32}{278} \times 100 = 11.51\%\)
Percentage of \(\text{H} = \frac{14}{278} \times 100 = 5.04\%\)
Percentage of \(\text{O} = \frac{176}{278} \times 100 = 63.31\%\)

ii. \(278\text{ kg}\) green vitriol \(= 56\text{ kg}\) iron
\(\therefore 4.54\text{ kg}\) green vitriol \(= x\)
\(\therefore x = \frac{56 \times 4.54}{278} = 0.915\text{ kg}\)
Mass of \(\text{7H}_2\text{O}\) in \(278\text{ kg}\) green vitriol \(= 7 \times 18 = 126\text{ kg}\)
\(\therefore 4.54\text{ kg}\) green vitriol \(= y\)
\(\therefore y = \frac{126 \times 4.54}{278} = 2.058\text{ kg}\)

Ans:
i. Mass percentage of Fe, S, H and O in \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) are \(20.14\%\), \(11.51\%\), \(5.04\%\) and \(63.31\%\) respectively.
ii. Mass of iron in \(4.54\text{ kg}\) green vitriol \(= 0.915\text{ kg}\)
Mass of water of crystallisation in \(4.54\text{ kg}\) green vitriol \(= 2.058\text{ kg}\). This calculation helps us understand the exact composition of the hydrated crystal.
In simple words: To find the percentage of each element, we divide its total mass in one molecule by the total molecular mass and multiply by 100. To find the actual mass in a larger sample, we use simple cross-multiplication based on these ratios.

🎯 Exam Tip: Always remember to include the water of crystallisation when calculating the total molar mass of a hydrated salt like green vitriol.

 

Question 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)
Answer:
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 \(= 55.84\text{ u}\)
\(\therefore\) Mass of 4 Fe atoms \(= 55.84 \times 4 = 223.36\text{ u} = 0.335\%\)
Let molecular weight of haemoglobin be \(x\).
Hence,
\(\frac{223.36}{x} \times 100 = 0.335\)

\(\implies x = \frac{223.36 \times 100}{0.335}\)

\(\implies x = 66674.6\text{ u}\)

Ans: The molecular weight of haemoglobin is approximately \(66674.6\text{ u}\) (or \(\text{g mol}^{-1}\)). This high molecular weight reflects the complex, large structure of the protein.
In simple words: Since 4 iron atoms make up only 0.335% of the entire haemoglobin molecule, we can find the total weight of the molecule by setting up a ratio where 223.36 units of mass equals 0.335% of the total.

🎯 Exam Tip: When solving percentage composition problems for large molecules, ensure you multiply the atomic mass of the element by the number of atoms present in one molecule before setting up the ratio.

 

Question 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).
Answer:
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 = \(\frac{\text{\% of Na}}{\text{Atomic mass of Na}} = \frac{43.4}{23} = 1.89\text{ mol}\)
Moles of C = \(\frac{\text{\% of C}}{\text{Atomic mass of C}} = \frac{11.3}{12} = 0.94\text{ mol}\)
Moles of O = \(\frac{\text{\% of O}}{\text{Atomic mass of O}} = \frac{45.3}{16} = 2.83\text{ mol}\)

Hence, the ratio of number of moles of Na:C:O is:
\(\frac{1.89}{0.94} = 2\), \(\frac{0.94}{0.94} = 1\) and \(\frac{2.83}{0.94} = 3\)

Hence, the empirical formula is \(\text{Na}_2\text{CO}_3\). This compound is commonly known as sodium carbonate or washing soda.
Ans: Empirical formula of the compound = \(\text{Na}_2\text{CO}_3\)
In simple words: To find the empirical formula, we calculate the number of moles of each element by dividing their percentage by their atomic mass. Then, we find the simplest whole-number ratio of these moles to get the formula.

🎯 Exam Tip: Always divide the mole values by the smallest mole value obtained to find the simplest ratio. Round off to the nearest whole number if the values are very close to integers.

 

Question I. Assuming the atomic weight of a metal M to be 56, find the empirical formula of its oxide containing 70.0% of M.
Answer:
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 = 100% - 70.0% = 30.0%
Atomic mass of O = 16 u

Moles of M = \(\frac{70.0}{56} = 1.25\text{ mol}\)
Moles of O = \(\frac{30.0}{16} = 1.875\text{ mol}\)

Simplest ratio of M to O:
M = \(\frac{1.25}{1.25} = 1\)
O = \(\frac{1.875}{1.25} = 1.5\)

To convert to whole numbers, multiply by 2:
M = \(1 \times 2 = 2\)
O = \(1.5 \times 2 = 3\)

Hence, the empirical formula of the metal oxide is \(\text{M}_2\text{O}_3\). This metal M is likely iron, and the oxide formed is iron(III) oxide, commonly known as rust.
Ans: Empirical formula of the compound = \(\text{M}_2\text{O}_3\)
In simple words: We find the percentage of oxygen by subtracting the metal's percentage from 100. Then, we calculate the moles of both metal and oxygen, and simplify their ratio to the nearest whole numbers to get the formula.

🎯 Exam Tip: If the simplest ratio contains a decimal like 1.5, multiply all numbers by 2 to get the simplest whole-number ratio instead of rounding 1.5 to 1 or 2.

 

Question 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)
Answer:
Given: Atomic mass of \( \text{Fe} = 56 \), \( \text{S} = 32 \), and \( \text{O} = 16 \)
Mass of iron, sulphur, oxygen and water = \( 0.2014\text{ g} \), \( 0.1153\text{ g} \), \( 0.2301\text{ g} \) and \( 0.4532\text{ g} \) respectively.
To find: The empirical formula of the compound
Calculation: Since the mass of crystal is \( 1\text{ g} \), the percentage of iron, sulphur, oxygen and water is \( 20.14\% \), \( 11.53\% \), \( 23.01\% \) and \( 45.32\% \) respectively. This percentage composition represents the mass of each component in a 100 g sample of the hydrated salt.

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

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

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

\( \text{Moles of water} = \frac{\% \text{ of water}}{\text{Molar mass of water}} = \frac{45.32}{18} = 2.518\text{ mol} \)

Hence, the ratio of number of moles of Fe:S:O:water is:
\( \frac{0.360}{0.360} = 1 \), \( \frac{0.360}{0.360} = 1 \), \( \frac{1.438}{0.360} = 4 \) and \( \frac{2.518}{0.360} = 7 \)

Hence, the empirical formula is \( \text{FeSO}_4\cdot7\text{H}_2\text{O} \).
Ans: Empirical formula of the compound = \( \text{FeSO}_4\cdot7\text{H}_2\text{O} \).
In simple words: To find the empirical formula, we calculate the number of moles of each element and water, and then divide each by the smallest value to get their simplest whole-number ratio.

🎯 Exam Tip: Always remember to use the molar mass of water (18) when calculating the moles of water of crystallisation, and divide all values by the smallest mole value to get the simplest ratio.

Question. A compound 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.
Answer:
Given: Percentage of carbon, hydrogen, nitrogen = 20%, 6.7%, 46.67% respectively.
Molar mass of the compound = \(60 \text{ g mol}^{-1}\)
To find: The molecular formula of the compound
Calculation:
\(\% \text{ carbon} + \% \text{ hydrogen} + \% \text{ nitrogen} = 20 + 6.7 + 46.67 = 73.37\%\)
This is less than 100%. Hence, the compound contains adequate oxygen so that the total percentage of elements is 100%.
Hence, \(\% \text{ of oxygen} = 100 - 73.37 = 26.63\%\)

Moles of C = \(\frac{\% \text{ of C}}{\text{Atomic mass of C}} = \frac{20}{12} = 1.667 \text{ mol}\)
Moles of H = \(\frac{\% \text{ of H}}{\text{Atomic mass of H}} = \frac{6.7}{1.0} = 6.700 \text{ mol}\)
Moles of N = \(\frac{\% \text{ of N}}{\text{Atomic mass of N}} = \frac{46.67}{14} = 3.334 \text{ mol}\)
Moles of O = \(\frac{\% \text{ of O}}{\text{Atomic mass of O}} = \frac{26.63}{16} = 1.664 \text{ mol}\)

Hence, the ratio of number of moles of C:H:N:O is:
\(\frac{1.667}{1.664} \approx 1\), \(\frac{6.700}{1.664} \approx 4\), \(\frac{3.334}{1.664} \approx 2\), and \(\frac{1.664}{1.664} = 1\)

Hence, empirical formula is \(\text{CH}_4\text{N}_2\text{O}\). This compound is commonly known as urea, which is a highly soluble organic compound.
Empirical formula mass = \(12 + 4 + 28 + 16 = 60 \text{ g mol}^{-1}\)
Hence,
Molar mass = Empirical formula mass
\(\therefore\) Molecular formula = Empirical formula = \(\text{CH}_4\text{N}_2\text{O}\)
Ans: Molecular formula of the compound = \(\text{CH}_4\text{N}_2\text{O}\)
In simple words: To find the molecular formula, we first calculate the percentage of oxygen to make the total 100%. Then, we find the ratio of the atoms of each element to get the simplest formula, which matches the molecular mass of 60.

🎯 Exam Tip: Always check if the sum of the given percentages is 100%. If it is less, the remaining percentage is always assumed to be oxygen.

 

Question 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.
Answer:
Given: Percentage of H, O, C = 9.09%, 36.36%, 54.55% respectively.
Molar mass of the compound = \(88 \text{ g mol}^{-1}\)
To find: The molecular formula of the compound
Calculation:
\(\% \text{ of H} + \% \text{ of O} + \% \text{ of C} = 9.09 + 36.36 + 54.55 = 100\%\)
Since the sum is 100%, no other element is present.

Moles of H = \(\frac{\% \text{ of H}}{\text{Atomic mass of H}} = \frac{9.09}{1} = 9.09 \text{ mol}\)
Moles of O = \(\frac{\% \text{ of O}}{\text{Atomic mass of O}} = \frac{36.36}{16} = 2.2725 \text{ mol}\)
Moles of C = \(\frac{\% \text{ of C}}{\text{Atomic mass of C}} = \frac{54.55}{12} = 4.545 \text{ mol}\)

Hence, the ratio of number of moles of C:H:O is:
C = \(\frac{4.545}{2.2725} = 2\)
H = \(\frac{9.09}{2.2725} = 4\)
O = \(\frac{2.2725}{2.2725} = 1\)

Hence, empirical formula is \(\text{C}_2\text{H}_4\text{O}\). This molecular formula corresponds to butyric acid, which gives butter its characteristic odor.
Empirical formula mass = \((2 \times 12) + (4 \times 1) + 16 = 24 + 4 + 16 = 44 \text{ g mol}^{-1}\)
We know that,
\(r = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{88}{44} = 2\)
\(\therefore\) Molecular formula = \(r \times \text{Empirical formula} = 2 \times (\text{C}_2\text{H}_4\text{O}) = \text{C}_4\text{H}_8\text{O}_2\)
Ans: Molecular formula of the compound = \(\text{C}_4\text{H}_8\text{O}_2\)
In simple words: We find the number of moles of each element by dividing their percentage by their atomic mass. Then, we find the simplest ratio to get the empirical formula and multiply it by 2 to match the given molar mass of 88.

🎯 Exam Tip: Remember to divide the molar mass by the empirical formula mass to find the multiplier 'r' before writing the final molecular formula.

 

Question 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 ?
Answer:
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.
\( \therefore \) Molar mass of water = 310 – 124 = 186 g
\( \text{Moles of C} = \frac{\text{Mass of C}}{\text{Atomic mass of C}} = \frac{124}{12} = 10.33 \text{ mol} \)
\( \text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{186}{18} = 10.33 \text{ mol} \)
The ratio of number of moles of C : water = C : \( \text{H}_2\text{O} \) = 1 : 1
Hence, empirical formula = \( \text{CH}_2\text{O} \)
Ans: Empirical formula of the carbohydrate = \( \text{CH}_2\text{O} \). This simple 1:1 ratio of carbon to water molecules is characteristic of many basic monosaccharides.
In simple words: Since the carbohydrate breaks down into carbon and water, we find the moles of both parts. Since they are in a 1:1 ratio, the empirical formula is simply CH₂O.

🎯 Exam Tip: Always write down the 'Given' and 'To find' steps clearly, as they carry step-wise marks in board exams.

 

Question N. Write each of the following in exponential notation :
a. 3,672,199
Answer:
a. \( 3.672199 \times 10^6 \)
To express a number in exponential notation, we move the decimal point to the left so that there is only one non-zero digit to its left, and multiply by 10 raised to the power of the number of places moved.
In simple words: To write a large number in scientific notation, place a decimal after the first digit and count how many places you moved the decimal to find the power of 10.

🎯 Exam Tip: Count the number of digits after the first non-zero digit carefully to determine the correct positive exponent for numbers greater than 1.

 

Question O. Write each of the following numbers in ordinary decimal form:
a. \( 3.49 \times 10^{-11} \)
b. \( 3.75 \times 10^{-1} \)
c. \( 5.16 \times 10^4 \)
d. \( 43.71 \times 10^{-4} \)
e. \( 0.011 \times 10^{-3} \)
f. \( 14.3 \times 10^{-2} \)
g. \( 0.00477 \times 10^5 \)
h. \( 5.00858585 \)
Answer:
a. \( 3.49 \times 10^{-11} = 0.0000000000349 \)
b. \( 3.75 \times 10^{-1} = 0.375 \)
c. \( 5.16 \times 10^4 = 51,600 \)
d. \( 43.71 \times 10^{-4} = 0.004371 \)
e. \( 0.011 \times 10^{-3} = 0.000011 \)
f. \( 14.3 \times 10^{-2} = 0.143 \)
g. \( 0.00477 \times 10^5 = 477 \)
h. \( 5.00858585 = 5.00858585 \)
Converting these numbers helps us visualize their actual scale in standard decimal notation.
In simple words: To write numbers with negative powers of 10 in ordinary form, move the decimal point to the left. For positive powers of 10, move the decimal point to the right.

🎯 Exam Tip: Count the number of places you move the decimal point carefully, especially when dealing with large negative exponents, to avoid missing any zeros.

 

Question P. Perform each of the following calculations. Round off your answers to two digits.
i. \( \frac{1}{3.40 \times 10^{24}} \)
ii. \( \frac{33}{9.00 \times 10^{-4}} \)
iii. \( \frac{1.4 \times 10^9}{(2.77 \times 10^3) \times (3.76 \times 10^5)} \)
iv. \( \frac{(4 \times 10^{-3}) \times (9.9 \times 10^{-7})}{(789) \times (1.002 \times 10^{-10}) \times (0.3 \times 10^2)} \)
Answer:
i. \( \frac{1}{3.40 \times 10^{24}} = 2.9 \times 10^{-25} \)
ii. \( \frac{33}{9.00 \times 10^{-4}} = 3.7 \times 10^4 \)
iii. \( \frac{1.4 \times 10^9}{(2.77 \times 10^3) \times (3.76 \times 10^5)} = 1.3 \)
iv. \( \frac{(4 \times 10^{-3}) \times (9.9 \times 10^{-7})}{(789) \times (1.002 \times 10^{-10}) \times (0.3 \times 10^2)} = 1.7 \times 10^{-3} \)
Rounding to two significant digits ensures that the precision of our final values matches the constraints of the problem.
In simple words: Solve the math step-by-step by grouping the regular numbers together and the powers of 10 together. Finally, round your answer so it only has two non-zero digits.

🎯 Exam Tip: When rounding to two digits, look at the third digit to decide whether to round up or down, and always keep track of your exponents during division.

 

Question Q. Perform each of the following calculations. Round off your answers to three digits.
a. \( (3.26 \times 10^4) (1.54 \times 10^6) \)
b. \( (8.39 \times 10^7) (4.53 \times 10^9) \)
c. \( \frac{8.94 \times 10^6}{4.35 \times 10^4} \)
d. \( \frac{(9.28 \times 10^9) \times (9.9 \times 10^{-7})}{(511) \times (2.98 \times 10^{-6})} \)
Answer:
i. \( (3.26 \times 10^4) (1.54 \times 10^6) = 5.0204 \times 10^{4+6} = 5.02 \times 10^{10} \)
ii. \( (8.39 \times 10^7) (4.53 \times 10^9) = 38.0067 \times 10^{7+9} = 38.0067 \times 10^{16} = 3.80 \times 10^{17} \)
iii. \( \frac{8.94 \times 10^6}{4.35 \times 10^4} = 2.055 \times 10^{6-4} = 2.06 \times 10^2 \)
iv. \( \frac{(9.28 \times 10^9) \times (9.9 \times 10^{-7})}{(511) \times (2.98 \times 10^{-6})} = 0.06033 \times 10^{9-7-(-6)} = 0.06033 \times 10^8 = 6.03 \times 10^6 \)
Rounding off to three significant figures ensures that the precision of the final result matches standard scientific conventions.
In simple words: To multiply numbers in scientific notation, multiply their decimal parts and add their exponents. For division, divide the decimal parts and subtract the bottom exponent from the top one, then round the final answer to three digits.

🎯 Exam Tip: When rounding to three digits, look at the fourth digit; if it is 5 or more, round up the third digit, otherwise keep it the same.

 

Question R. Perform the following operations :
a. \( 3.971 \times 10^7 + 1.98 \times 10^4 \)
b. \( 1.05 \times 10^{-4} - 9.7 \times 10^{-5} \)
c. \( 4.11 \times 10^{-3} + 8.1 \times 10^{-4} \)
d. \( 2.12 \times 10^6 - 3.5 \times 10^5 \)
Answer:
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.
a. \( 3.971 \times 10^7 + 1.98 \times 10^4 = 3.971 \times 10^7 + 0.00198 \times 10^7 = (3.971 + 0.00198) \times 10^7 = 3.97298 \times 10^7 \)
b. \( 1.05 \times 10^{-4} - 9.7 \times 10^{-5} = 1.05 \times 10^{-4} - 0.97 \times 10^{-4} = (1.05 - 0.97) \times 10^{-4} = 0.08 \times 10^{-4} = 8.0 \times 10^{-6} \)
c. \( 4.11 \times 10^{-3} + 8.1 \times 10^{-4} = 4.11 \times 10^{-3} + 0.81 \times 10^{-3} = (4.11 + 0.81) \times 10^{-3} = 4.92 \times 10^{-3} \)
d. \( 2.12 \times 10^6 - 3.5 \times 10^5 = 2.12 \times 10^6 - 0.35 \times 10^6 = (2.12 - 0.35) \times 10^6 = 1.77 \times 10^6 \)
Adjusting the decimal point allows us to align the place values of the coefficients before performing basic arithmetic.
In simple words: To add or subtract numbers in scientific notation, you must first make their exponents the same. Once the exponents match, simply add or subtract the front numbers and keep the exponent part as it is.

🎯 Exam Tip: Always convert the number with the smaller exponent to match the larger exponent, as this usually makes the final decimal adjustment easier and less prone to errors.

 

Question 5. 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?
Answer:
Precision:
To find the observed mass of acetone, we subtract the mass of the empty bottle from the mass of the acetone-filled bottle for each measurement:

Measurement	Mass of acetone observed (g)
1	38.7798 − 38.0015 = 0.7783
2	38.7795 − 38.0015 = 0.7780
3	38.7801 − 38.0015 = 0.7786

Now, we calculate the mean value of these measurements:
\( \text{Mean} = \frac{0.7783 + 0.7780 + 0.7786}{3} = 0.7783 \text{ g} \)

Next, we determine the absolute deviation for each measurement:

Measurement	Mass of acetone observed (g)	Absolute deviation (g) = | Observed value − Mean |
1	0.7783	0
2	0.7780	0.0003
3	0.7786	0.0003

Conclusion:
1. Precision: The individual measurements (0.7783 g, 0.7780 g, and 0.7786 g) are extremely close to each other, with a very small average absolute deviation of 0.0002 g. Therefore, the measurements have high precision.
2. Accuracy: The mean value of the measurements (0.7783 g) is slightly lower than the actual mass of the acetone (0.7791 g). Since there is a small difference between the mean value and the true value, the measurements are moderately accurate.
In simple words: Precision is about how close your measurements are to each other, while accuracy is about how close they are to the real value. Here, the measurements are very close to each other (high precision) but slightly off from the true weight of 0.7791 g (moderate accuracy).

🎯 Exam Tip: Always show the step-by-step subtraction to find the observed mass of the sample before calculating the mean and absolute deviation to secure full marks.

 

Question 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.
Answer:
a. The accepted measurement is 12.6 cm. 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 the metre stick is 0.1 cm. This means the measurement can only be accurate and reliable up to one decimal place when expressed in centimetres.
b. Reasons for rejecting the other measurements:
(i) 12.65 cm and 12.655 cm: These measurements have two and three decimal places respectively, which is beyond the least count (0.1 cm) of the metre stick.
(ii) 126.55 mm: Since the least count in millimetres is 1 mm, the measurement cannot have decimal fractions of a millimetre.
(iii) 12 cm: This measurement does not show the precision up to the least count of the instrument, as it should be written as 12.0 cm to show the correct precision.
In simple words: A standard metre stick can only measure as small as 1 millimetre (which is 0.1 centimetres). Any correct reading must match this limit, making 12.6 cm the only correct measurement because it has exactly one decimal place.

🎯 Exam Tip: Always identify the least count of the measuring device first. Your recorded values must always have the exact number of decimal places allowed by that least count to score full marks.

 

Question 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)
Answer:
Given: Mass of \( \text{CaCO}_3 \) consumed in reaction = \( 19.3\text{ g} \)
To find: Mass of \( \text{CaO} \) formed
Calculation: Calcium carbonate decomposes according to the balanced chemical equation:
\( \text{CaCO}_3 \xrightarrow{\Delta} \text{CaO} + \text{CO}_2\uparrow \)
Molar Mass of \( \text{CaCO}_3 = 40 + 12 + (3 \times 16) = 100\text{ g/mol} \) (100 parts)
Molar Mass of \( \text{CaO} = 40 + 16 = 56\text{ g/mol} \) (56 parts)
Molar Mass of \( \text{CO}_2 = 12 + (2 \times 16) = 44\text{ g/mol} \) (44 parts)
This thermal decomposition reaction demonstrates the law of conservation of mass in action.
So, \( 100\text{ g} \) of \( \text{CaCO}_3 \) produces \( 56\text{ g} \) of \( \text{CaO} \).
Therefore, \( 19.3\text{ g} \) of \( \text{CaCO}_3 \) will produce = \( \frac{56\text{ g}}{100\text{ g}} \times 19.3\text{ g} = 10.81\text{ g} \) of \( \text{CaO} \).
Ans: Mass of \( \text{CaO} \) formed = \( 10.81\text{ g} \)

[Calculation using log table:
\( 56 \times 0.193 \)
\( = \text{Antilog}_{10} [\log_{10}(56) + \log_{10}(0.193)] \)
\( = \text{Antilog}_{10} [1.7482 + \bar{1}.2856] \)
\( = \text{Antilog}_{10} [1.0338] = 10.81 \)]
In simple words: When we heat calcium carbonate, it breaks down into calcium oxide and carbon dioxide gas. By using their molecular weights, we can calculate that heating 19.3 grams of calcium carbonate will give us exactly 10.81 grams of calcium oxide.

🎯 Exam Tip: Always write down the balanced chemical equation first and calculate the molar masses carefully to avoid calculation errors in stoichiometry problems.

 

Question 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?
Answer:
Given:
Hourly requirement of sucrose = \( 34\text{ g} \)
Time period = \( 1\text{ day} = 24\text{ hours} \)
To find: Mass of oxygen (\( \text{O}_2 \)) required for one day
Calculation:
First, let us find the total mass of sucrose required for one day (24 hours):
Total sucrose required = \( 34\text{ g/hour} \times 24\text{ hours} = 816\text{ g} \)
The balanced chemical equation for the combustion of sucrose (\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)) is:
\( \text{C}_{12}\text{H}_{22}\text{O}_{11} + 12\text{O}_2 \rightarrow 12\text{CO}_2 + 11\text{H}_2\text{O} \)
Molar mass of sucrose (\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \)) = \( (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342\text{ g/mol} \)
Molar mass of \( 12 \) moles of \( \text{O}_2 = 12 \times (2 \times 16) = 12 \times 32 = 384\text{ g/mol} \)
This metabolic process provides the vital energy needed to sustain life functions in microgravity environments.
From the balanced equation, \( 342\text{ g} \) of sucrose requires \( 384\text{ g} \) of \( \text{O}_2 \).
Therefore, \( 816\text{ g} \) of sucrose will require = \( \frac{384\text{ g}}{342\text{ g}} \times 816\text{ g} = 916.21\text{ g} \) of \( \text{O}_2 \).
Ans: The mass of oxygen needed to be carried in the space capsule is \( 916.21\text{ g} \).
In simple words: An astronaut needs 34 grams of sugar (sucrose) every hour, which equals 816 grams for a whole day. By using the balanced chemical equation for burning sugar, we find that the astronaut will need 916.21 grams of oxygen to burn all that sugar and get energy for one day.

🎯 Exam Tip: Remember to convert the hourly requirement to a daily requirement (multiply by 24) before calculating the stoichiometric amount of oxygen needed.

 

Question. Calculate the mass of oxygen an astronaut needs to carry per day if 34 g of sucrose provides energy for an hour.
Answer: 34 g of sucrose provides energy for an hour.
Hence, for a day, the mass of sucrose needed = \( 34 \times 24 = 816\text{ g} \).
The balanced chemical equation is:
\[ \text{C}_{12}\text{H}_{22}\text{O}_{11(s)} + 12\text{O}_{2(g)} \longrightarrow 12\text{CO}_{2(g)} + 11\text{H}_2\text{O}_{(l)} \]
\( 342\text{ g} \) of sucrose requires \( 12 \times 32 = 384\text{ g} \) of oxygen.
Thus, 342 g of sucrose require 384 g of oxygen.

\( \implies \) \( 816\text{ g} \) of sucrose will require = \( \frac{816}{342} \times 384 = 916\text{ g of O}_2 \).
Therefore, the astronaut needs to carry 916 g of \( \text{O}_2 \) to sustain themselves for a full day.
In simple words: First, we calculate that the astronaut needs 816 grams of sucrose for a whole day. By using the balanced chemical equation, we find that 816 grams of sucrose reacts with exactly 916 grams of oxygen.

🎯 Exam Tip: Always write down the balanced chemical equation first and show the molar mass calculations clearly to secure full marks.