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Chapter 1 Some Basic Concepts Of Chemistry
Importance Of Chemistry
Chemistry is fundamentally the study of molecules and their transformations. It explores the vast number of molecules that can be created from a limited set of elements.
Science aims to organise knowledge to understand nature. We observe many chemical changes daily, such as milk turning into curd, sugarcane juice fermenting into vinegar, and iron rusting. Chemistry is a branch of science focused on the preparation, properties, structure, and reactions of material substances.
The field of chemistry has a rich history, although its modern form is relatively recent. Early studies, known as Alchemy and Iatrochemistry (around 1300-1600 CE), were often driven by the search for specific goals:
- A substance called the 'Philosopher’s stone', believed to turn common metals like iron and copper into gold.
- An 'Elixir of life' that was thought to grant immortality.
Ancient India had significant knowledge of chemical phenomena, applying it in areas like metallurgy, medicine, cosmetics, glass, and dyes, often referred to as Rasayan Shastra or Rasvidya. Archaeological evidence from sites like Mohenjodaro and Harappa indicates advanced practices, such as using baked bricks (an early chemical process involving mixing, moulding, and heating materials) and producing glazed pottery, faience (a type of glass for ornaments), and metal objects (melting, forging, and alloying copper with tin or arsenic for hardness). Texts like Rigveda mention leather tanning and cotton dyeing. Kautilya's Arthashastra described salt production.
Ancient Indian texts like Sushruta Samhita discussed the use of alkalies. Charaka Samhita mentioned the knowledge of preparing acids (sulphuric, nitric) and various metal oxides, sulphates, and carbonates. Rasopanishada and Tamil texts described gunpowder and fireworks preparation.
Notable Indian figures include Nagarjuna, a chemist and metallurgist whose work 'Rasratnakar' dealt with mercury compounds and metal extraction. 'Rsarnavam' (around 800 CE) described furnaces, crucibles, and methods for identifying metals by flame colour. Chakrapani is credited with discovering mercury sulphide and inventing soap using mustard oil and alkalies.
The lasting colours of Ajanta and Ellora paintings highlight the level of chemical science achieved. Varähmihir's Brihat Samhita mentioned materials for walls/roofs from plant extracts and resins, and later texts detailed dye stuffs and recipes for perfumes and cosmetics.
Paper making and ink use were also known in India centuries ago. Fermentation processes were well-understood, used for preparing various liquors and medicinal concoctions ('Asavas').
Ancient Indian philosophical thought included ideas about matter being made of indivisible particles. Acharya Kanda (around 600 BCE) proposed the concept of 'Paramãnu' (comparable to atoms), eternal, indestructible, and in motion, thousands of years before Dalton. He suggested different types of Paramãnu forming combinations, driven by unseen forces.
Charaka Samhita also touched upon reducing particle size of metals for medicinal use (bhasma), concepts now related to nanotechnology, as modern studies show bhasmas often contain metal nanoparticles.
Modern chemistry emerged in India in the late 19th century, influenced by European scientists, leading to a gradual decline in traditional indigenous techniques as foreign products and methods became prevalent.
Today, chemistry is crucial, interacting with many other sciences. It helps us understand and interact with the world around us, from natural phenomena like weather to technological advancements like computers and new materials. Chemistry contributes significantly to the economy through chemical industries producing essential goods like fertilisers, acids, bases, polymers, drugs, soaps, and metals.
It improves the quality of life by providing solutions for food production (fertilisers, pesticides), healthcare (life-saving drugs like cisplatin, taxol for cancer, AZT for AIDS), and materials with specific properties (superconductors, optical fibers). Chemistry also plays a vital role in addressing environmental challenges, such as developing safer alternatives to harmful substances like CFCs. However, major issues like managing greenhouse gases remain significant challenges for chemists.
Understanding the fundamental concepts of chemistry, starting with the nature of matter, is essential for tackling future challenges and contributing to national development, particularly for developing countries like India that need skilled chemists.
Nature Of Matter
Matter is defined as anything that has mass and occupies space. Everything we see and interact with, such as books, air, water, and living organisms, is composed of matter.
States Of Matter
Matter exists primarily in three physical states: solid, liquid, and gas.
The arrangement of constituent particles differs significantly in these states:
- In solids, particles are tightly packed in an ordered structure with very limited freedom to move.
- In liquids, particles are still close but can move past each other, resulting in a less rigid arrangement.
- In gases, particles are far apart and move randomly and rapidly throughout the available space.
This difference in particle arrangement leads to distinct characteristics:
- Solids: Have a definite volume and a definite shape.
- Liquids: Have a definite volume but take the shape of their container.
- Gases: Have neither a definite volume nor a definite shape, expanding to fill their container completely.
These three states can be interconverted by changing temperature and pressure. For instance, heating a solid usually forms a liquid (melting), and further heating produces a gas (boiling/vaporisation). Conversely, cooling a gas produces a liquid (condensation), and further cooling results in a solid (freezing).
Classification Of Matter
At a larger scale, matter can be classified as either a mixture or a pure substance.
A pure substance consists of particles that are all chemically identical. It has a fixed composition.
A mixture contains particles of two or more pure substances combined in any ratio. Therefore, mixtures have a variable composition. The pure substances forming a mixture are called its components, and they retain their individual properties within the mixture.
Examples of mixtures include sugar dissolved in water, air, and tea. Components of a mixture can often be separated using physical methods like filtration, distillation, or hand-picking.
Mixtures are further divided into two types:
- Homogeneous mixtures: Components are uniformly distributed throughout the mixture, and the composition is consistent everywhere. Examples are sugar solution and air.
- Heterogeneous mixtures: Composition is not uniform, and sometimes different components can be seen separately. Examples include mixtures of salt and sugar, or grains mixed with dirt.
Pure substances are classified into elements and compounds.
An element is composed of only one type of atom. These atoms can exist individually or bonded together to form molecules.
For example, sodium (Na) and copper (Cu) consist of individual atoms. Gases like hydrogen (H$_2$), nitrogen (N$_2$), and oxygen (O$_2$) exist as diatomic molecules (two atoms bonded together).
A compound is formed when atoms of different elements combine chemically in a definite fixed ratio. Unlike mixtures, the components of a compound cannot be separated by physical methods; chemical methods are required. A key characteristic is that the properties of a compound are completely different from the properties of its constituent elements.
For instance, water (H$_2$O) is a compound where two hydrogen atoms are bonded to one oxygen atom. Its properties (liquid at room temperature, used to extinguish fires) are very different from those of hydrogen (a flammable gas) and oxygen (a gas that supports combustion).
Properties Of Matter And Their Measurement
Every substance possesses unique characteristics or properties. These properties are categorised into two types:
Physical And Chemical Properties
- Physical properties are those that can be measured or observed without altering the substance's chemical identity or composition. Examples include colour, odour, melting point, boiling point, and density.
- Chemical properties are observed during or after a chemical reaction, meaning the substance's composition changes during the measurement. These include characteristics like combustibility, acidity, basicity, and reactivity with other substances.
Chemists rely on understanding and measuring both types of properties through careful experimentation to describe, interpret, and predict how substances will behave.
Measurement Of Physical Properties
Quantitative measurement is fundamental in scientific investigation. Properties like length, volume, and mass are quantitative. Any quantitative measurement consists of a numerical value followed by the unit in which it was measured (e.g., 6 meters, where 6 is the number and meter is the unit).
Historically, different measurement systems existed (e.g., English and Metric). The Metric System, based on decimals, gained popularity. The need for a globally consistent system led to the establishment of the International System of Units.
The International System Of Units (SI)
The SI system, officially 'Le Système International d’Unités', was established in 1960 by the General Conference on Weights and Measures (CGPM). It is the modern form of the metric system and is used worldwide in science, technology, and commerce.
The SI system is based on seven base units corresponding to seven fundamental physical quantities. All other physical quantities can be derived from these base units.
| Base Physical Quantity | Symbol for Quantity | Name of SI Unit | Symbol for SI Unit |
|---|---|---|---|
| Length | $l$ | metre | m |
| Mass | $m$ | kilogram | kg |
| Time | $t$ | second | s |
| Electric current | $I$ | ampere | A |
| Thermodynamic temperature | $T$ | kelvin | K |
| Amount of substance | $n$ | mole | mol |
| Luminous intensity | $I_v$ | candela | cd |
Each SI base unit has a precise definition, often based on fundamental physical constants. For example, the metre is defined based on the speed of light, the kilogram on the Planck constant, and the second on the frequency of caesium-133 atom's transition.
| Unit | Definition |
| Metre (m) | Defined by taking the fixed numerical value of the speed of light in vacuum ($c$) as $299792458$ when expressed in m s$^{-1}$, relative to the definition of the second. |
| Kilogram (kg) | Defined by taking the fixed numerical value of the Planck constant ($h$) as $6.62607015 \times 10^{-34}$ when expressed in J s (kg m$^2$ s$^{-1}$), relative to the definitions of the metre and the second. |
| Second (s) | Defined by taking the fixed numerical value of the caesium frequency ($\Delta v_{\text{Cs}}$), the unperturbed ground-state hyperfine transition frequency of the caesium-133 atom, as $9192631770$ when expressed in Hz (s$^{-1}$). |
| Ampere (A) | Defined by taking the fixed numerical value of the elementary charge ($e$) as $1.602176634 \times 10^{-19}$ when expressed in C (A s), relative to the definition of the second. |
| Kelvin (K) | Defined by taking the fixed numerical value of the Boltzmann constant ($k$) as $1.380649 \times 10^{-23}$ when expressed in J K$^{-1}$ (kg m$^2$ s$^{-2}$ K$^{-1}$), relative to the definitions of the kilogram, metre, and second. |
| Mole (mol) | Defined as containing exactly $6.02214076 \times 10^{23}$ elementary entities (Avogadro constant, $N_A$). Amount of substance ($n$) is a measure of the number of specified elementary entities (atom, molecule, ion, etc.). |
| Candela (cd) | Defined by taking the fixed numerical value of the luminous efficacy of monochromatic radiation of frequency $540 \times 10^{12}$ Hz, $K_{\text{cd}}$, to be $683$ when expressed in lm W$^{-1}$ (cd sr W$^{-1}$ or cd sr kg$^{-1}$ m$^{-2}$ s$^3$), relative to the definitions of the kilogram, metre, and second. |
Maintaining accurate measurement standards is crucial. National Metrology Institutes (NMIs) like India's National Physical Laboratory (NPL) are responsible for realising and maintaining these standards, comparing them internationally.
SI units use prefixes to denote multiples or submultiples of the base units, simplifying the representation of very large or small quantities.
| Multiple | Prefix | Symbol |
| $10^{-24}$ | yocto | y |
| $10^{-21}$ | zepto | z |
| $10^{-18}$ | atto | a |
| $10^{-15}$ | femto | f |
| $10^{-12}$ | pico | p |
| $10^{-9}$ | nano | n |
| $10^{-6}$ | micro | $\mu$ |
| $10^{-3}$ | milli | m |
| $10^{-2}$ | centi | c |
| $10^{-1}$ | deci | d |
| $10^{1}$ | deca | da |
| $10^{2}$ | hecto | h |
| $10^{3}$ | kilo | k |
| $10^{6}$ | mega | M |
| $10^{9}$ | giga | G |
| $10^{12}$ | tera | T |
| $10^{15}$ | peta | P |
| $10^{18}$ | exa | E |
| $10^{21}$ | zeta | Z |
| $10^{24}$ | yotta | Y |
Mass And Weight
Mass is an intrinsic property of an object, representing the quantity of matter it contains. Weight, on the other hand, is the force exerted on an object by gravity. Mass is constant regardless of location, while weight can vary depending on the local gravitational acceleration.
The SI unit for mass is the kilogram (kg). In chemistry labs, smaller quantities are often measured, making the gram (g) a more convenient unit ($1 \text{ kg} = 1000 \text{ g}$). Accurate mass measurements are typically done using an analytical balance.
Volume
Volume is the amount of three-dimensional space occupied by a substance. Its units are derived from length, typically (length)$^3$. The SI unit of volume is the cubic meter (m$^3$).
For laboratory work, smaller units are more practical, such as cubic centimeters (cm$^3$) or cubic decimeters (dm$^3$). A common unit for liquids is the litre (L), which is not an SI base unit but is widely used and defined as $1 \text{ L} = 1000 \text{ mL} = 1000 \text{ cm}^3 = 1 \text{ dm}^3$.
Various pieces of glassware are used in the lab to measure liquid volumes, including graduated cylinders, burettes, and pipettes. Volumetric flasks are used for preparing solutions of precise volumes.
Density
Density is a physical property that relates a substance's mass to its volume. It is defined as mass per unit volume:
$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$
The SI unit of density is kilograms per cubic meter (kg m$^{-3}$). More commonly in chemistry, grams per cubic centimeter (g cm$^{-3}$) or grams per milliliter (g mL$^{-1}$) are used.
Density provides insight into how tightly packed the particles of a substance are; a higher density indicates closer packing.
Temperature
Temperature is a measure of the average kinetic energy of the particles in a substance. It is commonly measured using three scales:
- Celsius ($^\circ$C): Water freezes at $0^\circ$C and boils at $100^\circ$C (at standard pressure).
- Fahrenheit ($^\circ$F): Water freezes at $32^\circ$F and boils at $212^\circ$F.
- Kelvin (K): This is the SI unit for thermodynamic temperature. The Kelvin scale is an absolute scale, meaning $0$ K (absolute zero) is the theoretically lowest possible temperature where particle motion effectively stops. Negative temperatures are possible in Celsius and Fahrenheit, but not in Kelvin.
Conversions between scales are done using these formulas:
$^\circ\text{F} = (\frac{9}{5} \times ^\circ\text{C}) + 32$
$\text{K} = ^\circ\text{C} + 273.15$
Uncertainty In Measurement
All experimental measurements have some degree of uncertainty. This is due to limitations in measuring instruments and the skill of the person taking the measurement. For example, a measurement of 9.4 g on a simple balance might be refined to 9.4213 g on a more precise analytical balance, indicating that the digit '4' in the first measurement was uncertain.
Uncertainty is typically represented by the concept of significant figures and using scientific notation for handling very large or small numbers precisely.
Scientific Notation
In chemistry, dealing with the masses and numbers of atoms and molecules involves extremely large or small numbers. For example, the mass of a hydrogen atom is about $0.00000000000000000000000166$ g, and the number of molecules in a few grams of hydrogen is enormous.
To manage these numbers conveniently, especially in calculations, scientific notation (or exponential notation) is used. A number in scientific notation is expressed in the form $N \times 10^n$, where:
- $N$ is the digit term, a number between $1.000...$ and $9.999...$.
- $n$ is the exponent, an integer (positive or negative).
To convert a number to scientific notation, move the decimal point until only one non-zero digit is to its left. The number of places the decimal point is moved determines the exponent ($n$). Moving left gives a positive exponent; moving right gives a negative exponent.
Examples:
- $232.508 = 2.32508 \times 10^2$ (decimal moved 2 places left, $n=2$)
- $0.00016 = 1.6 \times 10^{-4}$ (decimal moved 4 places right, $n=-4$)
Mathematical operations in scientific notation:
- Multiplication/Division: Multiply/divide the digit terms ($N$ values) and add/subtract the exponents of 10. * $(5.6 \times 10^5) \times (6.9 \times 10^8) = (5.6 \times 6.9) \times 10^{5+8} = 38.64 \times 10^{13} = 3.864 \times 10^{14}$ (adjust N to be between 1 and 10) * $(9.8 \times 10^{-2}) \div (2.5 \times 10^{-6}) = (9.8 \div 2.5) \times 10^{-2 - (-6)} = 3.92 \times 10^4$
- Addition/Subtraction: First, adjust the numbers so they have the same exponent ($10^n$). Then, add or subtract the digit terms ($N$ values). * $(6.65 \times 10^4) + (8.95 \times 10^3) = (6.65 \times 10^4) + (0.895 \times 10^4) = (6.65 + 0.895) \times 10^4 = 7.545 \times 10^4$ * $(2.5 \times 10^{-2}) - (4.8 \times 10^{-3}) = (2.5 \times 10^{-2}) - (0.48 \times 10^{-2}) = (2.5 - 0.48) \times 10^{-2} = 2.02 \times 10^{-2}$
Significant Figures
Significant figures are the meaningful digits in a measured or calculated quantity. They include all digits known with certainty plus one estimated or uncertain digit. The uncertainty is considered to be $\pm 1$ in the last significant digit unless otherwise stated.
Rules for determining the number of significant figures:
- All non-zero digits are significant. (e.g., 285 cm has 3 significant figures; 0.25 mL has 2).
- Zeros preceding the first non-zero digit are NOT significant. They only indicate the decimal point's position. (e.g., 0.03 has 1 significant figure; 0.0052 has 2).
- Zeros between two non-zero digits ARE significant. (e.g., 2.005 has 4 significant figures).
- Zeros at the end or to the right of a number ARE significant if they are to the right of the decimal point. (e.g., 0.200 g has 3 significant figures). If there is no decimal point, trailing zeros may or may not be significant; this ambiguity is avoided by using scientific notation. (e.g., 100 could have 1 significant figure ($1 \times 10^2$), 2 ($1.0 \times 10^2$), or 3 ($1.00 \times 10^2$)).
- Counting numbers or exact quantities have infinite significant figures. (e.g., 2 balls, 20 eggs, conversion factors like 1 m = 100 cm).
In numbers written in scientific notation ($N \times 10^n$), all digits in the digit term ($N$) are significant.
It is important to distinguish between precision and accuracy:
- Precision refers to the closeness of multiple measurements of the same quantity to each other (reproducibility).
- Accuracy refers to how close a measured value is to the true or accepted value.
Measurements can be precise but not accurate, accurate but not precise (less common), or both precise and accurate. Ideally, scientific measurements are both precise and accurate.
| Measurement 1 (g) | Measurement 2 (g) | Average (g) | |
|---|---|---|---|
| Student A | 1.95 | 1.93 | 1.940 |
| Student B | 1.94 | 2.05 | 1.995 |
| Student C | 2.01 | 1.99 | 2.000 |
Assuming the true value is 2.00 g: Student A's measurements are precise (close to each other) but not accurate (far from 2.00 g). Student B's are neither precise nor accurate. Student C's are both precise and accurate (close to each other and close to 2.00 g).
Rules for reporting results based on significant figures in calculations:
- Addition and Subtraction: The result should have no more decimal places than the number with the fewest decimal places among the original numbers. * Example: $12.11 + 18.0 + 1.012 = 31.122$. Since 18.0 has only one decimal place, the result is rounded to 31.1.
- Multiplication and Division: The result should have no more significant figures than the number with the fewest significant figures among the original numbers. * Example: $2.5 \times 1.25 = 3.125$. Since 2.5 has 2 significant figures (the fewest), the result is rounded to 3.1.
Rules for Rounding Off numbers to the required number of significant figures:
- If the digit to be removed is greater than 5, the preceding digit is increased by one. (e.g., 1.386 rounded to 3 significant figures becomes 1.39).
- If the digit to be removed is less than 5, the preceding digit is unchanged. (e.g., 4.334 rounded to 3 significant figures becomes 4.33).
- If the digit to be removed is exactly 5: * If the preceding digit is even, it is left unchanged. (e.g., 6.25 rounded to 2 significant figures becomes 6.2). * If the preceding digit is odd, it is increased by one. (e.g., 6.35 rounded to 2 significant figures becomes 6.4).
Dimensional Analysis
Dimensional analysis, also known as the factor label method or unit factor method, is a technique used to convert a quantity from one system of units to another. It is based on the principle of multiplying the given quantity by one or more "unit factors".
A unit factor is a ratio derived from an equivalence between two different units (e.g., 1 inch = 2.54 cm). This equivalence can be written as two unit factors: $\frac{1 \text{ in}}{2.54 \text{ cm}}$ and $\frac{2.54 \text{ cm}}{1 \text{ in}}$. Both ratios equal 1 because the numerator and denominator represent the same magnitude. Multiplying a quantity by a unit factor changes its units but not its value.
To perform a conversion, select the unit factor that cancels the original unit and introduces the desired unit. Units are treated algebraically, like numbers (they can be multiplied, divided, and cancelled).
Example 1. A piece of metal is 3 inch (represented by in) long. What is its length in cm?
Answer:
We know the equivalence: 1 in = 2.54 cm.
The unit factors are $\frac{1 \text{ in}}{2.54 \text{ cm}}$ and $\frac{2.54 \text{ cm}}{1 \text{ in}}$.
To convert inches to centimeters, we need the unit factor with cm in the numerator and in in the denominator to cancel the original 'in' unit:
Length in cm = $3 \text{ in} \times \frac{2.54 \text{ cm}}{1 \text{ in}}$
Cancel the 'in' units:
Length in cm = $3 \times 2.54 \text{ cm}$
Length in cm = $7.62 \text{ cm}$
Example 2. A jug contains 2 L of milk. Calculate the volume of the milk in m$^3$.
Answer:
We know the equivalences: 1 L = 1000 cm$^3$ and 1 m = 100 cm.
From 1 m = 100 cm, we get unit factors $\frac{1 \text{ m}}{100 \text{ cm}}$ and $\frac{100 \text{ cm}}{1 \text{ m}}$.
To convert cm$^3$ to m$^3$, we need to cube the unit factor: $(\frac{1 \text{ m}}{100 \text{ cm}})^3 = \frac{1^3 \text{ m}^3}{100^3 \text{ cm}^3} = \frac{1 \text{ m}^3}{1,000,000 \text{ cm}^3} = \frac{1 \text{ m}^3}{10^6 \text{ cm}^3}$.
First, convert L to cm$^3$:
Volume in cm$^3$ = $2 \text{ L} \times \frac{1000 \text{ cm}^3}{1 \text{ L}} = 2000 \text{ cm}^3$.
Now, convert cm$^3$ to m$^3$ using the cubed unit factor:
Volume in m$^3$ = $2000 \text{ cm}^3 \times \frac{1 \text{ m}^3}{10^6 \text{ cm}^3}$
Cancel the cm$^3$ units:
Volume in m$^3$ = $2000 \times \frac{1}{10^6} \text{ m}^3 = 2000 \times 10^{-6} \text{ m}^3 = 2 \times 10^3 \times 10^{-6} \text{ m}^3 = 2 \times 10^{-3} \text{ m}^3$.
Example 3. How many seconds are there in 2 days?
Answer:
We use a series of unit factors based on known equivalences:
- 1 day = 24 hours (h) $\implies \frac{24 \text{ h}}{1 \text{ day}}$
- 1 h = 60 minutes (min) $\implies \frac{60 \text{ min}}{1 \text{ h}}$
- 1 min = 60 seconds (s) $\implies \frac{60 \text{ s}}{1 \text{ min}}$
Start with the given quantity (2 days) and multiply by the unit factors sequentially, arranging them so units cancel:
Time in seconds = $2 \text{ days} \times \frac{24 \text{ h}}{1 \text{ day}} \times \frac{60 \text{ min}}{1 \text{ h}} \times \frac{60 \text{ s}}{1 \text{ min}}$
Cancel the units (days, h, min):
Time in seconds = $2 \times 24 \times 60 \times 60 \text{ s}$
Time in seconds = $172800 \text{ s}$.
Laws Of Chemical Combinations
The ways in which elements combine to form compounds are governed by several fundamental laws.
Law Of Conservation Of Mass
Proposed by Antoine Lavoisier in 1789 based on careful experiments, this law states:
Matter can neither be created nor destroyed in a chemical reaction.
This means the total mass of the reactants before a chemical reaction is always equal to the total mass of the products after the reaction. Lavoisier's use of precise mass measurements was crucial in establishing this law, which became a cornerstone for future chemical developments.
Law Of Definite Proportions
Formulated by French chemist Joseph Proust, this law states:
A given chemical compound always contains its component elements in a fixed ratio by mass, regardless of its source or method of preparation.
Proust demonstrated this by analyzing samples of cupric carbonate from different sources (natural and synthetic), finding the elemental composition by mass was identical in both cases.
| Sample | % of Copper | % of Carbon | % of Oxygen |
|---|---|---|---|
| Natural Sample | 51.35 | 9.74 | 38.91 |
| Synthetic Sample | 51.35 | 9.74 | 38.91 |
This law highlights that a specific compound always has a unique and consistent elemental composition by mass.
Law Of Multiple Proportions
This law was also proposed by John Dalton in 1803, building upon the previous laws. It states:
If two elements can combine to form more than one compound, then the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.
Consider hydrogen and oxygen, which can form water (H$_2$O) and hydrogen peroxide (H$_2$O$_2$).
- In water: 2 g of hydrogen combine with 16 g of oxygen.
- In hydrogen peroxide: 2 g of hydrogen combine with 32 g of oxygen.
Fixing the mass of hydrogen at 2 g, the masses of oxygen that combine are 16 g and 32 g. The ratio of these oxygen masses (16:32) simplifies to a small whole number ratio of 1:2. This supports the idea that elements combine in specific, often simple, proportions.
Gay Lussac’s Law Of Gaseous Volumes
Discovered by Joseph Louis Gay Lussac in 1808, this law specifically applies to reactions involving gases:
When gases react together, they do so in volumes that bear a simple whole number ratio to one another, and to the volume of the gaseous products (if any), provided that all volumes are measured at the same temperature and pressure.
For example, when hydrogen gas reacts with oxygen gas to form water vapour:
Hydrogen (g) + Oxygen (g) $\rightarrow$ Water (g)
100 mL of hydrogen reacts with 50 mL of oxygen to produce 100 mL of water vapour.
The volumes of hydrogen and oxygen that react (100 mL and 50 mL) are in a simple ratio of 2:1. Gay Lussac's observation of these integer volume ratios was a significant step, later explained by Avogadro's hypothesis.
Avogadro’s Law
Proposed by Amedeo Avogadro in 1811, this law provided a crucial link between the volume and the number of particles of gases:
Equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules.
Avogadro was the first to clearly distinguish between atoms and molecules. His law explained Gay Lussac's observations if one assumes that simple gases like hydrogen and oxygen exist as diatomic molecules (H$_2$, O$_2$) rather than individual atoms (H, O).
Applying Avogadro's Law to the reaction $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(g)}$:
If 2 volumes of H$_2$ react with 1 volume of O$_2$ to give 2 volumes of H$_2$O, then according to Avogadro's law, 2 molecules of H$_2$ react with 1 molecule of O$_2$ to give 2 molecules of H$_2$O. This supports the diatomic nature of H$_2$ and O$_2$.
Despite its correctness, Avogadro's hypothesis wasn't widely accepted until decades later when Stanislao Cannizaro presented it at a chemistry conference in 1860, highlighting its significance.
Dalton’s Atomic Theory
Building on the laws of chemical combination and the ancient Greek idea of indivisible particles ('atomos'), John Dalton proposed his atomic theory in 1808 in "A New System of Chemical Philosophy".
The main postulates of Dalton's atomic theory are:
- Matter is made up of very small, indivisible particles called atoms.
- All atoms of a specific element are identical in mass and properties. Atoms of different elements have different masses and properties.
- Compounds are formed when atoms of different elements combine in a fixed, simple whole-number ratio.
- Chemical reactions involve the reorganisation of atoms. Atoms are neither created nor destroyed during a chemical reaction (consistent with the Law of Conservation of Mass).
Dalton's theory successfully explained the Laws of Conservation of Mass, Definite Proportions, and Multiple Proportions. However, it had limitations, including its inability to explain Gay Lussac's Law of Gaseous Volumes and providing no reason for why atoms combine.
Atomic And Molecular Masses
Atoms and molecules are incredibly small, so their individual masses are also very small. While modern techniques like mass spectrometry allow for accurate determination of atomic masses, historical methods relied on comparing the masses of atoms relative to a chosen standard.
Atomic Mass
Initially, the lightest atom, hydrogen, was used as a reference. The current standard for atomic mass, adopted in 1961, is the carbon-12 ($^{12}$C) isotope. By convention, one atom of carbon-12 is assigned a mass of exactly 12 atomic mass units (amu).
The atomic mass unit (amu) is defined as exactly one-twelfth ($1/12$) the mass of one atom of carbon-12. All other atomic masses are determined relative to this standard.
$1 \text{ amu} = 1.66056 \times 10^{-24} \text{ g}$
For example, the mass of a hydrogen atom is approximately $1.6736 \times 10^{-24}$ g. In amu, this is $\frac{1.6736 \times 10^{-24} \text{ g}}{1.66056 \times 10^{-24} \text{ g/amu}} \approx 1.0080$ amu.
Today, 'amu' is often replaced by the symbol 'u', standing for unified mass.
When we use atomic masses in calculations for elements found naturally, we typically use the average atomic mass.
Average Atomic Mass
Many elements exist naturally as a mixture of isotopes (atoms of the same element with different numbers of neutrons, hence different masses). The average atomic mass of an element is a weighted average of the atomic masses of its naturally occurring isotopes, taking into account their relative abundances.
For example, carbon has three main isotopes: $^{12}$C, $^{13}$C, and $^{14}$C. Their natural abundances and masses are:
| Isotope | Relative Abundance (%) | Atomic Mass (amu) |
|---|---|---|
| $^{12}$C | 98.892 | 12.00000 |
| $^{13}$C | 1.108 | 13.00335 |
| $^{14}$C | $2 \times 10^{-10}$ | 14.00317 |
The average atomic mass of carbon is calculated as:
$(0.98892 \times 12.00000 \text{ u}) + (0.01108 \times 13.00335 \text{ u}) + (2 \times 10^{-12} \times 14.00317 \text{ u}) \approx 12.011 \text{ u}$.
The atomic masses listed in the periodic table are these average atomic masses.
Molecular Mass
For substances that exist as discrete molecules, the molecular mass is the sum of the atomic masses of all the atoms present in one molecule. It is calculated by adding the atomic masses of each element multiplied by the number of atoms of that element in the molecular formula.
Example: Molecular mass of methane (CH$_4$)
CH$_4$ contains 1 carbon atom and 4 hydrogen atoms.
Molecular mass of CH$_4$ = $(1 \times \text{Atomic mass of C}) + (4 \times \text{Atomic mass of H})$
= $(1 \times 12.011 \text{ u}) + (4 \times 1.008 \text{ u})$
= $12.011 \text{ u} + 4.032 \text{ u} = 16.043 \text{ u}$.
Example: Molecular mass of water (H$_2$O)
H$_2$O contains 2 hydrogen atoms and 1 oxygen atom.
Molecular mass of H$_2$O = $(2 \times \text{Atomic mass of H}) + (1 \times \text{Atomic mass of O})$
= $(2 \times 1.008 \text{ u}) + (1 \times 16.00 \text{ u})$
= $2.016 \text{ u} + 16.00 \text{ u} = 18.016 \text{ u} \approx 18.02 \text{ u}$.
Example 1.1. Calculate the molecular mass of glucose (C$_6$H$_{12}$O$_6$) molecule.
Answer:
The molecular formula for glucose is C$_6$H$_{12}$O$_6$.
To calculate the molecular mass, sum the atomic masses of 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
Atomic mass of C $\approx$ 12.011 u
Atomic mass of H $\approx$ 1.008 u
Atomic mass of O $\approx$ 16.00 u
Molecular mass of C$_6$H$_{12}$O$_6$ = $(6 \times \text{Atomic mass of C}) + (12 \times \text{Atomic mass of H}) + (6 \times \text{Atomic mass of O})$
= $(6 \times 12.011 \text{ u}) + (12 \times 1.008 \text{ u}) + (6 \times 16.00 \text{ u})$
= $72.066 \text{ u} + 12.096 \text{ u} + 96.00 \text{ u}$
= $180.162 \text{ u}$.
Formula Mass
Some substances, particularly ionic compounds like sodium chloride (NaCl), do not exist as discrete molecules. Instead, they consist of a repeating three-dimensional arrangement of positive and negative ions (e.g., Na$^+$ and Cl$^-$ ions).
In such cases, the formula (like NaCl) represents the simplest whole-number ratio of the ions in the structure. We calculate the formula mass instead of molecular mass by summing the atomic masses of the atoms in the formula unit.
Example: Formula mass of sodium chloride (NaCl)
Formula mass of NaCl = $\text{Atomic mass of Na} + \text{Atomic mass of Cl}$
= $22.99 \text{ u} + 35.45 \text{ u}$ (using rounded average atomic masses)
= $58.44 \text{ u}$. (The text uses 23.0 u and 35.5 u giving 58.5 u, which is also acceptable depending on precision needed).
Mole Concept And Molar Masses
Atoms and molecules are extremely small, and even a tiny amount of substance contains a vast number of these particles. To work with such large quantities in a convenient way, chemists use the concept of the mole.
Just as we use terms like 'dozen' (12) or 'gross' (144) to count everyday objects, the mole is a unit used to count entities at the microscopic level (atoms, molecules, ions, electrons, etc.).
In the SI system, the mole (symbol mol) is the seventh base quantity for the amount of substance.
The definition of the mole is tied to the Avogadro constant: One mole contains exactly $6.02214076 \times 10^{23}$ elementary entities. This number is the fixed numerical value of the Avogadro constant ($N_A$) when expressed in the unit mol$^{-1}$.
So, $N_A \approx 6.022 \times 10^{23} \text{ mol}^{-1}$.
To grasp the scale of this number, written out it is $602,214,076,000,000,000,000,000$.
Therefore:
- 1 mol of hydrogen atoms = $6.022 \times 10^{23}$ hydrogen atoms.
- 1 mol of water molecules = $6.022 \times 10^{23}$ water molecules.
- 1 mol of sodium chloride = $6.022 \times 10^{23}$ formula units of sodium chloride.
The number of entities in 1 mole is constant, regardless of the substance.
The value of the Avogadro constant was determined by experiments, such as measuring the mass of a carbon-12 atom ($1.992648 \times 10^{-23}$ g) and knowing that one mole of carbon-12 has a mass of 12 g. The number of atoms in 12 g of $^{12}$C is $\frac{12 \text{ g/mol}}{1.992648 \times 10^{-23} \text{ g/atom}} \approx 6.022 \times 10^{23} \text{ atoms/mol}$.
The molar mass of a substance is the mass of one mole of that substance, expressed in grams. Numerically, the molar mass in grams per mole (g/mol) is equal to the atomic mass, molecular mass, or formula mass in unified atomic mass units (u).
Examples:
- Molar mass of hydrogen atoms = 1.008 u $\implies$ Molar mass of H = 1.008 g/mol.
- Molecular mass of water = 18.02 u $\implies$ Molar mass of H$_2$O = 18.02 g/mol.
- Formula mass of sodium chloride = 58.44 u $\implies$ Molar mass of NaCl = 58.44 g/mol.
Molar mass provides a crucial link between the mass of a substance and the number of particles it contains (via the mole concept).
Percentage Composition
Often, it is useful to know the proportion of each element within a compound. This is expressed as the percentage composition by mass. Knowing the percentage composition can help determine the formula of an unknown compound or verify the purity of a known one.
The mass percentage of an element in a compound is calculated using the formula:
$\text{Mass \% of element} = \frac{\text{Mass of that element in one mole of the compound}}{\text{Molar mass of the compound}} \times 100$
Example: Percentage composition of water (H$_2$O).
Molar mass of H$_2$O = 18.02 g/mol.
Mass of hydrogen in 1 mol H$_2$O = 2 mol H $\times$ 1.008 g/mol = 2.016 g.
Mass of oxygen in 1 mol H$_2$O = 1 mol O $\times$ 16.00 g/mol = 16.00 g.
Mass % of hydrogen = $\frac{2.016 \text{ g}}{18.02 \text{ g}} \times 100 \approx 11.19\%$. (The text rounds differently, getting 11.18%)
Mass % of oxygen = $\frac{16.00 \text{ g}}{18.02 \text{ g}} \times 100 \approx 88.79\%$.
Example: Percentage composition of ethanol (C$_2$H$_5$OH).
Molecular formula is C$_2$H$_6$O.
Molar mass of ethanol = $(2 \times 12.011 \text{ g/mol}) + (6 \times 1.008 \text{ g/mol}) + (1 \times 16.00 \text{ g/mol}) = 24.022 + 6.048 + 16.00 = 46.070 \text{ g/mol}$. (The text uses 46.068 g)
Mass % of carbon = $\frac{(2 \times 12.011) \text{ g}}{46.070 \text{ g}} \times 100 = \frac{24.022 \text{ g}}{46.070 \text{ g}} \times 100 \approx 52.15\%$. (The text gets 52.14%)
Mass % of hydrogen = $\frac{(6 \times 1.008) \text{ g}}{46.070 \text{ g}} \times 100 = \frac{6.048 \text{ g}}{46.070 \text{ g}} \times 100 \approx 13.13\%$.
Mass % of oxygen = $\frac{16.00 \text{ g}}{46.070 \text{ g}} \times 100 \approx 34.73\%$.
(Sum of percentages should be close to 100%, differences due to rounding).
Empirical Formula For Molecular Formula
The empirical formula provides the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in a molecule of the compound.
If the mass percentage composition of a compound is known, its empirical formula can be determined. If the molar mass of the compound is also known, the molecular formula can then be calculated.
Problem 1.2. A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?
Answer:
Step 1: Convert mass per cent to grams.
Assume we have a 100 g sample of the compound. The masses of the elements are:
- Hydrogen: 4.07 g
- Carbon: 24.27 g
- Chlorine: 71.65 g
Step 2: Convert mass of each element to moles.
Divide the mass of each element by its atomic mass.
- Moles of Hydrogen: $\frac{4.07 \text{ g}}{1.008 \text{ g/mol}} \approx 4.04 \text{ mol}$
- Moles of Carbon: $\frac{24.27 \text{ g}}{12.01 \text{ g/mol}} \approx 2.021 \text{ mol}$
- Moles of Chlorine: $\frac{71.65 \text{ g}}{35.45 \text{ g/mol}} \approx 2.021 \text{ mol}$
Step 3: Determine the simplest whole-number ratio of moles.
Divide the number of moles of each element by the smallest number of moles calculated in Step 2 (which is 2.021 mol).
- Ratio of H: $\frac{4.04}{2.021} \approx 2.00$
- Ratio of C: $\frac{2.021}{2.021} = 1.00$
- Ratio of Cl: $\frac{2.021}{2.021} = 1.00$
The mole ratio H:C:Cl is approximately 2:1:1. If the ratios are not exact whole numbers, multiply by a small integer to get whole numbers.
Step 4: Write the empirical formula.
The empirical formula is CH$_2$Cl (using the whole-number ratios as subscripts).
Step 5: Determine the molecular formula.
(a) Calculate the empirical formula mass.
Empirical formula mass of CH$_2$Cl = $(1 \times \text{Atomic mass of C}) + (2 \times \text{Atomic mass of H}) + (1 \times \text{Atomic mass of Cl})$
= $(1 \times 12.01 \text{ u}) + (2 \times 1.008 \text{ u}) + (1 \times 35.45 \text{ u})$
= $12.01 + 2.016 + 35.45 = 49.476 \text{ u}$. (Text gets 49.48 g)
(b) Determine the factor ($n$) by which the empirical formula mass must be multiplied to get the molar mass.
$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$
$n = \frac{98.96 \text{ g}}{49.476 \text{ g}} \approx 2.000$
So, $n = 2$.
(c) Multiply the empirical formula by $n$ to get the molecular formula.
Molecular formula = (Empirical formula)$_n$ = (CH$_2$Cl)$_2$ = C$_2$H$_4$Cl$_2$.
Thus, the empirical formula is CH$_2$Cl and the molecular formula is C$_2$H$_4$Cl$_2$.
Stoichiometry And Stoichiometric Calculations
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. The word comes from Greek words meaning 'element' and 'measure'.
A balanced chemical equation is essential for stoichiometric calculations because it provides the relative number of moles (and molecules) of reactants and products involved in the reaction. The numbers in front of the chemical formulas are called stoichiometric coefficients.
For example, consider the combustion of methane:
CH$_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$
In this equation, CH$_4$ and O$_2$ are reactants, and CO$_2$ and H$_2$O are products. The coefficients are 1 for CH$_4$ and CO$_2$, and 2 for O$_2$ and H$_2$O. These coefficients represent the relative number of molecules or moles:
- 1 molecule of CH$_4$ reacts with 2 molecules of O$_2$ to produce 1 molecule of CO$_2$ and 2 molecules of H$_2$O.
- 1 mole of CH$_4$ reacts with 2 moles of O$_2$ to produce 1 mole of CO$_2$ and 2 moles of H$_2$O.
For gases at the same temperature and pressure, the coefficients also represent relative volumes (based on Avogadro's law):
- 1 volume of CH$_4$ reacts with 2 volumes of O$_2$ to produce 1 volume of CO$_2$ and 2 volumes of H$_2$O. (e.g., 22.7 L of CH$_4$ reacts with 45.4 L of O$_2$ to give 22.7 L of CO$_2$ and 45.4 L of H$_2$O at standard conditions).
Using molar masses (mass of 1 mole), we can also relate masses:
- 16 g of CH$_4$ (molar mass 16.04 g/mol) reacts with 2 $\times$ 32 g of O$_2$ (molar mass 31.998 g/mol) to give 44 g of CO$_2$ (molar mass 44.01 g/mol) and 2 $\times$ 18 g of H$_2$O (molar mass 18.016 g/mol). Note: Using approximate molar masses from the text gives 16g CH4, 2x32g O2, 44g CO2, 2x18g H2O. The sum of reactant masses ($16 + 64 = 80$ g) equals the sum of product masses ($44 + 36 = 80$ g), illustrating the conservation of mass.
Stoichiometric calculations use these relationships (mole-to-mole, mass-to-mass, mass-to-mole, mole-to-volume, etc.) to determine the amount of one substance given the amount of another in a reaction.
It's important that the chemical equation is balanced before performing stoichiometric calculations. Balancing ensures the number of atoms of each element is the same on both sides of the equation, satisfying the Law of Conservation of Mass. Balancing is typically done by adjusting the stoichiometric coefficients, not the subscripts within the chemical formulas.
Problem 1.3. Calculate the amount of water (g) produced by the combustion of 16 g of methane.
Answer:
The balanced equation for the combustion of methane is:
CH$_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$
From the periodic table (or given masses): Molar mass of CH$_4 \approx 16.04$ g/mol, Molar mass of H$_2$O $\approx 18.02$ g/mol.
Step 1: Convert the mass of methane to moles.
Moles of CH$_4 = \frac{\text{Mass of CH}_4}{\text{Molar mass of CH}_4} = \frac{16 \text{ g}}{16.04 \text{ g/mol}} \approx 0.9975 \text{ mol}$. (The text uses 16g for 1 mol CH4, implying using atomic mass 12 for C and 1 for H, making molar mass 16 g/mol. Let's follow the text's simplification for this problem).
Using Molar mass of CH$_4 = 16$ g/mol (as used in the text's example explanation):
Moles of CH$_4 = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mol}$.
Step 2: Use the stoichiometry from the balanced equation to find moles of water produced.
The equation shows that 1 mol of CH$_4$ produces 2 mol of H$_2$O.
Moles of H$_2$O produced = $1 \text{ mol CH}_4 \times \frac{2 \text{ mol H}_2\text{O}}{1 \text{ mol CH}_4} = 2 \text{ mol H}_2\text{O}$.
Step 3: Convert moles of water to grams.
Molar mass of H$_2$O $\approx 18.02$ g/mol. (Text uses 2+16=18 g/mol)
Using Molar mass of H$_2$O = 18 g/mol (as used in the text's example explanation):
Mass of H$_2$O = Moles of H$_2$O $\times$ Molar mass of H$_2$O
Mass of H$_2$O = $2 \text{ mol} \times 18 \text{ g/mol} = 36 \text{ g}$.
Therefore, 36 g of water is produced.
Problem 1.4. How many moles of methane are required to produce 22g CO$_2$ (g) after combustion?
Answer:
The balanced equation is:
CH$_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$
From the periodic table: Molar mass of CO$_2 = 12.01 + 2 \times 16.00 = 44.01$ g/mol. (Text uses 44g for 1 mol CO2, implying atomic masses 12 for C and 16 for O). We'll use 44 g/mol.
Step 1: Convert the mass of CO$_2$ to moles.
Moles of CO$_2 = \frac{\text{Mass of CO}_2}{\text{Molar mass of CO}_2} = \frac{22 \text{ g}}{44 \text{ g/mol}} = 0.5 \text{ mol}$.
Step 2: Use the stoichiometry from the balanced equation to find moles of methane required.
The equation shows that 1 mol of CH$_4$ produces 1 mol of CO$_2$.
Moles of CH$_4$ required = $0.5 \text{ mol CO}_2 \times \frac{1 \text{ mol CH}_4}{1 \text{ mol CO}_2} = 0.5 \text{ mol CH}_4$.
Therefore, 0.5 mol of methane is required.
Limiting Reagent
In real-world experiments, reactants are often not mixed in the exact stoichiometric ratios required by the balanced equation. In such cases, one reactant will be completely consumed before the others. This reactant is called the limiting reagent (or limiting reactant).
The limiting reagent determines the maximum amount of product that can be formed. The other reactants are present in excess.
Identifying the limiting reagent is crucial for accurately predicting product yields in reactions where reactants are not in stoichiometric proportions.
Problem 1.5. 50.0 kg of N$_2$ (g) and 10.0 kg of H$_2$ (g) are mixed to produce NH$_3$ (g). Calculate the amount of NH$_3$ (g) formed. Identify the limiting reagent in the production of NH$_3$ in this situation.
Answer:
The balanced equation for the synthesis of ammonia is:
N$_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$
Molar mass of N$_2 \approx 28.0$ g/mol (as used in the text). Molar mass of H$_2 \approx 2.016$ g/mol (as used in the text). Molar mass of NH$_3 \approx 17.0$ g/mol (as used in the text).
Step 1: Convert masses of reactants to moles.
- Moles of N$_2 = \frac{50.0 \text{ kg} \times 1000 \text{ g/kg}}{28.0 \text{ g/mol}} = \frac{50000 \text{ g}}{28.0 \text{ g/mol}} \approx 1785.7 \text{ mol} = 1.786 \times 10^3 \text{ mol}$
- Moles of H$_2 = \frac{10.0 \text{ kg} \times 1000 \text{ g/kg}}{2.016 \text{ g/mol}} = \frac{10000 \text{ g}}{2.016 \text{ g/mol}} \approx 4960.3 \text{ mol} = 4.960 \times 10^3 \text{ mol}$
Step 2: Identify the limiting reagent.
From the balanced equation, 1 mol of N$_2$ reacts with 3 mol of H$_2$.
To react completely with $1.786 \times 10^3$ mol of N$_2$, the required moles of H$_2$ would be:
Required H$_2$ = $1.786 \times 10^3 \text{ mol N}_2 \times \frac{3 \text{ mol H}_2}{1 \text{ mol N}_2} = 5.358 \times 10^3 \text{ mol H}_2$.
We only have $4.960 \times 10^3$ mol of H$_2$, which is less than the required amount ($5.358 \times 10^3$ mol). Therefore, H$_2$ is the limiting reagent, and N$_2$ is in excess.
Step 3: Calculate the amount of NH$_3$ formed based on the limiting reagent (H$_2$).
From the balanced equation, 3 mol of H$_2$ produces 2 mol of NH$_3$.
Moles of NH$_3$ produced = $4.960 \times 10^3 \text{ mol H}_2 \times \frac{2 \text{ mol NH}_3}{3 \text{ mol H}_2} \approx 3.307 \times 10^3 \text{ mol NH}_3$. (Text rounds to $3.30 \times 10^3$ mol)
Step 4: Convert moles of NH$_3$ to grams or kilograms.
Mass of NH$_3$ = Moles of NH$_3 \times$ Molar mass of NH$_3$
Mass of NH$_3 = 3.307 \times 10^3 \text{ mol} \times 17.0 \text{ g/mol} \approx 56220 \text{ g}$
Mass of NH$_3 \approx 56.22 \text{ kg}$. (Text gets 56.1 kg due to rounding)
Therefore, approximately 56.1 kg of NH$_3$ is formed, and H$_2$ is the limiting reagent.
To find the mass of the reactant in excess that remains unreacted:
Step 1: Calculate the moles of N$_2$ that reacted with the limiting reagent (H$_2$).
From the equation, 3 mol H$_2$ reacts with 1 mol N$_2$.
Moles of N$_2$ reacted = $4.960 \times 10^3 \text{ mol H}_2 \times \frac{1 \text{ mol N}_2}{3 \text{ mol H}_2} \approx 1.653 \times 10^3 \text{ mol N}_2$.
Step 2: Calculate the moles of N$_2$ remaining.
Moles of N$_2$ initial = $1.786 \times 10^3 \text{ mol}$.
Moles of N$_2$ remaining = Moles initial - Moles reacted
= $1.786 \times 10^3 \text{ mol} - 1.653 \times 10^3 \text{ mol} = 0.133 \times 10^3 \text{ mol} = 133 \text{ mol}$.
Step 3: Convert moles of N$_2$ remaining to grams.
Molar mass of N$_2 \approx 28.0$ g/mol.
Mass of N$_2$ remaining = $133 \text{ mol} \times 28.0 \text{ g/mol} \approx 3724 \text{ g} \approx 3.724 \text{ kg}$.
So, about 3.724 kg of dinitrogen remains unreacted.
Reactions In Solutions
Most chemical reactions in a laboratory setting are carried out in solutions. Therefore, it is important to express the amount of substance (solute) present in a given volume or mass of the solution (or solvent).
The concentration of a solution can be expressed in several ways:
1. Mass per cent or Weight per cent (w/w %)
This expresses the mass of the solute as a percentage of the total mass of the solution.
$\text{Mass \% of solute} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$
Note that the mass of the solution is the sum of the mass of the solute and the mass of the solvent.
Problem 1.6. A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute.
Answer:
Mass of solute (substance A) = 2 g
Mass of solvent (water) = 18 g
Mass of solution = Mass of solute + Mass of solvent = 2 g + 18 g = 20 g.
Mass per cent of solute = $\frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$
= $\frac{2 \text{ g}}{20 \text{ g}} \times 100 = \frac{1}{10} \times 100 = 10 \%$.
The mass per cent of the solute is 10%.
2. Mole Fraction
Mole fraction ($X$) is the ratio of the number of moles of a particular component (solute or solvent) to the total number of moles of all components in the solution.
If a solution contains component A (with $n_A$ moles) and component B (with $n_B$ moles), the mole fractions are:
$X_A = \frac{n_A}{n_A + n_B}$
$X_B = \frac{n_B}{n_A + n_B}$
The sum of the mole fractions of all components in a solution is always equal to 1 ($X_A + X_B = 1$). Mole fraction is a dimensionless quantity.
3. Molarity (M)
Molarity is one of the most common units of concentration. It is defined as the number of moles of solute dissolved in one litre of the solution.
$\text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$
The unit of molarity is mol/L or M.
Molarity is useful for reactions carried out in solution, as volumes are easily measured. However, molarity is temperature-dependent because the volume of a solution changes with temperature.
When preparing a solution of desired molarity from a more concentrated 'stock' solution, the dilution formula $M_1 V_1 = M_2 V_2$ is often used, where $M_1$ and $V_1$ are the molarity and volume of the stock solution, and $M_2$ and $V_2$ are the molarity and volume of the diluted solution. This formula is based on the fact that the number of moles of solute remains constant during dilution ($n = M \times V$).
Problem 1.7. Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution.
Answer:
Molar mass of NaOH = Atomic mass of Na + Atomic mass of O + Atomic mass of H
= 22.99 g/mol + 16.00 g/mol + 1.008 g/mol $\approx$ 40.00 g/mol. (Using approximate values, 23 + 16 + 1 = 40 g/mol)
Step 1: Calculate the number of moles of NaOH solute.
Moles of NaOH = $\frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{4 \text{ g}}{40.00 \text{ g/mol}} = 0.10 \text{ mol}$.
Step 2: Convert the volume of the solution to litres.
Volume of solution = 250 mL = $250 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.250 \text{ L}$.
Step 3: Calculate the molarity.
Molarity (M) = $\frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}} = \frac{0.10 \text{ mol}}{0.250 \text{ L}} = 0.40 \text{ M}$.
The molarity of the NaOH solution is 0.40 M.
4. Molality (m)
Molality is defined as the number of moles of solute dissolved in one kilogram of the solvent.
$\text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}$
The unit of molality is mol/kg or m.
Molality is advantageous because it is temperature-independent, as mass does not change with temperature, unlike volume.
Problem 1.8. The density of 3 M solution of NaCl is 1.25 g mL$^{-1}$. Calculate the molality of the solution.
Answer:
Given: Molarity (M) = 3 mol L$^{-1}$, Density of solution = 1.25 g mL$^{-1}$.
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol. (Text uses 58.5 g/mol).
Step 1: Find the mass of solute in a specific volume of solution (e.g., 1 L).
A 3 M NaCl solution contains 3 moles of NaCl per litre of solution.
Mass of NaCl in 1 L solution = Moles of NaCl $\times$ Molar mass of NaCl
= 3 mol $\times$ 58.5 g/mol = 175.5 g.
Step 2: Find the mass of the solution (1 L).
Volume of solution = 1 L = 1000 mL.
Mass of solution = Density $\times$ Volume = 1.25 g/mL $\times$ 1000 mL = 1250 g.
Step 3: Find the mass of the solvent (water).
Mass of solution = Mass of solute + Mass of solvent
Mass of solvent = Mass of solution - Mass of solute = 1250 g - 175.5 g = 1074.5 g.
Step 4: Convert the mass of the solvent to kilograms.
Mass of solvent = 1074.5 g $\times \frac{1 \text{ kg}}{1000 \text{ g}} = 1.0745 \text{ kg}$.
Step 5: Calculate the molality.
Molality (m) = $\frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{3 \text{ mol}}{1.0745 \text{ kg}} \approx 2.792 \text{ mol/kg}$.
Rounding to three significant figures as per typical density measurements might give 2.79 m.
The molality of the solution is approximately 2.79 m.
Exercises
Question 1.1 Calculate the molar mass of the following:
(i) $H_2O$
(ii) $CO_2$
(iii) $CH_4$
Answer:
Question 1.2 Calculate the mass per cent of different elements present in sodium sulphate ($Na_2SO_4$).
Answer:
Question 1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.
Answer:
Question 1.4 Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Answer:
Question 1.5 Calculate the mass of sodium acetate ($CH_3COONa$) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is $82.0245 \ g \ mol^{-1}$.
Answer:
Question 1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, $1.41 \ g \ mL^{-1}$ and the mass per cent of nitric acid in it being 69%.
Answer:
Question 1.7 How much copper can be obtained from 100 g of copper sulphate ($CuSO_4$)?
Answer:
Question 1.8 Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.
Answer:
Question 1.9 Calculate the atomic mass (average) of chlorine using the following data:
| Isotope | % Natural Abundance | Molar Mass |
|---|---|---|
| $^{35}Cl$ | 75.77 | 34.9689 |
| $^{37}Cl$ | 24.23 | 36.9659 |
Answer:
Question 1.10 In three moles of ethane ($C_2H_6$), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Answer:
Question 1.11 What is the concentration of sugar ($C_{12}H_{22}O_{11}$) in $mol \ L^{-1}$ if its 20 g are dissolved in enough water to make a final volume up to 2L?
Answer:
Question 1.12 If the density of methanol is $0.793 \ kg \ L^{-1}$, what is its volume needed for making 2.5 L of its 0.25 M solution?
Answer:
Question 1.13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
$1Pa = 1N \ m^{-2}$
If mass of air at sea level is $1034 \ g \ cm^{-2}$, calculate the pressure in pascal.
Answer:
Question 1.14 What is the SI unit of mass? How is it defined?
Answer:
Question 1.15 Match the following prefixes with their multiples:
| Prefixes | Multiples |
|---|---|
| (i) micro | $10^6$ |
| (ii) deca | $10^9$ |
| (iii) mega | $10^{-6}$ |
| (iv) giga | $10^{-15}$ |
| (v) femto | 10 |
Answer:
Question 1.16 What do you mean by significant figures?
Answer:
Question 1.17 A sample of drinking water was found to be severely contaminated with chloroform, $CHCl_3$, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in per cent by mass.
(ii) Determine the molality of chloroform in the water sample.
Answer:
Question 1.18 Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Answer:
Question 1.19 How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034
Answer:
Question 1.20 Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Answer:
Question 1.21 The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
| Mass of dinitrogen | Mass of dioxygen | |
|---|---|---|
| (i) | 14 g | 16 g |
| (ii) | 14 g | 32 g |
| (iii) | 28 g | 32 g |
| (iv) | 28 g | 80 g |
(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = ...................... mm = ...................... pm
(ii) 1 mg = ...................... kg = ...................... ng
(iii) 1 mL = ...................... L = ...................... $dm^3$
Answer:
Question 1.22 If the speed of light is $3.0 \times 10^8 \ m \ s^{-1}$, calculate the distance covered by light in 2.00 ns.
Answer:
Question 1.23 In a reaction
$A + B_2 \rightarrow AB_2$
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Answer:
Question 1.24 Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
$N_2 (g) + H_2 (g) \rightarrow 2NH_3 (g)$
(i) Calculate the mass of ammonia produced if $2.00 \times 10^3$ g dinitrogen reacts with $1.00 \times 10^3$ g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Answer:
Question 1.25 How are 0.50 mol $Na_2CO_3$ and 0.50 M $Na_2CO_3$ different?
Answer:
Question 1.26 If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:
Question 1.27 Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
Answer:
Question 1.28 Which one of the following will have the largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of $Cl_2(g)$
Answer:
Question 1.29 Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
Answer:
Question 1.30 What will be the mass of one $^{12}C$ atom in g?
Answer:
Question 1.31 How many significant figures should be present in the answer of the following calculations?
(i) $\frac{0.02856 \times 298.15 \times 0.112}{0.5785}$
(ii) $5 \times 5.364$
(iii) $0.0125 + 0.7864 + 0.0215$
Answer:
Question 1.32 Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes:
| Isotope | Isotopic molar mass | Abundance |
|---|---|---|
| $^{36}Ar$ | $35.96755 \ g \ mol^{-1}$ | 0.337% |
| $^{38}Ar$ | $37.96272 \ g \ mol^{-1}$ | 0.063% |
| $^{40}Ar$ | $39.9624 \ g \ mol^{-1}$ | 99.600% |
Answer:
Question 1.33 Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.
Answer:
Question 1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
Answer:
Question 1.35 Calcium carbonate reacts with aqueous HCl to give $CaCl_2$ and $CO_2$ according to the reaction, $CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2(g) + H_2O(l)$
What mass of $CaCO_3$ is required to react completely with 25 mL of 0.75 M HCl?
Answer:
Question 1.36 Chlorine is prepared in the laboratory by treating manganese dioxide ($MnO_2$) with aqueous hydrochloric acid according to the reaction
$4 HCl (aq) + MnO_2(s) \rightarrow 2H_2O (l) + MnCl_2(aq) + Cl_2 (g)$
How many grams of HCl react with 5.0 g of manganese dioxide?
Answer: