Chemistry Notes PDF

# Some Basic Concept Of Chemistry ## Matter - Anything that occupies space and has mass is called matter. - Matter can be classified as: - Solid - Liquid - Gas - **Solid:** Particles are closely arranged, so solids have a defined shape and volume. - **Liquid:** Particles are loosely ar...

# Some Basic Concept Of Chemistry ## Matter - Anything that occupies space and has mass is called matter. - Matter can be classified as: - Solid - Liquid - Gas - **Solid:** Particles are closely arranged, so solids have a defined shape and volume. - **Liquid:** Particles are loosely arranged. Liquids have a defined volume, but no defined shape. - **Gas:** Particles are freely arranged. Gases have no defined shape or volume. ## Classification of Matter (Chemically) | Matter | Mixture | Pure Substance | Element | Compound | |:---|:---|:---|:---|:---| | | Homogeneous Mixture | Heterogeneous Mixture | Element | Compound | | | Mixture | Pure Substance | | | ### Mixture - A mixture consists of two or more substances in any ratio. - **Homogeneous Mixture:** Substances are completely mixed with each other. - Example: Salt dissolved in water. - **Heterogeneous Mixture:** Substances are not completely mixed with each other. - Example: Sand with water. ### Pure Substance - A pure substance contains only one type of element or one type of compound. - **Element:** A pure substance made up of only one type of atom. - Examples: Hydrogen atom, oxygen atom. - **Compound:** A pure substance made up of different elements. - Examples: CH₄, NH₃, NaCl. ## Physical Quantity | Physical Quantity | Symbol | S.I Unit | Symbol For S.I Unit | |:---|:---|:---|:---| | Length | m | Metre | m/m | | Mass | m | Kilogram | Kg | | Time | t | Second | S | | Electric Current | T | Ampere | A | | Temperature | T | Kelvin | K | | Amount of Substance | n | mole | mol | | Luminuous Intensity | Iv | Candela | Cd | - **Density:** Mass of substance per unit volume. - S.I Unit= kgm³ - **Mass and Accuracy** - **Mass:** The amount of matter present in a substance. - **Weight:** The force exerted by gravity on a substance. - **Temperature:** Measured in °F, K, °C. - **Thermometer:** A device used to measure the temperature of bodies. - Kelvin scale is related to Celsius scale by the equation: K= °C+273. - °C = 270-K. - Fahrenheit is related to Celsius scale by the equation: F = 9/5[°C] + 32. - Convert the following °C into K: - 25°C -> K = °C + 273 =25 + 273 = 298K - 37°C -> K= 37 + 273, K = 310K - 110°C -> K = 110 + 273, K = 373K - 0°C -> K = 0 + 273, K = 273K - Convert the following into °C: - 100K -> °C= 273 -k = 273 - 100 = -173°C - 373K --> °C = 273 - k = 273 - 373 = -100°C - 400K -> °C= 273 - k = 273 - 400, °C = -127°C - Convert the following into °F: - 25°C -> °F= 9/5 * °C + 32 =9/5 * 25 + 32 = 77°F - 100°C -> °F = 9/5 * °C + 32 = 9/5 * 100 + 32 = 212°F ## Scientific Notation - Any number which is written in the form of N x 10ⁿ. - N = any number from 1 to 9. - n = power. - Write the following numbers in scientific notation. - 0,003 - N= 10³ -> 0.003 = 10³ x 10^-3 = 3 x 10^-3 - 0.00025 -> 2.5 x 10^-4 - 0.00000089 -> 8.9 x 10^-7 - 0.101 -> 1.01 x 10^-1 - 1000 -> 1 x 10³ - 265 -> 2.65 x 10² - 3625 -> 3.625 x 10³ - 465 -> 4.65 x 10² - 25 -> 2.5 x 10¹ ## Precisions and Accuracy - **Precision:** Closeness of various measurements for the same quantity. - **Accuracy:** Agreement of the particular value of the true value. - Example: - 83.4% is precision value. - 83% is accuracy value. ## Significant Figures - Meaningful digits which are known with certainty. - Rules for significant figures: 1. All non-zero numbers are significant. (Example: 283 -> 3 significant figures) 2. Zeroes which are present before the number, are not significant. (Examples: 0.003 -> 1 significant figure, 0.012 -> 2 significant figures) 3. If zeroes are present between numbers, they are significant figures. (Examples: 101 -> 3 significant figures, 320059 -> 6 significant figures) 4. If zeroes are present in the right side of the number, they are non-significant, but if a decimal point is present between the zeroes, then all the zeroes are significant. (Examples: 100 -> 1 significant figure, 100.0 -> 4 significant figures) - How many significant figures are in the following numbers: - 0.002 -> 1 significant figure. - 0.32 -> 2 significant figures. - 10101 -> 5 significant figures. - 239 -> 3 significant figures. - 100 -> 1 significant figure. - 100.0 -> 4 significant figures. - 2.60201 -> 6 significant figures. - 2.01 x 10⁵ -> 3 significant figures. - Round off the following numbers up to 3 significant figures: - 2.463 -> 2.46 - 3.167 -> 3.17 - 284.3 -> 284 - 4.645 -> 4.64 - 5.675 -> 5.67 - Round off the following numbers up to 2 significant figures: - 3.62 -> 3.6 - 4.28 -> 4.3 - 7.89 -> 7.9 - 5.25 -> 5.2 - 5.35 -> 5.3 ## Law of Chemical Combination 1. **Law of conservation of mass:** - Matter can neither be created or destroyed. This law was given by Antoine Lavoisier in 1789. 2. **Law of definite proportion:** - A given compound always contains the same proportion of elements by weight. 3. **Law of multiple proportion:** - If two elements combine to form more than one compound, the mass of one element that combines with a fixed mass of the other element is in a ratio of small numbers. - Example: Hydrogen combines with oxygen to get two combination mainly: 1. Water (H₂O) 2. Hydrogen peroxide (H₂O₂) - Hydrogen + Oxygen -> Water 2g + 16g -> 18g - Hydrogen + Oxygen -> Hydrogen peroxide 2g + 32g -> 34g 4. **Gay Lussac's law of gases, volume:** - When gases combine or produce in a chemical reaction, they do so in a simple ratio, provided all gases are at the same temperature and pressure. - Example: - H₂ + O₂ -> 2H₂O - 2 + 1 = 3 5. **Avogadro's Law:** - Equal volumes of gases at the same temperature and pressure should contain an equal number of molecules. ## Dallon's Atomic Theory - Matter consists of indivisible atoms. - All the atoms of a given element have the same property and same mass. Atoms of different elements have different properties and different masses. - Compounds are formed when atoms of different elements combine in a fixed ratio. - Chemical reactions involve reorganization of atoms. These are neither created or destroyed in a chemical reaction. ## Atomic Mass - The ratio of the average mass of an atom to 1/12 mass of an atom of carbon-12 isotope. - Atomic mass = Average mass of one atom / 1/12 the mass of an atom of C-12 isotope. - Its unit is 1 amu or 'U'. | Elements | Atomic Mass | |:---|:---| | Hydrogen (H) | 1 | | Carbon (C) | 12 | | Nitrogen (N) | 14 | | Oxygen (O) | 16 | | Sodium (Na) | 23 | | Magnesium (Mg) | 24 | | Aluminum (Al) | 27 | | Potassium (K) | 39 | | Sulfur (S) | 32 | | Chlorine (Cl) | 35.5 | | Calcium (Ca) | 40 | | Manganese (Mn) | 55 | | Iron (Fe) | 56 | | Copper (Cu) | 63.5 | | Silver (Ag) | 108 | ## Molecular Mass - The sum of the atomic masses of all the elements present in a molecule. - Calculate the molecular mass of the following: - H₂ -> 2 x Atomic mass of H = 2 x 1 = 2u - CH₄ -> 12 + 4 x 1 = 12 + 4 = 16u - H₂O -> 2 x 1 + 16 = 2 + 16 = 18u - NaCl -> 23 + 35.5 = 58.5u - H₂SO₄ -> 2 x 1 + 32 + (4 x 16) = 2 + 32 + 64 = 98u - C₆H₁₂O₆ -> (6 x 12) + (12 x 1) + (6 x 16) = 72 + 12 + 96 = 180u - CO₂ -> 12 + (2 x 16) = 12 + 32 = 44u - NH₃ -> 14 + (3 x 1) = 14 + 3 = 17u - Na₂CO₃ -> (23 x 2) + 12 + (3 x 16) = 46 + 12 + 48 = 106u - NaHCO₃ -> 23 + 1 + 12 + (3 x 16) = 23 + 1 + 12 + 48 = 84u - C₂H₂OH -> (2 x 12) + (2 x 1) + 16 + 1 = 24 + 2 + 16 + 1 = 46u - CH₃OH -> 12 + (3 x 1) + 16 + 1 = 12 + 3 + 16 + 1 = 32u - **Value of 1 amu = 1.66056 x 10^-24g (imp.)** - Amount of substance that contains particles or atoms present in exactly 12g of carbon-12 isotope. - The value of Avogadro constant is denoted by NA = 6.022 x 10²³ atoms and it is denoted by NA. - **Molar mass:** The mass of 1 mole of a substance in grams. - Its unit is grams. - Calculate the molar mass of: - H₂SO₄ -> (2 x 1) + 32 + (4 x 16) = 98g - Na₂CO₃ -> (23 x 2) + 12 + (3 x 16) = 106g - H₂O -> (2 x 1) + 16 = 18g ## Percentage composition - Formula: Mass % of an element = (Mass of the element/Molar mass of compound) x 100 - **Example:** What is the % composition of hydrogen and oxygen in water? - **Step 1:** Calculate molar mass of H₂O = (2 x 1) + 16 = 18g - Mass % of hydrogen = (2 x 1 / 18) x 100 = 11.11 % - Mass % of oxygen = (16/18) x 100 = 88.88% - **Example:** What is the % composition of nitrogen and hydrogen in ammonia? - **Step 1:** Molar mass of NH₃ = 14 + (3 x 1) = 17g - Mass % of nitrogen = (14 x 100 / 17) = 82.35% - Mass % of hydrogen = (3 x 1 x 100 / 17) = 17.64% - **Example:** What is the % composition of carbon, hydrogen and oxygen in ethanol? - **Step 1:** Molar mass of C₂H₅OH = (2 x 12) + (6 x 1) + 16 = 46g - Mass % of carbon = (2 x 12 x 100 / 46) = 52.17% - Mass % of hydrogen = (6 x 1 x 100 / 46) = 13.04% - Mass % of oxygen = (16 x 100 / 46) = 34.78% - **Example:** Calculate the % competion of sodium, oxygen and hydrogen in sodium hydroxide. - **Step 1:** Molar mass of NaOH = 23 + 16 + 1 = 40g - Mass % of sodium = (23 x 100 / 40) = 57.5% - Mass % of oxygen = (16 x 100/ 40) = 40% - Mass % of hydrogen = (1 x 100 / 40) = 2.5% - **Example:** Calculate the mass % of hydrogen, carbon and oxygen in CH₃OH. - **Step 1:** Molar mass of CH₃OH = 12 + (4 x 1) + 16 = 32g - Mass % of hydrogen = (4 x 1 x 100 / 32) = 12.5% - Mass % of carbon = (12 x 100 / 32) = 37.5% - Mass % of oxygen = (16 x 100 / 32) = 50% ## Empirical formula and Molecular Formula - **Empirical formula:** The simplest whole number ratio of various atoms present in a compound. - **Molecular formula:** The actual number of different atoms present in a molecule of compound. **Example:** A compound contains 4. 07% hydrogen, 24.27% carbon, and 77.65% chlorine; its molar mass is 98.96 g. What are its empirical formula and molecular formula? | Element | Mass % | Atomic Mass | No. of moles | Atomic mass/No. of moles | Smallest whole no ratio | |:---|:---|:---|:---|:---|:---| | Carbon | 24.27 | 12 | 2.02 | 2.02 | 1 | | Hydrogen | 4.07 | 1 | 4.07 | 4.07 | 2 | | Chlorine | 71.65 | 35.5 | 2.02 | 2.02 | 1 | - **Empirical Formula:** CH₂Cl - **Empirical Formula Mass:** CH₂Cl = 12 + (2 x 1) + 35.5 = 49.48g - **Molar mass:** 98.96g - **To calculate 'n':** n = (Molar mass / Empirical formula mass) = 98.96 / 49.48 = 2 - **Molecular formula:** (Empirical formula) x n = CH₂Cl x 2 = C₂H₄Cl₂ ## Example: - An organic compound contains 39.9% carbon, 6.7% hydrogen, and 53.4% oxygen. Molar mass is 90 g. Find its empirical formula and molecular formula. | Element | % of Atomic Mass | No. of moles | Atomic Mass/No. of moles | Atomic Mass/No. of moles - smallest whole no ratio | |:---|:---|:---|:---|:---|:---| | Carbon | 39.9 | 12 | 3.325 | 3.325 / 3.325 = 1 | | Hydrogen | 6.7 | 1 | 6.7 | 6.7 / 3.325 = 2 | | Oxygen | 53.4 | 16 | 3.33 | 3.33 / 3.325 = 1 | - **Empirical formula:** CH₂O - **Empirical formula mass:** CH₂O = 12 + (2 x 1) + 16 = 30g - **Molar mass:** 90g - **Calculate 'n':** n = (Molar mass / Empirical formula mass) = 90 / 30 = 3 - **Molecular Formula:** (Empirical formula) x n = CH₂O x 3 = C₃H₆O₃ ## Molar Volume - The volume occupied by 1 gram molecular mass of any substance. - 1 mole contains 6.022 x 10²³ molerul of any substance. - **Example:** Calculate the no. of molecules present in 10g of water. - Molar mass of H₂O = (2 x 1) + 16 = 18g - 18g of H₂O contains 6.022 x 10²³ molecules. - 10g of H₂O contains: - x = 10 x 6.022 x 10²³ / 18 = 3.345 x 10²³ molecules. - **Example:** Calculate the no. of moles present in 20 gram of NaOH. - Molar mass of NaOH = 23 +16 + 1 = 40g - 40g of NaOH contain 6.022 x 10²³ molecules. - 20g of NaOH contain: - x = 20 x 6.022 x 10²³ / 40 = 3.011 x 10²³ molecules. - **Example:** Calculate the no. of moles present in 40 grams of CH₃OH. - Molar mass of CH₃OH = (12 x 1) + (4 x 1) + 16 = 32 grams - We know that 32g of CH₃OH contain 6.022 x 10²³ molecules. - 40g of CH₃OH contain: - x = 40 x 6.022 x 10²³ / 32 = 7.338 x 10²³ molecules. - **Example:** Calculate the no. of moles present in 18 grams of Na₂CO3. - Molar mass of Na₂CO₃ = (23 x 2) + 12 + (3 x 16) = 106g - We know that 106g of Na₂CO₃ contain 6.022 x 10²³ molecules - 18g of Na₂CO₃ contain: - x = 18 x 6.022 x 10²³ / 106 = 1.022 x 10²³ molecules. ## Stoichiometric Calculation 1. **Combustion of methane:** - CH₄ + 2O₂ → CO₂ + 2H₂O 2. **Decomposition of CaCO₃:** - CaCO₃ → CaO + CO₂ 3. **Combustion of magnesium oxide:** - 2Mg + O₂ → 2MgO - **Example:** Calculate the amount of water produced by the combustion of 18g of methane. - **Combustion of methane:** - CH₄ + 2O₂ → CO₂ + 2H₂O - **Molar mass of CH₄:** 12 + (4 x 1) = 16g - **Molar mass of 2H₂O:** 2 x [(2 x 1) + 16] = 36g - **16g of CH₄ → 36g of H₂O** - **18g of CH₄ -> x** - x = 18 x 36 / 16 = 40.5 g - 40.5 g of water is produced by the combustion of 18g of methane. - **Example:** Calculate the amount of water produced by combustion of 11g of methane. - **Combustion of methane:** - CH₄ + 2O₂ → CO₂ + 2H₂O - **Molar mass of CH₄:** 12 + (4 x 1) = 16g - **Molar mass of 2H₂O:** 2 x [(2 x 1) + 16] = 36g - **16g of CH₄ → 36g of H₂O** - **11g of CH₄ → x** - x = 11 x 36 / 16 = 24.75 g - 24.75 g of water is produced by the combustion of 11g of methane.