IB SL Chemistry Unit 1

IB SL CHEMISTRY UNIT ONE Matter as a particulate and basic background Elements form chemical compounds. These can come in forms of ionic or covalent bonds. <--(Ionic Bond) (Covalent bond) 1. Types of mixtures. a. When two compounds combine non-chemically, you can form either a homogeneous or heterogeneous mixture. b. When you have two gases that form a mixture, they can be separated through fractional distillation. This relies on exploiting their boiling points to separate the gases. 2. Fractal Distillation of Crude Oil. a. This is the first example of IB’s specific examples. b. Crude Oil is a combination of many hydrocarbons. Hydrocarbons are molecules that are composed only of carbon and hydrogen atoms. These hydrocarbons will vary in boiling points, which will be explained later, so by boiling the crude oil to certain temperatures, you can isolate certain oils. These oils have many different applications within our life. 3. Solutions. a. Composed of a solute and solvent. i. The solute is the solid dissolved in the solvent. ii. The solvent is the liquid in which the solute is dissolved in. b. You can find the Molar concentration of the solute in the solvent by solving for its stoichiometric relationship. c. Concentration is measured in moles of solute divided by liters of solvent. You will see many different units that describe this calculation, eg: i. ii. M 𝑀𝑜𝑙 𝐿 −1 iii. 𝑀𝑜𝑙 ∗ 𝐿 iv. 𝑀𝑜𝑙 ∗ 𝑑𝑚−3 ----- This is the form most commonly used by IB. 1. For example, if you’re given 2 moles of a NaOH and one liter of Water, a. 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 = 2 𝑀𝑜𝑙𝑒𝑠 1 𝐿𝑖𝑡𝑒𝑟 =2M 2. If you’re given 3 Moles of NaCl, and 0.5 L of water, your concentration of NaCl will be a. 3 𝑀𝑜𝑙𝑒𝑠 = 6 M. 0.5 𝐿𝑖𝑡𝑒𝑟𝑠 d. Parts per million i. Found by dividing mass of solute by mass of solvent then multiplying by 106 e. Standard Solution i. A solution with an accurately known concentration ii. Primary standard vs secondary standard 1. Primary a. High purity, low molar mass, low reactivity, doesn’t react with air 2. Secondary a. A solution that has been standardized against a primary solution. 4. Chemical Reactions a. In chemistry, we produce products by taking two reactants and adding them together. b. These reactions will always obey the laws of conservation of both energy and mass. That means on both sides of the reaction, the mass of the products and reactants will be the same. The energy on each side will also be the same. c. Chemicals may go through many different types of reactions, which we will explore later on. These reactions include i. Single placement reactions ii. Double placement reactions ( precipitation reactions) iii. Redox reactions iv. Acid-base reactions v. Organic reactions vi. Combustion vii. Synthesis d. Writing Chemical Reactions. i. Writing chemical reactions requires two main things. 1. The reactant- on the left side 2. The product- on the right side 3. An arrow or equilibrium arrow, that shows 4. the direction the reaction occurs. e. Balancing Chemical reactions i. According to the law of conservation of mass, there will be the same amount of each atom/ element on each side of the reaction. ii. The balanced chemical reaction is when both sides are equal in mass. Shown above is the reaction between HCl and NaOH into table salt and water. The chemical reaction is NaOH + HCl →NaCl + H2O This reaction is already balanced. If we count the amount of each element on each side, we’ll find that they're equal. ● Reactant Side ○ 1 - Na ○ 1-O ○ 1 - Cl ○ 2-H ● Product Side ○ 1 - Na ○ 1-O ○ 1 - Cl ○ 2-H Understanding the Mole/ intro to stoichiometry ● One mole of any substance is equal to 6.0223 *1023 - named Avagadro’s number ● For example ○ 1 mol of Oxygen is 6.0223 * 1023 atoms. ● Molar Mass ○ How many grams of an element or compound are in one mole of that element/compound. Measured in Grams/Moles. ○ Example: Find molar mass of substance x if it contains 0.573 moles and has a mass of 156 grams. ○ 156 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑋 0.573 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑋 = 232.14 𝐺𝑟𝑎𝑚𝑠/𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑋 ● Percent Composition ○ Percentage of any element/ isotope in any given chemical. ■ Example: ● One mole of Carbon Dioxide has a mass of 44.01 gram ● To find the percentage of carbon in CO2 a. Calculate the molar amount of Carbon in CO2. In this case, there is one mole of Carbon in Carbon dioxide. b. Next, convert moles of carbon into grams of carbon. c. 1 𝑚𝑜𝑙𝑒 𝑐𝑎𝑟𝑏𝑜𝑛 ∗ 12.01 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝐶𝑎𝑟𝑏𝑜𝑛 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝑎𝑟𝑏𝑜𝑛 = 12.01 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑐𝑎𝑟𝑏𝑜𝑛. d. Finally, divide mass of carbon by mass of carbon dioxide and convert to percent e. 12.01 𝑔𝑟𝑎𝑚𝑠 𝐶𝑎𝑟𝑏𝑜𝑛 44.01 𝑔𝑟𝑎𝑚𝑠 𝐶𝑂2 ∗ 100 = 27.39 % ● Going from mass- moles - molar mass ○ Calculate the mass of 3 moles of Ethanol ( C2H5OH) ■ Mass of One mol Ethanol = 46.06 grams/mol ■ 3 moles = 46.06 𝑔𝑟𝑎𝑚𝑠 𝐸𝑡ℎ𝑎𝑛𝑜𝑙 1 𝑀𝑜𝑙 𝐸𝑡ℎ𝑎𝑛𝑜𝑙 138.19 𝑔𝑟𝑎𝑚𝑠 𝐸𝑡ℎ𝑎𝑛𝑜𝑙 ∗ 3 𝑚𝑜𝑙𝑒𝑠 𝐸𝑡ℎ𝑎𝑛𝑜𝑙 = ○ Calculate the amount of atoms in 3.5 grams of propane ( C3H8) ■ Propane molar mass = 44.11 grams/moles ■ 3.5 𝐺𝑟𝑎𝑚𝑠 𝑃𝑟𝑜𝑝𝑎𝑛𝑒 ∗ 1 𝑚𝑜𝑙𝑒 𝑃𝑟𝑜𝑝𝑎𝑛𝑒 44.11 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑃𝑟𝑜𝑝𝑎𝑛𝑒 ■ 0.079 𝑚𝑜𝑙𝑒𝑠 𝑃𝑟𝑜𝑝𝑎𝑛𝑒 ∗ 6.0223 ∗1023 𝑎𝑡𝑜𝑚𝑠 1 𝑚𝑜𝑙𝑒 𝑝𝑟𝑜𝑝𝑎𝑛𝑒 = 0.079 𝑚𝑜𝑙𝑒𝑠 𝑝𝑟𝑜𝑝𝑎𝑛𝑒 = 4.778 𝑎𝑡𝑜𝑚𝑠 𝑜𝑓 𝑝𝑟𝑜𝑝𝑎𝑛𝑒 ● Finding the molecular and empirical formula ○ Empirical formula- the formula of the simplest form of the compound ■ Example ( C2H6) ○ Molecular formula - the actual formula of the compound ■ Example (C6H18) ○ Finding Empirical Formula ■ Steps ● First, convert percentage mass into grams by multiplying by 100; assuming that there are 100 grams of the compound. ● Then Find the number of moles of each element in the compound. Divide each molar concentration by the smallest. ■ Example ● Given unknown compounds of C H and N, with mass percentages of 24.02, 32, and 43.98, respectively. ● Step 1: Converting to moles a. 24.02 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑐𝑎𝑟𝑏𝑜𝑛 ∗ 1 𝑚𝑜𝑙 𝐶𝑎𝑟𝑏𝑜𝑛 12.01 𝑔𝑟𝑎𝑚𝑠 𝐶𝑎𝑟𝑏𝑜𝑛 2 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝑟𝑏𝑜𝑛 1 𝑚𝑜𝑙 𝐻 b. 32 𝑔𝑟𝑎𝑚𝑠 ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 ∗ 1.01 𝑔𝑟𝑎𝑚𝑠 𝐻 = 31.68 𝑚𝑜𝑙𝑒𝑠 𝐻𝑦𝑑𝑟𝑜𝑔𝑒𝑛 c. 44 𝑔𝑟𝑎𝑚𝑠 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛 ∗ 1 𝑚𝑜𝑙 𝑁𝑖𝑡𝑟𝑜𝑔𝑒𝑛 14 𝑔𝑟𝑎𝑚𝑠 𝑁𝑖𝑡𝑟𝑜𝑔𝑒𝑛 = 3.14 𝑚𝑜𝑙𝑒𝑠 𝑁𝑖𝑡𝑟𝑜𝑔𝑒𝑛 ● Step 2: Dividing moles a. b. 31.68 𝑚𝑜𝑙𝑒𝑠 𝐻𝑦𝑑𝑟𝑜𝑔𝑒𝑛 2 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝑟𝑏𝑜𝑛 3.14 𝑚𝑜𝑙𝑒𝑠 𝑁𝑖𝑡𝑟𝑜𝑔𝑒𝑛 2 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝑟𝑏𝑜𝑛 = 15.84 ≃ 16 𝐻𝑦𝑑𝑟𝑜𝑔𝑒𝑛 = 1.57 ≃ 2 𝑁𝑖𝑡𝑟𝑜𝑔𝑒𝑛 = c. 2 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝑟𝑏𝑜𝑛 2 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝑟𝑏𝑜𝑛 = 1 𝐶𝑎𝑟𝑏𝑜𝑛. ● Therefore, the empirical formula of the compound is CH16N2. ○ Finding Molecular formula ■ Given the Molar mass of CH16N2 is 168.51, find the molecular formula. ● Step 1: Divide the molecular mass by the empirical mass. ● 168.51 𝑔𝑟𝑎𝑚𝑠 56.17 𝑔𝑟𝑎𝑚𝑠 = 3 ■ Therefore, the molecular formula is three times greater than the empirical formula. In this case, the molecular formula would be C3H48N6. ● Limiting and Excess reactants ○ The amount of a product made is dictated by the limiting reactant. ○ The limiting reactant is the reactant that will be depleted first. ○ Example: ■ 2Fe2O3 (s) + 3C (s) → 4Fe (s) + 3CO2 (g) ■ 250 grams of iron oxide are reacted with 250 grams of Carbon. ■ ■ 250 𝑔𝑟𝑎𝑚𝑠 𝐼𝑟𝑜𝑛 𝑂𝑥𝑖𝑑𝑒 159.7 𝑔𝑟𝑎𝑚𝑠/𝑚𝑜𝑙𝑒 𝐼𝑟𝑜𝑛 𝑂𝑥𝑖𝑑𝑒 250 𝑔𝑟𝑎𝑚𝑠 𝐶𝑎𝑟𝑏𝑜𝑛 12.01 𝑔𝑟𝑎𝑚𝑠/𝑚𝑜𝑙𝑒 𝐶𝑎𝑟𝑏𝑜𝑛 = 1.56 𝑚𝑜𝑙𝑒𝑠 𝐼𝑟𝑜𝑛 𝑂𝑥𝑖𝑑𝑒 = 20.82 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝑟𝑏𝑜𝑛 ○ To find the limiting reactant, divide the concentrations by their coefficient. ■ 1.56 𝑀𝑜𝑙𝑠 𝐼𝑟𝑜𝑛 𝑂𝑥𝑖𝑑𝑒 ■ 20.82 𝑀𝑜𝑙𝑠 𝐶𝑎𝑟𝑏𝑜𝑛 2 3 = 0.78 𝑀𝑜𝑙𝑒𝑠 𝐼𝑟𝑜𝑛 𝑂𝑥𝑖𝑑𝑒 = 6.94 𝑀𝑜𝑙𝑒𝑠 𝐶𝑎𝑟𝑏𝑜𝑛. ○ Iron Oxide is the limiting reactant. ● Theoretical and experimental yield. ○ Chemistry is by no means a perfect science. We will always have errors in the lab, whether it’s random or systematic. We judge how effective our reaction was through the calculation of the theoretical yield. ○ The theoretical yield is how much product should be made under perfect conditions. ○ Example - synthesis of aspirin through Salicylic Acid and Ethanoic Acid Anhydride. Aspirin is C9H8O4 ■ C7H6O3 (s) + C4H6O3 (l) → C9H8O4 (s) + C2H4O2 (l) ○ Find how much aspirin is theoretically produced when 2 grams of Salicylic Acid and Ethanoic Acid react. ○ Convert grams to moles ■ ■ 2 𝑔𝑟𝑎𝑚𝑠 𝑆𝑎𝑙𝑖𝑐𝑦𝑐𝑙𝑖𝑐 138.12 𝑔𝑟𝑎𝑚𝑠/ 𝑚𝑜𝑙𝑒𝑠 𝑆𝑎𝑙𝑖𝑐𝑦𝑙𝑖𝑐 2 𝑔𝑟𝑎𝑚𝑠 𝐸𝑡ℎ𝑎𝑛𝑜𝑖𝑐 102.09 𝑔𝑟𝑎𝑚𝑠/𝑚𝑜𝑙𝑒𝑠 𝐸𝑡ℎ𝑎𝑛𝑜𝑖𝑐 = 0.0144 𝑀𝑜𝑙𝑒𝑠 𝑆𝑎𝑙𝑖𝑐𝑦𝑙𝑖𝑐 = 0.0196 𝑀𝑜𝑙𝑒𝑠 𝐸𝑡ℎ𝑎𝑛𝑜𝑖𝑐 ○ Find the limiting reactant by dividing each by their coefficient. ■ Salicylic is the limiting reactant. ○ Convert Salicyclic to Aspirin. ■ 0.0144 𝑚𝑜𝑙 𝑆𝑎𝑙𝑖𝑐𝑦𝑐𝑙𝑖𝑐 1 1 𝑚𝑜𝑙 𝐴𝑠𝑝𝑖𝑟𝑖𝑛 ∗ 1 𝑚𝑜𝑙 𝑆𝑎𝑙𝑖𝑐𝑦𝑐𝑙𝑖𝑐 = 0.0144 𝑚𝑜𝑙 𝐴𝑠𝑝𝑖𝑟𝑖𝑛 ○ Convert mol to grams. ■ 0.0144 𝑚𝑜𝑙 𝐴𝑠𝑝𝑖𝑟𝑖𝑛 ∗ 180.16 𝑔𝑟𝑎𝑚𝑠 𝐴𝑠𝑝𝑖𝑟𝑖𝑛 1 𝑚𝑜𝑙 𝐴𝑠𝑝𝑖𝑟𝑖𝑛 = 2.609 𝑔𝑟𝑎𝑚𝑠 𝐴𝑠𝑝𝑖𝑟𝑖𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 ● Percentage yield ○ Found by dividing experimental over theoretical yield, then multiplying by 100. ○ Example ■ 2.102 grams of aspirin formed. What is the percent yield? ■ 2.102 2.609 ∗ 100 = 81% 𝑦𝑖𝑒𝑙𝑑. ● Avogadro and gas laws ○ Kinetic Molecular Theory has 5 postulates ● The particles in a gas are in constant, random, straight-line motion. ● There are negligible forces of attraction (intermolecular forces) between the particles. ● Collisions between particles or with the walls of the container are perfectly elastic (no energy is lost). ● The distance between the particles is much greater than the size of the particles, therefore, gas particles have negligible volume. ● The kinetic energy of the particles in a gas is directly proportional to the absolute temperature (in kelvin). ○ Standard Pressure and Temperature ■ Baseline for the molarity of a gas ■ At STP, one mol of a gas take up 22.7 dm3 ■ Calculating moles of gas ● 𝑚𝑜𝑙𝑒𝑠 (𝑛) = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑚3 𝑣𝑜𝑙𝑢𝑚𝑒 𝑎𝑡 𝑠𝑡𝑝 ● Gas Laws ○ Boyle’s law ■ Named after Robert William Boyle ■ Explores relation between pressure and volume of a system ■ Discovered pressure and volume are inversely related ■ PV=k ■ K is the gas constant ■ As pressure increases, volume decreases ○ Charles Law ■ Named after Jacques Charles ■ Volume and Temperature are directly proportional ■ 𝑉 𝑇 =𝑘 ○ Gay-Lussac’s law ■ Named after Joseph Louis Gay-Lussac ■ Pressure and temperature are directly proportional ○ General Gas equation ■ When the three equations are combined, you make the equation ■ 𝑃1𝑉1 𝑇1 = 𝑃2𝑉2 𝑇2 ● Ideal Gas Law ○ If you combine general gas with the concentration of a gas equation derived by Avagadro, you get ○ 𝑃𝑉 = 𝑛𝑅𝑇 ■ Where ■ P is pressure in pascals ■ V is volume in m3 ■ N is amount of moles of a gas ■ R is universal gas constant - 8.314 Joules mol-1k-1 ■ T is temperature in kelvin ○ You can move the equation around to solve for all parts