Chemistry Lecture 3 PDF

Chemistry Dr Ramin Boroujerdi PhD | MSc | BSc | FHEA [email protected] Lecture 3 More on Stoichiometry, Temperatures, Solutions and Dilutions Stoichiometric Steps in Converting between Masses of Reactant and Product Calculation 1. Convert the mass of one substance (substance A) to t...

Chemistry Dr Ramin Boroujerdi PhD | MSc | BSc | FHEA [email protected] Lecture 3 More on Stoichiometry, Temperatures, Solutions and Dilutions Stoichiometric Steps in Converting between Masses of Reactant and Product Calculation 1. Convert the mass of one substance (substance A) to the corresponding number of moles using its molar mass. 2. From the balanced chemical equation, obtain the number of moles of another substance (B) from the number of moles of substance A using the appropriate mole ratio (the ratio of their coefficients). 3. Convert the number of moles of substance B to mass using its molar mass. It is important to remember that some species are present in excess by virtue of the reaction conditions. For example, if a substance reacts with the oxygen in air, then oxygen is in obvious (but unstated) excess. Converting amounts of substances to moles—and vice versa—is the key to all stoichiometry problems, whether the amounts are given in units of mass (grams or kilograms), weight (pounds or tons), or volume (liters or gallons). The US Space Shuttle Example The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb). Strategy Step A Write the balanced chemical equation for the reaction. Step B Convert mass of oxygen to moles. From the mole ratio in the balanced chemical equation, determine the number of moles of hydrogen required. Then convert the moles of hydrogen to the equivalent mass in tons. The US space shuttle Discovery during liftoff. The 1 lb = 453.6 large cylinder in the middle contains the oxygen and hydrogen that fueled the shuttle’s main engine.  The US Space Shuttle Example The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb). Step A We use the same general strategy for solving stoichiometric calculations as in the preceding example. Because the amount of oxygen is given in tons rather than grams, however, we also need to convert tons to units of mass in grams. Another conversion is needed at the end to report the final answer in tons. We know the identity of both the reactants and the product, we can write the reaction and then balance it to have equal number of atoms on both sides of the reaction: 2H2(g)+O2(g)→2H2O(g) Using the balanced equation, we can say 2 mol of H 2 react with 1 mol of O2 to produce 2 mol of H2O.  The US Space Shuttle Example The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb). Step B To convert tons of oxygen to units of mass in grams, we multiply by the appropriate conversion factors: Using the molar mass of O2 (32.00 g/mol, to four significant figures), we can calculate the number of moles of O2 contained in this mass of O2:  The US Space Shuttle Example The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb). Now use the coefficients in the balanced chemical equation to obtain the number of moles of H2 needed to react with this number of moles of O2: The molar mass of H2 (2.016 g/mol) allows us to calculate the corresponding mass of H2: 2H2(g)+O2(g)→2H2O(g)  The US Space Shuttle Example The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb). Finally, convert the mass of H2 to the desired units (tons) by using the appropriate conversion factors: The space shuttle had to be designed to carry 0.126 tn of H2 for each 1.00 tn of O2. Even though 2 mol of H2 are needed to react with each mole of O2, the molar mass of H2 is so much smaller than that of O2 that only a relatively small mass of H2 is needed compared to the mass of O2. Solutions Solutions are homogenous mixtures. Solutions may exist in any of the three states of matter; that is, they may be gases, liquids, or solids. The terms solute and solvent refer to the components of a solution. The solute, in the case of a solution of a gas or solid dissolved in a liquid, is the gas or solid; in other cases, the solute is the component in smaller amount. The solvent, in a solution of a gas or solid dissolved in a liquid, is the liquid; in other cases, the solvent is the component in greater amount. Solutions The solubility of a solute in a solvent (that is, the extent of the mixing of the solute and solvent species) depends on a balance between the natural tendency for the solute and solvent species to mix and the tendency for a system to have the lowest energy possible. For a solution, the amount of solute dissolved in a unit volume of solution (or a unit amount of solvent) is the concentration of the solute. If an excess of solute (more than will normally dissolve) is added to a quantity of liquid solvent, an equilibrium is established between the pure solute and the dissolved solute. Solutepure ⇌ Solutedissolved The Solution Process London forces are the only intermolecular forces between nonpolar covalent molecules. The intermolecular attractions between polar covalent molecules are due to dipole-dipole forces as well as London forces. In substances in which there is hydrogen bonding (an exceptionally strong type of dipole-dipole attraction), the intramolecular forces are usually strong. Nonpolar substances and polar substances are generally immiscible.  The Solution Process Like dissolves like! Polar solvents (liquids) can mix and dissolve in each other. Polar compounds (solids) are more readily dissolved in polar solvents. Nonpolar solvents (liquids) can mix and dissolve in each other. Nonpolar compounds (solids), are more readily dissolved in nonpolar solvents.  The Solution Process Polar liquids (water, in particular) can function as solvents for many ionic compounds. The ions of the solute are electrostatically attracted by the polar solvent molecules – negative ions by the positive poles of the polar solvent molecules – positive ions by the negative poles of the solvent molecules. These ion-dipole attractions can be relatively strong. Cell membrane Why do phospholipids form a bilayer in the cell membrane? Heat of Solution When a solute dissolves in a solvent, energy is absorbed or evolved, the actual quantity of energy per mole of solute depending upon the concentration of final solution. The heat effect observed when a solution is prepared, which is the net result of the energy required to break apart certain chemical bonds or attractions (solute-solute and solvent- solvent) and the energy released by the formation of new ones (solute- solvent). Heat of Solution Changes in enthalpy (heat) for a process between initial and final states is showed by ΔH. Hess's law states that the overall change in enthalpy for a chemical reaction is equal to the sum of the enthalpy changes for each step. When two substances mix to form a solution, heat is either evolved (an exothermic process) or absorbed (an endothermic process); only in the special case of an ideal solution do substances mix without any heat effect. Heat of Solution Molarity The general term concentration refers to the quantity of solute in a standard quantity of solution. Concentration sometimes showed by “[ ]”, so we can say [X] means the concertation of molecule, ion or element X. Qualitatively, we say that a solution is dilute when the solute concentration is low and concentrated when the solute concentration is high. Usually, these terms are used in a comparative sense and do not refer to a specific concentration. We say that one solution is more dilute, or less concentrated, than another. However, for commercially available solutions, the term concentrated refers to the maximum, or near maximum, concentration available. Molar concentration, or molarity (M), is defined as the moles of solute dissolved in one liter (cubic decimeter) of solution. Molarity Example A sample of NaNO3 weighing 0.38 g is placed in a 50.0 mL volumetric flask. The flask is then filled with water to the mark on the neck, dissolving the solid. What is the molarity of the resulting solution? To calculate the molarity, you need the moles of solute. Therefore, you first convert grams NaNO3 to moles. The molarity equals the moles of solute divided by the liters of solution. You find that 0.38 g NaNO3 is 4.47×10-3 mol NaNO3; the last significant figure is underlined. The volume of solution is 50.0 mL, or 50.0×10-3 L, so the molarity is: Calculating Moles from Volume An expanded version of the flowchart for stoichiometric calculations illustrated in previous Lecture (2) you can see that it is possible to use the balanced chemical equation for the reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products to determine the amounts of other species. Calculating Moles from Volume Example Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN) 2]− ion. Gold is then recovered by reduction with metallic zinc according to the following equation: Zn (s) + 2[Au(CN)2]− (aq) → [Zn(CN)4]2− (aq) + 2Au (s) What mass of gold would you expect to recover from 400.0 L of a 3.30 × 10 −4 M solution of [Au(CN)2]−? Strategy: Step A Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN) 2]− present by multiplying the volume of the solution by its concentration. Step B From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass. Calculating Moles from Volume Example Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN) 2]− ion. Gold is then recovered by reduction with metallic zinc according to the following equation: Zn (s) + 2[Au(CN)2]− (aq) → [Zn(CN)4]2− (aq) + 2Au (s) What mass of gold would you expect to recover from 400.0 L of a 3.30 × 10 −4 M solution of [Au(CN)2]−? Step A The equation is balanced as written, so we can proceed to the stoichiometric calculation. Calculating Moles from Volume Example Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN) 2]− ion. Gold is then recovered by reduction with metallic zinc according to the following equation: Zn (s) + 2[Au(CN)2]− (aq) → [Zn(CN)4]2− (aq) + 2Au (s) What mass of gold would you expect to recover from 400.0 L of a 3.30 × 10 −4 M solution of [Au(CN)2]−? Step A As indicated in the strategy, we start by calculating the number of moles of [Au(CN)2]− present in the solution from the volume and concentration of the [Au(CN) 2]− solution: Calculating Moles from Volume Example Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN) 2]− ion. Gold is then recovered by reduction with metallic zinc according to the following equation: Zn (s) + 2[Au(CN)2]− (aq) → [Zn(CN)4]2− (aq) + 2Au (s) What mass of gold would you expect to recover from 400.0 L of a 3.30 × 10 −4 M solution of [Au(CN)2]−? Step B Because the coefficients of gold and the [Au(CN) 2]− ion are the same in the balanced chemical equation, if we assume that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN) 2]− we started with (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so we need to convert the number of moles of gold to the corresponding mass using the molar mass of gold:  Calculating Moles from Volume Example Determine the molar concentrations of Na+ and PO43- in a 1.50 M Na3PO4 solution. Molar mass of Na, P and O are 22.98 g/mol, 30.97 g/mol and 15.99 g/mol, respectively. You are given the concentration of an ionic solution and asked to find the concentrations of the component ions. A formula unit of Na3PO4 contains 3 Na+ ions (as indicated by the subscript), so the concentration of Na+ is three times the concentration of Na3PO4. Since the same formula unit contains one PO43- ion, the concentration of PO43- is equal to the concentration of Na3PO4. 1.50 M solution of Na3PO4 molarity of Na+ = 3×(1.50 M) = 4.50 M molarity of PO43- = 1×(1.50 M) = 1.50 M *Na3PO4 : trisodium phosphate Dilution You must know the relationship between the molarity of the solution before dilution (the initial molarity) and that after dilution (the final molarity). To obtain this relationship, first recall the equation defining molarity: You can rearrange this to give: Dilution The product of molarity and the volume (in liters) gives the moles of solute in the solution. Writing Mi for the initial molar concentration and Vi for the initial volume of solution, you get: When the solution is diluted by adding more water, the concentration and volume change to Mf (the final molar concentration) and Vf (the final volume), and the moles of solute equals: Because the moles of solute has not changed during the dilution, you can write: Dilution Example How to make 5.00 L of a 1.50 M KCl solution from a 12.0 M stock solution?  Dilution You might see MiVi=MfVf is also written as C1V1=C2V2, which is technically the same equation. M only refers to molarity, mol/L, but C refers to concertation, which is a more general term, so it can be any form/unit of concertation such as mol/L, mg/mL, nmol/mL, etc. You can use any volume units, but both V1 and V2 must be in the same units. You can use any concentration units, but both C1 and C2 must be in the same units. Dilution Example You are given a solution of 14.8 M NH3. How many milliliters of this solution do you require to give 100.0 mL of 1.00 M NH3 when diluted? You know the final volume (100.0 mL), final concentration (1.00 M), and initial concentration (14.8 M). You write the dilution formula and rearrange it to give the initial volume; then you substitute the known values into the right side of the equation.  Dilution Example Consider the reaction of sulfuric acid, H2SO4, with sodium hydroxide, NaOH. H2SO4(aq) + 2NaOH(aq) → 2H2O(l) + Na2SO4(aq) Suppose a beaker contains 35.0 mL of 0.175 M H2SO4. How many milliliters of 0.250 M NaOH must be added to react completely with the sulfuric acid? Strategy This is a stoichiometry problem that involves the reaction of sulfuric acid and sodium hydroxide. You have a known volume and concentration of H 2SO4 in the beaker, and you want to determine what volume of a known concentration of NaOH is required for complete reaction. If you can determine the number of moles of H 2SO4 that are contained in the beaker, you can then use the balanced chemical reaction to determine the number of moles of NaOH required to react completely with the H2SO4. Finally, you can use the concentration of the NaOH to determine the required volume of NaOH. Following this strategy, you convert from 35.0 mL (or 35.0×10-3 L) H2SO4 solution to moles H2SO4 (using the molarity of H2SO4), then to moles NaOH (from the chemical equation). Finally, you convert this to volume of NaOH Dilution Example Consider the reaction of sulfuric acid, H2SO4, with sodium hydroxide, NaOH. H2SO4(aq) + 2NaOH(aq) → 2H2O(l) + Na2SO4(aq) Suppose a beaker contains 35.0 mL of 0.175 M H2SO4. How many milliliters of 0.250 M NaOH must be added to react completely with the sulfuric acid? Thus, 35.0 mL of 0.175 M sulfuric acid solution reacts with exactly 49.0 mL of 0.250 M sodium hydroxide solution. Dilution Types of dilution Stock dilution Different volumes of the same stock solution are transferred directly to other containers to generate various concentration of the solution. Serial dilution A stepwise dilution of a solution to generate various concentrations of the solution. Usually, the dilution factor at each step is constant.  Dilution Example A student tries to dilute a Mauveine (purple dye) solution using serial solution. Suppose the concentration of the mother (stock) solution is 500 μmol/L, which is diluted into three different solutions using 100 mL volumetric flasks. As shown in the figure. 10 mL of the stock solution is accurately measured and added to the first volumetric flask, and water is added precisely to the mark. Similarly, 10 mL of the first solution was added to the second volumetric flask, and water was added to the mark. Finally, 10 mL of the second solution was added to the third volumetric flask, and water was added to the mark. What is the concentration (molarity, in μmol/L) of solutions 1, 2 and 3? C1V1=C2V2 C1V1=C2V2 C1V1=C2V2 50 μmol/L × 10 mL = C2 × 100 5 μmol/L × 10 mL = C2 × 100 mL 500 μmol/L × 10 mL = C2 × 100 mL C2 = 0.5 μmol/L (solution 3) mL C2 = 5 μmol/L (solution 2) C2 = 50 μmol/L (solution 1)  Titration An important method for determining the amount of a particular substance is based on measuring the volume of reactant solution. Suppose substance A reacts in solution with substance B. If you know the volume and concentration of a solution of B that just reacts with substance A in a sample, you can determine the amount of A. Titration is a procedure for determining the amount of substance A by adding a carefully measured volume of a solution with known concentration of B until the reaction of A and B is just complete. Volumetric analysis is a method of analysis based on titration. Titration Example A flask contains a solution with an unknown amount of HCl. This solution is titrated with 0.207 M NaOH. It takes 4.47 mL NaOH to complete the reaction. What is the mass of the HCl? NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) If you use the numerical information from the problem to determine the moles of NaOH added to the solution, you then can use the stoichiometry of the reaction to determine the moles of HCl that reacted. Once you know the moles of HCl, you can use the molar mass of HCl to calculate the mass of HCl. Employing this strategy, you convert the volume of NaOH (4.47×10 -3 L NaOH solution) to moles NaOH (from the molarity of NaOH). Then you convert moles NaOH to moles HCl (from the chemical equation). Finally, you convert moles HCl to grams HCl. Titration Example Consider the reaction for the neutralization of sulfuric acid. How much 0.125 M NaOH solution do we need to completely neutralize 0.225 L of 0.175 M H2SO4 solution? H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) Titration Example Consider the reaction for the neutralization of sulfuric acid. How much 0.125 M NaOH solution do we need to completely neutralize 0.225 L of 0.175 M H2SO4 solution? H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) It will take 0.630 L of the NaOH solution to completely neutralize the H2SO4 Temperature: Random Motion of Molecules and Atoms The atoms and molecules that compose matter are in constant random motion—they contain thermal energy. The temperature of a substance is a measure of its thermal energy. The hotter an object, the greater the random motion of the atoms and molecules that compose it, and the higher its temperature. We must be careful to not confuse temperature with heat. Heat, which has units of energy, is the transfer or exchange of thermal energy caused by a temperature difference. For example, when a cold ice cube is dropped into a warm cup of water, energy is transferred as heat from the water to the ice, resulting in the cooling of the water. Temperature, by contrast, is a measure of the thermal energy of matter (not the exchange of thermal energy).  Temperature: Random Motion of Molecules and Atoms Three different temperature scales are in common use: The scale scientists use is the Celsius (°C) scale. On this scale, water freezes at 0 °C and boils at 100 °C. Room temperature is approximately 22 °C. The Fahrenheit and Celsius scales differ in both the size of their respective degrees and the temperature each calls “zero”. The most familiar/common in the United States is the Fahrenheit (°F) scale. On the Fahrenheit scale, water freezes at 32 °F and boils at 212 °F. Room temperature is approximately 72 °F. The Fahrenheit scale was initially set up by assigning 0 °F to the freezing point of a concentrated saltwater solution and 96 °F to normal body temperature (although body temperature is now known to be 98.6 °F). Both the Fahrenheit and Celsius scales contain negative temperatures. A third temperature scale, called the Kelvin (K) scale, avoids negative temperatures by assigning 0 K to the coldest temperature possible, absolute zero. Absolute zero (-273.15 °C or -459.7 °F) is the temperature at which molecular motion virtually stops. There is no lower temperature. The kelvin degree, or kelvin (K), is the same size as the Celsius degree—the only difference is the temperature that each scale designates as zero. Temperature: Random Motion of Molecules and Atoms We can convert between these temperature scales using the following formulas: Temperature: Random Motion of Molecules and Atoms Example Convert 310 K to Fahrenheit:  Osmosis Osmosis is the process where solvent molecules move through a semipermeable membrane from a dilute solution into a more concentrated solution (which becomes more dilute). The solvent may be a liquid or even a gas. In biology, osmosis is the movement of water molecules from a solution with a high concentration of water molecules to a solution with a lower concentration of water molecules, through a cell's partially permeable membrane.  Seawater is a Thirsty Solution Why drinking seawater or any other salt-water causes dehydration?  Osmosis Example A solution was produced by dissolving 10 grams NaOH in 250 mL water and added to the left side of a semipermeable membrane. If on the right side of the membrane, we already had 250 mL of a 1 M solution of NaOH, which side will hold more water as a result of osmosis? Molar mass of NaOH is about 40 g/mol. Let’s use grams of NaOH on the left side and calculate the molarity on the left side: The concentration of NaOH on both sides is similar, so both sides will maintain same amount of water.  Osmosis Osmotic pressure is a colligative property of a solution equal to the pressure that, when applied to the solution, just stops osmosis. The osmotic pressure, π, of a solution is related to the molar concentration of solute, M: Here R is a constant (0.082 L.atm/K.mol) and T is the absolute temperature. Osmosis Example The formula for low-molecular-mass starch is (C6H10O5)n, where n averages 2.00×102. When 0.798 g of starch is dissolved in 100.0 mL of water solution, what is the osmotic pressure, at 25 oC? Strateg y Calculations of osmotic pressure involve the formula π=MRT. In order to employ this equation here, we need to determine the molarity (M) of the solution. If we obtain the molecular mass of the starch, we can determine the number of moles of starch in the 0.798 g sample used to prepare the solution. Then, knowing the moles of starch and the volume of the solution (100.0 mL), we can calculate the molarity. Then we can use the formula to determine osmotic pressure. Osmosis Example The formula for low-molecular-mass starch is (C6H10O5)n, where n averages 2.00×102. When 0.798 g of starch is dissolved in 100.0 mL of water solution, what is the osmotic pressure, at 25 oC? The molecular mass of (C6H10O5)200 is 32,400 amu. The number of moles in 0.798 g of starch is: The molarity of the solution is:  Osmosis Example The formula for low-molecular-mass starch is (C6H10O5)n, where n averages 2.00×102. When 0.798 g of starch is dissolved in 100.0 mL of water solution, what is the osmotic pressure, at 25 oC? and the osmotic pressure at 25 oC is: *Absolute temperature scales are Kelvin and Rankine.