Topic 1: Quantitative Chemistry (Stoichiometry)12.5hr

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1 Topic 1: Quantitative Chemistry (Stoichiometry)12.5hr 1.1 The Mole, Mass, & Avogadro s constant (number) Introduction mole the amount of a substance that contains the same number of particles as the number of atoms in 12 g of carbon-12 = SI unit for amount of a substance Avogadro s number the # of particles in one mole of a pure substance Avogadro s # = x This can mean x atoms (elements) molecules (covalent compounds) formula units (ionic compounds) 1 mole of carbon-12 = x atoms of carbon-12 = 12.0 g of carbon-12 1 mole of glucose = x molecules of C6H12O6 = g of C6H12O6 1 mole of NaCl = x formula units of NaCl = 58.5 g of NaCl There are several terms for mass with which you should be familiar: molar mass (M) the mass (in grams) of one mole of an element or compound If it s a compounds, just add up the masses of all the elements in the compound. molar mass of H2O = 18.0g (more appropriately 18.0 g mol -1 ) molar mass of C6H12O6 = 180g (180 g mol -1 ) formula mass sum of the masses of all elements in the formula of a compound molar mass, in grams (g) = formula mass, in atomic mass units (u) formula mass of H2O = 18.0u formula mass of C6H12O6 = 180u atomic mass the mass of one mole of an atom average atomic mass the average mass of all the isotopes of an element relative atomic mass the mass of an atom relative to carbon 12 the average atomic mass of all oxygen isotopes = u the relative atomic mass of oxygen 16 = u molecular mass the mass of one mole of a molecule Here again, just add up the masses of all the elements in the molecule: molecular mass of H2O = 18.0g molecular mass of C6H12O6 = 180g The big picture: the term molar mass (M) can be applied to all the other three (formula mass, atomic mass, and molecular mass) *Note that Ebbing uses molecular weight (MW) and formula weight (FW) instead of molecular mass and formula mass (pg 87) IB prefers the terms mass Calculate the following : M NaCl M C3H8 M NH3 2 other terms you should be familiar with: AR relative atomic mass (1.2.1) The weighted average of all the naturally occurring isotopes of an element, relative to carbon-12. This explains why the relative atomic masses given for the elements on the periodic table are not whole numbers. MR relative molecular mass / relative formula mass(1.2.1) The mass of an average molecule of a compound relative to 1/12 of the mass of an atom of carbon-12. The term relative molecular mass is used for covalent compounds; for ionic compounds, the term relative formula mass is used. The units of molar mass are g mol -1, but relative molar masses have no units: the Mr of glucose (C6H12O6) is

2 1.1.2 Practice converting Between Particles and moles Converting Moles to Particles Converting Particles to Mole Number of x moles 6.02 x representative particles 1 mole Number of x 1 mole Particles 6.02 x representative particles 1. Determine the number of atoms in 2.5 mol Zn 3. How many moles contain each of 2. Given 3.25 mol AgNO3, determine the number of the following? formula units a) atoms Al b) 3.75 x molecules CO2 c) 3.58 x formula units ZnCl2 Mass and Mole Converting moles to Mass and to numbers of particles is not very difficult using the following diagram 4. Determine the mass of 3.25 moles of sulfuric acid (H2SO4) 5. What is the mass of 4.35 x 10-2 moles of zinc chloride (ZnCl2) 6. Determine the number of moles present in ach of the following a) 22.6 g AgNO3 b) 6.50 g ZnSO4 c)35.0g HCl 1.2 Formulas Percent Composition 1.2.5(pg 328) percent composition: % by mass of each element in a compound For example even though water s molecular formula is H2O, 2 parts H: 1 part O it s percent composition is: ~89% oxygen and 11% hydrogen Explain this: ex 1: Find % composition of Cu2S ex 2: Find % composition of PbCl2 ex 3: Find % composition of Ba(NO3)2

3 Simplest (Empirical) Formula and Molecular Formula( ) Simplest (Empirical) Formula formula showing the smallest whole # ratio of atoms in a Compound Molecular Formula the simplest (empirical) formula may not be the correct formula! For example: most sugars have the simplest formula CH2O; but glucose is C6H12O6 (same ratio, but not exact same formula: CH2O is the simplest formula; C6H12O6 is the molecular formula) Determine the empirical formula from the percentage composition ex: You have a compound composed of boron & hydrogen. The compound is 78.0% boron & 22.0 % hydrogen Find the empirical formula 1 st change % to grams 78.0g B, 22.0g H 2 nd convert grams to moles so, the mole ratio is 7.22 mol B : 21.8 mol H but, you can t have B7.22H21.8!! 3 rd divide both #s by the smallest # now, you ve got 1 mol B : 3.02 mol H Sometimes, you get whole #s, sometimes (due to rounding or experimental error) you get almost whole #s (for example, here you have 3.02 moles of boron). ROUND!! simplest (empirical) formula = BH3 ex 2: 26.56% K, 35.41% Cr, 38.03% O Find the simplest formula ex 3: Find the simplest (empirical) formula for a compound that is 63.52% Fe & 36.48% S. Molecular Formula the simplest (empirical) formula may not be the correct formula! For example: most sugars have the simplest formula CH2O; but glucose is C6H12O6 (same ratio, but not exact same formula: CH2O is the simplest formula; C6H12O6 is the molecular formula) How do you determine the molecular formula? relationship: (simplest formula)x = (molecular formula) x = whole # multiple of simplest formula What would x be for glucose? ex 1: Go back to BH3 empirical formula. After an experiment you discover that the formula mass of the true compound is u. What is its molecular formula? You may want to set up a chart to solve:

4 Formula mass e.f. BH u m.f. B2H u (BH3)2 B2H6 ex 2: The simplest formula of a compound of P & O is P2O5. You discover the formula mass of the compound is u. What is the molecular formula? Mass of element Mass of compound x 100 = % by mass 1.3 Chemical Equations Before doing any stoichiometry with chemical equations, make sure that they re balanced you must have the same number of each atom on the reactant and product sides of the equation Balance the following equations (1.3.1): KClO3(s) KCl(s) + O2(g) Pb(NO3)2(aq) + KI(aq) PbI2(s) + KNO3(aq) Al(OH)3(aq) + HCl(g) AlCl3(aq) + H2O(l) C5H12(g) + O2(g) CO2(g) + H2O(g) C4H10(g) + O2(g) CO2(g) + H2O(g) Symbols Used In Chemical Equations Symbol (s) (l) or (l) (aq) (g) Meaning reacts to yield reversible reaction reactant or product is a solid reactant or product is a liquid reactant or product is an aqueous solution reactant or product is a gas reactants are heated temperature at which the reaction is carried out pressure at which the reaction is carried out pressure is greater than 1 atm formula of the catalyst used to alter reaction rate Several symbols may be used in the same reaction. For example: C2H4(g) + H2(g) C2H6(g)

5 1.4.1 Stoichiometry g A mol A mol B g B mole-mole calculations 6H2O(l) + 6CO2(g) C6H12O6(s) + 6O2(g) If 6.00 moles of water react with carbon dioxide, how many moles of glucose are produced? If moles of oxygen were produced, how many moles of carbon dioxide were needed to react? If 3.53 moles of Carbon dioxide react with excess water how many moles of each product will be produced? mole-mass / mass-mole calculations 6H2O(l) + 6CO2(g) C6H12O6(s) + 6O2(g) What mass, in grams, of carbon dioxide is needed to react with 3.00 mol of water in this reaction? 5C(s) + 2SO2(g) CS2(l) + 4CO(g) If 8.00g of SO2 reacts, how many moles of each product are formed? mass/mass calculations 5C(s) + 2SO2(g) CS2(l) + 4CO(g) How many grams of carbon monoxide are produced from the reaction of 16.0 g of SO2 with carbon? Work on IB Chemistry Stoichiometric Conversions Worksheet Limiting Reactant limiting reactant controls the amount of product that is formed in a chemical reaction How do you calculate limiting reactant? 1) If you start with moles: You have 2 reactants: A & B.

6 Starting with your given moles of A, calculate how many moles of B you would need to completely react. Compare your given moles of B with your needed moles of B. If your given is more than your need, A is the limiting reactant. If your given is less than your need, B is the limiting reactant. (because you don t have enough B) 2) If you start with grams: You have 2 reactants: A & B. Starting with your given grams of A, calculate how many grams of B you would need to completely react. Compare your given grams of B with your needed grams of B. If your given is more than your need, A is the limiting reactant. If your given is less than your need, B is the limiting reactant. (because you don t have enough B) SiO2(s) + 4HF(g) SiF4(g) + 2H2O(l) If 2.0 mol of HF is combined with 4.5 mol SiO2, which is the limiting reactant? N2H4(l) + 2H2O2(l) N2(g) + 4H2O(g) (propellants used in some rockets) What is the limiting reactant when 24.0 g of N2H4 reacts with 17.0 g of H2O2? How many grams of excess reactant remain? How many grams of water vapor are produced? Percent Yield % yield = x 100 actual yield is also known as experimental yield (the amount you actually get when you do the experiment) C6H6(l) + Cl2(g) C6H5Cl(s) + HCl(g) When 36.8 g of C6H6 react with excess Cl2, 38.8g of C6H5Cl(s) are produced (the actual yield). What is the percent yield of C6H5Cl(s)? CO(g) + 2H2(g) CH3OH(l)

7 If 75.0g of carbon monoxide reacts with excess hydrogen gas to produce 68.4g of methanol, what is the percent yield of methanol? The percent yield of a reaction could never be 100%. What are factors that lower the percent yield? 1) Some product may be lost when trying to recover it (separate it from the reaction mixture). 2) There may be other competing reactions occurring simultaneously. 3) Many reactions stop before they reach completion (some portion of the reactants may simply not react, giving a mixture of reactants and products) mass/volume calculations Avogardro s Law equal volumes of gases at the same temperature & pressure contain an equal number of particles (and thus an equal # of moles) For example, this means that 22.4 liters of helium and 22.4 liters of radon (assuming both are at 1 atm pressure and 0 C) would both contain x atoms of gas. There are two important implications of Avogadro s Law: All gases show the same physical behavior. A gas with a larger volume must consist of a greater number of particles This is also why, at STP, the volume occupied by one mole of ANY gas is always 22.4 L (22.4 dm 3 ) (This is called standard molar volume). Of course, if the temperature changes, so will standard molar volume. At 1 atm pressure and ambient (room) temperature (25 C, or 293 K), the volume of one mole of any gas = 24.8 L (24.8 dm 3 ). These conditions are called Standard Ambient Temperature and Pressure, or SATP A chemical rxn. produced L of sulfur dioxide (SO2) gas at STP. What was the mass in g of the gas produced? How many molecules of SO2 were produced? Find the volume (in dm 3 ) of 3.50 g of CO at STP. Propane burns according to the following equation: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) Assuming all volume measurements are made at the same temperature and pressure: (a) What volume (dm 3 ) of oxygen is required for the combustion of dm 3 of propane at STP? (b) What volume of CO2 will be produced in the reaction?

8 Empirical Gas Laws (Boyle s Law and Charles Law) Consider the effects of pressure, volume, and temperature changes on a fixed mass of an ideal gas (the fixed mass part is important you must assume no gas can enter or escape): As volume, pressure (when T is constant) Boyle s Law As the V decreases, concentration of the particles increases, resulting in more collisions with the container walls, and thus greater pressure. Since pressure and volume are inversely proportional, decreasing the volume by one-half would double the pressure. As temperature, pressure (when V is constant) At higher temperatures, the particles have a higher average velocity, and thus collide with container with greater force (i.e. each collision is more energetic). Since particles are moving faster, if the volume is held constant, the collisions will also occur more frequently. Temperature and pressure are directly proportional, so doubling the temperature would double the pressure of the gas. As temperature, volume (when P is constant) Charle s Law As we just discussed, higher temperatures higher average velocity of gas particles, and thus more energetic collisions between gas particles and container walls. To maintain constant pressure, there must be fewer collisions per unit area. Therefore, volume of the gas must increase. Temperature and volume are directly proportional, so doubling the temperature would double the volume of the gas. Combined Gas Law (we can mathematically derive a combined equation) When the amount of a gas is fixed: P1V1 T1 P2V2 T2 A helium-filled balloon has a volume of 50.0 L at 25.0 C and 820. mm Hg. What volume will it occupy at 650. mm Hg and 10.0 C? A helium-filled balloon is under a pressure of 1.55 atm at 25.0 C. What pressure would the gases in balloon exert 20. C at constant volume? When you read a problem and one of these ingredients (P, V, or T) is missing, just assume that it was held constant and therefore would just cancel out (divide out in the equation) The Ideal Gas Law PV = nrt

9 P = pressure V = volume n = # of moles of gas R = a constant (ideal gas constant) T = temperature (K) 3 values for R: = 62.4 = Obviously, the only difference in units for each R value is the unit for pressure. The purpose of the units for R is to cancel out all the other units in the problem. Therefore, V must be in liters (L), because R has liters in it, which must cancel out. What is the volume in liters occupied by mol of oxygen at 20.0 C and 740 mm Hg pressure? What mass in grams of chlorine (Cl2) is contained in a 10.0 L tank at 27 C and 3.50 atm of pressure? What pressure in atmospheres is exerted by 12.2 g of hydrogen gas in a 4.08 L container at 35.0 C? How do you know when to use the Combined Gas Law and when to use the Ideal Gas Law? Combined Gas Law you will be given at least 2 of P, V, or T (i.e. 2 temperatures, 2 pressures, and/or 2 volumes) Ideal Gas Law you will see some mention of either moles or grams (and only one P, V, and T) 1.5 Solutions solution homogeneous (uniform) mixture of two or more substances in a single phase (ex: sweet tea sugar evenly distributed throughout the tea) soluble capable of being dissolved (ex: sugar in sweet tea) solvent does the dissolving (ex: water) solute is dissolved (ex: sugar) electrolyte a substance that conducts electricity when dissolved in water (in solution) Any soluble ionic compound (ex: NaCl) is an electrolyte. nonelectrolyte does not conduct an electric current in solution How can you increase the rate of dissolving? (e.g.: How can you make a cube of sugar dissolve faster in tea?) 1. increase surface area of the solute (crush up the sugar cube) 2. agitate the solution (stir the tea) 3. heat the solvent (heat up the tea) saturated solution contains all the solute that can possibly be dissolved in the solution No more solute will dissolve; add any more and it just falls to the bottom. Factors affecting solubility:

10 1.types of solvents and solutes like dissolves like ex: water is polar & dissolves polar molecules; hexane is nonpolar & dissolves non-polar molecules (but water cannot dissolve non-polar molecules, etc.) immiscible substances are not soluble in each other 2.pressure gases are more soluble in liquids at higher pressures 3.temperature most substances are more soluble in solution at higher temperatures concentration how much solute is present in a given amount of solution One way to measure concentration is molarity. molarity - # of moles of solute per liter of solution (mol / liter) symbol = M or M Molarity = one molar solution of sodium hydroxide means one mole of NaOH in every liter of solution (NaOH + water) What is the molarity of 3.50 L of solution that contains 90.0 g of sodium chloride? How many liters of water must be used to dissolve 14.6g of HCl to produce a M HCl solution? Liters are units of volume; cubic measurements such as cm 3 or dm 3 are also units of volume Thus, there are other ways to measure concentration, such as g dm -3 or mol dm -3. What is the concentration (in mol dm -3 ) of 245.0g of C6H12O6 dissolved in 2.30dm 3 of CCl4? How many cubic decimeters of water must be used to dissolve 55.8g CaCO3 to produce a mol dm -3 CaCO3 solution? Given the following balanced reaction: 3KOH(aq) + H3PO4(aq) K3PO4(aq) + 3H2O(l) How many grams of potassium phosphate are produced from the reaction of 125mL of 0.500M KOH? Given the following balanced reaction: 3KOH(aq) + H3PO4(aq) K3PO4(aq) + 3H2O(l) How many grams of water are produced from the reaction of 1.30dm 3 of 0.350mol dm 3 phosphoric acid?

11 Remember that concentration is often expressed using brackets [ ]. Use the previous reaction once more: dm 3 of 0.475mol dm 3 H3PO4 were needed to neutralize dm 3 of KOH. What is the [KOH]? Keep in mind: on the IB exam, they may ask for the concentration in: (1) mol L -1 (2) mol dm -1 (3) g dm -1 Recall also how to write overall and net ionic equations and identify spectator ions. Consider this reaction: KCl(aq) + AgNO3(aq) KNO3(aq) + AgCl(s) Look at the ions: overall ionic equation: K + (aq) + Cl - (aq) + Ag + (aq) + NO3 - (aq) AgCl(s) + K + (aq) + NO3 - (aq) K + (aq) & NO3 - (aq) remain aqueous ions on both sides of the equation they don t change! these ions that do not take part in a chemical reaction, but remain aqueous ions as both reactants & products are called spectator ions Cl - (aq) & Ag + (aq) react to produce AgCl(s), which is effectively insoluble this is a chemical change! net ionic equation shows only the compounds and ions that undergo a chemical change: Ag + (aq) + Cl - (aq) AgCl(s) Write the overall and net ionic equations for the following balanced chemical reaction and identify the spectator ions: Zn(NO3)2(aq) + (NH4)2S(aq) ZnS(s) + 2NH4NO3(aq) overall ionic equation: Zn +2 (aq) + 2NO3 - (aq) + 2NH4 + (aq) + S -2 (aq) ZnS(s) + 2NH4 + (aq) + 2NO3 - (aq) net ionic equation: Zn +2 (aq) + S -2 (aq) ZnS(s) spectator ions: 2NO3 - (aq) & 2NH4 + (aq) Write the overall and net ionic equations for the following balanced chemical reaction and identify the spectator ions: K2SO4(aq) + Ca(ClO3)2(aq) CaSO4(s) + 2KClO3(aq) overall ionic equation: 2K + (aq) + SO4-2 (aq) + Ca 2+ (aq) + 2ClO3 - (aq) CaSO4(s) + 2K + (aq) + 2ClO3 - (aq) net ionic equation: Ca +2 (aq) + SO4-2 (aq) CaSO4(s) spectator ions: 2K + (aq) & 2ClO3 - (aq)

1.1 Introduction to the Particulate Nature of Matter and Chemical Change MATTER. Homogeneous (SOLUTIONS)

TOPIC 1: STOICHIOMETRIC RELATIONS 1.1 Introduction to the Particulate Nature of Matter and Chemical Change MATTER Mass Volume Particles Particles in constant motion MATTER Pure Matters Mixtures ELEMENTS