Solutions Chemistry’s Counting Unit: The Mole

Compound Composition & Solutions Chemistry’s Counting Unit: The Mole • A mole of anything has 6.02 x 1023 units. • Analogy: one dozen implies 12 units. • 6.02214179 x 1023 units is known as Avogadro’s number after Amedeo Avogadro. • Example of what a mole of water molecules represents is illustrated in this picture. • The number of molecules in a single droplet of water is roughly 100 billion times greater than the number of people on earth. • Different elements have different 1 mole masses, even though they contain the same number of units. • Example: 1 dozen apples (12 apples) do NOT have the same mass as 1 dozen watermelons (12 watermelons). Atomic Mass Unit (amu) & the Mole • Definition of a mole: • It is the amount of a substance containing the same number of discrete entities (such as atoms, molecules, or ions) as the number of atoms in a sample of pure carbon-12 weighing exactly 12.000 g. 1 mol of 12C = 12.000 g of C 1 mol of 12C = 6.022 x 1023 atoms of C 12.000 g of 12C has an atomic mass unit • The element carbon (C) has an averaged atomic mass of 12.011 grams when considering all of the isotopes of carbon. • One mole of all carbon atoms has a mass of 12.011 grams. • The atomic mass of one mole of carbon atoms is 12.011 grams. • The mole provides a link between the mass of a sample and the number of atoms, molecules, or ions in that substance. Problem: How long would it take to spend is a mole of pennies? You have inherited ¼ mole of pennies. If you spent 1 x 1012 dollars every second how many decades would it take for you to spend your inheritance? Mole Calculations Strategies • The mole is the grand central station when converting between mass to units or units to mass. • Calculation Strategy: • Mass (element or compound) → MOLE → number (atoms or molecules). 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑) mass (g) x (𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑜𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠 mass (g) x ( 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡) ) 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑔 𝑔 )x( 6.02 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 ) 1 𝑚𝑜𝑙𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 6.02 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 ) 1 𝑚𝑜𝑙𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 x( = #molecules = #atoms • Number (atoms or molecules) → MOLE → mass (atoms or molecules). #molecules x ( 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 ) 6.02 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔) ) 1 𝑚𝑜𝑙𝑒 x( 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 (𝑔) ) 1 𝑚𝑜𝑙𝑒 #atoms x (6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠) x ( = mass of molecules = mass of atoms Problem: How many atoms of Americium (Am) are in a 5.00 mg sample? Useful information: atomic mass of Am is 243.1 g/mol Terminology • Atomic mass is the average mass of an atom, expressed in atomic mass unit (where 1 amu is ~ 1 mole of element atoms). • Molecular or Molar mass is the average mass of a molecule, expressed in grams/mole. • Formula mass is the average mass of a formula unit of any type of substance (i.e., element or ionic compound), expressed in grams/mole. Molecular Compound/Molecular Mass In one mole of sucrose, C12H22O11 there are: 12 moles of Carbon atoms 22 moles of Hydrogen atoms 11 moles of Oxygen atoms • There are 6.02 x 1023 molecules of C12H22O11 in one mole of sucrose. • One mole of sucrose has a mass of 342 grams (342.0 amu). • Example: • C: 12.0 amu x 12 • H: 1.0 amu x 22 • 0: 16.0 amu x 11 = = = Total: 144.0 amu 22.0 amu 176.0 amu 342.0 amu • 342 amu is the molecular mass of sucrose • 342 g/mol Formula Mass of Ionic Compounds • Given the formula of an ionic compound, its formula mass can be computed from the formula’s elements atomic masses (amu). • Ionic substances (i.e., compounds formed from the combination of a metal & nonmetal elements) are composed of discrete cations and anions combined in ratios to yield electrically neutral matter. • NOTE: Ionic compounds do NOT exist as “molecules”, therefore the formula mass for an ionic compound technically should NOT be referred to as its molecular mass. • The average atomic masses of the ions can be approximated to be the same as the average atomic masses of the neutral atoms. • Example: PbI2 (lead (II) iodide) formula mass is 461.0 amu Pb: I: 207.0 amu x 1 127.0 amu x 2 = = Total: 207.0 amu 254.0 amu 461.0 amu Problem: Determine the formula mass (amu) for the following: a. (NH4)2S b. C2H5O2N c. BaCl2∙2H2O Problem: A packet of an artificial sweetener contains 0.040 g of saccharin (C7H5NO3S), which has the structural formula: Reference credit: OpenStax2017 Given that saccharin has a molar mass of 183.18 g/mol. 1. How many saccharin molecules are in a 0.040 g sample of saccharin? 2. How many carbon atoms are in the same sample? Problem: 1. How many moles of NaCl are in 0.30 mg? 2. How many atoms are in 0.30 mg of salt? Percent Composition • A pure compound always consists of the same elements combined in the same proportions by weight. • Therefore, we can express molecular composition as a PERCENT BY WEIGHT or PERCENT COMPOSITION. • Example: A 10.0 g sample of a compound is determined to contain 2.5 g hydrogen and 7.5 g carbon. 2.5 g H %H= ´ 100 = 25% 10.0 g compound 7.5 g C %C = ´ 100 = 75% 10.0 g compound Problem: Determine the percent composition of glucose (C6H12O6). Problem Strategy: Determine the percent composition of glucose (C6H12O6). 1. Determine the molecular mass of glucose. 2. Determine the % contribution for each element in the compound. • (# atoms x atomic mass)/molecular mass of compound. 3. Sum of all % = 100% Empirical Formula: Atom to Atom Ratio • An empirical formula is the ratio of atom to atom in a molecule. • Example: Glucose • Molecular formula for glucose is C6H12O6. • A molecular formula represents the actual number of atoms in the molecule. • The empirical formula for glucose is (CH2O)n where n = 6 • The subscript n represents the number of (CH2O) units in a glucose molecule. • An empirical formula represents the simplest whole-number ratio of the number of atoms (or ions) in the compound. • Using percent by weight of each element present in the compound an empirical formula can be derived. Empirical Formulas from Percent Composition • An EMPIRICAL formula is the ratio of atom to atom in the compound. • From a chemical analysis we determine the % by weight of each element in a given amount of pure compound and can derive the EMPIRICAL or SIMPLEST formula representing that compound. • A compound’s empirical formula can be determined from the masses of its constituent elements. Reference credit: OpenStax2017 Problem: Determine empirical formulas given the following. a. BO1.5 b. C6H6 c. Na0.5P0.25O d. C3H2.67O1.33 Guide to Converting Decimal Subscripts to Whole Numbers Decimal Ending Fractional Equivalent To Convert to a Whole Number, Multiply All Subscripts by .20 ⅕ 5 .25 ¼ 4 .33 ⅓ 3 .40 ⅖ 5 .50 ½ 2 .67 ⅔ 3 .75 ¾ 4 Strategy for Determining an Empirical Formula & its Molecular Formula by % Composition Empirical formula from % Composition Step 1: • To determine the EMPIRICAL FORMULA • • • • % to mass Mass to moles Divide by the smallest And if not Whole (number) multiply until whole. • This strategy gives you the EMPIRICAL formula of the compound Problem: The bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29% C and 72.71% O. What is the empirical formula for this gas? Strategy for Determining Molecular Formula Knowing its Empirical Formula Empirical Formula to Molecular Formula Step 2: • To determine the MOLECULAR FORMULA • MUST be given the molecular mass of the compound in the problem. • Calculate the “unit” mass of the empirical formula • This is the mass (in moles) of one unit of the empirical formula • Divide this amount into the molecular mass given. • This is the number (n) units of the empirical formula that make up the molecular formula • Example: if empirical formula is (CH)n & n = 4 then molecular formula is C4H4 Problem: • A compound has an empirical formula of CH2O (empirical formula mass = 30 amu) and a molecular mass of 180 amu. Problem: A compound is composed of B atoms (81.10%) & H atoms. It has a molecular mass of 53.3 g/mol. Determine this compound’s molecular formula? Introduction to Solution Chemistry Solutions can be described by the relative amounts of their components, solute & solvent, to each other. • Dilute and concentrated are terms that qualitatively describe the amount of solute for a given amount of solvent. • A dilute solution has a larger solvent component relative to its solute component. • Unsaturated, Saturated & Super saturated. • A solution is unsaturated if the amount of solute dissolved is just below the maximum for a given amount of solvent at that temperature. • A saturated solution contains the maximum quantity of solute that can be dissolved at that temperature for a given quantity of solvent. • Supersaturated solutions contain more solute than is possible for a given quantity of solvent at that temperature. • They may appear clear or have solid solute present in solution. • Supersaturated solutions are unstable. They have been prepared by “forcing” (i.e. pressure, heating) more solute into solution if under normal conditions. Concentration Units: Molarity • Molarity (M) is defined as the moles of solute per one liter of solution. M= 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐭𝐮𝐞 𝐀 𝐯𝐨𝐥𝐮𝐦𝐞 𝐋 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐀 • Representations or Symbols for molarity • Capital letter M • mol/L • Shorthand symbol for representing molarity is [ ] • [H+] is 0.25 indicates that the molarity of H+ ion in solution is 0.25 M. Molar Concentrations • Molecular compounds: • A 2.0 M solution of glucose means that there is one mole of glucose (solute) per liter of glucose solution. • Ionic compounds: • In a 1.0 M solution of copper (II) chloride you have one mole of CuCl2 dissolved in one liter of solution. • But CuCl2 ionizes in water, so one mole of CuCl2 will produce one mole of Cu2+ ions & two moles of Cl- ions. • Or 1.0 molar (M) CuCl2 solution yields 1.0 M Cu2+ ions & 2.0 M Cl• So, if [CuCl2] = 0.30 M, then [Cu2+] = 0.30 M & [Cl-] = 2 x 0.30 M = 0.60 M Molarity Calculations • Making solution from a solid solute: MA (mol/L) x VA (L) = moles A where moles A = • Making a solution from a concentrated (liquid) solution or a dilution: M1V1 = M2V2 • • • • M1 is the original (stock or concentrated solution) M2 is the dilute solution concentration V1 is the amount of stock solution used V2 is the new volume of the diluted solution • Finding moles of a substance when given molarity and volume of substance. MA x VA = moles of solute A 𝐬𝐚𝐦𝐩𝐥𝐞 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐀 (𝐠) 𝐠 𝐟𝐨𝐫𝐦𝐮𝐥𝐚 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐀 (𝐦𝐨𝐥) Problems: 1. Determine the molarity of a solution when 6.75 grams nickel (II) chloride hexahydrate (NiCl2 6H20) is dissolved in 500.0 mL of water. 2. Calculate the amount of solid KOH required to make 250.0 mL of 0.175 M solution. 3. A 355.0 mL soft drink sample contains 47.88 grams of sucrose (table sugar; mol. mass = 360.0 g/mol). What is the molar concentration of sucrose in the beverage? Making a Solution from a Solution: A Dilution • A dilution is making a solution from a more concentrated (e.g. stock) to make a less concentrated solution. • Formula to use: M1V1 = M2V2 if molarity is the concentration unit or C1V1 = C2V2 (general). • MOLES of SOLUTE DO NOT CHANGE when a solution is diluted. • The moles of solute in the DILUTED (new) solution is EQUAL to the moles of solute in the ORIGINAL (undiluted) solution. • However, the CONCENTRATION of the solution DOES CHANGE!! Why Do Solute Moles Remain Constant in a Dilution? • Flask A (original) contains 50.0 mL of 3.00 M KOH • (3.00 M) x (0.0500 L) = 0.150 moles KOH • If 50.0 mL of the 3.00 M is diluted to 300.0 mL then: • 3.00 M x 0.0500 L = Mdilute x 0.3000 L • Mdilute = 0.500 M KOH • Flask B (dilute solution) • (0.500 mol/L) x 0.3000L = 0.150 mole KOH Reference credit: www.slideshare.net Problems: 1. If 0.850 L of a 5.00M solution of copper nitrate, Cu(NO3)2, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution? 2. 25.0 mL of 0.75 M LiOH solution is diluted to make 0.25 M LiOH. Determine the volume of the diluted solution. Serial Dilutions: An Illustration • The illustration (below) is of a 10X or ten-fold dilution. • Each dilution step results in the next concentration being 1/10 less than the previous. Reference credit: biology.kenyon.edu Dilution: A different look Problem: • Make a 1:5 dilution of 50.0 mLs of 0.30 M CH3COOH solution • What does this mean? • A 1:5 dilution of 0.30 M CH3COOH means that the new concentration will be 0.060 M Or • 1/5 x 0.30 M HCl = 0.060 M HCl • The volume of the diluted solution will be 5x that of the original volume or 250.0 mLs