IB Chemistry S1.4 PDF

Counting Particles by mass: The Mole How do we quantify matter on the atomic scale?  Table of contents 01 02 03 Amounts in Relative atomic and Molar Mass Chemistry formula mass 04...

Counting Particles by mass: The Mole How do we quantify matter on the atomic scale?  Table of contents 01 02 03 Amounts in Relative atomic and Molar Mass Chemistry formula mass 04 05 06 Empirical Formulas IA Practice Empirical Formula Practice 07 08 09 Molar Concentration Standard Solutions Avogadro’s Law (molarity)  Prior Knowledge What is the ratio of elements in each of the following chemical formulas? 1. MgO i. 1 Mg: 1 O 2. H2O i. 2 H: 1 O 3. Al2O3 i. 2 Al: 3 O 4. Li3PO4 i. 3 Li: 1 P: 4 O 5. Mg(NO3)2 i. 1 Mg: 2 N: 6 O Amounts in Chemistry I can calculate the number of moles or particles using dimensional analysis  Amounts in Chemistry Units for amounts 1. If you were shopping for 12 eggs, what would you ask for? 2. If you wanted the three Lord of the Rings films, what would you ask for? 3. How do we describe 2 goals scored by the same person in football? 4. How many pieces of paper are in a ream? 5. If you wanted 6.02 x 1023 atoms, what would you ask for?  Amounts in Chemistry In 18.02 cm3 of distilled water how many molecules are there? 602,000,000,000,000,000,000,000 Is it practical to talk in terms of the number of atoms? No, so we use scientific notation: 6.02 x 1023 Amounts in Chemistry  Amounts in Chemistry Avogadro's number (constant) is the number to which the mass of an atom must be multiplied to give a mass in grams numerically equal to its relative atomic mass Ex: Hydrogen has a relative atomic mass of 1 therefore 6.02 x 1023 hydrogen atoms have a mass of 1g Carbon has a relative mass of 12 therefore 6.02 x 1023 carbon atoms have a mass of 12g Magnesium has a relative atomic mass of 24 therefore 6.02 x 1023 magnesium atoms have a mass of 24g  Amounts in Chemistry 1. The amount of any substance containing an Avogadro number of particles of that substance is called a mole 2. One mole of any substance has a mass equal to its relative mass expressed in grams Ex: 1 mole of magnesium contains 6.02 x 1023 magnesium atoms 1 mole of magnesium has a mass of 24g 12g of magnesium is equivalent to 1/2 moles = 0.5 moles of magnesium 12g of magnesium contains 1/2 moles of magnesium atoms = 0.5 x 6.02 x 1023 = 3.01 x 1023 magnesium atoms  Amounts in Chemistry Avogadro’s Constant, Na Avogadro’s constant Na = 6.02 x 1023 mol-1 Complete the sentence for Avogadro’s constant: For every 1 mole there are… 6.02 x 1023 particles  Amounts in Chemistry Quick Check How many atoms would be found in a. 1 mol of atoms? i. 6.02 x 1023 atoms b. 2 mol of atoms? i. 1.204 x 1024 atoms c. 10 mol of atoms? i. 6.02 x 1024 atoms d. 0.1 mol of atoms? i. 6.02 x 1022  Amounts in Chemistry The amount of a substance: number of particles/ 6.02 x 1023, this can be written as n(mol)= where n or Na is 6.02 x 1023 mol-1 It is important to note that particles can include atoms, molecules, ions, protons, neutrons, or electrons  Amounts in Chemistry Quick Check How many atoms of hydrogen are there in 2 moles of H2O? 2 mol x 6.02 x 1023 mol-1 x 2 H atoms = 24.08 x 1023 H atoms = 2.408 x 1024 in scientific notation Given x Wanted/Given  Amounts in Chemistry Quick Check How many atoms of hydrogen are there in…? 1. 0.5 mol H2O a. 6.02 x 1023 atoms H 2. 0.2 mol NH4Cl a. 4.816 x 1023 atoms H 3. 1 mol Al(HCO3)3 a. 1.806 x 1024 atoms H Quick Check Amounts in Chemistry Try to answer the following questions WITHOUT a calculator. 1. Calculate the total number of protons and electrons in one mole of hydrogen gas, H2. a. 2.408 x 1024 2. Calculate the number of atoms of nitrogen in 0.50 mol of (NH4)2CO3. a. 6.02 x 1023 3. Calculate the total number of atoms in 0.100 mol of [Pt(NH3)2Cl2]. a. 6.62 x 1023 4. Calculate the number of ions in 0.20 mol of (NH4)3PO4. a. 4.8 x 1023 Relative Atomic Mass and Formula Mass I can solve for relative atomic mass and formula mass  Relative Atomic Mass and Formula Mass Why does relative mass not have any units? Dog Mass (kg) 2.5 2.0 13.2 Relative mass 2.5 2.0 13.2 Relative Atomic Mass and Formula Mass Relative Atomic Mass, Ar, is the average mass of naturally occurring atoms of an element on a scale where the 12C atom has a mass of exactly 12 units Relative Molecular Mass, Mr, is the sum of the relative atomic masses The numerical value of Ar is the same as the mass number of the element Ex: Ar of copper - Cu is 64 Ar of helium - He is 4 To find Mr, take the Ar of each element and multiply by the number of atoms of the respective element found in the substance and add all the values together Ex: Mr of Cl2 : 35.5 x 2 = 71 Mr of NaCl : (23 x 1) + (35.5 x 1) = 58.5 Mr of H2O : (2 x 1) + 16 = 18 Mr CH4 : 12 + (4 x 1)= 16  Relative Atomic Mass and Formula Mass What is the relative atomic What is the relative formula mass, Ar , of these atoms? mass, Mr , of these molecules? a. C c. O2 b. O d. CO2 Relative Atomic Mass and Formula Mass Quick Check 1. What are the symbols for: a. Relative atomic mass b. Relative formula mass 2. Calculate the relative formula mass for: a. H2 b. H2O c. NaCl d. Mg(NO3)2 e. CuSO4ᐧ 5H2O  Relative Atomic Mass and Formula Mass Find percentage of an element in a compound using Ar or Mr 1. Consider the Ar of a particular element 2. Multiply Ar but the number of atoms of the element 3. Divide by total Mr of the substance 4. Multiply by 100 Ex: Find the percentage of sodium (Na) in NaCl 1. 23 2. 23 x 1= 23 3. 23/58.5 = 0.393 4. 0.393 x 100 = 39.3% Relative Atomic Mass and Formula Mass Quick Check Find the percentage of hydrogen in water 1. 1 2. 1x2=2 3. 2/18 = 0.111 4. 0.111 x 100 = 11.1%  Relative Atomic Mass and Formula Mass Find mass of an element in a compound using Ar or Mr 1. Consider the Ar of a particular element 2. Multiply Ar by the number of atoms of the element 3. Divide by total Mr of the substance 4. Multiply by total mass Ex: Find the mass of carbon in 20 g of methane 1. 12 2. (12 x 1) = 12 3. 12/16 = 0.75 4. 0.75 x 20 g = 15.0 g  Relative Atomic Mass and Formula Mass Quick Check Find the mass of oxygen in 30 g of Al2O3 1. 16 2. 3 x 16 = 48 3. 48/102 = 0.471 4. 0.471 x 30 g = 14.1 g Molar Mass I can solve for relative atomic mass and formula mass Molar Mass At 4 oC, 0.500 mol of H2O molecules have a volume of 9.01 cm3. If the density of water at 4 oC is 1.00 g cm-3, what is the mass of 1 mole of H2O? Molar Mass Relative atomic mass is used to define the relative mass of one atom of a metallic element, or non-metallic element, which may be considered to consist of unique atoms for the purposes of calculations.  Inert gases Relative atomic mass Other non-metallic elements Relative atomic mass Neon - Ne 20 Phosphorus - P 31 Argon - Ar 40 sulfur - S 32 Giant covalent elements Relative atomic mass Metals Relative atomic mass Boron - B 10.8 Sodium - Na 23 Silicon - Si 28 Magnesium - Mg 24 Carbon - C 12 Iron - Fe 56 Uranium - U 238 Molar Mass Relative molecular mass is used to define the relative mass of one molecule of a covalent substance, whether element or compound. Molecular elements Relative molecular mass simple covalent compounds Relative molecular mass Hydrogen - H2 2 Ammonia - NH3 17 Nitrogen - N2 28 Methane - CH4 16 Oxygen - O2 32 Water - H2O 18 Fluorine - F2 38 Hydrogen chloride - HCl 36.5 Chlorine - Cl2 71 Glucose - C6H12O6 180 Bromine - Br2 160 Carbon dioxide- CO2 44 Iodine - I2 254 Molar Mass Relative formula mass is used to define the mass of the simplest ratio of particles in ionic and giant covalent compounds.  Ionic compounds Relative formula mass sodium chloride - NaCl 58.5 magnesium oxide - MgO 40 potassium nitrate - KNO3 101 calcium carbonate - CaCO3 100 Giant covalent compounds Relative formula mass Silicon dioxide - SiO2 60  Molar Mass Complete the sentence for molar mass for water (M = 18.02 g mol-1): For every mole of H2O there is ______ g of water  Molar Mass Substance Ar or Mr (no units) M (g mol-1) H2 2.02 C H2O CuSO4ᐧ 5H2O  Molar Mass If M is equal to molar mass (g mol-1), n = amount (mol), and m = mass (g) then complete the following equation: M=  Molar Mass Remember the relationship between amount (mol) and mass (g) —> Molar Mass Quick Check 1. 0.250 mol of an element has a mass of 4.75 g. State the name of the element a. M = 4.75 g / 0.250 mol = 19.0 g mol -1 —> Fluorine 2. Calculate the amount (mol) in 13.49 g of aluminum powder a. n = 13.49 g/ 26.98 g mol-1 = 0.5000 mol 3. Calculate the mass of 0.10 mol of carbon dioxide gas a. m = 44.01 g mol-1 x 0.10 mol = 4.4 g *In multiplication and division calculations, we provide our answer to the lowest SF given the data (ignore molar masses from the PTE) ** n = m/M is provided in the data booklet, but you need to know how to use it Molar Mass  Molar Mass Retrieval Practice: Significant Figures Measured value SFs 1.01 230 83040 0.005 0.0720 1.20 x 104 Molar Mass Putting It Together Fe + S → FeS The stoichiometry of the equation shows us that one atom of iron is needed to react with each atom of sulfur. Extending this idea we can see that the same number of iron and sulfur atoms are always needed for a complete reaction. Therefore the moles of iron are always equal to the moles of sulfur in this reaction. If we are told the mass of iron that we start with is 5.6g then we can calculate the mass of sulfur needed. The calculation proceeds via the number of moles. [Relative atomic mass of Fe=56, S=32] Moles of iron = mass /RAM = 5.6/56 = 0.1 moles Therefore moles of sulfur = 0.1 moles RAM of sulfur = 32 Therefore mass of sulfur needed = moles x RAM = 0.1 x 32 = 3.2g  Exit Ticket Molar Mass 1. Calculate the following for hydrated iron(II) sulfate, FeSO4ᐧ 7H2O powder. a. the molar mass b. the amount (mol) in 5.00 g c. the number of oxygen atoms in the amount calculated in b 2. A diatomic molecule contains two different group 17 elements. 0.150 mol of this compound has a mass of 8.18 g. a. Identify the formula of the the compound. b. Calculate the total number of protons in the 0.150 mol. Empirical Formulas I can solve for relative atomic mass and formula mass Empirical Formulas The empirical formula is the simplest ratio of atoms within a chemical compound. This stems from the historic development of chemistry, from the times when determination of structure was an extremely important activity, as many new materials were being isolated and their formulae determined. An unknown compound would be broken down into simpler substances and these measured quantitatively. By finding the percentage of the elements within the original compound it became possible to determine the empirical and, with further evidence, the molecular formula of the unknown compound  Empirical Formulas Calculating the empirical formula from molecular formulas. What is the empirical formula of these molecular formulas? 1. C2H6 1. C1H3 2. H2O 2. H2O 3. C6H6 3. CH 4. C4H10O2 4. C2H5O 5. C2HF4Cl 5. C2HF4Cl  Empirical Formulas Finding the Empirical Formula from formula mass Indicate the lowest whole number ratio of the atoms in a compound: 1. Determine moles of each element present in the compound 2. Divide molar amounts by the smallest number of moles present 3. Obtain whole numbers by multiplying by integers as necessary  Empirical Formulas Ex: A compound is found to contain 2.199 g Copper and 0.277 g Oxygen. Calculate its empirical formula. 1. 2.199 g Cu x (1 mole Cu/63.5 g Cu) = 0.03560 moles Cu 0.277 g O x (1 mole O/16.0 g O) = 0.01731 moles O 2. 0.03460 mol Cu/0.0173 mol = 1.99 Cu → 2 Cu 0.0173 mol O/0.0173 mol = 1.00 O 3. Cu2O  Empirical Formulas Quick Check: 1. A compound is analyzed and found to contain 79.8 g C and 20.2 g H. Determine the empirical formula. 2. What is the empirical formula of a compound that is 25.9 g Nitrogen and 74.1 g Oxygen. (Hint: subscripts MUST be whole numbers)  Empirical Formulas Finding the Molecular Formula from the Empirical Formula Indicates the actual formula for a compound Will be a multiple of the empirical formula 1. Find the formula mass 2. Divide given mass by formula mass 3. Multiply variable by empirical formula Note: Subscripts MUST be whole numbers. If your resulting factor is off by more than 0.1, multiply all ratios by a whole number to get a whole number subscript Ex: Glucose Empirical formula: CH2O Molecular formula: C6H12O6  Empirical Formulas Quick Check: Calculate the molecular formula of the compound whose molar mass is 60.0 g and the empirical formula is CH4N. CH4N’s molar mass can be calculated by adding together each atomic mass 12 + 1(4) + 14 = 30.0 g According to the question, the actual molar mass is 60.0 g Find the difference and multiply by that variable 60.0 / 30.0 = 2 2 (CH4N) = C2H8N2  Empirical Formulas Starting from Percent composition 1. Convert your grams to moles 2. Divide all moles by the smallest calculated mole This provides the empirical formula 3. Calculate the molecular weight of the empirical formula 4. Divide the molecular formula/empirical formula 5. Multiply the resulting factor by the empirical formula Note: Subscripts MUST be whole numbers. If your resulting factor is off by more than 0.1, multiply all ratios by a whole number to get a whole number subscript  Empirical Formulas Starting from Percent composition A material is found to be composed of 38.7% Carbon, 51.6% Oxygen, and 9.7% Hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol. Calculate the empirical and molecular formula for the compound. (Hint: If you assume a sample weight of 100 grams, then your percents= your grams) 1. 38.7 g C x (1 mol/12.0 g) = 3.23 mol 51.6 g O x (1 mol/16.0 g) = 3.23 mol 9.7 g H x (1 mol/1.0 g) = 9.7 mol  Empirical Formulas 2. 3.23 mol C/ 3.23 mol = 1 3.23 mol O/3.23 mol = 1 9.7 mol H/3.23 mol = 3 empirical formula: CH3O 3. C - 12 x 1 = 12 g/mol H - 1 x 3 = 3 g/mol O - 16 x 1 = 16 g/mol Total = 31 g/mol 4. 62 g/mol / 31 g/mol = 2 5. 2 x (CH3O) = C2H6O2 Molar Concentrations (molarity) I can calculate the concentration of a solution and the mass of the compounds in it.  Entrance Ticket: Empirical Practice A sample of peroxyacylnitrate (PAN) contains the following: element C N O H % by mass 20.2 11.4 65.9 Determine the empirical formula of PAN.  Molar Concentrations Which of the following solutions contains the most particles per cm3?  Molar Concentrations A solute is dissolved in a solvent to form a solution (homogenous mixture) ○ Solutions can involve a range of physical states Liquid and liquid ex: vodka water (ethanol dissolved in H2O) Solid and solid ex: steel is an alloy (C dissolved in Fe) Gas and liquid ex: soda (CO2 dissolved in H2O) Gas and gas ex: Air (O2+ dissolved in N2)  Molar Concentrations The concentration of a solution is the quantity of solute that is contains per unit volume ○ This may be given in grams per 100 cm3 or grams per litre, but it is usually given in terms of molarity as this gives a direct measure of the number of solute particles contained by the solution. Molarity is the need to know the amount of solute present in a solution in moles. One mole of any substance contains Avogadro’s number of particles of that substance 1 molar solution contains 1 mole of solute dissolved in 1 liter of solution ***Note that the definition is not per 1 liter of solvent, but 1 liter of SOLUTION Molarity = number of moles of solute per liter of solution (1 liter = 1000 cm3)  Molar Concentrations Molarity, sometimes called molar concentration, is denoted by the capital letter C, and given the units mol dm-3 or M. Molarity can be calculated as n= CV where V is the volume of the solution in dm3 Example: Calculate the molarity of a solution containing 0.15 moles of potassium nitrate in 100cm3 of solution. n = CV Convert 100 cm3 to dm3 (0.1 dm3) 0.15 moles = C (0.1 dm3) C = 1.5 M  Molar Concentrations The molarity of a specific ion within an ionic solution may also be considered separately. ○ In a 1M solution of copper sulfate (CuSO4) the copper ions are separate from the sulfate ions. The solution may be said to be both 1 molar in terms of copper 2+ ions and 1 molar in terms of sulfate 2- ions CuSO4(aq)→Cu2+(aq)+SO42−(aq) ○ A 1 molar (1M) solution of copper nitrate (Cu(NO3)2), however, is 1 molar with respect to copper 2+ ions but 2M with respect to nitrate ions. Cu(NO3)2(aq)→Cu2+(aq)+2NO3−(aq) Example: Calculate the molarity of hydrogen ions in a 0.15 molar solution of sulfuric acid (H2SO4). ○ Hint: Sulfuric acid dissociates in solution according to the following equation H2SO4 ⇆ 2H+ + SO42- If a solution is 1 molar in sulfuric acid, it must be double that in Hydrogen ions C = 0.15 M, therefore molarity in hydrogen atoms is 0.15 M x 2 or 0.3 mol/dm3  Molar Concentrations Using the above equations, we can calculate the mass needed to prepare solutions or the mass contained in solutions of a known concentration. Example: Calculate the mass of sulfuric acid in 100 cm3 of 0.1 mol dm-3 solution. The formula of sulfuric acid is FeSO4. ○ Relative formula mass = 152 Number of moles in 100cm3 of 0.1 mol dm-3 solution (Convert 100 cm3 to dm3) ○ n=CV= 0.1 mol dm-3 x 0.1 dm3 = 0.01 moles Therefore mass of iron (II) sulfate = moles x relative formula mass ○ 0.01 x 152 = 1.52 g  Molar Concentrations Entrance Ticket 1. Calculate the molar concentration of a solution containing 0.0250 mol solute and a total volume of 25.0 cm3. 2. Calculate the number of moles of solute in 40.0 cm3 of a 0.500 mol/dm3 solution 3. Calculate the volume, in cm3 of a 1.00 mol dm-3 solution that would contain 2.50 x 10-3 mol of solute Answers 1. 1.00 mol dm-3 2. 0.0200 mol 3. 2.50 cm3  Molar Concentrations Mol dm-3 x molar mass = g dm-3  Molar Concentrations How might we convert a 1.00 mol dm-3 solution of glucose, C6H12O6 , into units of grams per decimeter cubed (g dm-3)? 1. Find the molar mass of glucose. 2. Convert the concentration from mol/dm3 to g/dm3 using the equation below: Mass (in g) = Concentration (in mol/dm3)x Molar mass (in g/mol)  Molar Concentrations Exit Ticket 1. 3.00 g NaCl is dissolved in distilled water to make a 150.0 cm3 solution. Calculate the concentration of this solution in g/dm3 and mol/dm3. a. 20.0 g/dm3 b. 0.342 mol/dm3 2. The molar concentration of a sodium carbonate solution, Na2CO3(aq) = 0.100 mol dm-3. Convert this concentration in g cm-3. a. 10.6 g/dm3 b. 0.0106 g/cm3 3. Calculate the mass of lithium fluoride, LiF, required to make a 300.0 cm3 solution with a concentration of 2.00 mol dm-3. a. 15.6 g LiF Standard Solutions I can calculate the materials needed to prepare a standard solution  Standard Solutions Name these pieces of equipment  Standard Solutions In the laboratory, specially designed pieces of glassware are used to ensure accuracy, when dealing with solutions. These are: Pipette The volumetric flask also has a Burette circle around the upper narrow Volumetric (graduated) flask part of the flask and when the lower part of the liquid meniscus A pipette is used in conjunction with a pipette filler. The filler is level with the line, the draws the solution into the glassware by means of suction, volumetric flask holds the until the liquid reaches the correct level, indicated by a small correct amount. circular line etched (drawn) around the upper thin part of the pipette. The pipette is emptied by allowing the liquid to run freely into its final receptacle and then, when it has all drained out, touching the tip of the pipette on the surface of the liquid to allow the surface tension of the aqueous liquid surface to remove the final part.  Standard Solutions A solution requires 0.5 g of solute and a total volume of 20 cm3. Complete each table and identify the most precise piece of equipment. Absolute % uncertainty Absolute % uncertainty Measurement Measurement of uncertainty in uncertainty in of mass solution volume (g) measurement (cm3) measurement Standard ±0.01 25 cm3 beaker ±5 cm3 balance Precision 20 cm3 measuring ±0.001 ±0.5 cm3 balance cylinder 20 cm3 volumetric ±0.01 cm3 pipette  Standard Solutions If we wish to make up a solution of known concentration, then the starting material must also be 'pure'. In reality, it is very difficult to obtain chemical compounds in a very pure state and, even if pure when bought, reaction with moisture or carbon dioxide in the air, light sensitivity or decomposition on standing can all reduce the level of purity. Reactions in solution are often carried out for analytical and determination/estimation purposes. If the compound being used cannot be guaranteed to be pure, then it first must be standardised against a solution whose concentration can be determined to a high degree of accuracy. Such substances are called primary standards or standard solutions.  Standard Solutions A suitable primary standards must have the following characteristics: obtainable in a pure state stable in the air readily soluble react with the reagents of choice high relative molecular mass ○ The high relative molecular mass is simply to reduce percentage errors in the weighing stage.  Standard Solutions Procedure: 1. The desired quantity of solute is weighed out on an accurate balance 2. The solute is then dissolved in a small quantity of water and transferred to a volumetric flask a. Rinse any “stuck” droplets into the volumetric flask using distilled water 3. The volumetric flask is then filled up to the mark using distilled water and shaken a. Shake the volumetric flask by repeatedly inverting it.  Standard Solutions Procedure: Example: To prepare 250cm3 of 0.1M sodium hydroxide solution. 0.1 M means that 1 litre of solution contains 0.1 moles of solute. The solute in this case is sodium hydroxide. Sodium hydroxide has a relative formula mass of 40 (Na=23, O=16, H=1). 1 mole of NaOH = 40g/mol therefore 0.1 moles of NaOH = 4.0g However, we don't require 1 litre (1000cm3) of solution, only 250cm3 therefore the mass of sodium hydroxide needed = 250/1000 x 4.0g = 1.0g  Standard Solutions Dilution A calculated quantity of solution is transferred into a graduated (volumetric) flask using a pipette, then the flask is filled up to the mark using distilled water. Can be calculated using the equation C1V1=C2V2 Practice Problem If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution. C1V1=C2V2 (1.6 mol/dm3)(0.175 L)= (C2)(1 L) C2= 280 M Dilution Standard Solutions Example: To prepare 250cm3 of 0.1 mol dm-3 hydrochloric acid using 2.0 mol dm-3 standardised acid. In this case it is better to proceed by calculating the number of moles required in the final solution. Then, the volume of standardized solution that contains this number of moles can be calculated. Final solution required: 250cm3 of 0.1 M hydrochloric acid Moles = molarity x volume (litres) This contains 0.1 mol dm-3 x 0.25 dm3 of HCl = 0.025 moles We wish to obtain 0.025 moles from the 2.0 M solution volume (litres) = moles/molarity Volume = 0.025/2 = 0.0125 dm3 Volume required = 12.5 cm3 Procedure: Measure 12.5cm3 of the standardised 2.0 M HCl into a volumetric flask using a burette. Add distilled water up to the mark mixing carefully.  Standard Solutions Parts per million When expressing extremely low concentrations a unit that can be used is parts per million or ppm. This is useful when giving the concentration of a pollutant in water or the air when the absolute amount is tiny compared the the volume of water or air. 1 ppm is defined as a mass of 1 mg dissolved in 1 dm3 of water Since 1 dm3 weighs 1 kg we can also say it is a mass of 1 mg dissolved in 1 kg of water, or 10-3 g in 103 g which is the same as saying the concentration is 1 in 106 or 1 in a million.  Standard Solutions Parts per million Example: The concentration of chlorine in a swimming pool should between between 1 and 3 ppm. Calculate the maximum mass, in kg, of chlorine that should be present in an olympic swimming pool of size 2.5 million litres. Begin by calculating the total mass in mg assuming 3ppm = 3 mg/L (1 litre is the same as 1 dm3) m= CV = 3mg/L x 2.5 x 106 L = 7.5 x 106 mg Convert the mass into kilograms (1 mg = 10-6 kg) 7.5 x 106mg x 1 kg = 7.5 kg 6 10 mg Avogadro’s Law I can solve problems involving the mole ratio of reactants and/or products and the volume of gases Exactly 1 mole of 3 different gases are placed inside 3 balloons. Which balloon would occupy the largest volume? Avogadro’s Law STP (Standard Temperature and Pressure) One mole of any gas has a volume of 22.7 dm3 The units are usually written as dm3 mol-1 The conditions of STP are: ○ A temperature of 0oC (273 K) ○ A pressure of 100 kPa  Avogadro’s Law French chemist Gay Lussac investigated the reactions of gases and came to the conclusion that when gases combine chemically they do so in volumes that have a simple ratio to one another, and to any gaseous product, provided that all gases are measured at the same temperature and pressure. This became known as Gay Lussac's law. P1/T1=P2/T2 where P is pressure and T is temperature ** Recall that the conversion from Celsius to Kelvin is oC + 273 = K Avogadro’s Law k constant Proportionality constant The volume of a gas is directly proportional to the number of moles of that gas at a constant temperature and pressure k>1 : reaction favors products k