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Atoms and reactions

Atoms and reactions. Revision for resit F321. Aims . Atomic structure Isotopes and relative masses The mole Calculations using the mole Acids and bases Reactions of acids and bases. THE STRUCTURE OF ATOMS. Atoms consist of a number of fundamental particles, the most important are .

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Atoms and reactions

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  1. Atoms and reactions Revision for resit F321

  2. Aims • Atomic structure • Isotopes and relative masses • The mole • Calculations using the mole • Acids and bases • Reactions of acids and bases

  3. THE STRUCTURE OF ATOMS Atoms consist of a number of fundamental particles, the most important are ... 1 +1 1.672 x 10-27 1.602 x 10-19 1 0 1.675 x 10-27 0 1 1836 -1 9.109 x 10-31 1.602 x 10-19 Calculate the mass of a carbon-12 atom; it has 6 protons, 6 neutrons and 6 electrons 6 x 1.672 x 10-27+6 x 1.675 x 10-27+6 x 9.109 x 10-31 = 2.0089 x 10-26 kg

  4. 23 Na 11 MASS NUMBER AND ATOMIC NUMBER Atomic Number (Z) Number of protons in the nucleus of an atom Mass Number (A) Sum of the protons and neutrons in the nucleus Mass Number (A) PROTONS + NEUTRONS Atomic Number (Z) PROTONS

  5. RELATIVE MASSES Relative Atomic Mass (Ar) The mass of an atom relative to the 12C isotope having a value of 12.000 Ar = average mass per atom of an element x 12 mass of one atom of carbon-12 Relative Isotopic Mass Similar, but uses the mass of an isotope 238U Relative Molecular Mass (Mr) Similar, but uses the mass of a molecule CO2, N2 Relative Formula Mass Used for any formula of a species or ion NaCl, OH¯

  6. Atomic structure summary • The nucleus contains protons (positively charged) and neutrons (neutrally charged i.e. no charge). • The atomic number (proton number) is equal to the number of protons in the atom’s nucleus. • The mass number is the total number of protons and neutrons in the nucleus. • Ions do not have the same number of electrons as protons, and so have an overall charge.

  7. Isotopes and relative masses • Isotopes are atoms having the same number of protons but different numbers of neutrons. • The relative atomic mass is the weighted mean mass of an atom relative to 12C, so that carbon is exactly 12 on this scale. • The average relative atomic mass is equal to the sum of each isotope’s mass for an element x its relative abundance. • The relative formula mass of a compound is equal to the sum of the individual relative atomic masses.

  8. Definitions so far: • Isotopes • Atomic number • Mass number • Ion • Relative isotopic mass • Relative atomic mass • Relative molecular mass • Relative formula mass

  9. THE MOLE – AN OVERVIEW WHAT IS IT?The standard unit of amount of a substance - just as the standard unit of length is a METRE It is just a number, a very big number It is also a way of saying a number in words like DOZEN for 12 GROSS for 144 HOW BIG IS IT ?602200000000000000000000(approx) - THAT’S BIG !!! It is a lot easier to write it as 6.022 x 1023 And anyway it doesn’t matter what the number is as long as everybody sticks to the same value ! WHY USE IT ?Atoms and molecules don’t weigh much so it is easier to count large numbers of them. In fact it is easier to weigh substances. Using moles tells you :-how many particles you get in a certain mass the mass of a certain number of particles

  10. Empirical formula Analysis showed that 0.6075g of Mg combines with 3.995g of bromine to form a compound. Find the empirical formula: Ar Mg 24.3 Br 79.9 Mg: Br : 0.025: 0.050 1:2 MgBr2

  11. Molecular formula A compound has an empirical formula of CH2 and a relative molecular mass, Mr of 56.0. What is its molecular formula? • Empirical formula mass of CH2: = 12 + (1x2) = 14.0 • Number of CH2 units in the molecule: • Molecular formula: (4xCH2) = C4H8

  12. The mole summary • A mole is the S.I. unit for amount of substance and has units of mol. • One mole of a substance is simply the relative formula mass for a compound, or relative atomic mass for an element in grams. • The empirical formula is the simplest whole-number ratio of atoms of each element present in a compound. • The molecular formula is the actual number of atoms of each element in a molecule.

  13. MOLES = MASS MOLAR MASS THE MOLE CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE moles = mass / molar mass mass = moles x molar mass molar mass = mass / moles UNITS mass g or kg molar mass g mol-1or kg mol-1 MASS MOLES x MOLAR MASS COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED

  14. MOLES OF A SINGLE SUBSTANCE • 1. Calculate the number of moles of oxygen molecules in 4g • oxygen molecules have the formula O2 • relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1 • moles = mass = 4g = 0.125 mol • molar mass 32g mol-1 • What is the mass of 0.25 mol of Na2CO3 ? • Relative Molecular Mass of Na2CO3 = (2x23) + 12 + (3x16) = 106 • Molar mass of Na2CO3 = 106g mol-1 • mass = moles x molar mass = 0.25 x 106 =26.5g

  15. REACTING MASS CALCULATIONS CaCO3 + 2HCl ———> CaCl2 + CO2 + H2O 1. What is the relative formula mass of CaCO3? 40 + 12 + (3 x 16) = 100 2. What is the mass of 1 mole of CaCO3 100 g 3. How many moles of HCl react with 1 mole of CaCO3? 2 moles 4. What is the relative formula mass of HCl? 35.5 + 1 = 36.5 5. What is the mass of 1 mole of HCl? 36.5 g 6. What mass of HCl will react with 1 mole of CaCO3 ? 2 x 36.5g = 73g 7. What mass of CO2 is produced ? moles of CO2 = moles of CaCO3 moles of CO2 = 0.001 moles mass of CO2 = 0.001 x 44 = 0.044g

  16. MOLES CONC x VOLUME MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3) 1000 COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION UNITSconcentration mol dm-3 volume dm3 BUT IF... concentration mol dm-3 volume cm3 MOLES = CONCENTRATION x VOLUME MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)

  17. THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.100 mol dm-3 volume pipetted out into the conical flask= 25.00 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of0.100 mol dm-3 This means that there are 0.100 mols of solute in every 1 dm3 (1000 cm3) of solution Take out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles moles in 1dm3 (1000cm3) = 0.100 moles in 1cm3 = 0.100/1000 moles in 25cm3 = 25 x 0.100/1000 = 2.5 x 10-3 mol

  18. MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME 1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = conc x volume in cm3 1000 = 2 mol dm-3 x 25cm3 = 0.05 moles 1000 2 What volume of 0.1M H2SO4 contains 0.002 moles ? volume = 1000 x moles (re-arrangement of above) (in cm3) conc = 1000 x 0.002 = 20 cm3 0.1 mol dm-3

  19. STANDARD SOLUTION ‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’ 4.240g of Na2CO3 was placed in a cleanThe solution was transferred beaker and dissolved in de-ionised water quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm-3 ? mass of Na2CO3 in a 250cm3 solution = 4.240g molar mass of Na2CO3 = 106g mol-1 no. of moles in a 250cm3 solution = 4.240g / 106g mol-1 = 0.04 mol Concentration is normally expressed as moles per dm3 of solution Therefore, as it is in 250cm3, the value is scaled up by a factor of 4 no. of moles in 1000cm3 (1dm3) = 4 x 0.04 = 0.16 molANS.0.16 mol dm-3

  20. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? = 0.100 mol dm-3 How many moles will be in 1 dm3 ? = 0.100 mol How many moles will be in 250cm3 ? = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate? = Na2CO3 What is the relative formula mass? = 106 What is the molar mass? = 106g mol -1 What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650g of Na2CO3 ? (mass = moles x molar mass) ANS. The chemist will have to weigh out 2.650g, dissolve it in water and then make the solution up to 250cm3 in a graduated flask.

  21. VOLUMETRIC CALCULATIONS Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake. 2NaOH + H2SO4 ——> Na2SO4 + 2H2O you need 2 moles of NaOH to react with every 1 mole of H2SO4 i.emoles of NaOH = 2 x moles of H2SO4 or moles of H2SO4 = moles of NaOH 2 REMEMBER... IT IS NOT A MATHEMATICAL EQUATION 2moles of NaOH DO NOT EQUAL 1 mole of H2SO4 More examples follow

  22. VOLUMETRIC CALCULATIONS Calculate the volume of sodium hydroxide (concentration 0.100 mol dm-3) required to neutralise 20cm3 of sulphuric acid of concentration 0.120 mol dm-3. 2NaOH + H2SO4 ——> Na2SO4 + 2H2O you need 2 moles of NaOH to react with every 1 mole of H2SO4 therefore moles of NaOH = 2 x moles of H2SO4 moles of H2SO4 = 0.120 x 20/1000 (i) moles ofNaOH = 0.100 x V/1000 (ii) where V is the volume of alkali in cm3 substitute numbers moles of NaOH = 2 x moles of H2SO4 0.100 x V/1000 = 2 x 0.120 x 20/1000 cancel the 1000’s0.100 x V = 2 x 0.120 x 20 re-arrange Volume of NaOH(V) = 2 x 0.120 x 20 = 48.00 cm3 0.100

  23. MOLAR VOLUME ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3at stp 1. Calculate the volume occupied by 0.25 mols of carbon dioxide at stp 1 mol of carbon dioxide will occupy a volume of 24 dm3 at stp 0.25 mol of carbon dioxide will occupy a volume of 24 x 0.25 dm3 at stp 0.25 mol of carbon dioxide will occupy a volume of 6 dm3 at stp 2. Calculate the volume occupied by 0.08g of methane (CH4) at stp Relative Molecular Mass of CH4 = 12 + (4x1) = 16 Molar Mass of CH4 = 16g mol-1 Moles = mass/molar mass 0.08g / 16g mol-1 = 0.005 mols 1 mol of methane will occupy a volume of 24 dm3 at stp 0.005 mol of carbon dioxide will occupy a volume of 24 x 0.005 dm3 at stp 0.005 mol of carbon dioxide will occupy a volume of 0.12 dm3 at stp stp = standard temperature and pressure (298K and 105 Pa) ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure

  24. Calculations using the mole summary • Mass calculations: calculate the amount of substance and then use the chemical equation to deduce the moles of required substance. • Gas calculations: 1 mol of any gas occupies 24  000 cm3 or 24 dm3 at room temperature. • Solution calculations: the amount of substance dissolved is equal to the concentration x the volume of solution (in dm3). • A dilute solution consists of a small amount of dissolved solute. A concentrated solution consists of a large amount of solute

  25. Some more definitions • Amount of substance • Avogadro constant • The mole • Molar mass • Empirical formula • Molecule • Molecular formula • Molar volume • Concentration • Standard solution • Stoichiometry

  26. Acids and Bases 3 common acids: • Sulfuric acid H2SO4 • Hydrochloric acid HCl • Nitric acid HNO3 Acids are proton donors Weaker acids: Ethanoic acid CH3COOH Methanoic acid HCOOH Citric acid C6H8O7 All acids contain H+ ions this is the active ingredient in acids and it is responsible for acid reactions. They have a pH of less than 7

  27. Bases and Alkalis A base is a proton acceptor. An alkali is a base that dissolves is water forming OH- ions Common Alkalis Sodium hydroxide NaOH Potassium hydroxide KOH Ammonia NH3 Common bases: Metal oxides: MgO, CuO Metal hydroxides: NaOH, Mg(OH)2 Ammonia: NH3 H+(aq) + OH-(aq) H2O(l)

  28. Acid base summary • An acid is a hydrogen ion (H+) or proton donor in solution, whereas a base is a hydrogen ion or proton acceptor in solution. • Hydrochloric acid (HCl), sulfuric acid (H2SO4) and nitric acid (HNO3) are common acids. • Bases include metal oxides (e.g. MgO), metal hydroxides (e.g. NaOH) and ammonia (NH3). • Alkalis are soluble bases and form hydroxide ions, OH-, in solution.

  29. Acid base reactions • Acid + carbonate  salt + CO2 + H2O • Acid + base  salt + water • Acid + alkali  salt + water • Metal + acid  salt + hydrogen You must be able to write balanced equations for all of these reactions.

  30. Water of crystallisation Or how much water is in a compound How would you calculate the water of crystallisation? Mass of hydrated salt Mass of anhydrous salt Mass of water that was in the Hydrated salt

  31. Determine the formula of hydrated magnesium sulfate Mass of hydrated salt: 4.312g Mass of anhydrous salt: 2.107g So mass of H2O in MgSO4.xH2O = 2.205 Calculate the amount in mol of anhydrous MgSO4 MgSO4 = 24.3 + 32.1+ (16.0 x 4) = 120.4gmol-1 n(MgSO4) = = 0.0175 mol Calculate the amount in mol of water n(H2O) = MgSO4 : H2O 0.0175: 0.1225 (divide by the smallest number) 1: 7 MgSO4. 7 H2O

  32. Acid base reactions summary • Salts are formed when a hydrogen ion from the acid is replaced by a metal ion, or an ammonium ion. • Acids react with bases to form a salt and water only; they react with metal carbonates to form a salt, water and carbon dioxide gas. • Metals react with acids to form a salt and hydrogen gas. • Salts may chemically combine with water as water of crystallisation in hydrated salts. (Without water in anhydrous salts.)

  33. Final definitions • Salt • Hydrated • Anhydrous • Water of crystallisation

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