1 / 95

Quantitative Chemistry

Quantitative Chemistry. Activity:. Relative atomic mass Relative molecular mass Relative formula mass Avogadro’s number Element Compound Molecule Mole Isotope Diatomic. Print out the keywords worksheet and add the following words, once you know the meanings fill them in.

elom
Download Presentation

Quantitative Chemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Quantitative Chemistry

  2. Activity: • Relative atomic mass • Relative molecular mass • Relative formula mass • Avogadro’s number • Element • Compound • Molecule • Mole • Isotope • Diatomic Print out the keywords worksheet and add the following words, once you know the meanings fill them in.

  3. Chemical Formulas and the Mole concept • Elements • Compounds • Mole concept and Avogadro's constant • Isotopes • Formulas of compounds • Empirical formula • Molecular formula • Structural formula

  4. Elements • All substances are made up of one or more __________. • An element ___________be broken down by any chemical process into ___________substances. • There are just over _____ known elements. • The smallest part of an element is called an __________. cannot 100 elements atom simpler

  5. Molecules and compounds • Some substances are made up of a single element although there may be more than one atom of the element in a particle of the substance. • For example, Oxygen is diatomic, that is, a molecule of oxygen containing two oxygen atoms. • A compound contains more than one element. • For example; a molecule of water contains two hydrogen atoms and one oxygen atom. • Water is a compoundnot an element because it canbe broken down chemically in to it’s constituent elements: hydrogen and oxygen

  6. Mole concept and Avogadro’s constant • A single atom of an element has an extremely small mass. For example, an atom of carbon–12 has a mass of 1.993 x 10-23 g. • This is far too small to weigh • Instead of weighing just one atom we can weigh a mole of atoms. • 1 mole contains 6.02 x 10 23 particles • 6.02 x 10 23atoms of carbon–12 = ? • This number is known as Avogadro’s constant(NA or L) • 1 mole of Carbon = 12g

  7. Mole concept and Avogadro’s constant • Chemists measure amounts of substances in moles. • A mole is the amount that contains L particles of that substance. • The mass of one mole of any substance is known as the molar mass and has the symbol M. For example, Hydrogen atoms have one 12th the mass of carbon–12 atoms so one mole of hydrogen atoms contains 6.02 x 10 23hydrogen atoms and has a mass of 1.01 g. • For diatomicmolecules e.g. H2 there are 6.02 x 1023 molecules of hydrogen and therefore 12.04 x 1023 atoms.

  8. Relative Atomic mass • The actual atomic mass of an individual atom of an element is so small that we use a relative atomic mass as seen on the periodic table. • Since hydrogen is the lightest atom (it has an actual mass of 8.275 x 10-25) we say that hydrogen has a relative atomic mass of 1. • Experiments have shown that an atom of carbon weighs 12 times as much as an atom of hydrogen. So the relative atomic mass of carbon is 12. • Relative atomic mass can be abbreviated to Ar or RAM and is found on the periodic table.It has no units.

  9. Activity • Complete the worksheet IB 1.1 – Using Avogadro’s Number • There are 20 questions to complete • Show all working

  10. Isotopes and RMM/RFM • In reality elements are made up of a mixture of isotopes. • The relative atomic mass of an element Ar is the weighted mean of all the naturally occurring isotopes of the element relative to carbon–12. • This explains why the relative atomic masses given for elements on the periodic table are not whole numbers • The units of molar mass are g mol-1(this means grams per mole) but relative molar masses Mr have no units. • For molecules relative molecular mass is used (RMM). For example the Relative Molecule Mass of glucose, C6H12O6 = (6 x 12.01) + (12 x 1.01) + (6 x 16.00) = 180.18

  11. Calculating RFM • This is very simple but there are one or two rules to remember: • Numbers in subscript only refer to the number directly in front of it. E.g. SO4 there is one Sulphur atom and 4 Oxygen atoms RFM = 96 in total • Everything inside a bracket is multiplied by the small number directly outside of it on the right hand side. E.g. (NH4)2SO4 So the RFM of NH4 must be multiplied by 2 RFM = 132 in total • When a . is placed into a formula it means another compound is attached to the formula and must be included in the RFM e.g. CuSO4.5H2O means there are 5 water molecules added to this compound so you must include them in the total RFM = 249.5 in total

  12. Activity • Complete the worksheet IB 1.2 - Calculation of the Molar Mass of Compounds • There are 60 questions to complete • Don’t forget to refer to the previously stated rules • Show all working

  13. Using Molar masses • As we have seen 1g of Hydrogen contains many more atoms than 1g of carbon. • Calculate the number of atoms in 1g of hydrogen and 1g of carbon. • Suggest why we cannot use mass as a means of keeping a fair test during an experiment i.e. The reaction of substances with oxygen which would give out more energy 1g of hydrogen or 1g of carbon? • To keep a test fair we use molar quantities.

  14. Calculating molar quantities • As we have seen a mole is a fixed number of particles, 6 x 1023, we can find the molar mass of an element from the periodic table (mass number). The molar mass of a compound is just the sum of all the elements in the formula in their correct proportions. • E.g. • 1 mole of Na = 23g • 1 mole of NaOH = 23 + 16 + 1 = 40g • 1 mole of Na2SO4 = (2 x 23) + 32 + (4 x 16) = 142g

  15. Finding the number of moles • It is very rare to use exactly one mole of a substance in a reaction. So it is important to be able to find the number of moles of elements and compounds in different reactions. • We can do this using the following equation triangle. • You must learn this!!!!!! Mass ÷ RAM or RFM Number of moles x

  16. Example 1; • Find the number of moles in 46g of sodium Mass ÷ Mass 46 Number of moles ________ = 2 RAM or RFM Number of moles x 23 RAM or RFM

  17. Example 2; • Find the mass of 0.2 moles of sodium Mass Number of moles RAM or RFM x 4.6 g 0.2 23 ÷ Mass = RAM or RFM Number of moles x

  18. Activity • Complete the worksheet IB 1.3– Molar calculationsand IB 1.4 Molar calculations (II) • There are 60 questions to complete • Don’t forget to refer to the previously stated rules • Show all working

  19. Formula’s of compounds • Compounds can be described by different chemical formulas: • Empirical • Molecular • Structural

  20. Empirical Formula • Literally this is the formula obtained by experiment. • This shows the simplest whole number ratio of atoms of each element in a particle of the substance • It can be obtained by either knowing the mass of each element in the compound or from the percentage composition by mass of the compound. • The percentage composition can be converted directly into mass by assuming 100g of the compound are taken.

  21. Formula Triangle • You can use a formula triangle to help you rearrange the equation: Mass ÷ Mass Number of moles ________ = RAM or RFM Number of moles RAM or RFM x

  22. Example A compound contains 40.00% carbon, 6.73% hydrogen and 53.27% oxygen by mass, determine the empirical formula. 40.00/12.01 = 3.33 1 6.73/1.01 = 6.66 2 53.27/16.00 = 3.33 1 Empirical formula is therefore – CH2O

  23. Worked example An organic compound contains 60.00% carbon and 4.46% hydrogen by mass. Calculate the empirical formula of the compound. • As the percentages do not add up to 100% it can be assumed that the compound also contains 35.54% oxygen by mass. • Assume that there are 100g of compound, and calculate the amount of each element by dividing the mass of the element by its atomic mass Ar • Find the simplest ratio by dividing through by the smallest amount and then converting any decimals to give the simplest whole number ratio • Write the empirical formula, using subscripts for the numbers in the ratio. • The empirical formula of the compound is C9H8O4

  24. Activity; • Read through the examples 16, 17 and 18 on pages 31 and 32 in ‘Calculations in AS/A Level Chemistry’ – Jim Clark • Complete problems 15 and 16 part a only. • Answers are at the back of the book page 279 • Complete worksheet 1.5 Formulae and Percentage composition

  25. Molecular formula • For molecules this is much more useful as it shows the actual number of atoms of each element in a molecule of the substance. • It can be obtained from the empirical formula if the molar mass of the compound is known. • Methanal CH2O (Mr = 30), ethanoic acid C2H4O2 (Mr = 60) and glucose C6H12O6 (Mr = 180) are different substances with different molecular formulas but all with the same empirical formula CH2O. • Note that the subscripts are used to show the number of atoms of each element in the compound.

  26. Activity; • Using your answers from part a (from last lesson)complete problems 15 and 16 part b only. • Answers are at the back of the book page 279 • Complete worksheet; Worksheet IB1.6 Section B-Molecular Formulae

  27. Structural formula • This shows the arrangement of atoms and bonds within a molecule and is particularly useful in organic chemistry. • The three different formulas can be illustrated using ethene: CH2 C2H4 CH2CH2 Empirical Molecular Structural

  28. Structural formula • Two compounds with the same molecular formula but different structural formulas are known as isomers. • For example, both 1,1-dichloroethane and 1,2-dichloroethane have the same empirical formula, CH2Cl, and the same molecular formula, C2H4Cl2, but their structural formulas are different. • In 1,1-dichloroethane both the chlorine atoms are bonded to the same carbon atom, whereas in 1,2-dichloroethane each of the two carbon atoms has one chlorine atom bonded to it Cl H Cl Cl H Cl C C C C H H H H H H 1,1-dichloroethane 1,2-dichloroethane

  29. Structural formulas • The use of structural formulas is particularly important in organic chemistry. • Compounds with the same molecular formula but with different structural formulas are called structural isomers. • Structural isomers often have very different physical and chemical properties. For example, methoxymethane, CH3OCH3, and ethanol, C2H5OH, both have the same molecular formula, C2H6O but, unlike methoxymethane, ethanol is completely miscible with water and is generally much more reactive chemically. H H H H O H H C C C O C H H methoxymethane H H H H ethanol

  30. Quick Questions • What is the empirical formula of the following compounds? You could try to name them too. • A liquid containing 2.0g of hydrogen, 32.1 g sulphur, and 64g oxygen. • A white solid containing 0.9g beryllium, 3.2g oxygen, and 0.2g hydrogen. • A white solid containing 0.234g magnesium and 0.710g chlorine. • 3.888g magnesium ribbon was burnt completely in air and 6.488g of magnesium oxide were produced. • How many moles of magnesium and of oxygen are present in 6.488g of magnesium oxide? • What is the empirical formula of magnesium oxide? More to follow…

  31. Just a few more! • What are the empirical formulae of the following molecules? • Cyclohexane C6H12 • Dichloroethene C2H2Cl2 • Benzene C6H6 • Mr for ethane-1,2-diol is 62.0. It is composed of carbon, hydrogen and oxygen in the ratio by moles of 1:3:1. What is its molecular formula?

  32. Answers • 1 • H2SO4sulphuric acid • BeO2H2 which is Be(OH)2, beryllium hydroxide • MgCl2 magnesium chloride • 1 • 0.16 moles of each • MgO • 1 • CH2 • CHCl • CH • C2H6O2

  33. Working out formulae for ionic compounds • You can’t write equations until you can write formulae. • People tend to remember the formulae for common covalent substances like water or carbon dioxide or methane, and will rarely need to work them out. • That is not true of ionic compounds. You need to know the symbols and charges of the common ions and how to combine them into a formula.

  34. The need for equal numbers of “pluses” and “minuses” • Ions are atoms or groups of atoms which carry electrical charges, either positive or negative. • Compounds are electrically neutral. • In an ionic compound there must therefore be the right number of each sort of ion so that the total positive charge is exactly the same as the total negative charge. • Obviously, then, if you are going to work out the formula, you need to know the charges on the ions.

  35. Some charges you may remember! Can you put the correct charges into the table? +3 +1 -1 +2 -2 -3

  36. Cases where the name tells you the charge on an ion • A name like lead(II)oxide tells you that the charge on the lead (a metal) is +2. • Iron(III)chloride contains 3+ iron ion. • Copper(II) sulphate contains a Cu2+ion • Notice that all metals form positive ions. Ions that need to be learnt: Zn2+ • NO3- Ag+ • OH- • H+ • HCO3- • NH4+ • CO32- • SO42- Pb2+ • PO43-

  37. Confusing endings • Do not confuse ions like sulphate with sulphide. • A name like sodium sulphide means that it contains sodium and sulphur only. • Once you have an “ate” ending it means that there is something else there as well – often, but not always, oxygen • Try naming the following compounds: • Mg3N2 • Mg(NO3)2 • CaC2 • CaCO3

  38. Activity • Fill in the worksheet; Worksheet IB1.6 Important ions for IB calculationshow many can you complete without looking them up? • Click here for answers.

  39. Working out the formula of an ionic compound Example 1. • To find the formula for sodium oxide, first find the charges on the ions • Sodium is in Group 1, so the ion is Na+ • Oxygen is in Group 6, so the ion is O2- • To have equal numbers of positive and negative charges, you would need 2 sodium ions for each oxide ion • The formula is therefore Na2O

  40. Working out the formula of an ionic compound Example 2. • To find the formula for barium nitrate, first find the charges on the ions • Barium is in Group 2, so the ion is Ba2+ • Nitrate ions are NO3- • To have equal numbers of positive and negative charges, you would need 2 nitrate ions for each barium ion • The formula is therefore Ba(NO3)2 Notice the brackets around the nitrate group. Brackets must be written if you have more than one of these complex ions (ions containing more than one type of atom).

  41. Working out the formula of an ionic compound Example 3. • To find the formula for iron(III)sulphate, first find the charges on the ions • Iron(III) tells us that the ion is Fe3+ • Sulphate ions are SO42- • To have equal numbers of positive and negative charges, you would need 2 iron (III) ions for every 3 sulphate ions • The formula is therefore Fe2(SO4)3

  42. Activity • Complete the worksheet; Worksheet IB1.7 Writing formulae from names • For answers click here

  43. Chemical Reactions and equations Properties of chemical reactions Chemical equations State symbols One way or reversible reactions Ionic equations

  44. Properties of chemical reactions • Once the correct formulas of all the reactants and products are known, it is possible to write a chemical equation to describe a reaction taking place. • Newsubstances are formed. • Bonds in the reactants are broken and bonds in the products are formed, resulting in an energy change between the reacting system and the surroundings. • There is a fixed relationship between the number of particles of reactants and products, resulting in nooverall change in mass.

  45. Chemical Equations • In a chemical equation the reactants are written on the left hand side, and the products are written on the right hand side. • As there is no overall change in mass, the total amount of each element must be the same on the two side of the equation. • For example, consider the reaction between zinc metal and hydrochloric acid to produce zinc chloride and hydrogen gas. • The correct formulas for all the reacting species and products are first written down. Zn + HCl → ZnCl2 + H2 REACTANTS PRODUCTS

  46. Chemical Equations • The equation is then balanced by adding the correct coefficients • When the correct coefficients are in place, the reaction is said to be stoichiometricallybalanced. • The stoichiometry tells us that in this reaction two moles of hydrochloric acid react with one mole of zinc to form one mole of zinc chloride and one mole of hydrogen. Zn + 2 HCl → ZnCl2 + H2 REACTANTS PRODUCTS

  47. State symbols • The physical state that the reactants and products are in can affect both the rate of the reaction and the overall energy change so it is good practice to include the state symbols in the equation. (s) solid (aq) In aqueous solution (l) liquid (g) gas Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g) REACTANTS PRODUCTS

  48. Activity • Complete the worksheet; Worksheet IB1.8 Balancing equations • For answers click here

  49. One way or reversible • A single arrow → is used if the reaction goes to completion. • Sometimes the reaction conditions are written on the arrow: • Reversible arrows are used for reactions where both the reactants and products are present in the equilibrium mixture: Ni Catalyst, 180oC + C2H4(g) H2(g) C2H6(g)

  50. Ionic equations • Ionic compounds are completely dissociated in solution so it is sometimes better to use ionic equations to describe their reactions. • For example, when silver nitrate solution is added to sodium chloride solution a precipitate of silver chloride is formed. • Na +(aq) and NO3– (aq) are spectator ions and do not take part in the reaction. So the ionic equation becomes: + AgNO3(aq) NaCl(aq) + NaNO3(aq) AgCl(s) + Ag+(aq) Cl-(aq) AgCl(s)

More Related