Friday 18 September 2015

The molar gas volume in calculations, moles, gas volumes and Avogadro's Lawstudy examples carefully

9. The molar gas volume in calculations, moles, gas volumes and Avogadro's Lawstudy examples carefully
  • Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules.
    • So the volumes have equal moles of separate particles (molecules or individual atoms) in them.
    • Therefore one mole of any gas (formula mass in g), at the same temperature and pressure occupies the same volume .
    • This is 24dm3 (24 litres) or 24000 cm3, at room temperature of 25oC/298K and normal pressure of 101.3 kPa/1 atmosphere (such conditions are often referred to as RTP).
    • The molar volume for s.t.p is 22.4 dm3 (22.4 litres) at 0oC and 1atmosphere pressure.
    • Historically, s.t.p unfortunately stands for standard temperature and pressure, but these days 25oC/298K is usually considered the standard temperature (RTP).
  • Some handy relationships for substance Z below:
moles Z = mass of Z gas (g) / atomic or formula mass of gas Z (g/mol)
  • mass of Z in g = moles of Z x atomic or formula mass of Z

  • atomic or formula mass of Z = mass of Z / moles of Z
  • 1 mole = formula mass of Z  in g.
gas volume of Z = moles of Z x volume of 1 mole
  • rearranging this equation gives ...
  • moles of Z = gas volume of Z / volume of 1 mole
  • moles = V(dm3) / 24   (at RTP)
The latter form of the equation can be used to calculate molecular mass from experimental data because
  • moles = mass / molecular mass = gas volume / volume of 1 mole
  • mass / molecular mass = gas volume / volume of 1 mole
  • molecular mass = mass x volume of 1 mole/volume
  • therefore at RTP: Mr = mass(g) x 24 / V(dm3)
  • so, if you know the mass of a gas and its volume, you can work out moles of gas and then work out molecular mass.
  • This has been done experimentally in the past, but these days, molecular mass is readily done very accurately in a mass spectrometer.
  • Note (i): In the following examples, assume you are dealing with room temperature and pressure i.e. 25oC and 1 atmosphere pressure so the molar volume is 24dm3 or 24000cm3.

  • Note (ii):
    • Apart from solving the problems using the mole concept (method (a) below, and reading any equations involved in a 'molar way' ...
    • It is also possible to solve them without using the mole concept (method (b) below). You still use the molar volume itself, but you think of it as the volume occupied by the formula mass of the gas in g and never think about moles!





    Methods of measuring how much gas is formed (volume can be compared with theoretical prediction!)
    • (a) methods of gas preparation - apparatus, chemicals and equation (c) doc b You can collect the gases in a calibrated gas syringe.
      • You must make sure too much gas isn't produced and too fast!
      • A gas syringe is more accurate than collecting the gas in an inverted measuring cylinder under water shown below, but its still only accurate to the nearest cm3.
      • You can collect any gas by this method.
    • (b) The gas is collected in a measuring cylinder filled with water and inverted over a trough of water.Methods of measuring how much gas is formed
    (volume can be compared with theoretical prediction!)


      You can get a more accurate result by using an inverted burette instead of a measuring cylinder.
    • However, this method is no good if the gas is soluble in water!
    • Burettes are calibrated in 0.10 cm3 intervals. measuring cylinders to the nearest cm3 or worse!
    • In both methods the reaction is carried out in conical flask fitted with a sealing rubber bung, but a tube enabling the gas evolved to be collected in some suitable container.
    • (c) A third method is to measure the gas loss by carrying out the reaction in a flask set up on an accurate one-pan electronic balance.
    • You need to put a cotton wool plug in the neck of the conical flask in case you lose any of the solution in a spray as the gas bubbles up - effervescence can produce an aerosol.
    • This method can be used for any reaction that produces a gas, but the gas is released into the laboratory, ok if its harmless.
    • It is potentially the most accurate method, BUT, the mass loss may be quite small especially hydrogen [Mr(H2) = 2], better for the 'heavier' gas carbon dioxide [MrCO2) = 44]
    • Molar

    Introduction to MOLES

    Introducing 'moles' and their the connection with mass and formula massstudy the mole examples carefully

    (a) WHAT IS THE MOLE CONCEPT? and WHAT IS ONE MOLE OF A SUBSTANCE?
    The 2nd part of the heading is 'easy', the first part is a bit more 'abstract' to get your head round!
    The mole concept is an invaluable way of solving many quantitative problems in chemistry!
    Its a very important way of doing chemical calculations!
    The theoretical basis is explained in section (b).
    The mole is most simply expressed as the relative 'formula mass in g' or the 'molecular mass in g' of the defined chemical 'species', and that is how it is used in most chemical calculations. The mass of one mole of a substance is sometimes referred to as the molar mass.
    The atomic/formula mass in grams = one mole of the defined substance.
    If your are dealing with individual atoms, one mole of equals the relative atomic mass in grams.
    This can be expressed as a simple formula ...
    moles of species = (actual mass of species in g) / (atomic/formula mass of species)
    therefore (using triangle on right if necessary)
     mass of species in g = moles species x atomic/formula mass of species
    atomic/formula mass of species = mass of species in g / moles of species
    Note these equations are for either an element or a compound,
    but, whatever, you must clearly define the chemical species you mean for any mole calculation e.g.
    Al metal element atom, H2O covalent molecule, an element O2 molecule, Na+Cl- ionic compound or just any compound formula like CuSO4 etc. etc.
    This specificity cannot be overemphasised.
     
    Mr is 'shorthand' for relative formula mass or molecular mass in amu (atomic mass units) and you must be able to work these out correctly from a given formula (Calculating relative formula/molecular mass of a compound or element molecule).
    The term relative molecular mass (sum of the atomic masses of the atoms in a single molecule of the substance) is usually applied to definite molecular species.
    Using the following atomic masses: H = 1, O = 16, N = 14, C = 12, Na = 23, Cl = 35.5, S = 32
    and the three formulae above relating moles, mass and formula mass ...
    molecular mass 18 for the water molecule H2O, 17 for the ammonia molecule NH3
    so 1 mole of water is 18g, 0.333 mole = 0.333 x 18 = 6g
    for ammonia 1mol = 17g, 34g = 34/17 = 2 mols ammonia
     
    16 for the methane molecule CH4 and 180 for the glucose sugar molecule C6H12O6
    so 0.5 mol methane = 0.5 x 16 8g, 72g = 72/16 = 4.5 mols methane
    for glucose 18g = 18/180 = 0.10 moles, 0.05 mole = 0.05 x 180 = 9.0g glucose
    the element nitrogen consists of N2 molecules, molar mass = 28g, 0.25 moles = 0.25 x 28 = 7.0g
    Relative atomic mass of iron is 56, 7g = 7/56 = 0.125 mol Fe (Relative atomic mass explained)
    So, these calculations are quite simple, but they are often just one part of solving a more complex problem.
     
    The term relative formula mass (sum of the atomic masses of the atoms in a specified formula) can be used for ANY specified formula of ANY chemical substance, though it is most often applied to ionic substances.
    e.g. mass of 1 mole of ionic sodium chloride NaCl or Na+Cl- is 58.5g (from 23 + 35.5)
    mass of 1 mole of ammonium sulfate (ionic salt) (NH4)2SO4 or (NH4+)2(SO42-) =  130g
     

    Thursday 17 September 2015

    Empirical formula and formula mass from reacting masses (easy start, no moles!)

     Empirical formula and formula mass from reacting masses (easy start, no moles!)study examples carefully


    The EMPIRICAL FORMULA of a compound can be worked out by knowing the exact masses of the elements that combine to form a given mass of a compound.
    The empirical formula of a compound is the simplest whole number ratio of atoms present in a compound.


     Here the word 'empirical' means from experimental data.
    Do not confuse with molecular formula which depicts the actual total numbers of each atom in a molecule.
    The molecular formula and empirical formula can be different or the same.
    They are the same if the molecular formula cannot be simplified on a whole number basis.
    Examples where molecular formula = empirical formula
    e.g. for sodium sulfate Na2SO4  and  propane C3H8
    You cannot simplify the atomic ratios 2 : 1 : 4 or 3 : 8 to smaller whole number (integer) ratios
    Examples of where molecular formula and empirical formula are different e.g.
    butane molecular formula C4H10, empirical formula C2H5
    numerically, the empirical formula of butane is 'half' of its molecular formula
    4 : 10 ==> 2 : 5
    glucose molecular formula C6H12O6, empirical formula CH2O
    numerically, the empirical formula of glucose is '1/6th' of the full molecular formula
    6 : 12 : 6 ==> 1 : 2 : 1


    The following examples illustrate the ideas using numbers more easily appreciated than in real experiments.

    In real laboratory experiments only a fraction of a gram or a few grams of elements would be used, and a more 'tricky' mole calculation method is required than shown here

    However the examples below show in principal how formulae are worked out from experiments.

    Any calculation method must take into account the different relative atomic masses of the elements in order to get to the actual ratio of the atoms in the formula.

    For example, just because 10g of X combines with 20g of Y, it does not mean that the formula of the compound is XY2 !

    If you divide the mass of each element by its atomic mass, you actually get the atomic ratio.

    • Empirical formula calculation Example 5.1 The compound formed between lead and sulfur
      • It is found that 207g of lead combined with 32g of sulphur to form 239g of lead sulphide.
      • From the data work out the formula of lead sulphide. (Relative atomic masses: Pb = 207 and S = 32)
      • In this case it easy to see that by the atomic mass ratio, 239 splits on a 1 to 1 basis of 1 atom of lead to 1 atom of sulphur (1 x 207 to 1 x 32 by mass)
      • so the formula is simply PbS
      • You can set out the calculation in a simple table format, in this case the numbers are very easy to deal with!
      •  



    RATIOS ...

    lead (Ar = 207)

    sulphur S (Ar = 32)

    Comments and tips

    Reacting mass

    207g

    32g

    not the real atom ratio

    atom ratio from mass / atomic mass values

    207/207 = 1

    32/32 = 1

    work out the simplest whole number ratio

    simplest whole number atom ratio by trial & error

    1

    1

    therefore the integer simplest ratio of 1 : 1 gives the empirical formula for lead sulphide as PbS

      • -
    • Empirical formula calculation Example 5.2 The empirical formula of a lead oxide
      • It is found that 207g of lead combined with oxygen to form 239g of a lead oxide.
      • From the data work out the formula of the lead oxide. (Relative atomic masses: Pb = 207 and O = 16)
      • In this case, you first have to work out the amount of oxygen combined with the lead.
      • By simple logic from the law of conservation of mass, this is 239 - 207 = 32g
      • In atomic ratio terms, the 207 is equivalent to 1 atom of lead and the 32 is equivalent to 2 atoms of oxygen (1 x 207 to 2 x 16),
      • so the formula is simply PbO2
      • Note: The mass of oxygen combined with the lead is deduced by subtracting the original mass of lead from final total mass of lead oxide.
     



    RATIOS ...

    lead (Ar = 207)

    oxygen O (Ar = 16)

    Comments and tips

    Reacting mass

    207g

    239-207 = 32g

    not the real atom ratio

    atom ratio from mass / atomic mass values

    207/207 = 1

    32/16 = 2

    work out the simplest whole number ratio

    simplest whole number atom ratio by trial & error

    1

    2

    therefore the simplest whole number ratio of 1 : 2 gives the empirical formula for this lead oxide as PbO2

    Its actually called lead(IV) oxide

    Credit to Docbrown



     
     

     
    Check out what is available? Study the different examples then try the Quizzes!
    CALCULATION of EMPIRICAL FORMULA
    Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations
    study examples carefully5. Simple empirical formula and formula mass from reacting masses
    or % composition (easy start, no moles involved !)study examples carefully

    Quantitative Chemistry calculations online Help for problem solving in doing empirical formula calculations, using experiment data, making predictions. Practice revision questions on empirical formula from reacting masses or % composition by mass of a compound. This page describes and explains, with fully worked out examples, how to work out the empirical formula of a compound. The empirical formula of a compound is defined and explained. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses
    Spotted any careless error? EMAIL query?comment or request a type of GCSE calculation not covered?
     Self-assessment Quizzes: type in answer click me for QUIZ!for F and H  or  multiple choice click me for QUIZ!for F and H
    study examples carefully5. Empirical formula and formula mass from reacting masses (easy start, no moles!)study examples carefully
    The EMPIRICAL FORMULA of a compound can be worked out by knowing the exact masses of the elements that combine to form a given mass of a compound.
    The empirical formula of a compound is the simplest whole number ratio of atoms present in a compound. (see section 3. for some simpler examples). Here the word 'empirical' means from experimental data.
    Do not confuse with molecular formula which depicts the actual total numbers of each atom in a molecule.
    The molecular formula and empirical formula can be different or the same.
    They are the same if the molecular formula cannot be simplified on a whole number basis.
    Examples where molecular formula = empirical formula
    e.g. for sodium sulfate Na2SO4  and  propane C3H8
    You cannot simplify the atomic ratios 2 : 1 : 4 or 3 : 8 to smaller whole number (integer) ratios
    Examples of where molecular formula and empirical formula are different e.g.
    butane molecular formula C4H10, empirical formula C2H5
    numerically, the empirical formula of butane is 'half' of its molecular formula
    4 : 10 ==> 2 : 5
    glucose molecular formula C6H12O6, empirical formula CH2O
    numerically, the empirical formula of glucose is '1/6th' of the full molecular formula
    6 : 12 : 6 ==> 1 : 2 : 1
    Where the empirical formula and molecular formula are different, you need extra information to deduce the molecular formula from the empirical formula (see link below).
    This page is only concerned with calculating empirical formula.
    For more advanced students see Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)
    The following examples illustrate the ideas using numbers more easily appreciated than in real experiments.
    In real laboratory experiments only a fraction of a gram or a few grams of elements would be used, and a more 'tricky' mole calculation method is required than shown here (dealt with later for higher students in section 8).
    However the examples below show in principal how formulae are worked out from experiments.
    Any calculation method must take into account the different relative atomic masses of the elements in order to get to the actual ratio of the atoms in the formula.
    For example, just because 10g of X combines with 20g of Y, it does not mean that the formula of the compound is XY2 !
    If you divide the mass of each element by its atomic mass, you actually get the atomic ratio.
    • Empirical formula calculation Example 5.1 The compound formed between lead and sulfur
      • It is found that 207g of lead combined with 32g of sulphur to form 239g of lead sulphide.
      • From the data work out the formula of lead sulphide. (Relative atomic masses: Pb = 207 and S = 32)
      • In this case it easy to see that by the atomic mass ratio, 239 splits on a 1 to 1 basis of 1 atom of lead to 1 atom of sulphur (1 x 207 to 1 x 32 by mass)
      • so the formula is simply PbS
      • You can set out the calculation in a simple table format, in this case the numbers are very easy to deal with!
      • RATIOS ...lead (Ar = 207)sulphur S (Ar = 32)Comments and tips
        Reacting mass207g32gnot the real atom ratio
        atom ratio from mass / atomic mass values207/207 = 132/32 = 1work out the simplest whole number ratio
        simplest whole number atom ratio by trial & error11
        therefore the integer simplest ratio of 1 : 1 gives the empirical formula for lead sulphide as PbS
      • -
    • Empirical formula calculation Example 5.2 The empirical formula of a lead oxide
      • It is found that 207g of lead combined with oxygen to form 239g of a lead oxide.
      • From the data work out the formula of the lead oxide. (Relative atomic masses: Pb = 207 and O = 16)
      • In this case, you first have to work out the amount of oxygen combined with the lead.
      • By simple logic from the law of conservation of mass, this is 239 - 207 = 32g
      • In atomic ratio terms, the 207 is equivalent to 1 atom of lead and the 32 is equivalent to 2 atoms of oxygen (1 x 207 to 2 x 16),
      • so the formula is simply PbO2
      • Note: The mass of oxygen combined with the lead is deduced by subtracting the original mass of lead from final total mass of lead oxide.
      • RATIOS ...lead (Ar = 207)oxygen O (Ar = 16)Comments and tips
        Reacting mass207g239-207 = 32gnot the real atom ratio
        atom ratio from mass / atomic mass values207/207 = 132/16 = 2work out the simplest whole number ratio
        simplest whole number atom ratio by trial & error12
        therefore the simplest whole number ratio of 1 : 2 gives the empirical formula for this lead oxide as PbO2 Its actually called lead(IV) oxide
      • -
    • Empirical formula calculation Example 5.3 The empirical formula of aluminium sulfide
      • It is found that 54g of aluminium forms 150g of aluminium sulphide.
      • Work out the formula of aluminium sulphide. (Relative atomic masses: Al = 27 and S = 32).
      • Amount of sulphur combined with the aluminium = 150 - 54 = 96g
      • By atomic ratio, the 54 of aluminium is equivalent to 2 atoms of aluminium and the 96 of sulphur is equivalent to 3 atoms of sulphur.
      • Therefore the atomic ratio is 2 to 3,
      • so the formula of aluminium sulphide is Al2S3
      • RATIOS ...aluminium (Ar = 27)sulfur S (Ar = 32)Comments and tips
        Reacting mass54g150-54 = 96gnot the real atom ratio
        atom ratio from mass / atomic mass values54/2796/32work out the simplest whole number ratio
        simplest whole number atom ratio by trial & error23
        therefore the simplest integer ratio of 2 : 3 gives the empirical formula for aluminium sulphide as Al2S3
      • -
    • Empirical formula calculation Example 5.4 From now on, questions just using the table method to work out empirical formula from more awkward numbers! In this case a compound formed between copper and chlorine.
      • A compound of copper contained 47.4% copper and 52.6% chlorine.
      • The atomic masses are: Cu = 64 and Cl = 35.5
      • Think of the percentages as masses in grams to solve the empirical formula problem.
      • RATIOS ...CuClComments and tips
        Reacting mass47.452.6not the real atom ratio
        atom ratio from mass / atomic mass values47.4/64 = 0.7452.6/35.5 = 1.48work out the simplest whole number ratio
        simplest whole number atom ratio by trial & error0.74/0.74 = 1.01.48/0.74 = 2.0
        therefore the simplest whole number ratio of 1 : 2 gives the empirical formula for copper chloride as CuCl2 Its actually called copper(II) chloride
      • -
    • Empirical formula calculation Example 5.5 The empirical formula of a compound of carbon and chlorine
      • It was found that 0.39 g of carbon was combined with 4.61g of chlorine.
      • Atomic masses: C = 12 and Cl = 35.5
      • RATIOS ...CClComments and tips
        Reacting mass0.394.61not the real atom ratio
        atom ratio from mass / atomic mass values0.39/12 = 0.03254.61/35.5 = 0.130work out the simplest whole number ratio
        simplest whole number atom ratio by trial & error0.0325/0.0325 = 1.00.130/0.0325 = 4.0
        therefore the simplest ratio of 1 : 4 gives the empirical formula is CCl4 Its actually called tetrachloromethane
      • -
    • Empirical formula calculation Example 5.6 The formula of a hydrocarbon.
      • It was that 0.75g of carbon was combined with 0.25g of hydrogen.
      • Atomic masses: C = 12 and H = 1
      • Calculate the empirical formula of the hydrocarbon
      • RATIOS ...CHComments and tips
        Reacting mass0.75g0.25gnot the real atom ratio
        atom ratio from mass / atomic mass values0.75/12 = 0.06250.25/1 = 0.25work out the simplest whole number ratio
        simplest whole number atom ratio by trial & error0.0625/0.0625 = 1.00.25/0.0625 = 4.0
        therefore the simplest ratio gives the empirical formula for the hydrocarbon = 1 : 4, so formula is CH4 This is the simplest hydrocarbon molecule called methane (main constituent in natural gas)
      • -
    • Empirical formula calculation Example 5.7 The analysis of sodium sulfate, calculating its empirical formula from the % composition by mass.
      • On analysis, the salt sodium sulfate was found to contain 32.4% sodium, 22.5% sulfur and 45.1% oxygen.
      • Atomic masses: Na = 23, S = 32 and O = 16
      • Calculate the empirical formula of sodium sulfate
      • RATIOS ...NaSOComments and tips
        Reacting mass32.422.545.1not the real atom ratio
        atom ratio from mass / atomic mass values32.4/23 = 1.4122.5/32 = 0.7045.1/16 = 2.82work out the simplest whole number ratio, in this case you have to make a reasonable judgement as to the values of the integers
        simplest whole number atom ratio by trial & error1.41/7 = 2.01 ~2.00.70/0.70 = 1.02.82/0.70 = 4.03 ~4.0
        the simplest ratio gives the empirical formula for sodium sulfate = 2 : 1 : 4, formula is Na2SO4
      • -
    • Empirical formula calculation Example 5.8 The formula of a hydrocarbon.
      • Analysis of hydrocarbon showed it consisted of 83.3% carbon and 16.7% hydrogen.
      • Atomic masses: C = 12 and H = 1
      • Calculate the empirical formula of the hydrocarbon (just think of it as 83.3g C combined with 16.7g H)
      • RATIOS ...CHComments and tips
        Reacting mass83.316.7not the real atom ratio
        atom ratio from mass / atomic mass values83.3/12 = 6.9416.7/1 = 16.7work out the simplest whole number ratioIn this case from the 1 2.4 to the 5:12 ratio, you have to multiply the 2.4 up until you get a whole number, x2, x3 and x4 don't work, but x5 does!
        simplest whole number atom ratio by trial & error6.94/6.94 = 1.016.7/6.94 = 2.4
        This is a bit awkward!1.0 x 5 = 5.02.4 x 5 = 12.0
        therefore the simplest ratio gives the empirical formula for the hydrocarbon = 5 : 12, so formula is C5H12 This is the simplest hydrocarbon molecule called pentane
      • -
    • -
    Self-assessment Quizzes
    [emp] type in answer click me for QUIZ!for F and H  or  multiple choice click me for QUIZ!for F and H
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