Maths and Chemistry for Biologists

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Maths and Chemistry for Biologists
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Chemistry 2 Solutions, Moles and Concentrations

• This section of the course covers
• why some molecules dissolve in water to form
  solutions
• how we measure the amounts of atoms and molecules
  - the mole
• The relative molar mass of molecules
• how the concentrations of solutions are described
  and how solutions of a given concentration are
  made

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The bonds in some molecules are polarised

• C H N H O H Cl H

increasing electronegativity of the heavy atom
As electronegativity increases the pair of
electrons in the bond is displaced towards the
heavy atom and hence the bond is
polarised Electrons spend a larger proportion of
their time near the electronegative atom. Shown
by e.g. ?- O H? The ? indicates a fraction
of a charge
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Water is a bent, polar molecule

• The oxygen in a water molecule has two
• lone pairs of electrons which take up
• space. The two bonds and two lone
• pairs point to the corners of a tetrahedron
• hence molecule is bent. The bonds are
• polar so the molecule can be drawn as
• In water the value of ? is about 0.3

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Consequences of the structure of water (1)

• It forms hydrogen bonds electrostatic
  interaction
• between ? on H and ?- on O of adjacent molecules
• O H? ?-O H
• hydrogen bond
• H-bonds are weak (about 1/20th strength of
  covalent
• O H bond) and long (about 0.176 nm compared
• with O H bond) but are enormously important for
• the properties of water.
• H-bonds also occur in compounds containing N H

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Consequences of the structure of water (2)
Water dissolves ionic compounds such as NaCl
because it interacts electrostatically with the
ions
Water dissolves polar molecules such as ethanol
because it interacts electrostatically with the
partial charges
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Consequences of the structure of water (3)
In liquid water molecules stick together to give
a partially ordered structure
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The need to weigh atoms and molecules

• Consider the reaction Na F ? NaF
• One meaning of this is that 1 atom of sodium
  reacts with 1 atom of fluorine to give 1 molecule
  of NaF
• Cant weigh atoms so how can we know how much
  sodium to take to react exactly with a given
  amount of fluorine?

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Weighing atoms continued
Na has a mass number of 23 (it contains 11
protons and 12 neutrons) so its relative atomic
mass is 23 F has a mass number of 19 (9 protons
and 10 neutrons) so its relative atomic mass is
19 So 23 g of sodium will react exactly with 19
g of fluorine to give 42 g of sodium fluoride
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The mole

• 23 g of Na and 19 g of F contain the same numbers
  of atoms
• 23 g of Na is a mole of sodium
• 19 g of F is a mole of fluorine
• Definition- a mole is the amount of a substance
  that contains the same number of particles as
  there are atom in 12 g of the isotope 12C
• This is the Avogadro constant equal to 6.022 x
  1023

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The mole continued

• Na F ? NaF
• Also means that 1 mol of Na reacts with 1 mol of
  F to give 1 mol of NaF
• (mol is the abbreviation of mole)
• What the mol gives us the weights of atoms that
  will exactly react together to give molecules

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One for you to do

• Relative atomic mass of fluorine 19
• What mass of F is 0.1 mol?
• How many mol of F is 38 g?

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Answers

• 1 mol 19 g ? 0.1 mol 1.9 g
• 19 g 1 mol ? 38 g 2 mol

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Relative molar mass (Mr) of compounds

• Calculate these by adding the relative atomic
  masses (Ar) of the constituent atoms
• H2O 1 x 2 16 18
• Urea (NH2CONH2) (2 x 14) (4 x 1) 12 16
  60
• Hence 18 g of water is 1 mol
• 1.8 g of water is 0.1 mol
• 18 ?g of water is 18 ?mol
• 60 g of urea is 1 mol

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Molecular weight

• You will see this term used sometimes. It is
  numerically equal to the relative molar mass but
  it needs units. The correct unit is the Dalton
  (Da) which is 1/12th of the mass of an atom of
  12C
• Better to stick to relative molar mass

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One for you to do

• C O2 ? CO2
• (means that 1 mol of carbon reacts with 1 mol of
  oxygen molecules to give 1 mol of carbon dioxide)
• What weight of O2 is required to react completely
  with 1.2 g of C?

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Answer

• 1.2 g of C is 0.1 mol.
• Hence we require 0.1 mol of O2
• Mr of 02 is 2 x 16 32
• Hence we need 3.2 g of O2
• Product is 4.4 g (0.1 mol) of CO2

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Concentrations of solutions

• Expressed in terms of MOLARITY the number of
  moles per litre
• A 1 MOLAR solution contains 1 MOLE in 1 LITRE
• or a 1 M solution contains 1 mol/L
• NB the abbreviations for mole (mol) and molar (M)
  NEVER confuse them. One is an amount of
  substance (mol) the other is a concentration (M)

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Concentrations of solutions (contd)

• A 1 MOLAR solution contains 1 MOLE in 1 LITRE
• or 1 mol in 1 L gives a 1 M solution
• 1 mol/L solution is 1M
• 0.1 mol/L solution is 0.1 M
• 0.1 mol/100 ml solution is 1 M
• 1 mol/100ml solution is 10 M
• 1 mmol/ml solution is 1M

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Examples of how to make solutions

• How much NaOH is needed to make 500 ml of a 0.2 M
  solution?
• Mr of NaOH is 23 16 1 40. 1L of a 0.2 M
  solution contains 0.2 mol. Hence 500 ml contains
  0.1 mol. 0.1 mol of NaOH 4 g
• How much KCl is there in 100 ml of a 2 M soln?
• Mr of KCl is 39 35.5 74.5. 1L of 2M soln
  contains 2 mol hence 100 mol contains 0.2 mol.
• 0.2 mol of KCl is 0.2 x 74.5 14.9 g

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More examples

• What is the conc of a solution of NaH2PO4
  containing 2g/L?
• Mr is 23 2 31 64 120. Hence 2 g is 2/120
  of a mol or 0.0167 mol. This is contained in 1 L
  so conc 0.0167 M (or 1.67 x 10-2 M or 16.7 mM)
• How many mol is 1 mg of a substance with Mr
  250?
• 1 mg 10-3 g. Hence 10-3/250 4 x 10-6 mol or
• 4 ?mol

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One for you to do

• What is the concentration of a solution of NaOH
  containing 20 g in 500 ml? (Mr 40)
• How much NaOH is required to make 10 ml of a 1M
  solution?

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Answer

• Solution has 20 g in 500 ml and hence 40 g in 1
  L. 40 g is 1 mol and hence conc 1 M
• For 1 M need 40 g / L
• 10 ml 0.01 L. Hence need 0.01 x 40 g 0.4 g
• (or 400 mg)

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Dilutions

• If you dilute a solution (add more solvent) the
  amount of solute remains the same but the
  concentration decreases
• Take 250 ml of a 0.1 M solution and dilute to 1L
• New conc 0.1 x 0.025 M
• Amount of solute remains as 0.025 mol

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Example

• How would you make 50 ml of a 10 mM solution by
  dilution of a 1M stock solution?
• New solution has conc 1/100 that of stock. So
  need 50/100 0.5 ml of stock solution plus
• 49.5 ml water
• or 1 L of 10mM solution contains 10 mmol of
  solute so 50 ml (1/20 L) contains 0.5 mmol.
• Stock has 1 mol/L or 1 mmol/ml. So need 0.5 ml
  of stock plus 49.5 ml water as above.