Unimelb CHEM10003 ILT Notes

Chemistry ILT notes ILT 1 y When naming compounds, tert means tertiary, sec means secondary, and these apply to alkyl groups. Remember alkyl groups end in yl. -Trivial names we need to know: According to IUPAC nomenclature, "isobutyl", "sec-butyl", and "tert-butyl" are all retained trivial names. skeletal formula common name IUPAC name systematic name alternate notation n-butyl butyl butyl butan-1-yl isobutyl isobutyl 2-methylpropyl 2-methylpropan-1-yl sec-butyl sec-butyl 1-methylpropyl butan-2-yl tert-butyl tert-butyl 1,1-dimethylethyl 2-methylpropan-2-yl Butyl is the largest substituent for which trivial names are commonly used for all isomers. Note that the R group in the table is the rest of the molecule and the line structures are the side chains that correspond to the names shown. y y Note that isobutene is the equivalent of 2-methylpropane. Another trivial name we need to know is tert-butanol: The proper name for this compound is 2-methyl-2-propanol. y y y y y y IL 2 y The average atomic mass of Re is 186.207 amu. It is calculated from the two naturally occurring isotopes 185Re and 187Re. If the isotopic abundance of 185Re is 37.4% and the isotopic mass of 185Re is 184.953 amu calculate the mass of 187Re. , you need to use the percentages to work it out. So therefore, (184.953 × 0.374) + (X × (1-0.374)) = 186.207 X = 186.956amu. Remember the equations n= and C= and their applications. Remember the importance of mole ratios and calculations. Remember how to calculate empirical formulae and molecular formulae. Remember how to do limiting reagent problems. For more practice on the above 4 points, do the questions again on ILT -1. Remember how to do significant figures. The rules for identifying significant digits when writing or interpreting numbers are as follows: y y y y All non-zero digits are considered significant. For example, 91 has tw o significant figures (9 and 1), while 123.45 has five significant figures (1, 2, 3, 4 and 5). Zeros appearing anywhere between two non -zero digits are significant. Example: 101.12 has five significant figures: 1, 0, 1, 1 and 2. Leading zeros are not significant. For example, 0.00052 has two significant figures: 5 and 2. Trailing zeros in a number containing a decimal point are significant. For example, 12.2300 has six significant figures: 1, 2, 2, 3, 0 and 0. The number 0.000122300 still has only six significant figures (the zeros before the 1 are not significant). In addition, 120.00 has five significant figures. This convention clarifies the precision of such numbers; for example, if a result accurate to four decimal places is given as 12.23 then it might be understood that only two decimal places of accuracy are available. Stating the result as 12.2300 makes clear that it is accurate to four decimal places. The significance of trailing zeros in a number not containing a decimal point can be ambiguous. For example, it may not always be clear if a number like 1300 is accurate to the nearest unit (and just happens coincidentally to be an exact multiple of a hundred) or if it is only shown to the nearest hundred due to rounding or uncertainty. Various conventions exist to address this issue: y 4 3 y Isotopic abundance the atomic mass given on the periodic table is an AVERAGE. This is because in nature atoms exist as multiple isotopes, and there are differing proportions of these isotopes. So if the question as ed: 0 0 '     & % ! " y y y Av s e s 6 02 × 1023 Av s e ees e e c es 1 e s s ce So if they as e you how many moles of carbon were present in 1 000 000 carbon atoms it would be 1000000 ÷ 6 02 × 1023 Also, the atomic masses in the periodic table given are in grams PER MOLE. So if they as ed you the mass of a SINGLE carbon atom, it would be 12.01 ÷ 6.02 × 1023. 1 1 0 ) ( & % $    #        " ! " " !    © ¤ ¨§¦¥  ¤£¢ ¡  2 5 y y y y A bar may be placed over the last significant digit; any trailing zeros following this are insignificant. For example, has three significant figures (and hence indicates that the number is accurate to the nearest ten). The last significant figure of a number may be underlined; for example, "2 0000" has two significant figures. A decimal point may be placed after the number; for example "100." indicates specifically that three significant figures are meant. However, these conventions are not universally used, and it is often necessary to determine from context whether such trailing zeros are intended to be significant. If all else fails, the level of rounding can be specified explicitly. The abbreviation s.f. is sometimes used, for example "20 000 to 2 s.f." or "20 000 (2 sf)". Alternatively, the uncertainty can be stated separately and explicitly, as in 20 000 ± 1%, so that significant-figures rules do not apply. y y y y y y y y IL 3 y Remember that in a calculation, the answer should not have a greater accuracy than the original data, i.e. the significant figures of the answer should be the same as the lowest significant figure of the data given. Measurements should not have a greater number of significant figures than the equipment precision can support. Converting cm3 into m3: Since 100cm is 1m, 1000000cm 3 is 1m3. Converting dm 3 into m3: Since 10dm is 1m, 1000dm 3 is 1m 3. 1m = 1,000mm = 1,000,000 m = 1,000,000,000nm = 10,000 ,000,000Å (angstrom) = 1,000,000,000,000pm (picometre) 1L = 1dm3 1,000L = 1m 3 1GL = 1,000,000L y y y y 7 6 The pressure that must be exerted on a solution to prevent the passage of water/solvent molecules into the solution when the solution and solvent/water are separated by a semi-permeable membrane is called the os otic pressure. The higher the concentration of particles in the solution, the greater it attracts the water into it by osmosis, hence a greater pressure is required to stop the water from moving into the solution. Therefore, a higher concentration of solutes means a higher osmotic pressure. Isotonic/isosmotic solutions have the same osmotic pressure. Cells rupture when they are bathed in a hypotonic solution. Crenation is the name of the phenomenon whe re cells shrivel up after being placed in a hypertonic solution. When a liquid is in a closed container, the amount of liquid at first decreases but eventually becomes constant. The decrease occurs because there is an initial net transfer of molecules from the liquid to the vapour phase. However, as the number of vapour molecules increases, so does the rate of return of these molecules to the liquid. The process by which vapour y molecules re-form a liquid is called condensation. Eventually, enough vapour molecules are present above the liquid so that the rate of condensaton i equals the rate of evaporation at which point no further net change in the amount of liquid or vapour occurs because the two processes are at equilibrium. Hence, the pressure e erted by the vapour at this equilibrium is called the equilibrium vapour pressure, or more commonly known as the vapour pressure of the liquid. Solutes can affect the vapour pressure of a solution. If the added solute is non-volatile, i.e. does not evaporate easily, then the vapour pressure of the new solution will be lowered. Raoult's Law decribes the relationship between the vapour pressure of a solution and the amount of solute present, for an "ideal" solution. The equation describing Raoult s law is: Psoln = XsolventPosolvent Psoln is the observed vapour pressure of the solution, Xsolvent is the mole fraction of solvent, and Posolvent is the vapour pressure of the pure solvent. To calculate the mole fraction of the solvent: Xsolvent =  Colligative properties are the collective term that is used to describe the properties of solutions that are dependent on the number (not the type) of solute particles. Free ing-point depression, boiling-point elevation, and osmotic pressure are called colligative properties. Molality (not molarity) is equal to the number of moles of solute per kilogram of solvent and it has the unit m. @ 9 8 y y y y The Van't Hoff factor (i) is the relationship between the moles of solute dissolved and the resulting number of moles of particles in solution. It is a measure of the effect of a solute upon colligative properties such asvapor pressure, osmotic pressure and free ing point depression. The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved, and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van' t Hoff factor is essentially 1. For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance. For e ample, NaCl has a Van't Hoff factor (i) of 2, while Fe3(PO4)2 has i=5, while glucose has i=1. The normal boiling point of a liquid is the temperature at which the vapour pressure is equal to 1atm. We have seen in Raoult s law, that a non-volatile solute lowers the vapour pressure of the solvent. Therefore, such a solution must be heated to a higher temperature than the boiling point of the pure solvent to reach a vapour pressure of 1atm and boil . Hence, a non-volatile solute elevates and changes the boiling point of the solvent. The magnitude of the boiling-point elevation depends on the concentration of the solute, and can be represented by the equation: T=Kbmsolute C C B A y y y y y y y y Where the change in boiling point equals the molal boiling point elevation constant multiplied by the molality of the solution. Obviously a common solvent is water which as a molal boiling point elevation constant of 0.51oCkg/mol. Note pure water has a boiling point of 100 oC. Solutes can also lower the freezing-point, known as the freezing point depression. It is similar to the boiling point elevation in terms of the equation where: T= fmsolute Where the change in freezing point equals the molal freezing point depression constant multiplied by the molality of the solution. Obviously a common solvent is water which as a molal freezing point elevation constant of 1.86oCkg/mol. Note pure water has a freezing point of 0 oC. Remember in both boiling point elevation and freezing point depression, if the solute is able to dissociate (i.e. has a Van t Hoff factor of i>1 such as NaCl which has i=2) you need to multiply the right hand side of the equation by i, as you are increasing the number of particles in the solution that you actually dissolved. Since it is a colligative property (i.e. doesn t matter what the particles are), you need to multiply by i. If it says aqueous you can usually assume that its dissolved in water and f=1.86 and b=0.51. Another colligative property other than boiling point elevation and freezing point depression is osmotic pressure. We have already discussed what osmotic pressure is. It can be illustrated by the equation showing its dependence on the solution concentration: = MRT where is the osmotic pressure in atmospheres, M is the molarity of the solution, R is the gas law constant, and T is the temperature in elvin. The gas constant is 0.08206L atm 1 mol 1 in these calculations. It is 8.314J 1 mol 1 when used in the equations PV=nRT (pressure is also in kPa, not atm in these calculations). To convert between atm and torr , 1atm=760torr. To convert between atm and kPa, 1atm=101. 3kPa Density is mass divided by volume, e.g. g/L. It has an import ant relationship with molality and molarity (concentration of mol/L). You can obtain molality using density and molarity as illustrated in this example question (note the kg part of molality ONLY REFERS TO THE SOLVENT, not the solvent + the solute. It is only measuring moles of solute per kg of solvent, not per k g of the whole solution itself): The molality (m) of a 1.37 molar solution of citric acid (H 3C6H5O7) with a density of 1.10 g/cm 3 is : The density in g/L = 1100 g/L Therefore 1L has mass 1100 g = mass (H 3C6H5O7) + mass (H2O) Mass of H3C 6H5O7 in 1L = 1.37 x molar mass (H 3C 6H5O7) Mass of H2O = 1100-263 = 837 g Molality = 1.37moles / 0.837 kg = 1.64 m . For practice questions on all 4 of these calculations, refer to the textbook Zumdahl and also the ILT itself. F D F E F E