AS Chemistry Unit 1 PDF
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This document appears to be notes on AS Chemistry, covering topics like formulae, equations, and amounts of substance. It includes examples and definitions related to chemical concepts.
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AS CHEMISTRY UNIT 1 1.1 Formulae, Equations and Amounts of Substance Writing and Balancing Chemical Equations Molecular Ions. Ions with “ate” have the ION NAME FORMULAE...
AS CHEMISTRY UNIT 1 1.1 Formulae, Equations and Amounts of Substance Writing and Balancing Chemical Equations Molecular Ions. Ions with “ate” have the ION NAME FORMULAE highest number of oxygens. Sulfate SO42- Ions with “ite” have a lower number of oxygens. Sulfite SO32- (e.g. Sulfate / Sulfite, Nitrate / Thiosulfate S2 O32 - Nitrite) Hydrogensulfate HSO4- Hydrogencarbonate HCO3- Carbonate CO32 - Nitrate NO3- Nitrite NO2- Phosphate PO43- Chlorate ClO3- Hypochlorite ClO- Hydroxide OH- Dichromate Cr2O 2- 7 Chromate CrO42 - Permanganate MnO4- Ammonium NH4+ roup 7 elements are all diatomic. G FINCHBRO F - Fluorine Br - Bromine I - Iodine O - Oxygen N - Nitrogen C - Chlorine H - Hydrogen 1 ompounds sometimes use prefixes to show the amount of a specific element it C contains; - “mono-”= one - “di-”= two - “tri-”= three - “tetra-”= four - “penta”= five Writing Ionic Equations . S 1 tart with an ordinary equation. 2. Write all aqueous ionic compounds (salts, acids and bases) with ions separated. 3. Write all insoluble ionic compounds and the covalent compounds in the usual way. 4. Cross out spectator ions (ions that appear on both sides of equation). Example (1.)CuSO4(aq) + Mg(s) → MgSO4(aq) + Cu(s) (2. & 3.)Cu2+(aq)+ SO42 -(aq) + Mg(s) → Mg2+(aq)+ SO42 -(aq)+ Cu(s) (4.)Cu2+(aq)+SO42 -(aq) + Mg(s) → Mg2+(aq)+SO42 -(aq)+ Cu(s) Final Eq.Cu2+(aq)+ Mg(s)→ Mg2+(aq)+ Cu(s) RAM, RMM, RFM and Molar Mass AM; Relative atomic mass, used for singular elements. R RMM; Relative molecular mass, used for covalentlybonded compounds. RFM;Relative formula mass, used for ionic compounds. Molar Mass;MMis the mass of one mole of a substance. elative formula mass is calculated by adding up all relative atomic masses of each R atom within a compound. (e.g. NaCl would have a RFM of 58.5, because Na has an RAM of 23, and Cl has an RAM of 35.5.) The Mole Avogadro’s constant- defined as “the number of atomsin 12.000g of carbon-12.” mole of a substance is the amount of a substance that contains the Avogadro A constant (6.02x1023) number of atoms, molecules orgroups of ions. 2 ( e.g. “Cu + S → CuS” Can be read as 1 mole of copper atoms reacts with 1 mole of sulfur atoms to form 1 mole of copper (II) sulfide. This is also due to the ratios, of which are 1:1:1.) Calculating the amount of moles in a solid. For solid matter, this triangle is used. o find the amount of moles in a substance, the T equation; 𝑚𝑎𝑠𝑠(𝑔) 𝑛 = 𝑅𝐹𝑀 he expression can be rearranged to find the T Mass or RFM of a substance instead. Example; - Calculate the amount, in moles, present in 2.71g of carbon dioxide. Give your Q answer to three significant figures. 𝑚𝑎𝑠𝑠(𝑔) 2.71 A -𝑛 = 𝑅𝐹𝑀 = 44 = 0. 0616𝑚𝑜𝑙 (See workbook for other examples) Calculating the amount of moles in a solution. For solutions, this triangle is used. 𝑣𝑜𝑙𝑢𝑚𝑒( 𝑐𝑚3)𝑥 𝑐 𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛( 𝑚𝑜𝑙/𝑑 𝑚3) 𝑛 = 1000 ince volume is typically measured in “cm3” , the S equation is divided by 1000 to get it into “dm3”. dm3 = to 1L, chemists prefer to use 1dm3. 1cm3 is 1 the same as 1ml, there are 1000cm3in 1dm3. Question may ask for the answer to a certain umber of significant figures or decimal places but if not it is usually best to give 3 n significant figures if the answer is usually best to give 3 significant figures if the 3 nswer is less than 1 and give the answer to 2 decimal places if the answer is a greater than 1. Example; - Calculate amount, in moles, of sodium hydroxide present in 15.0cm3 of a solution Q of concentration 1.25 mol dm-3. Give your answer to3 sig. figs. 𝑣𝑜𝑙𝑢𝑚𝑒( 𝑐𝑚3)𝑥 𝑐 𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛( 𝑚𝑜𝑙/𝑑 𝑚3) 15.0𝑥1.25 A -𝑛 = 1000 = 1000 = 0. 0188𝑚𝑜𝑙 sing the Avogadro’s constant U Avogadro’s constant (NA or L) is used in calculations which involve number of particles or the mass of a certain number of particles. Avogadro’s constant is equal to 6.02x1023 mol-1. Example; Q - Calculate mass of 100 zinc atoms. 100 −22 A -𝑛 = 23 = 1. 66 × 10 𝑚𝑜𝑙 6.02×10 −22 −20 𝑚 = 1. 66 × 10 × 65 = 1. 08 × 10 𝑔 eacting Mass Questions R Balanced symbol equation is key to these calculations. 1. Calculate number of moles for reactant whose mass is given in question. 2. Use balancing numbers to calculate number of moles for the desired substance. 3. Convert moles of this substance to mass (or volume required). Example; 4 Limiting Reactant and Excess Reactant imiting reactantis reactant in a chemical reactionthat limits amount of product L that can be formed. Reaction ceases when limiting reactant is consumed. xcess reactantis reactant that remains when a reactionstops due to the limiting E reactant being consumed. he bike frames and wheels on the left represent the reactants. Completed bikes are T the products. Bike frames are the limiting reactant, wheels are in excess. The bike frames limit how much product is obtained and the wheels left over at the end are in excess. In a chemical reaction, limiting and excess reactant can be determined from the number of moles and using the balanced equation. Example; 5 Water of Crystallisation any salts (formed from acids) when solid are hydrated (a salt which contains water M of crystallisation). Water of crystallisation is water chemically bonded within a crystal structure. If hydrated salt heated to constant mass in open container (so water vapour can escape) all water of crystallisation is removed, leaving behind an anhydrous salt (a salt that contains no water of crystallisation). xam questions often ask how E you carry out heating a hydrated salt to constant mass. You would heat and weigh and repeat this process until the mass no longer changes. uestions often ask what initial Q weighings you would make; should weigh the empty evaporating basin, and the mass of the basin with the hydrated salt inside of it. etermining the degree of hydration D If asked to find the degree of hydration using mass data or percentage data, use the following flow chart; . 1 ind the mass of the anhydrous salt and mass of water lost; F 2. Find the moles of the anhydrous salt (Divide by RFM); 3. Find the moles of the water lost (Divide by water’s RFM); 4. Divide the moles of the anhydrous salt and the moles of water by the moles of the anhydrous salt; . This gives the ratio ofanhydrous salt : water, thevalue of water is the 5 degree of hydration. Example: 6 See booklet for more examples. 1.2 Atomic Structure Atomic Structure toms are composed of three subatomic particles; protons, neutrons and electrons. A “Relative”, used to compare particles as the masses and charges of these particles are so small: Subatomic Particle Relative Mass Relative Charge Location in Atom Proton 1 +1 Nucleus Neutron 1 0 Nucleus Electron 1/1840 -1 Shells Definitions; 7 Atomic number Number of protons in (the nucleus of) an atom. Mass Number otal number of protons and neutrons in (the nucleus T of) an atom. Relative Atomic Mass (RAM) he average (weighted mean) mass of an atom of an T element relative to one-twelfth of the mass of an atom of carbon-12. Relative Isotopic Mass (RIM) ass of an atom of an isotope of an element relative to M one-twelfth of the mass of an atom of carbon-12. tomic number (Z), often known as proton number. Atomic A number is always the same. ass number (A), is sum of protons and neutrons, so M subtracting the atomic number from the mass number gives the number of neutrons. Relative Formula Mass (RFM) verage (weighted mean) mass of a formula unit A relative to one-twelfth of the mass of an atom of carbon-12. Relative Molecular Mass (RMM) verage (weighted mean) mass of a molecule A relative to one-twelfth of the mass of an atom of carbon-12. elative molecular mass is calculated from total of all relative atomic masses in a R single molecule. MM is used for molecular covalent elements and compounds. RFM is used for R everything else. toms electrically neutral (same number of protons and electrons). Simple ions are A charged particles which form when atoms gain or lose electrons. Charge on an ion can be determined by subtracting number of electrons from number of protons. Relative Atomic Mass (RAM) Chlorine has two isotopes: chlorine-35 and chlorine-37. 8 he relative atomic mass of chlorine is the average mass of the atoms taking into T account the proportions in which they occur. Σ(𝑚𝑎𝑠𝑠𝑜𝑓𝑖𝑠𝑜𝑡𝑜𝑝𝑒×𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑎𝑏𝑢𝑛𝑑𝑎𝑛𝑐𝑒) 𝑅𝐴𝑀 = Σ𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒𝑎𝑏𝑢𝑛𝑑𝑎𝑛𝑐𝑒 Isotopes are atoms which have the same atomic number as an element, but have a different mass number. (same no. of protons, diff. no. of neutrons). xample; E Q. Chlorine exists as two isotopes,35Cl and37Cl,which occur in the relative proportions 75% and 25% respectively. Calculate the relative atomic mass of chlorine. (75×35)+(25×37) 3550 A.𝑅𝐴𝑀 = (75+25) = 100 = 35. 5 Isotopesare atoms which have the same atomic numberbut different mass number (same number number of protons but different number of neutrons). Mass Spectrometry A mass spectrometer is used to determine the mass of atoms and molecules. A mass spectrometer atomises (turns into gas) and ionises a sample, producing ions with a single positive charge. Assumed all ions in a mass spectrometer have a single positive charge. If a mass spectrometer is used on an element, it’ll show a spectrum with the asses and relative abundances for all isotopes of the element. m Mass Spectrum Graphs eaks are also given along the relative P abundances. Horizontal axisis always the measurement of mass. Vertical axisis usuallyrelative abundance or percentage abundance. an be asked to identify the species that C causes a peak at a certain mass value. Include mass number, and all species in a mass spectrum are assumed to have a single positive charge. 9 Mass Spectrum of a Diatomic Element For a diatomic element, there would be 5 peaks in the spectrum. ○ 35 + 37 + 35 Cl , Cl , [ Cl-35Cl]+, [37Cl-37Cl]+, and [35Cl-37Cl]+ m/z Species 35 35 Cl+ 37 37 Cl+ 70 [3 5Cl-35Cl]+ 72 [3 5Cl-37Cl]+ 74 [3 7Cl-37Cl]+ 35 land37Cl naturally occur in a 3:1 ratio, C explains why peak at 35 is 3x larger than the peak at 37. Peaks at 70, 72, and 74 are in a ratio of 9:6:1. Bromine exists as79Br and81Br. Its isotopes arein a 1:1 ratio, and the relative abundance of the peaks at 158, 160 and 162 are in a ratio of 1:2:1. Mass Spectrum of a Compound Mass spectrum for a compound is more complicated because molecule breaks up during the process. Thelast major peakin the mass spectrum of a compoundis called the molecular ion peak. Molecular ion peak has same RMM as the compound. In mass spectrum for ethanol (CH3C H2O H). The peaks seen below the molecularion peak are the fragmentation pattern, they’re unique to each compound.Can be used alongside RMM to identify a compound. 10 + S pecies responsible for the peak at 46 is CH3C H2O H . The peak with the highest relative abundance in any mass spectrum is called the base peak. Electronic Configuration E lectrons arranged in energy levels in which the energy of electrons increase as their distance from the nucleus increases. Energy levels are labelled “n=1” (closest to nucleus), “n=2”, “n=3”, etc. Energy levels are subdivided in subshells, which are made up of orbitals. Anorbitalis a region within an atom that can holdup to two electrons with opposite spin. ○ An “s subshell” is made up ofones orbital; ○ A “p subshell” is made up ofthreep orbitals; ○ A “d subshell” is made up offived orbitals. S -orbitalsare spherical in shape, can hold 2 electrons. P-orbitalsare dumbbell shaped, come in set of 3.Since each orbital holds 2 electrons, the three p-orbitals can hold a maximum of 6 electrons. D-orbitalscome as a set of 5. Each d-orbital holds2 electrons, in total the five d-orbitals can hold a maximum of 10 electrons. Type Shape tart at which S umber of this N aximum M energy level type of orbital number of in a subshell electrons s Spherical 1 1 2 p Dumbell 2 3 6 d Not required 3 5 10 f Not required 4 7 14 A t n=1 there is only an s-subshell; At n=2 there is an s-subshell and a p-subshell; At n=3 there is an s-subshell, p-subshell, and d-subshell. 11 E ach individual orbital holds two electrons which spin in opposite directions to minimise repulsion. Opposite spins often represented as ⇅ in “electrons-in-box” diagrams. Should be written in the following order: 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p… ○ In terms of energy, it would be 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p… When losing electrons, subshells lose electrons from the highest numbered orbitals first. Transition metals have their outermost electrons in the 3d subshell, but lose their 4s electrons first. Electrons fill in order of energy levels and orbitals closest to the nucleus. Ground state is a term which describes the electronic configuration when all the electrons are in the lowest available energy levels. Atoms in SPD Notation (With new orbitals) Atomic Number Element SPD Notation 1 Hydrogen 1s1 3 Lithium 1s22s1 5 Boron 1s22s22p1 11 Sodium 1s22s22p63 s1 13 Aluminium 1s22s22p63 2 s 3p1 19 Potassium 1s22s22p63 2 s 3p64 s1 21 Scandium 1s22s22p63 2 s 3p64 2 s 3d1 24 Chromium 1s22s22p63 2 s 3p64 1 s 3d5 29 Copper 1s22s22p63 2 s 3p64 1 s 3d10 31 Gallium 1s22s22p63 2 s 3p64 2 s 3d104p1 C hromium and copper are exceptions, they only have 1 electron in their 4s orbital, so that the 3d orbitals can be half filled or completely filled. ○ Half filled or completely filled p or d-orbitals are more stable. T o determine the electronic configuration of an atom and/or ion, write the atom’s normal configuration, then add / remove electrons to make the ion. ○ When working out the electronic configuration, first write it in order of power, then write it in numerical order. ○ e.g.1s22 2 s 2p63 2 s 3p64 2 s 3d1→ 1s22s22 p63 2 s 3p63 d14 2 s 12 Blocks of the Periodic Table If outer electrons in s-subshell, element is in the s-block. If outer electrons in p-subshell, element is in the p-block. If outer electrons in d-subshell, element is in the d-block. Electrons-in-boxes diagrams Electronic configuration can be asked for in “electrons-in-box” form. Electrons only pair when no other space is available in the subshell. Sometimes you’ll have to identify and label the subshells 1s, 2s, 2p, etc. S subshell only has one orbital, P subshell has three orbitals, and the D subshell has five orbitals. Can be asked up to4p. When filling p and d subshells, there should only be one electron in each orbital until each is half filled, then start pairing electrons. Draw electrons as arrows pointing up and down, opposite directions represent different directions of spin. Charge Electrons gained / lost + Lost electrons - Gained electrons 13 Ionisation Energy F irst ionisation energyis the energy required toconvert 1 mol of gaseous atoms into gaseous ions with a single positive charge. X(g) → X+( g) + e- S econd ionisation energyis the energy required toconvert 1 mol of gaseous ions with a single positive charge into ions with a double positive charge. X+ ( g) → X2+(g) + e- T hird ionisation energyis the energy required toconvert 1 mol of gaseous ions with a double positive charge into ions with a triple positive charge. X2 +(g) → X3+(g) + e- Could be asked to write equation for 1st, 2nd, or 3rd, etc ionisation energy. V alues of ionisation energy are always endothermic, always measured in “ kJ mol-1”; Atoms and ions must be gaseous; Only 1 mol of electrons removed each time; Electrons do not need state symbols. First Ionisation Energy for Elements 1-36 Graph shows change in first ionisation energy from hydrogen (atomic no. 1) to krypton (atomic no. 36). 14 Factors to Explain Changes in Ionisation Energy 1. Nuclear Charge; 2. Atomic Radius; 3. Shielding (By inner electrons); a. Inner shell electrons block attraction that nucleus has for outermost electrons. 4. Stability of half-filled and filled subshells. Trends across the graph of first ionisation energy Across a period: first ionisation energy generallyincreases (some decreases). ○ Why?Nuclear charge increases across a group, andatomic radius decreases, shielding is not important here, because across a period, atoms have the same number of inner electrons. Down a group: First ionisation decreases. ○ Why? Atomic radius increases and shielding increases.Elements in Groups II, V and 0 have higher than expected first ionisation energy values due to stability of half-filled and filled subshells. W hy is there a decrease in first ionisation energy from Be to B in period 2? ○ Beryllium has configuration of 1s2, 2s2, whereas Boronhas configuration of 1s2, 2s2, 2p1. Boron’s outer electronis in a 2p orbital, which is higher in energy and further from nucleus, so it’s easier to remove. Could also be explained that outer 2s electrons of beryllium are in a filled subshell, so they’re more stable and require more energy to remove. W hy is there a decrease in first ionisation energy between nitrogen and oxygen in period 2? ○ Nitrogen atom-1s2, 2s2, 2p3 - has a half filled2p subshell, which is more stable. ○ Oxygen atom- 1s2, 2s2, 2p4 - more than half-filled2p subshell, repulsion between two electrons in same subshell, making it easier to remove one. Successive Ionisation Energies L arge break occurs in successive ionisation energies of an element when moving from one energy level to another that is closer to the nucleus. 15 D iagram shows successive ionisation energies values for sodium. Plotted as Log (ionisation energy) as there is large difference in the ionisation energy values and couldn’t be plotted on a conventional scale. ○ For instance; 1st ionisation energy is +500 kJ mol-1,but 11th ionisation energy is +158700 kJ mol-1 A question may give successive ionisation energies for some elements, and ask to determine the group to which the element belongs. Look for large jump in successive ionisation energy as this occurs when all outer electrons have been removed, and indicate number of electrons in outer shell. 1.3 Bonding Bonding Looking at bonding of; ○ Metals; ○ Ionic compounds; ○ Covalent compounds. Semimetals (metalloids) have properties of both metals and non–metals. Metals M etals contain metallic bonding (electrostatic attraction between layers of positive ions (cations) and delocalised electrons). This structure described as metallic lattice (regular arrangement of atoms / ions). 16 Physical Property xplanation of physical property in terms of structure and E bonding Hardness trong attraction between positive ions and negative electrons, S and a regular structure. High melting point arge amount of energy is required to break the bonds, which L are strong attractions between positive ions and negative electrons. ood electrical G elocalised electrons can move and carry charge through the D conductivity metal. alleability and M ayers of positive ions can slide over each other without L ductility disrupting the bonding. Metallic bonds are stronger when there are more delocalised electrons. ○ Sodium has M.P. of 98OC , and magnesium has M.P. of625OC . Mg has two electrons in its outer shell, so can delocalise two electrons per atom, whereas Na can only delocalise one electron per atom. More electrons that are delocalised, the stronger the metallic bond. D-block metals have many outer electrons, so have higher melting points. Ionic Compounds Ionic bonding is the electrostatic attraction between oppositely charged ions in a regular ionic lattice, the structure of an ionic compound is described as an ionic lattice. Generally contain a metal (group I or II particularly) and non-metal (particularly from group VI or VII) Positive ions calledcations, negative ions calledanions ○ Cat-ions are paws-itive. When cations form from atoms, name stays same as parent atom. Simple anions have an -ide suffix. (ox-ide [O2-]). Positive molecular ions end in -onium. (ammonium [NH4+ ]) Negative molecular ions usually end in -ate or -ite. (sulfate [SO42 -]), (hypochlorite [OCl-] ). Unusual ones end in -ide. (hydroxide[OH-] ). 17
