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# Moles P1

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A set of slides created to teach Moles P1 to learners at Bishops Diocesan College in Cape Town.

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### Moles P1

1. 1. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY K Warne
2. 2. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Atomic Weights You must be able to… · Describe the mole as the SI unit for amount of substance · Relate amount of substance to relative atomic mass · Describe relationship between the mole and Avogadro’s number · Conceptualise the magnitude of Avogadro’s number · Describe the relationship between molar mass and relative molecular mass · Calculate the molar mass of a substance given its formula
3. 3. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY THE NEUTRAL ATOM • The atom consists of a ……………. containing protons and neutrons surrounded by a cloud of …………………. • Atomic Number (Z) Number of ……………. in the Nucleus (= number of electrons in a neutral atom.) • Mass number (A) = Number of protons + neutrons. Notation Z A X………….. Number (smaller) …………. …………. Number (bigger) …………. symbol Neutrons = ………………………
4. 4. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY THE NEUTRAL ATOM • The atom consists of a nucleus containing protons and neutrons surrounded by a cloud of electrons. • Atomic Number Z - Number of protons in the Nucleus = number of electrons in a neutral atom. • Mass number A = Number of protons + neutrons. Notation Z A XAtomic Number (smaller) = PROTONS Mass Number (bigger) = P + N symbol Neutrons = Mass number – Atomic Number = A - Z
5. 5. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Relative Mass Atomic • Certain products, such as paper for example, are sold by the ream. A ream is 500 sheets. Since it is impractical to actually count out 500 sheets, the weight (mass) of 500 sheets is determined; then each ream is packaged according to this mass. • Atoms are even smaller than paper, so it is not possible to actually count them. However, it is possible to know the mass of an atom in respect to the mass of another atom. • The Relative mass of an object is expressed by comparing it mathematically to the mass of another object. So the relative mass of an orange in relation to a grapefruit is .6. The relative mass of the grapefruit in relation to a grapefruit is 1.0. • Atoms are compared to the lightest atom (hydrogen) which is 12 times lighter (1/12 of the mass of) one carbon atom. • THE RELATIVE ATOMIC MASS IS THE NUMBER OF TIMES AN ATOM IS HEAVIER THAN 1/12 OF A C12 ATOM.
6. 6. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY The Mole The mole is defined as, “the amount of ………….. with the same number of ……………………… particles as ….. grams of carbon 12”. (n used as symbol for moles) 602 300 000 000 000 000 000 000 Six hundred and two thousand, three hundred, billion billion ! 6.023x1023 particles 12.00 g C Symbol (….) Number of particles = no of moles x no. particles in a mole Particles = ……………..
7. 7. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY The Mole The mole is defined as, “the amount of matter with the same number of elementary particles as 12 grams of carbon 12”. (n used as symbol for moles) 602 300 000 000 000 000 000 000 Six hundred and two thousand, three hundred, billion billion ! 6.023x1023 particles 12.00 g C Symbol (L) Number of particles = no of moles x no. particles in a mole Particles = n x L
8. 8. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY The Mole and Mass The mole is defined in such a way that the MASS NUMBER (A) of an element is equal to the relative atomic mass mass of one mole of the substance. (in grams) - THE MOLAR MASS • Eg Na = 23g/mol, water(H2O)=18g/mol Z A XAtomic Number (smaller) Mass Number (bigger) protons + neutrons Periodic Table Symbol Relative atomic mass or mass(g) of one mole
9. 9. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Relative Masses • Relative atomic(Ar) - The mass of the atom relative to 1/12 of the mass of a C12 atom. (Number of times heavier than…) O - 16one atom of oxygen is 16 times heavier than 1/12 of the mass of a C12 atom, Na - 23 one atom of sodium… , H - 1 etc. • Formula mass (Mr) - The sum of all the atomic masses of the atoms in a molecule. Water H2O one molecule of water has a relative mass of (2x(1)+16) = 18 - that is the molecular or formula mass of water. Mr(H2O) = 18 (Times heavier than…)
10. 10. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Atomic Structure Symbol Notation Name Protons Neutrons Mass Number Electrons Calcium 31 15P 14 30 13 6 Mg
11. 11. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY ISOTOPES Symbol PROTONS ELECTRONS NEUTRONS Carbon 12 12 6C Carbon 13 13 6C Boron 10 10 5B Boron 11 11 5B Hydrogen 1 Hydrogen 2 Chlorine 35 Chlorine 37
12. 12. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Relative Masses - examples Calculate the Formula (Molecular) masses of: • O2 (oxygen gas) Mr (O2) = 2x16 = 32 • Cl2 (chlorine gas) Mr (Cl2) = 2x35. 5 = 71.0 • NaCl (sodium chloride - table salt) Mr (NaCl) = 23+35.5 = 58.5 • CaCO3 (calcium carbonate) Mr (CaCO3) = 40.1+12+(3x16) = 100.1 • (NH4)2Cr2O7 (ammonium dichromate) Mr ((NH4)2Cr2O7 ) = 2(14+4)+2(52)+7(16) = 252
13. 13. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Isotopes Chlorine has two isotopes 37 17Cl & 35 17Cl Cl(35) has 35-17=18neutrons Cl(37) has 20 neutrons! • 37Cl (25%) & 35Cl (75%) - exist in the ratio 1:3 Calculate the average mass of a Cl atom. (Two methods) In 100 atoms – 25 have a mass of 37 and 75 have mass 35! Average Ar(Cl)= total mass = (37x25)+(35x75) = 35.50 no of atoms 100 Or 4 atoms – 3 are 35 and one is 37! Av Ar(Cl) = (37x1)+(35x3) = 35.50 4
14. 14. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Relative Atomic Mass Z A XAtomic Number (smaller) Mass Number (bigger) protons + neutrons Relative atomic mass or mass(g) of one mole Periodic Table Symbol Calculate: The mass in grams - 1. of one mole of copper chloride (CuCl2) 2. one mole of carbon dioxide (CO2) 3. One and a half moles of oxygen (O2) 4. TWO moles of methane (CH4) 5. Four moles of water. m = n x Mr mass of substance = number of moles x mass of 1 mole
15. 15. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY The Mole - moles --> Mass m = n x Mr Calculate the mass of • 2 moles of copper oxide (CuO) • 0.5 moles of copper (II) sulphate (CuSO4) • 0.01 moles of calcium carbonate • 5 moles of ammonium carbonate mass = moles x relative mass
16. 16. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY The Mole - moles --> Mass m = n x Mr Calculate the mass of • 2 moles of copper oxide (CuO) m = nxMr = 2x(63.5+16) = 159 g • 0.5 moles of copper (II) sulphate m (CuSO4) = n x Mr = 0.5 x ( (63.5) + 32.1 + 4(16) ) = 79.8 g • 0.01 moles of calcium carbonate Mr (CaCO3) = n x Mr = 0.01 x ( 40 + 12 + 3(16) ) = 1 g • 5 moles of ammonium carbonate m(NH4)2CO3 = n x Mr = 5 x ( 2(14+4)+12+3(16) ) = 5 x (96) = 480g
17. 17. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY The Mole - mass calculations C + O2  CO2 Carbon reacts with oxygen to form carbon dioxide as shown. If 0.12g of carbon are reacted with excess oxygen what mass of carbon dioxide would be formed? 1. Balance the reaction 2. Work out moles of reactant(mass given). 3. Go through the equation to find out the number of moles being formed 4. 4. Work out quantity asked for.
18. 18. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY The Mole - mass calculations C + O2  CO2 Carbon reacts with oxygen to form carbon dioxide as shown. If 0.12g of carbon are reacted with excess oxygen what mass of carbon dioxide would be formed? 1. Balance the reaction 2. Work out moles of reactant(mass given). n(C) = m/Ar = 0.12/12 = 0.01 mol 3. Go through the equation to find out the number of moles being formed - the molar ratio: C:CO2 1:1 - => n(CO2) = 0.01 mol 4. Work out quantity asked for. m(CO2) = nxMr = 0.01 x (12+2(16)) = 0.01 x 44 = 0.44 g
19. 19. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Limiting reagent example Ammonia gas is made by reacting ammonium chloride with calcium hydroxide according to: NH4Cl + Ca(OH)2  NH3 + CaCl2 + H2O If 32.1 g of ammonium chloride reacts with 7.5 g calcium hydroxide in solution, Show by calculation; which is the limiting reagent and what mass of ammonia is produced. 1. Balance the reaction. 2. Calculate moles given (both). 3. Work through molar ratio to decide which is limiting reagent. 4. Use limiting reagent to calculate quantity asked as before.
20. 20. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Limiting reagent example Ammonia gas is made by reacting ammonium chloride with calcium hydroxide according to: NH4Cl + Ca(OH)2  NH3 + CaCl2 + H2O If 32.1 g of ammonium chloride reacts with 7.5 g calcium hydroxide in solution, Show by calculation; which is the limiting reagent and what mass of ammonia is produced.
21. 21. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Percentage Composition Analysis of a compound by mass makes it possible to work out the % mass of each element. eg Table salt: NaCl mass analysis: One mole of NaCl would have a mass of 23 + 35.5 = 58.5g • The % composition can be found using the formula: Mass element X x100 Total Mass Compound • %Na = […../ (…..) ]x100 = …………..% (by mass) • %Cl = (…../ (…….) )x100 = …………% % Mass Element X =
22. 22. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Percentage Composition from mass. Eg2 Calculate the % of oxygen in water. Mr (H2O) = (m(O)/Mr(H2O))x100 = (16/18)x100 = 88.9%
23. 23. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Empirical and Molecular Formula. A compound consists of carbon, hydrogen and oxygen only. The % by mass are Carbon 40.0% and 6.7% hydrogen. Calculate the empirical and molecular formula of the compound if Mr = 60g·mol-1 %(O) = 100 – (40+6.7) = 53.3 C H O In 100g: 40.0g 6.7g 53.3g n=m/Mr: 40/12 6.7/1 53.3/16 3.33 6.7 3.33 3.33 3.33 3.33 Simplest: 1 2.01 1 Empirical Formulae: CH2O (12+2+16 = 30) Molecular Formula: 2(CH2O) C2H4O2 (Mr = 2x30)
24. 24. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Concentration - Molarity The concentration of a solution is defined as the ………………. of ……………………… per ………………. (dm3) of …………………. solute solute Final volume of …………….. 500cm3 =+ Concentration = Amount of ……… (……….) Volume of ……………… 30g of NaCl C = n v
25. 25. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Volume Conversions 1 dm = ….. cm 1 dm3 = ………… cm3 1 m3 = …………….. dm3 = ………………….cm3 (10….) 1cm3 1 dm3 (1 litre) 10 cm3 10 cm3 10 cm3
26. 26. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Decimal Conversions King Henry Died a miserable death called measles Kilo Hecta Decca m(unit) deci centi milli 1000 100 10 1 1/10 1/100 1/1000 1m ?km 0. 0 0 1km 1 m = 0.001 km 1 km = 1000 m
27. 27. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Conversions 1 cm = 0.1 dm 1 cm2 = (0.1)2 dm2 1 cm2 = 0.01 dm2 1 cm3 = (0.1)3 dm3 1 cm3 = 0.001 dm3 1 dm3= 1000 cm3 25 cm3 = 0.025 dm3
28. 28. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Molar Volumes One mole of an ideal (ANY) gas occupies a volume of 22,4dm3 at standard temperature and pressure. (STP) STP: T= 0ºC, 273K P =1 atmosphere (101,3kPa) Fe2O3 + 3H2  2Fe + 3H2O What volume of hydrogen reacts with 50g of Fe2O3 Fe2O3 : H2 1 : 3 n(H2) =3n(Fe2O3) = 3(0.3125) = 0.9375 v(H2) = nxMv = 0.9375x22.4 = 21dm3 n(Fe2O3) = m/Mr = 50/(2(56)+3(16)) = 0.3125mol moles = volume/molar volume ==> n = v/Mv 22.4 dm3
29. 29. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY ASKEDGIVEN Mole Calculations MOLES MOLES MASS MASS VOLUME VOLUME CONCENTRATION CONCENTRATION MOLAR RATIO Number Of particles Number Of particles
30. 30. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Volume - Volume Calculations H2 + N2 --> NH3 If 3.00 dm3 of nitrogen are reacted to produce ammonia, what volume of hydrogen will be required? (At STP) 1. 3H2 + N2 --> 2NH3 2. n(N2) = v/Mv = 3/22.4 = 0.134mol 3. N2 : H2 1:3 n(H2) = 3(N2) 4. n(H2) = 3(0.13) = 0.401mol 5. v(H2) = n(H2)Mv = 0.401(22.4) = 8.98dm3
31. 31. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Mass Volume Calculations 1. Balance the equation - 3H2 + N2 --> 2NH3 2. Calculate the moles of the substance given. n(NH3) = v/Mv = 46/22.4 = 2.05 mol 3. Work through the molar ratio to find out the moles of the substance asked. N2 : NH3 as 1 : 2 n(N2) = 1/2(n(NH3)) = 1/2(2.05) = 1.03 mol 4. Calculate the quantity asked for. m(N2) = n(N2)Mr = 1.03(28) = 28.84 g 2. H2 + N2 --> NH3 What mass of nitrogen (in dm3) would be needed to produce 46dm3 of ammonia?
32. 32. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Standard Solution A standard solution is one for which the concentration is precisely known. Since c = n(solute)/v(solvent) = m/Mr V • The number of moles of solute (Mass) • The volume of solution. These values must be accurately determined. 2.45g Mass is determined accurately using an electronic balance. • Possible accuracies of 0.1 - 0.0001g KMnO4 Volume is measured using a volumetric flask. • 250 cm3 • 100 cm3, 200 cm3,
33. 33. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Weighing Technique Procedure - Weighing by difference. 1. Zero scales and clean the pan. 2. Weigh the weighing container. 3. Add (approximately) the required amount of salt. Take care not to drop any salt onto the pan. 4. Transfer the salt to a clean beaker. 5. Reweigh the weighing container. 6. Subtract the final mass of the container from the mass of salt and container to give the mass of salt transferred to the beaker. Mass is determined accurately using an balance (electronic or triple beam). • Possible accuracies of 0.1 - 0.0001g 2.45g KMnO4 Results: Mass salt + container: ………… Final Mass container: ………… Mass salt transferred:
34. 34. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Volumetric Flask Making a standard solution. 1. Rinse a clean & dry 100 cm3 beaker with a little distilled water. 2. Transfer the correctly weighed amount of salt to the beaker. Ensure NO SALT IS LOST. 3. Add 50 - 80 cm3 water the salt and stir gently with a glass rod until all salt is dissolved. DO NOT REMOVE THE ROD FROM THE SOLUTION NOR ALLOW ANY DROPS OF SOLUTION TO ESCAPE. 4. Add ALL the solution to volumetric flask via funnel. Ensure glass rod and beaker are thoroughly rinsed. (Include rinsings.) 5. Add enough solvent to bring the level up to the mark.
35. 35. For FULL presentation click HERE >> www.warnescience.net SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY Hi - This is a SAMPLE presentation only. My FULL presentations, which contain a lot more more slides and other resources, are freely available on my resource sharing website: www.warnescience.net (click on link or logo) Have a look and enjoy! WarneScience
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