Understanding Atomic Masses and Moles in Chemistry, Lecture notes of Chemistry

Download Understanding Atomic Masses and Moles in Chemistry and more Lecture notes Chemistry in PDF only on Docsity! 1B – The Mole Atomic Masses Protons, electrons and neutrons  So an atom of one element must have a different mass from another element, we call this the Mass Number.  The number of protons determines which element an atom is and the bottom number tell us this, we call this the Atomic number. Examples:- 1) Lithium  The top number is the Mass number. This means that the total number of protons and neutrons are 7.  The bottom number is the Atomic number. This is the number of protons.  Because an atom is neutral, this means that this is also the number of electrons. This atom has 3 protons and 3 electrons.  If we take the Atomic number (Z) from the Mass number (A) we get the number of neutrons. 7-3=4 neutrons. 2) Nitrogen  An atom of nitrogen is twice as heavy as an atom of Lithium.  The top number is the mass number. This means that the total number of protons and neutrons are 14.  The bottom number is the Atomic number. This is the number of protons.  Because an atom is neutral, this means that this is also the number of electrons. This atom has 7 protons and 7 electrons. 7 Li 3 14 N 7  From GCSE this table shows that only protons and neutrons have a mass.  Since different elements have different atoms.  These atoms have different numbers of protons and neutrons (and electrons).  This means that different elements atoms must have different masses. Sub-Atomic Atomic Atomic working Particle Mass Charge it out Proton 1 +1 bottom Electron 1/2000 -1 bottom Neutron 1 0 top - bottom  If we take the Atomic number (Z) from the Mass number (A) we get the number of neutrons. 14-7=7 neutrons. Isotopes An atom of the same element that has the same number of protons and electrons but a different number of neutrons.  As the atom has the same number of protons and electrons it will have the same chemical properties.  They are all hydrogen atoms because they all have the same number of protons  Hydrogen can be used as an example:- Measurement of relative masses  Because Chemistry is about reacting ratios we have to measure the amount of reacting particles of each reactant.  We now know that each atom (and therefore molecules) have different masses (because they have different numbers of protons and neutrons), we have to be able to weigh out a number of particles of reactants to react with each other.  Since atoms are so small we give them a mass scale of their own.  This scale is called: Unified atomic mass unit, u: 1u = 1.66 x 10-24g  This is basically the masses of a proton / neutron. The mass of a carbon - 12 atom = 12u The mass of 1/12th of a carbon - 12 atom = 1u  We have to state the Atomic mass number of the atom as elements usually have isotopes: There is a carbon - 13 atom containing an extra neutron, this would have an mass of 13u!!  This is not the only thing we have to be careful of, there are 4 different types of masses and we have to use the correct one depending what we are referring to: 1) Relative isotopic mass:  We use this one when we are only referring to one isotope of an element 16O has a mass of 16u P P N P N N E E E Hydrogen – 0 Neutrons Hydrogen – 1 Neutron (deuterium) Hydrogen – 2 Neutrons (tritium) Types of Formula Empirical Formula is the simplest ratio of atoms in a molecule. Molecular formulae is the actual ratio of atoms in a molecule  This can be calculated using moles from percentage composition:- Example 1 A sample of iron oxide was found to have 11.2g of iron and 4.8g of oxygen. Calculate the formula of this compound Element Fe O Masses 11.2 4.8 Divide by Ar 11.2 / 55.8 4.8 / 16 Moles 0.2 : 0.3 Divide by smallest 0.2 / 0.2 : 0.3 / 0.2 Ratio 1 : 1.5 Whole No Ratio 2 : 3 Empirical formula Fe2O3 Example 2 A sample of hydrocarbon was found to have 1.20g of carbon and 0.25g of hydrogen. Calculate the Empirical formula of this compound. Then find out the molecular formula if the Mr = 58 Element C H Masses 1.20 0.25 Divide by Ar 1.20 / 12 0.25 / 1 Moles 0.10 : 0.25 Divide by smallest 0.10 / 0.10 : 2.5 / 0.10 Ratio 1 : 2.5 Whole No Ratio 2 : 5 Empirical formula C2H5 (29 x 2 = 58) Molecular formula C4H10 Questions 1,2 p13 / 6,7 p35 Moles and gas volumes Avogadro's hypothesis Equal volumes of gases will have the same number of atoms / molecules  This makes gases particularly easy to calculate 1 mole of any gas, no matter what it is, occupies the same volume (at RTP) 1 mole of any gas occupies 24dm3 (24000cm3) at room temperature and pressure.  Again the numbers are not always simple so a general formula will help:- Questions 1-3 p15 / 8 p35 / 3 p36 Moles and solution Concentration:  So far when we have looked at reacting quantities we have dealt with the mole.  Many reactions in chemistry involve solutions.  A solution is expressed as a number of moles in 1dm3 (1 litre or 1000cm3) of solvent (usually water).  Concentration is the number of moles of specified entities in 1 dm3 of solution.  We use square brackets to denote concentration [X]. (also known as molarity, M). No. Moles = Volume (dm3) 24dm3 Moles = V (dm3) 24  For example – a solution of sodium hydroxide has a concentration of 1.0 Mol dm-3 (1.0M) [NaOH(aq)] = 1.0M This means there is 1 mole of sodium hydroxide dissolved in 1dm3 of water. [NaCl(aq)] = 2.0M This means there is 2 moles of sodium chloride dissolved in 1dm3 of water. [KOH(aq)] = 0.5M This means there is 0.5 mole of potassium hydroxide dissolved in 1dm3 of water.  This is fine so long as we keep making up 1dm3 solutions.  Most lab experiments only use a 100cm3 or so and making up 1000cm3 would be waste so we need to scale down:  For example – 500cm3 solution of sodium hydroxide has a concentration of 1.0 Mol dm-3 (1.0M) 500cm3 [NaOH(aq)] = 1.0M: 0.5 moles of sodium hydroxide dissolved in 0.5dm3 water. 2000cm3 [NaCl(aq)] = 2.0M: 4 moles of sodium chloride dissolved in 2dm3 of water. 100cm3 [KOH(aq)] = 0.5M: 0.05 moles of sodium hydroxide dissolved in 0.1 dm3 of water.  The values are not usually as nice as this so we can use the following formula:- Concentration = No. of moles C = n (mol dm-3) Volume(dm3) V (dm3)  Basically the number of moles per volume of solution.  We usually like our formula to have n, number of moles at the start: No. of moles = Concentration (Mol dm-3) x Volume (dm3) n = C x V (dm3)  However this formula assumes we are working in dm3 and we usually work in cm3.  dm3 and cm3 are related by a factor of 1000 – you must convert into dm3 by dividing cm3 by 1000 STEP4: Check the question/ state symbol to decide whether to convert it to mass / concentration / volume - (s) = mass Gas / mole calculations: Example: 2.3g sodium reacts with excess sulphuric acid to form sodium sulphate and hydrogen gas. Calculate the volume of hydrogen made: STEP1: Write a balanced chemical equation and add the amounts given and question mark what you are asked to work out: STEP2: Check the state symbol of your starting mass to decide which moles equation you will use - (s) - means you use Moles = mass / Ar STEP3: Use the reacting ratios to work out how many moles you have made (or need): STEP4: Check the question/ state symbol to decide whether to convert it to mass / concentration / volume - (g) = volume Concentration / mole calculations: Example: 2.3g sodium reacts with 250cm3 sulphuric acid to form sodium sulphate and hydrogen gas. Calculate the concentration of sodium sulphate solution made: STEP1: Write a balanced chemical equation and add the amounts given and question mark what you are asked to work out: STEP2: Check the state symbol of your starting mass to decide which moles equation you will use - (s) - means you use Moles = mass / Ar STEP3: Use the reacting ratios to work out how many moles you have made (or need): Titrations - Volumetric analysis  This technique can be used to find: Concentration Mr Formula Water of crystalisation  To do this you react a certain volume of a solution with an unknown concentration with a solution of known concentration.  Using moles and reacting ratios, you can calculate the concentration of this solution. This is known as titration.  The only requirement is that you can tell when one solution has completely reacted with the other.  Between acids and alkalis, we use indicators to let us know when the resulting solution is neutral.  An indicator will change colour at the ‘end point’ (neutral).  Common indicators are: Indicator Acidic colour Base colour End point colour Methyl orange Red Yellow Orange Bromothymol Blue Yellow Blue Green Phenylphthalein colourless Pink Pale pink Technique/procedure  Rinse the burette with distilled water then acid.  Fill the burette to the graduation mark ensuring the air bubble is removed from the tap.  Rinse a pipette with alkali, fill and transfer a known volume to a clean, dry conical flask.  Add some indicator.  Run the acid into the alkali and stop when the colour changes. This is your ‘range finder’.  Wash the conical flask with pure water.  Repeat the previous steps but stop a few cm3 before the colour change.  Add drop wise until the first drop changes the colour of the indicator.  Repeat until you get 2 results that agree within 0.10cm3.  Record results in a table like the one below: Range finder / titration / cm3 Run 1 / cm3 Run 2 / cm3 2nd burette reading 1st burette reading Volume added Calculation – example 25cm3 of 0.01M sulphuric acid was added to exactly neutralize 50cm3 of sodium hydroxide. Calculate the concentration of the sulphuric acid: 1 Write a balanced equation 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H20(l) 2 Calculate the number of moles of acid added from the burette No. of moles = Concentration (Mol dm3) x Volume (cm3) (H2SO4) No. of moles = 0.01 Mol dm3 x 25 cm3 (H2SO4) 1000 No. of moles = 25 x 10-3 Moles (H2SO4) 3 Use the ratio to work out the number of moles in the sample of alkali 2 : 1 NaOH : H2SO4 2x the number of moles for H2SO4 No. of moles = 25 x 10-3 x 2 (NaOH) No. of moles = 50 x 10-3 Mol (NaOH) 4 Calculate the concentration. (Convert to g dm3 if necessary) Concentration (Mol dm3) = No Moles Volume (dm3) Concentration (Mol dm3) = 50 x 10-3 0.05 (dm3) Concentration (Mol dm3) = 1 Mol dm-3 5 Convert to g dm3 if required (usually for solubility) Mass = No. moles x Mr (NaOH) Mass = 1 x 40 Concentration = 40 g dm-3 Questions 1-2 p29 / 17 p35 Questions 1-2 p33 / Remaining questions p35 - 37