IGCSE Double Award Extended Coordinated Science

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Christiana Zoe Carr
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1 IGCSE Double Award Extended Coordinated Science Chemistry The Mole Concept The Atomic Mass Unit You need to know the atomic mass unit and the relative atomic mass. In Unit C3.3, 1 atomic mass unit (amu) was mentioned as the mass of one nucleon. 1 atomic mass unit is defined as 1/12th the mass of a carbon-12 atom. - Which makes sense, because carbon-12 has 12 nucleons (6 protons and 6 neutrons). Meaning all the masses on the periodic table is relative to carbon-12 s mass Also in Unit C3.3, the term relative atomic mass was also defined as - the average masses of all the isotopes of an element, according to their percentage abundances. Relative atomic mass is sometimes written as A r (A for atomic, r for relative mass) and calculated by: A r = (isotopic mass x percentage abundance) isotope 1 + (isotopic mass x percentage abundance) isotope The Mole You need to know what the mole, the Avogadro s constant, and the relative molecular mass is The mole (unit symbol mol ) is a way of counting particles (atoms and molecules) - Since particles are so small and there are so many of them, we count them in groups called moles. - Like we can count socks in pairs, or count eggs in dozens, we can count particles in moles. - A pair of something is 2, a dozen of something is 12, a mole of something is 6.02 x That number is also called the Avogadro s Constant ( 6.02 x ) A mole is also defined as: the number of particles in 12 grams of carbon-12 (which is 6.02 x particles) - So we know that 1 mol = number of carbon atoms in 12 grams of carbon-12 There is also a value called the relative molecular mass M r (M for molecular mass, r for relative mass) - This is simply the mass of the molecule in atomic mass units - It is calculated by adding up the A r of all the atoms of a molecule - sodium chloride, NaCl - A r of sodium + A r of chlorine = M r of sodium chloride = sulphuric acid, H 2 SO 4-1x x4 = 98 - calcium nitrate, Ca(NO 3 ) ( x3) = 164

(#sorrynotsorry) For example, to find out how the number of moles in 16 grams of oxygen, we can: 1.

2 Mole Calculations You need to know the importance and of calculations involving moles, and the three mole formulae. The mass formula allows the conversion between moles and grams: - mass = moles x molar mass (A r or M r ) - g = mol x g/mol - You can remember it as g = Mr x mol - Which sounds like Gee! Mr. Mole! (#sorrynotsorry) For example, to find out how the number of moles in 16 grams of oxygen, we can: 1. Use the formula g = Mr x mol - We need to find mol, so we substitute the values we know for grams and M r - The M r of oxygen is NOT 16, but 32, because oxygen exists as O 2 (diatomic) - (other elements like hydrogen, nitrogen, chlorine, fluorine, iodine are also diatomic) 2. 16g = 32g/mol x moles - 16g 32g/mol = 0.5mol = moles - Solving the equation gives us that in 16 grams of oxygen, we have 0.5mol. Using the formula we can also find out other values, for example: - Q. What is the mass of 3 moles of sodium carbonate? 1. Use the formula: g = Mr x mol - Sodium carbonate, Na 2 CO 3 s M r is ( 23x x3 ) = 106 g/mol - Number of moles is 3 2. mass = 106 g/mol x 3 mol = 318 g: 3 moles of sodium carbonate is 318 grams. The gas formula allows us to find moles in gases: - If the substance given is a gas, it is hard to calculate moles using the mass formula. - So instead we use another formula relating gas volume and moles Avogadro found that: - At room temperature (20C) and pressure (1 atm) one mole of ANY gas occupies 24dm 3 ( 1L ) Note that the type of gas does not matter, - so 1 mol of hydrogen (Mr = 2) and 1 mol of xenon (Mr = 131) will both occupy 24dm 3 - So we know that in the gas formula, the Mr is irrelevant. From this, we can form a formula that links the volume of gas The conversion for decimetre cubed (dm 3 ) is : - Volume of gas in dm 3 = moles x 24 1 dm 3 = 1000 cm 3 = 1 litre - dm 3 = mol x dm 3 /mol Remember the formula as : vol = mol x 24 which sounds like voldemort 24 Q : How much space does 2 moles of helium occupy at room temperature and pressure? 1. Use the formula vol = mol x 24 (remember the Mr is irrelevant here) - 2 mol x 24 = volume = 48dm 3. Easy! Q. If a gas X occupies 600cm 3 at RTP, how many moles are there in the gas? 1. vol = mol x 24 - But the volume must be in dm 3 and not cm 3, so we convert by dividing by a thousand cm 3 / 1000 = 0.6dm = mol x = mol = 0.025mol of gas X

3 The molarity formula allows us to find moles in solutions It is hard to measure mass or volumes in solutions, because think about it: Let s say we have 1 mole of sodium chloride (salt) dissolved in water to make 1 litre of solution. If we add another 1 litre water, only the volume increases, the amount of sodium chloride is still 1 mole. So it makes no sense to calculate using the total volume or the just the moles of the solute. - So we have to use the ratio of the two values, mass of solute and the total solution volume, concentration. Molarity (M) is the solution concentration, calculated by moles solution volume - Or, rearranged to: - moles = concentration (molarity) x volume of solution - mol = M (mol/litre) x litre - Or remember it as : mol = con x vol (I can t think of a pun for this one - sorry) Q : How many moles of calcium hydroxide are there in 2 litres of 3M solution of calcium hydroxide? 1. Use the mol = con x vol formula - Mol = 3M x 2L = 6 moles. Easy! Q : What is the volume of solution if the concentration is 5M and there is 1 mole of solute? 1. Use the mol = con x vol formula - 1 mol = 5M x vol = vol = 0.2 litres or 200ml. You need to know how to manipulate these formulae for more complex questions (use more than one formulae) More complex question might require you to use more than one formulae from above Q: How much space would 0.07 grams of nitrogen gas occupy at RTP? - For the volume of gas, we need to use the gas formula, vol = mol x However we are not given the value of the moles for gas, so we cannot use this formula directly - But since we are given the mass of the gas and the identity of the gas (hence the Mr) - We can calculate the moles by the mass formula, g = Mr x mol 1. Nitrogen is diatomic, so the Mr would be = Using the mass formula, 0.07g / 28 = mol. - We now know that the gas contains moles. 2. We can use that result in the gas formula. - Vol = x 24 = 0.06dm 3, or 60cm 3.

4 Stoichiometry You need to know that chemical equations obey the law of conservation of matter CH 4 + 2O 2 O Balanced equations show the conservation of matter in chemical reactions - Where no matter is lost or gained by the reaction - If it was not balanced, CH 4 + O 2 + H 2 O, - it would mean the reaction somehow produced another atom of oxygen, and this is not possible Balanced equations also show the ratio of molecules reacting. - CH 4 + 2O 2 O - x : 2x : x : 2x - The coefficients are the number of molecules, so - For every 1 molecule of methane, twice the amount of oxygen are needed to react and form: - 1 molecule of carbon dioxide and twice the amount of molecules of water. - Since we can also count number of molecules in moles, - We can think the same: just looking at the coefficients. - Thinking it algebraically might make it easier. - x moles of CH 4 reacts with 2x moles of oxygen, producing x moles of CO 2 and 2x moles of water. - So if there were 2 (x) moles of methane, it would produce 4 (2x) moles of water You need to know how to calculate using balanced chemical equations. Q : From Ca(OH) 2 + 2HCl -> CaCl 2 O, if 5 moles of HCl was used how many moles of CaCl 2 are produced? 1. The stoichiometric ratios show that every 2x moles of HCl produces x moles of CaCl 2, so: - 2x = 5 mol, x = 2.5 mol: so 2.5 mol of CaCl 2 is produced. Q: 6HNO 3 + 2Fe -> 2Fe(NO 3 ) 3 + 3H 2, if 7 moles of hydrogen is produced. How many moles of nitric acid is used? 1. The ratios are 6x : 2x : 2x : 3x. For every 3x moles of hydrogen produced, 6x moles of nitric acid is used. - 3x = 7 mol, meaning x = 7/3 mol - Then 6x = 7/3 x 6 = 42/3 = 14 mol of nitric acid was used. You need to know how to calculate, even using the mole formulae. This question will be the most difficult type of stoichiometry question you will ever get. (most will be easier) Try this question without the guide, if you can solve it, you are very ready. Q: For the complete combustion of propane, C 3 H 8 + 5O 2 -> 4H 2 O + 3CO 2, - if 32 grams of oxygen is consumed, what is the volume of the produced carbon dioxide under STP? A: Since the stoichiometric ratios are ratios of moles and not masses, we need to convert the mass into moles - g = mr x mol, for oxygen: 32g / 32 mr = 1 mol of oxygen was used. - Now using the ratios, we know that 5x mols of oxygen produces 3x moles of carbon dioxide. - 5x = 1 mol, meaning x = 0.2 mol, and therefore 3x = 0.6 mol of carbon dioxide. - vol = mol x 24, for carbon dioxide: vol = 0.6mol x 24 = 14.4dm 3 of carbon dioxide. Q: During the synthesis of ammonia, 3H 2 + N 2 -> 2NH 3, - If 384cm 3 of ammonia was produced in the process, what was the total mass of reactants used?

5 Limiting and Excess Reagents You need to know the idea of limiting reagents and excess reagents, (or reactants) So far, all the reactions were perfectly prepared, meaning there were no excess reactants - All the reactants were used up in the chemical reaction. - Sometimes, one of the reactants may be left over. Reaction CH 4 + 2O 2 O molar ratio x : 2x : x : 2x From this, we know that if we have 1 mol of methane, we need 2 mol of oxygen. - What if we have 3 mol of oxygen instead? - The 1 mol of methane will react with 2 mol of oxygen, and there will be 1 mol of oxygen left over. - This is not a complete reaction, because not all the reactants were used. - The leftover reactant is called the excess reagent. - In this case, the amount of methane is limiting the reaction from completion. - In this reaction, methane is the limiting reagent. - Consequently, the products formed will only be produced according to the limiting reagent. - 1 mol of oxygen is not reacted, and does not produce anything. You need to know know how to find limiting reagents from reaction data - And calculate from reactions with a limiting reagent If a question gives you a data set with a reaction, you might have to check for limiting reagents. For example, 3H 2 + N 2 -> 2NH 3, - Q: 9 mol of hydrogen and 5 mol of nitrogen is reacted, how many moles of ammonia is produced? - The ratios are 3x mol of H 2 to x mol of N 2. - If we used up all the nitrogen, it would require 15 mol of hydrogen, this is not possible. - If we used up all the hydrogen, it would require 3 mol of nitrogen, this is possible - This leaves 2 mol excess nitrogen. Nitrogen is the excess reagent - Meaning that hydrogen is the limiting reagent - The amount of the limiting reagent determines the amount of products, so - Limiting reagent : hydrogen with 3x = 9 mol. - x = 3 mol which means for the product ammonia, the moles would be 2x = 6 mol. - A: 6 mol of ammonia is produced. Similar types of questions might be asked, but you might have to use the other mole formulae as well. Q: 3H 2 + N 2 -> 2NH 3, 7g hydrogen and 36dm 3 nitrogen is reacted, what is the volume of the produced ammonia? - First convert all data into useful moles. - g = mr x mol for hydrogen gives us 11/2 = 5.5 mols hydrogen - vol = mol x 24 for nitrogen gives us 36 / 24 = 1.5 mols of nitrogen - Using the mole ratio and idea of limiting reagent, we can find that - 5.5mol hydrogen requires 1.83mol of nitrogen, and there is not enough nitrogen - So nitrogen is the limiting reagent - So, we have to use the mole value of the limiting reagent - For nitrogen, x mol = 1.5 mol, so for ammonia, 2x = 3mol. - A: 3 mol of ammonia would occupy (3 x 24 = 72) 72dm 3 at STP.

6 The syllabus says you should be able to, (SO check if you can): - Define relative atomic mass, Ar - Define relative molecular mass, Mr, as the sum of the relative atomic masses (relative formula mass or Mr - will be used for ionic compounds). - Define the mole in terms of a specific number of particles called Avogadro s constant. (Questions requiring recall of Avogadro s constant will not be set.) - Use the molar gas volume, taken as 24dm 3 at room temperature and pressure. - Calculate stoichiometric reacting masses and reacting volumes of solutions; solution concentrations will be expressed in mol/dm 3. (Calculations involving the idea of limiting reactants may be set.)

The Mole. Relative Atomic Mass Ar

The Mole. Relative Atomic Mass Ar STOICHIOMETRY The Mole Relative Atomic Mass Ar Relative Molecular Mass Mr Defined as mass of one atom of the element when compared with 1/12 of an atom of carbon-12 Some Ar values are not whole numbers