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Published byAutumn Phillips Modified over 5 years ago

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**Aim: to revise key concepts about molar calculations**

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Calculations 6.02 x 1023 atoms in a mole of any substance

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Calculations Relative atomic mass (RAM, Ar) is a measure of the mass of an atom relative to the mass of 1 gram of carbon-12 1 mole of carbon-12 atoms has a mass of 12 grams So e.g. for an element whose atoms are twice as heavy as carbon-12, the mass of one mole of atoms would be 24 grams

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Calculations The RAM is calculated by looking at all the isotopes of the atoms for that element E.g. Chlorine has two isotopes Cl-35 and Cl-37 75% of one mole of atoms are Cl-35 25% of one mole of atoms are Cl-37 RAM = (75 x 35) + (25 x 37)/100 = 35.5

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Calculations Relative formula mass (RFM, Mr) is the sum of the RAM of individual atoms in the compound E.g. C6H12O6 has a RFM of: RAM C = 12 H = 1 O = 16 (6 x 12) + (12 x 1) + (6 x 16) = 180

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Calculations We can find out the % of an element in a compound (analogy: % girls in the room). This is useful for finding out the amount of active ingredient in medicines. % element in compound = RAM element x 100 RFM compound e.g. What % of C is in C2H6? = 2 x x 100 (2 x 12) + (6 x 1) = 80%

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**Calculations Moles of element = mass of element RAM**

e.g. How many moles of Mg are in 48g? moles = = 2 moles 24 Moles of compound = mass of compound RFM e.g. How many moles of MgO are in 20g? moles = = 0.5 moles

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Calculations We can measure the amount of product made and calculate it as a % of the total mass of expected product. We want to get as high a % yield as possible. % yield = actual amount of product made x 100 (actual yield) maximum amount of product that could be made (theoretical yield)

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Calculations In a reaction of carbon and oxygen the maximum yield of carbon dioxide would be 1100 grams. When this was carried out in the school laboratory, 539 grams of carbon dioxide was obtained. What is the % yield of carbon dioxide? % yield = x = 49% 1100

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Calculations We can measure the atom economy of a reaction which gives us a measure of how much of the reactant has been converted into useful product. The higher the atom economy, the less waste and better use of resources has occurred in a reaction. % atom economy = RFM of useful product x 100 (desired product) Total RFM of ALL products

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**Calculations CH4 + 2O2 CO2 + 2H2O**

This reaction produced 88 grams CO2 and 72 grams H2O. We are trying to make water and need to know the usefulness of this reaction. What is the % atom economy for water? % atom economy = x = 45%

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Calculations The molecular formula of a compound is the actual number of each type of element present in the compound, e.g. C2H6 The empirical formula of a compound is the simplest ratio of atoms in the compound, e.g. CH3. We use this to help us identify formulae of compounds and unknown substances

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Calculations To calculate the empirical formulae these are the steps to follow: Calculate moles of each element from % or grams given for each element Divide all the figures by the smallest answer in (1) to give a ratio of elements Convert ratio into whole numbers by multiplying answer in (2)

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**Calculations Therefore CH3 x2 = C2H6**

Find the empirical formula and molecular formula of a compound with 80%C and 20%H that has a mass of 30. Empirical: C = H = moles = = ratio = = = 3 whole numbers CH3 If the answer to H had been approximately 2.5 then the answer would need to be multiplied up, e.g. C2H5 Molecular: CH3 has a mass of 15, so x 2 = 30 Therefore CH3 x2 = C2H6

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Calculations We can work out how much reactant we need or how much product we can make in a reaction by doing reacting mass calculations The steps to follow: Calculate the moles of known substance Work out the ratio of each substance by using the big numbers in the balanced symbol equation Find out the mass of “unknown” substance

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Calculations What mass of sodium is needed to make grams of sodium oxide when it is burnt in air? 4Na + O Na2O Moles Na2O = = (23x2) + 16 Ratio Na:Na2O is 4:2 i.e. 2:1 so 2x(1.745):1.745 = moles Na Mass Na = 3.49 x 23 = grams

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Chemical Quantities Chapter 9

Chemical Quantities Chapter 9

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