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Calculations What you need to know: Relative formula mass Empirical formula % composition by mass Use balanced equations to calculate masses of reactants.

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Presentation on theme: "Calculations What you need to know: Relative formula mass Empirical formula % composition by mass Use balanced equations to calculate masses of reactants."— Presentation transcript:

1 Calculations What you need to know: Relative formula mass Empirical formula % composition by mass Use balanced equations to calculate masses of reactants and products Yields

2 Mass number – atomic number = number of neutrons E.g. Sodium 23 – 11 = 12 Isotopes Same number of protons Different number of neutrons

3 Relative Molecular Mass (M r ) M r is calculated by adding together all the relative atomic masses (A r ) in a molecule/formula For example NaCl - 1 sodium + 1 Chlorine = (1 x 23) + (1 x 35.5) = 58.5 ZnCl 2 – 1 Zinc + 2 Chlorine = (1 x 65) + (2 x 25.5) = 136

4 Number of Moles Number of moles = mass of X of substance XA r or M r of X For example 7g of Nitrogen (A r 14) = 7/14 = 0.5 moles 4g of silicon (A r 28) = 4/28 = 0.14 moles 10g of CaO (M r 56) = 10/56 = 0.18 moles

5 Relative atomic masses (A r ) Mass of atom compared to 12 C e.g. Na = 23, Cl = 35.5 Relative formula masses (M r ) Mass of a compound found by adding A r of each element e.g. NaCl = = 58.5 Moles A mole of any substance always contains same number of particles - Relative atomic mass in grams - Relative formula mass in grams

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7 Percentage Composition To calculate the percentage of one element in a compound % composition = A r of Xx 100 of X in compoundM r of compound For example % of Na (A r 23) in NaCl (M r 58.5) = (23/58.5) x 100 = 39%

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9 Percentage mass % = mass of element total mass of compound Percentage composition / empirical formula AlCl Mass935.5 Ar Moles(9/27) = 0.33(35.5/35.5) = 1 Simplest ratio (divide by smallest number of moles) (0.33 / 0.33) = 1(1 / 0.33) = 3 FormulaAlCl 3

10 Maximum Yield Maximum yield is the maximum mass of a substance that could be produced in a chemical reaction X + YXY Maximum yield = (mass of X) x (M r of XY) A r of X For example – what is the maximum mass of zinc oxide (M r 81) when 20g of Zn (A r 65) reacts? maximum yield = (20/65) x 81 = 24.9g

11 Percentage Yield Percentage yield is how much you actually get compared to what is theoretically possible (maximum yield) % yield of X = mass of X obtained x 100 maximum yield For example – what is the percentage yield if you get 14 g of ZnO when the maximum yield is 24.9g? Maximum yield = (14/24.9) x 100 = 48%

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13 Empirical Formula The empirical formula is the simplest form of a formula, for example C 6 H 12 would be CH 2 There are 3 steps to calculating the empirical formula Step 1 – calculate the number of moles of each substance Step 2 – divide the numbers by the smallest number Step 3 - write it down as a formula

14 Empirical Formula An example A substance contains 2g hydrogen and 16g oxygen Step 1 – 2g H (A r 1) = 2/1 = 2 moles of H 16g O (A r 16) = 16/16 = 1 mole O Step 2 – 2/1 = 2 ‘H’ 1/1 = 1 ‘O’ Step 3 – H 2 O

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16 Very few chemical reactions have a yield of 100% because: Reaction is reversible Some reactants produce unexpected products Some products are left behind in apparatus Reactants may not be completely pure More than one product is produced and it may be difficult to separate the product we want

17 Percentage yield % yield = amount of product produced (g) x 100% max. amount of product possible (g)


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