IC5.3.4 Percentage masses from formulae © Oxford University Press Percentage masses from formulae.

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Presentation transcript:

IC5.3.4 Percentage masses from formulae © Oxford University Press Percentage masses from formulae

IC5.3.4 Percentage masses from formulae © Oxford University Press The formula of a compound, and the relative atomic masses of the elements it contains, can be used to calculate percentage masses.

IC5.3.4 Percentage masses from formulae © Oxford University Press Ammonium nitrate, NH 4 NO 3, is used as an artificial fertiliser. Suppose you wanted to calculate the percentage by mass of nitrogen in ammonium nitrate. Follow these three steps to see how it’s done…

IC5.3.4 Percentage masses from formulae © Oxford University Press NH 4 NO 3 A r of N = 14, A r of H = 1, A r of O = 16 1.Calculate the relative formula mass of the compound using the relative atomic masses.  M r of NH 4 NO 3 = 14 + (4 × 1) (3 × 16) = 80

IC5.3.4 Percentage masses from formulae © Oxford University Press NH 4 NO 3 A r of N = 14, A r of H = 1, A r of O = 16 2.Multiply the relative atomic mass of the element in the question by the number of its atoms in the compound.  Mass of N = 2 × 14 = 28 (notice that N appears twice in the formula)

IC5.3.4 Percentage masses from formulae © Oxford University Press NH 4 NO 3 A r of N = 14, A r of H = 1, A r of O = 16 3.Work out the percentage by mass of nitrogen in ammonium nitrate using the numbers from steps 1 and 2.  Percentage of N in NH 4 NO 3 = 28/80 × 100% = 35%

IC5.3.4 Percentage masses from formulae © Oxford University Press Now use these A r values to help you to answer the questions: H = 1, O = 16, Na = 23, S = 32, Cu = Calculate the percentage by mass of sodium in sodium hydroxide, NaOH.  M r of NaOH = = 40  A r of Na = 23  Percentage of Na in NaOH = 23/40 × 100% = 57.5%

IC5.3.4 Percentage masses from formulae © Oxford University Press A r values : H = 1, O = 16, Na = 23, S = 32, Cu = Calculate the percentage by mass of oxygen in copper sulfate, CuSO 4.  M r of CuSO 4 = (4 × 16) =  Mass of O = (4 × 16) = 64  Percentage of O in CuSO 4 = 64/159.5 × 100% = 40%