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Empirical formulae

Empirical formulae. An empirical formula is the smallest whole number ratio of atoms in a compound. Eg; Hexane has a molecular formula of C 6 H 14 . This gives a C : H ratio of 6 : 14 , which can be simplified to; 3 : 7 This cannot be further simplified whilst remaining whole numbers.

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Empirical formulae

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  1. Empirical formulae An empirical formula is the smallest whole number ratio of atoms in a compound.

  2. Eg; Hexanehas a molecular formula ofC6H14 . • This gives a C:H ratio of 6:14, which can be simplified to; • 3:7 • This cannot be further simplified whilst remaining whole numbers. • Therefore the empirical formula of hexane is C3H7 .

  3. For some compounds the molecular formula is the same as the empirical formula. • Eg; Water, the molecular formula is H2O • This cannot be expressed more simply using whole numbers. • So the empirical formula of water is also H2O

  4. Eg; A compound with an empirical formula of CH2 could have a molecular formula which is any multiple of CH2. • It might be 2(CH2)=C2H4, • 3(CH2)=C3H6, • 4(CH2)=C4H8, • 5(CH2)=C5H10 • To find the molecular formula you must also know the molecular mass.

  5. Eg; The empirical formula is CH2 andRMM 56. • CH2 has a mass of 12 + 2 = 14. • Divide the RMM by 14. • 56/14 = 4 • Therefore the molecular formula is 4x CH2 • C4H8

  6. How to calculate an empirical formula. • First divide the masses of the elements by their RAM. • Then divide each by the smallest number of moles. • If this is a whole number ratio it represents the empirical formula.

  7. If the ratii are only slightly more or less than whole numbers ignore it. • If one is off by .5 multiply the ratii by 2. • If it is off by .333 or .666 multiply by 3. • If it is off by .25 or .75 multiply by 4.

  8. What is the empirical formula of a compound that has 10g of Ca, 3g of C and 12g of O? • nCa = 10/40 = 0.25 • nC = 3/12 = 0.25 • nO = 12/16 = 0.75 • The smallest number of moles is 0.25. • nCa/0.25 = 0.25/0.25 = 1 • nC/0.25 = 0.25/0.25 = 1 • nO/0.25 = 0.75/0.25 = 3 • The empirical formula is CaCO3

  9. If the composition is given as a percentage simply assume that the sample has a mass of 100g. • Eg A compound is made up of 43.66% P and 56.34% O. • Assume that there are 43.66g P and 56.34g O then proceed as before. • nP = 43.66/31 = 1.41 • nO = 56.34/16 = 3.52. • nP/1.41 = 1.41/1.41 = 1 • nO/1.41 = 3.52/1.41= 2.5 • The ratio, P1:O2.5 is not a whole number and so must be multiplied by 2. • Empirical formula = P2O5

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