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Empirical formula of substance

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Empirical formula of substance

1. 1. Experiment 4 : Empirical Formula of a CompoundSource: (2008) Introductory Chemistry Laboratory. Minneapolis Community and Technical College. Retrieved fromhome.minneapolis.edu/~naughtwe July 2010.Objectives: • To prepare a compound from a weighed quantity of metal. • To determine the empirical formula of the compound.Introduction:Chemical compounds are composed of atoms of two or more elements chemically combined in definiteproportions. The atoms in a compound are held together by chemical bonds. The total mass of eachelement in the compound depends on the number of atoms involved, and therefore, the elements are indefinite proportions by mass. The ratio of moles of the constituent elements in the compound is nearlyalways a ratio of small, whole numbers. The formula containing the lowest possible whole number ratiois known as the empirical formula.To find the empirical formula we must combine the elements to generate the compound under conditionsthat allow us to determine the mass of each element. From these data the moles of each element may bedetermined. By dividing the moles obtained for each element by the smallest number of moles, we obtainquotients that are in a simple ratio of integers, or are easily changed to such a ratio. An example of thistype of calculation is shown below.Sample calculations:A strip of aluminum weighing 0.69 g is ignited yielding an oxide that weighs 1.30 g. Calculate theempirical formula of the compound formed.A. Calculate the mass of oxygen that reacted to combine with the aluminium: Mass of oxide = mass of aluminium + mass of oxygen Therefore: Mass oxygen = mass oxide – mass aluminium = 1.30g – 0.69g = 0.61gB. Now that you have the masses, you can calculate the moles of each element: Moles = mass/ molar mass Moles aluminium = 0.69g/ 27.0gmole-1 = 0.026 mole Moles oxygen = 0.61g/16.0gmol-1 = 0.038 moleC. Obtain the ratio of atoms by dividing the moles of each by the smallest number of moles obtained from step B. In this case the moles of aluminium (0.026 mole) is smaller than the moles of oxygen (0.038mole). Al:O = (0.026/0.026) : (0.038/0.026) = 1 : 1.5 The ratios need to be in whole numbers. Multiply out the values to get rid of any fraction and get the least whole number ratio. Al:O = 2:3 Use the ratio to obtain the formula. Formula = Al2O3