Csec Chemistry Notes 3

The mole concept

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Relative atomic mass, molecular mass and formula mass are measured in atomic mass units 12 (amu), where 1amu is 1/12 of the mass of a carbon-12 atom, C. The relative atomic mass of an element, Ar, is the average mass in amu of all of its atoms. These values are generally given in tables. The relative molecular mass of a compound, Mr, is the average mass of all of its molecules in amu, while the relative formula mass, Mr, is the average mass of the formula units and applies specifically to ionic compounds. Molar mass is the mass of one mole of the substance. The values of Ar and Mr stated in grams are the molar masses. Avogadro's Law states that equal volumes of all gases measured at the same temperature and pressure contain equal numbers of molecules. The temperatures and pressures usually used are standard temperature and pressure (STP), which is 00C and 1 atmosphere pressure and room temperature (RTP) which is 200C and 1 atmosphere pressure. 3 1 dm of any gas at STP contains the same number of molecules as 1 dm3 of any other gas at STP. 3 23 22.4 dm of any gas contains L molecules (6.0 * 10 ) at STP. This is one mole of the gas and is called the molar volume. 3 At RTP the molar volume is 24 dm . The formula of a compound shows how many atoms of each element are present in a molecule or formula unit. The empirical formula is the simplest formula, which represents the composition of the compound. The actual formula is called the molecular formula. It is generally a multiple of the empirical formula and is calculated from the molar mass.

Question 1 (i). Write a balanced equation to represent the reaction between calcium carbonate and dilute hydrochloric acid. (ii) Calculate the number of moles of calcium carbonate in 20 g of calcium carbonate. 3

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(iii) Calculate the number of moles of hydrochloric acid in 40 cm of 2 mol dm hydrochloric acid. (iv) Identify the limiting reagent. (v) Calculate the (a) excess in mass of the second reagent. (b) Theoretical volume of carbon dioxide that could be obtained from this reaction at rtp. Answers (i) CaCO3 (s) + 2HCl (aq) ==== CaCl2 (aq) + CO2 (g) + H2O (g) (ii) # mols calcium carbonate = mass/ Mr = 20/ 100 = 0.20 mols

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(iii) 2 mol dm HCl means that there are 2 mol HCl in 1 dm Since 1 dm == 1000 cm 2 mol HCl in 3 3 1000cm Thus, x mol HCl in 40 cm x = (40*2)/ 1000 = 0.080 mol HCl (iv) The limiting reagent is usually present in the smaller quantity (least number of moles). HCl is the limiting reagent (v) a. Based on the reaction CaCO3 react with HCl in a 1:2 mol ratio Hence since # mol HCl = 0.080 mol, # mol CaCO33 = 0.080/2 = 0.040 mol If 20 g of calcium carbonate contains 0.20 mol Excess mol CaCO3 = (initial # mol - #mol react) = 0.20 - 0.040 = 0.16 mol. Mass of 0.16 mol excess CaCO3 = 0.16*100 = 16 g b. Based on the reaction #mol CaCO3 used = #mol CO2 produced #mol CO2 = 0.040 mol At rtp (room temperature and pressure) volume 1 mol gas = 24 dm 3 Volume of 0.040 mol CO2 = 0.040*24 = 0.96 dm3