The mole concept
Francine Taylor-Campbell, Contributor

MAIN POINTS

• Relative atomic mass, molecular mass and formula mass are measured in atomic mass units (amu), where 1amu is 1/12 of the mass of a carbon-12 atom, ¹²C.
• 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.
• 1 dm³ of any gas at STP contains the same number of molecules as 1 dm3 of any other gas at STP.
• 22.4 dm³ of any gas contains L molecules (6.0 * 10²³) at STP. This is one mole of the gas and is called the molar volume.
• At RTP the molar volume is 24 dm³.
• 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.

(iii) Calculate the number of moles of hydrochloric acid in 40 cm³ of 2 mol dm^-3 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/ M_r = 20/ 100 = 0.20 mols

(iii) 2 mol dm^-3 HCl means that there are 2 mol HCl in 1 dm³ Since 1 dm³ == 1000 cm³ 2 mol HCl in 1000cm³ Thus, x mol HCl in 40 cm³ 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 dm₃

Volume of 0.040 mol CO₂ = 0.040*24 = 0.96 dm₃

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Francine Taylor-Campbell is an independent contributor.