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Chemistry

Chemistry. The mole. The mole . Very important concept Used in almost all calculation in chemistry What is a mole? A mole is the amount of substance that contains 6x10 23 elementary units (ions, molecules and atoms). 1 mole of sodium chloride ions would contain 6x10 23 ions

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Chemistry

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  1. Chemistry The mole

  2. The mole • Very important concept • Used in almost all calculation in chemistry • What is a mole? • A mole is the amount of substance that contains 6x1023 elementary units (ions, molecules and atoms). • 1 mole of sodium chloride ions would contain 6x1023 ions • 1 mole of carbon dioxide would contain 6x1023 molecules

  3. The mole • Mole = given mass/ R.F.M • R.F.M- Relative Formula Mass • R.F.M= Sum of the Relative Atomic Mass (R.A.M) • R.F.M of NaCl= 23+35.5= 58.5 • R.F.M of ammonium sulphate- (NH4)2SO4 = 2x(14+4) + 32+ 4(16)= 132

  4. Calculating the number of moles • Number of moles in 13.2g (NH4)2SO4 = given mass/ RFM= 13.2/132=0.1 moles

  5. The mole and the equation • Calculate how many grams of Calcium Chloride will be formed when 2 grams of calcium is mixed with an excess of chlorine gas. • To solve this problem, we must first write a balanced equation of the reaction taking place. Then determine the mole ratio. • This is shown on the next slide.

  6. Solution • Ca + Cl2 −> CaCl2 • There is a 1:1 ratio of Calcium and Calcium Chloride • This means that there will be a 1:1 mole ratio of calcium reacted to calcium chloride produced

  7. Solution continued • No. moles calcium= given mass/ molar mass • = 2g/40= 0.05 moles • Since there is a 1:1 mole ration • 0.05 moles of Calcium Chloride will be formed • Since moles = mass/R.F.M • mass= R.F.M x moles= (40+ 2(35.5) x 0.05 • =111 x 0.05 = 5.55g CaCl2 • Answer= 5.55g CaCl2

  8. Useful points to note • 1 mole of gas, occupies 22.4 dm3 (This will be important when calculating volumes of gases produced). Reactivity points to note • Reactive metals (e.g. Group 1,2 and 3 metals and metals such as Zn, Fe and Cu) react with dilute acids to form a salt and hydrogen gas e.g. Zn(s) + H2SO4(aq) −> ZnSO4(aq) + H2(g)

  9. Useful points to note • More reactive metals will displace less reactive metals from solution according to order of reactivity. • The order of reactivity is shown below for some of the common metals you will come across during your CSEC studies. • K>Na>Ca>Mg>Al>Zn>Fe>Cu>Ag

  10. Useful points to note • You should know most of these symbols however you may not be familiar with Zn, Fe, Cu and Ag. These are defined below: Zn-Zinc Fe- Iron Cu- Copper Ag-Silver

  11. Useful Points to note • Fe(s) + CuSO4 (aq ) −> FeSO4(aq) + Cu(s) • Iron will displace copper from copper sulphate .

  12. Problem 1 • Calculate the volume of hydrogen produced when 6.5g of Zn is reacted with excess dilute sulphuric acid

  13. Solution • Equation • Zn(s) + H2SO4(aq) −> ZnSO4(aq) + H2(g) • No moles of Zn reacted = given mass/R.F.M =6.5/65 • No. moles of Zn reacted = 0.1 Moles • What is the mole ratio of Zn reacted to H2 liberated

  14. Solution • 1:1 ratio • From 1:1 mole ration, the no. moles of hydrogen gas is liberated = 0.1moles • Since 1 mole occupies 22.4 dm3 , then 0.1 moles should occupy 2.24 dm3( 0.1 x 22.4). Answer = 2.24 dm3

  15. Solution • 1:1 ration • From 1:1 mole ration, the no. moles of hydrogen gas is liberated = 0.1moles • Since 1 mole occupies 22.4 dm3 , then 0.1 moles should occupy 2.24 dm3( 0.1 x 22.4). Answer = 2.24 dm3

  16. Problem 2 • Calculate the volume of oxygen gas formed during the electrolysis of 18 moles of water. Equation is shown below: • 2H20(l) −> O2(g) + 2H2(g)

  17. Solution • Mole ratio • 2 mole of water: 1 moles of oxygen gas • Hence 18 moles of water would produce 9 moles of oxygen gas. • We know that 1 mole of gas occupies 22.4 dm3 • Then 9 moles of oxygen will occupy 9 x 22.4 = 201.6 dm3

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