Moles Number of atoms/molecules ; Relative Molecular Mass ; Volume occupied. https://www.youtube.com/wa tch?v=PvT51M0ek5c
Avogadro’s Number of particles = 6.0x1023 I mole of a substance is the amount of that substance which contains particles of that 6.0 x 1023 substance. https://www.youtube.com/watch?v=TEl4jeET Vmg Avogadro’s Number of particles = 6.0x1023
1 mole of oxygen = 16g
Definitions RMM in grams – Relative Molecular Mass The RMM of a compound = Sum of the molecular masses of the atoms in the compound.
RMM in Grams What is the RMM of each of the following H2O = (2 x 1) + 16 = 18 (NH4)2SO4 = (14+1*4)*2+32+16*4 = 132 What is the RMM of each of the following H3PO4 ; Al2(SO4)3 ; CuSO4.5H2O? H3PO4 = 1 * 3 + 31 + 16 * 4 = 98 Al2(SO4)3 = 27 * 2 + (32 +16*4)*3 = 342 CuSO4.5H2O = 64 + 32 +16*4+ 5(1*2 +16) = 250
Questions H3PO4 ; Al2(SO4)3 ; CuSO4.5H2O How many moles in 10 grams of each? What mass of each is needed to make 0.75 mole? H3PO4 = Al2(SO4)3 = CuSO4.5H2O =
Questions H3PO4 ; Al2(SO4)3 ; CuSO4.5H2O How many moles in 10 grams of each? What mass of each is needed to make 0.75 mole? H3PO4 = 10/98 = 0.1 mole Al2(SO4)3 = 10/342 = 0.03 mole CuSO4.5H2O 10/250 = 0.04 mole H3PO4 = 0.75 of 98 = 73.5 g Al2(SO4)3 = 0.75 of 342 = 256.5 g CuSO4.5H2O = 0.75 of 250 = 187.5 g
Ammonium nitrate [NH4NO3] Calcium carbonate [CaCO3] Avogadro’s Number = 6 x 1023 How many molecules are there in 5 g of each of the following? Hydrogen gas [H2] Methane [CH4] Water [H2O] Ammonium nitrate [NH4NO3] Calcium carbonate [CaCO3]
Grams H2 1 mole 2 g 1 g 5 g Molecules 6 * 1023 6 * 1023 / 2 5 * 6 * 1023/ 2 1.5 * 1024
Ammonium nitrate [NH4NO3] Calcium carbonate [CaCO3] Avogadro’s Number = 6 * 1023 How many atoms are there in 5 g of each of the following? Hydrogen gas [H2] Methane [CH4] Water [H2O] Ammonium nitrate [NH4NO3] Calcium carbonate [CaCO3]
H2 = 1 * 2 = 2g 5/2 = 2.5 moles = 2.5 * 6 * 1023 = 15 * 1023 * 2 = 3.0 *1024 CH4 = 12 + (1*4) = 16 g5/16 = 0.31 mole = 0.31 * 6 * 1023 = 1.88 * 1023 * 5 = 9.4 * 1023 H2O = (1*2) + 16 = 18g 5/18 = 0.28 mole = 0.28 * 6 * 1023 = 1.68 * 1023 * 3 = 5.04 * 1023 NH4NO3 = 14 + (1*4) + 14 + (16*3) = 80 g 5/80 = 0.0625 mole = 0.0625 * 6 * 1023 = 0.375 * 1023 = 3.75 * 1022 * 9 = 3.38 * 1023 CaCO3 = 40+12+(16*3) =100 = 5/100 = 0.05 mole 0.05*6*1023 = 0.3*1023 = 3.0*1022*5 = 1.5 * 1023
22.4 L of any gas at STP Volume occupied by each of the following 8g of Oxygen 32g methane 20g of neon 7.1g of chlorine 0.01g of hydrogen
O2 = 16 * 2 = 32 g 8/32 = 0.25 moles 0.25 of 22.4 L = 5.6 L CH4 = 12+ (1*4) = 16g 32/16 = 2 mole = 2 * 22.4 l = 44.8 L Ne = 20 = 20g 20/20 = 1 mole = 22.4 L Cl2 = 35.5 * 2 = 71 g 7.1/71 = 0.1 moles 0.1 of 22.4 L = 2.24 L
General rule for questions Write down the two units Put the one you are given the amount of first and the one you are asked the amount of second Give the value of 1 mole of each Bring the one you know the amount of down to 1 unit and divide the other by the molar amount Multiply both by the amount given Watch out for the number of atoms / molecules
Example How many atoms in 2.24L of methane at STP? 22.4L = 6 * 1023 molecules 1 L = 6 * 1023 / 22.4 molecules 2.24L = 2.24 * 6 * 1023 / 22.4 Methane is CH4 so 5 atoms per molecule Atoms = 5 * 2.24 * 6 * 1023 / 22.4 = 3 * 1023
Race Find the number of grams of: 3.1 moles of sodium chloride 2.5 moles of MgSO4.7H2O 1.5mol. of chlorine gas (remember that chlorine is diatomic) 0.25mol. of sodium carbonate
Christmas Test Revision! Structure = 10 Short Questions 3 Long Questions Exam = 90% Essay = 5% Presentation = 5% Atomic Structure – Calculating Protons, Neutrons and Electrons Balancing Chemical Equations Physical and Chemical Change Isotopes – Calculations Electronic Configuration Mole Calculation