Recap – Chemical Equations Chem 1001 Lecture 4 Recap – Chemical Equations Formula equation – gives overview, balance for numbers and types of atoms eg Mg + 2HCl MgCl2 + H2 Molecular equation – include states of reactants and products eg C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) Net ionic equation – focus on just what is involved and balance charges eg Ca2+(aq) + CO32-(aq) CaCO3(s)
Masses of Atoms 1H atom 1p 1 nucleon 4He atom 2p + 2n 4 nucleons Chem 1001 Lecture 11 Masses of Atoms 1H atom 1p 1 nucleon 4He atom 2p + 2n 4 nucleons 4 times the mass of H 7Li atom 3p + 4n 7 nucleons 7 times the mass of H 238U atom 92p + 146n 238 nucleons 238 times the mass of H Dr Chiara Neto
Isotopes and Average Mass Chem 1001 Lecture 11 Isotopes and Average Mass 1H atom 2H atom 1p 1p + 1n 99.99% 0.01% 1.008 6Li atom 7Li atom 3p + 3n 3p + 4n 7.5% 92.5% (6 x 7.5%) + (7 x 92.5%) = 6.9 12C atom 13C atom 6p + 6n 6p + 7n 98.9% 1.1% (12 x 98.9%) + (13 x 1.1%) = 12.01 35Cl atom 37Cl atom 17p + 18n 17p + 20n 75.8% 24.2% (35 x 75.8%) + (37 x 24.2%)= 35.5 Dr Chiara Neto
Avogadro Number, NA NA = 6.022 1023 Chem 1001 Lecture 11 Avogadro Number, NA Actual mass of H is 1.67 x 10-24 g Use a scaling factor: the Avogadro number, NA NA = 6.022 1023 Definition: The Avogadro number is the number of 12C atoms present in exactly 12 g of 12C. One ‘mole’ of anything contains the Avogadro number, NA, of items Dr Chiara Neto
Avogadro Number, NA Atomic mass mass of NA atoms ‘Molar mass’ Hydrogen Chem 1001 Lecture 11 Avogadro Number, NA Atomic mass mass of NA atoms ‘Molar mass’ Hydrogen 1.008 amu 1.008 g Lithium 6.94 amu 6.94 g 12C 12.00 amu 12.00 g Carbon 12.01 amu 12.01 g Chlorine 35.45 amu 35.45 g Dr Chiara Neto
Chem 1001 Lecture 12 Molar Masses Relative masses of molecules is simply the sum of the atomic masses of the component atoms. eg H2 2 x 1.01 = 2.02 g mol-1 H2O (2 x 1.01) + 16.0 = 18.02 g mol-1 C6H12O6 (6 x 12.01) + (12 x 1.01) + (6 x 16.0) = 180.18 g mol-1 For ionic solids, no molecules, use the empirical formula. eg NaCl 23.0 + 35.5 = 58.5 g mol-1 Dr Toby Hudson
The Mole no. of moles (n) = mass (m) / molar mass (M) or Chem 1001 Lecture 12 The Mole So now we can relate the number of grams of a substance we weigh out in the lab to the number of moles, and thus the number of particles of the substance that we have: no. of moles (n) = mass (m) / molar mass (M) or no. of moles (n) = no. of items / 6.022 x 1023 Dr Toby Hudson
The Mole Q: How many atoms present in 1.0 g of silver? Chem 1001 Lecture 12 n = mass / molar mass The Mole n = no. of items / NA Q: How many atoms present in 1.0 g of silver? Moles = mass / molar mass = 1.0 / 107.9 = 0.0093 mol No. atoms = moles NA = 0.0093 6.022 1023 = 5.6 1021 atoms Dr Toby Hudson
Chem 1001 Lecture 12 n = mass / molar mass The Mole n = no. of items / NA Q: Calculate the no of moles of water in 1000 g water. Molar mass = 18.02. Moles = mass / molar mass = 1000 / 18.02 = 55.49 moles Q: How many water molecules would be present? No. molecules = moles NA = 55.49 6.022 1023 = 3.342 1025 molecules Dr Toby Hudson
Learning Outcomes: By the end of this lecture, you should: recognise the mass reported is a weighted average of the mass of the individual isotopes of an element be able to calculate an average mass from isotope data understand that NA is a scaling factor understand the difference between atomic mass and molar mass be able to calculate the molar mass of a compound be able to convert between mass, moles and numbers of atoms/ions/molecules be able to complete the worksheet (if you haven’t already done so…)
Questions to complete for next lecture: Chem 1001 Lecture 4 Questions to complete for next lecture: Give the definition of a mole of substance. How many atoms are there in 0.25 mole of silver? Silver contains two isotopes: 107Ag (51.8%) and 109Ag(48.2%). What is the average atomic mass of silver? Table sugar (sucrose) has the formula C12H22O11. What is the molar mass of sucrose? How many moles of sucrose are present in a sugar lump with mass 5.0 g? How many molecules of sucrose are present in the sugar lump?