This document discusses relative atomic mass (Ar) and relative molecular mass (Mr). It defines these terms and provides examples of calculating them. Ar is a ratio of the average mass of a given atom to 1/12 the mass of one carbon-12 atom. It allows easy comparison of atomic masses. Mr is the sum of Ar values for all atoms in a molecule or polyatomic ion. Examples are given for calculating Mr of common substances like CO2, glucose, and ionic compounds. The document also addresses why some Ar values are not whole numbers due to isotope mixtures.

1. Chapter 7 Relative Masses of Atoms and Molecules

4. Introduction Atoms are soooo small and their masses are so small! Is there a way to measure them conveniently??? Since atoms are so small, let’s compare them to each other on the atomic scale.

5. We can compare every element to H or 1/12 of C-12

6. relative atomic mass The lightest atom is the hydrogen atom. As one nitrogen atom is 14 times heavier than a hydrogen atom, nitrogen is said to have a relative atomic mass of 14. N H H H H H H H H H H H H H H

10. relative atomic mass of an element Relative atomic masses of common elements: relative atomic mass Si 28 silicon O 16 oxygen N 14 nitrogen Mg 24 magnesium Na 23 sodium Al 27 aluminium P 31 phosphorus He 4 helium S 32 sulphur carbon hydrogen Elements 12 1 Relative Atomic Mass C H Symbol

11. relative atomic mass of an element Relative atomic masses of common elements: relative atomic mass Cu 64 copper Zn 65 zinc Br 80 bromine I 127 iodine K 39 potassium Ca 40 calcium C l 35.5 chlorine Fe 56 iron lead Elements 207 Relative Atomic Mass Pb Symbol

18. Relative molecular mass of some molecules 1 + 35.5 = 36.5 1H; 1Cl HCl Hydrogen chloride 32 + 16 x 2 = 64 1S; 2O SO 2 Sulfur dioxide 12 + 16 x 2 = 44 1C; 2O CO 2 Carbon dioxide 1 x 2 + 16 = 18 1O; 2H H 2 O Water 14 x 2 = 28 2N N 2 Nitrogen Calculating M r Atoms in formula Formula Substance

21. Relative formula mass of ionic compounds 40 + 12 + 16 x 3 = 100 1Ca;1C;3O CaCO 3 Calcium carbonate 23 x 3 + 31 + 16 x 4 = 164 3Na;1P;4O Na 3 PO 4 Sodium phosphate 56 + 32 + 16 x 4 + 7(1 x 2 + 16) = 278 1Fe;1S;4O;7H 2 O FeSO 4 .7H 2 O Iron (II) sulphate crystals 56 + 3 x 35.5 = 162.5 1Fe;3Cl FeCl 3 Iron (III) chloride 24 + 32 + 4 x 16 = 120 1Mg; 1S;4O MgSO 4 Magnesium sulphate Calculating M r Atoms present Formula Name