The
electrochemistry of a Lead acid cell at each plate is such that two electrons are transferred per atom of Lead, but as both plates are needed, to store two electrons it needs PbO
2+ Pb +2H
2SO
4. Adding up the atomic masses:
Pb: 207.2 *2 = 414.4
O: 16 *10 = 160
H: 1 *4 = 4
S: 32 *2 = 64
Total: 642.4 or 321.2 per single electron.
Additionally, the electrolyte of a fully charged battery is 5.2 molar, so for every 5.2 moles of acid there is 1 litre of water. That's approximately 10.7 molecules of H
2O per molecule of acid so add 192.3 for the molecular mass of the required electrolyte, giving a result of 513.5 per electron. Divide by Faraday's constant and multiply by 3600, and you get 19.5 g/AH. Multiply by 6 for a 12V battery, and by 7 for 7AH, and you get just under 805g for the chemically active parts of the battery. (unless I've FUBARed the maths)
However, you need a lead matrix to support the active electrodes, more electrolyte to saturate the separator and surround the electrodes + the case etc. so 19.5 g/AH/cell is only the theoretical limit.
Additionally to the problems with cell internal resistance mentioned above, the lead matrix of the plates tends to corrode during the life of the battery, loosening the electrode material in its pockets so it sheds, reducing the capacity and also increasing the internal resistance, so the expected life of a battery (optimised for a particular application) is directly related to the thickness of lead in its plates and thus its mass.