i.e. mass of 1 mole of glucose, C6H12O6 = (6 × 12.01 + 12 × 1.01 + 6 × 16.00) g. = 180.18 g ... Analysis gives 38.7 % carbon and 9.8 % hydrogen by mass.
It uses elimination againLet A be 15% juice and B is 5% juice A+B = 100.15A + 0.05B = 0.11*10 = 1.1Multiply 2nd equation by 100 to get rid of decimals A+B = 1015A + 5B = 110
