how many aluminum ions are present?how many chloride ions are present?what is the mass in grams of one molecule of aluminum chloride?help! i have no clue how to do this i need to know how for my quiz!!!
You have to remember Avogadro's number: 6.02 x 10^23 Read about it in your text book. It is the number of molecules of any atom that make up 1 mole. 1 mole of a molecule is equal to it's atomic weight in grams. So, the atomic weight of aluminum is 26.9. So, 26.9 grams of aluminum contains exactly 6.02 x 10^23 molecules of aluminum (or 1 mole). What's the molecular weight of aluminum chloride? Well, 26.9 + 3(35.5) = 133.4 the 35.5 is the molecular weight of chloride and there are 3 of them. So, 133.4 grams of aluminum chloride would be 6.02 x 10^23 molecules. We only have 37.2 g. 37.2 g/ 133.4 g per mole = 0.279 moles. 0.279 moles x (6.02 x 10^23 molecules/mole) = 1.68 x 10^23 molecules of aluminum chloride. So, since there is one aluminum per molecule, you have your answer. Just multiply by three to find out how many molecules of chloride are present. Finally, if 133.4g = 6.02 x 10^23 molecules, then 1 molecule = 133.4 / 6.02x10^23 = 22.16 x 10^-23 or 2.216 x 10^-22 Make sure that the periodic table you use for class has the same values (some tables are more simplified then others). Also, make sure that you use the correct number of scientific digits when calculating your answer.
