Calculate the empirical formula of aluminum chloride produced
ConstanceP > 01-15-2020, 08:47 AM
I have the following homework question:quot;A sample of 1.27 g of aluminum reacted with chlorine gas to give 6.28 g of aluminum chloridePlease calculate the empirical formula of the aluminum chloride that was produced.quot;If I know that aluminum chloride is AlCl3 is there no calculation to be done, as this is already in the simplest form or am I not misunderstanding the question? Thanks.You\'re not understanding the intent of the questionIt is asking you to prove that the formula is AlCl3 using the data givenIt goes this way: (1.27 g Al) / (26.98154 g Al/mol) 0.047069 mol Al (6.28 g - 1.27 g) Cl / (35.4532 g Cl/mol) 0.14131 mol Cl Divide by the smaller number of moles: (0.047069 mol Al) / 0.047069 mol 1.00 (0.14131 mol Cl) / 0.047069 mol 3.00 Round to the nearest whole numbers to find the empirical formula: AlCl3You\'re not understanding the intent of the questionIt is asking you to prove that the formula is AlCl3 using the data givenIt goes this way: (1.27 g Al) / (26.98154 g Al/mol) 0.047069 mol Al (6.28 g - 1.27 g) Cl / (35.4532 g Cl/mol) 0.14131 mol Cl Divide by the smaller number of moles: (0.047069 mol Al) / 0.047069 mol 1.00 (0.14131 mol Cl) / 0.047069 mol 3.00 Round to the nearest whole numbers to find the empirical formula: AlCl3Other related question