Mass of A: 10 g Mass of B: 90 g Molar mass of A: 30 g/mol Molar mass of B: 100 g/mol Balanced equation: nA + mB --> AnBm What is the empirical formula forAnBm? Here is what I have so far: 10 g/mol / 30 g/mol= 0.33 mol of A 90 g/mol / 100 g/mol= 0.90 mol of B .33/.90= 0.370 ratio So where do I go from here, If I am wrong so far pleasecorrect me. Mass of A: 10 g Mass of B: 90 g Molar mass of A: 30 g/mol Molar mass of B: 100 g/mol Balanced equation: nA + mB --> AnBm What is the empirical formula forAnBm? Here is what I have so far: 10 g/mol / 30 g/mol= 0.33 mol of A 90 g/mol / 100 g/mol= 0.90 mol of B .33/.90= 0.370 ratio So where do I go from here, If I am wrong so far pleasecorrect me. What is the empirical formula forAnBm? Here is what I have so far: 10 g/mol / 30 g/mol= 0.33 mol of A 90 g/mol / 100 g/mol= 0.90 mol of B .33/.90= 0.370 ratio So where do I go from here, If I am wrong so far pleasecorrect me.
Explanation / Answer
no. of moles of A = mass / Molar mass = 10 g / 30 g / mol = 0.3333 mol No. of moles of B = mass / Molar mass =90 g / 100 g /mol = 0.9 mol ratio of no. of moles of A & B is 0.3333 : 0.9 1: 3.7 ~ 1 : 4 So, the emperical formula is A 1 B4