Answer:
CuCl2- emperical formula
CuCl4- molecular formula
Explanation:
to get empirical formula: convert the 55% into grams
so 55g CuCl over 1 times 1 mol CuCl over 35.453g equals 2 (rounded off)
so the empirical formula is CuCl2
To get molecular formula: multiply 2 to 35.453g ( 2 is from earlier which is the product of 55 devided by 35.453g) (while 35.453g is the mass of Chloride) so the product of 2 times 35.453 equals ( CuCl2=70.906 )
so CuClx - 127 divided by 70.906 = 1.79 (round off so 2)
so the molecular formula is CuCl4
its CuCl4 since we multiplied the empirical's product and the molecular's solution
- sorry if there's some grammatical errors
