1.00 g of a compound of copper & chlorine contained 0.524 g chlorine.i need 2 calculate the empirical formula?

help please

Answer:
Atomic mass Cl = 35.453 g/mol
0.524 / 35.453 = 0.0148 moles Cl
Atomic mass Cu = 63.54
1.00 - 0.524 = 0.476 g Cu
0.476 / 63.54 = 0.00749
We have 0.0148 moles Cl and 0.00749 mole Cu. We divide for the smallest number
0.0148 / 0.00749 = 2
The ratio between Cu and Cl is 1 : 2
The formula is CuCl2

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