Question

the composition of copper chloride is as follows :

Copper = 47.3%, Chlorine = 52.7%. Calculate the empirical formula of the compound.

Atomic mass of copper = 63.5 and Atomic mass of chlorine = 35.5.​

Answer 1

Explanation:

$\huge{\underline{\bold{\pink{Answer:⤵}}}} \\ \huge \colorbox{lightgreen}{attached \: answer } \:$

please mark as brainliest answer please thank my answer please friend ❣️✌️❤️ God bless you

Answer 2

:

When a chemical formula that represents proportions of elements depicted in a compound is known as an empirical formula.

Since, it is given that in four gram copper chloride there is 1.890 g of copper and 2.110 g of chlorine. So, calculate the number of moles present in copper and chlorine as follows.

$No. of moles in copper = \frac{mass}{molar mass of copper} </p><p>molarmassofcopper</p><p>mass</p><p> </p><p> </p><p></p><p> = \frac{1.890 g}{63.54 g/mol} </p><p>63.54g/mol</p><p>1.890g</p><p> </p><p> </p><p></p><p> = 0.0297 g/mol</p><p></p><p> No. of moles in chlorine = \frac{mass}{molar mass of chlorine} </p><p>molarmassofchlorine</p><p>mass</p><p> </p><p> </p><p></p><p> = \frac{2.110 g}{35.45 g/mol} </p><p>35.45g/mol</p><p>2.110g</p><p> </p><p> </p><p></p><p> = 0.0595 g/mol</p><p></p><p>Therefore, molar ratio will be calculated as follows.</p><p></p><p> Molar ratio for copper = \frac{0.0595 g/mol}{0.0595 g/mol} </p><p>0.0595g/mol</p><p>0.0595g/mol</p><p> </p><p> </p><p></p><p> = 1</p><p></p><p> Molar ratio for chlorine = \frac{0.0595 g/mol}{0.0297 g/mol} </p><p>0.0297g/mol</p><p>0.0595g/mol</p><p> </p><p> </p><p></p><p> = 2</p><p></p><p>Thus, we can conclude that empirical formula of given compound is CuCl_{2}CuCl </p><p>2</p><p>$

.