Question

A metal 'M' having an atomic weight of 197yeilds a chloride containing 35.1% of chlorine. Determine the emprical formula of the compound.​

Answer 1

automic mass of a use goes to 197. the rest will be.Now, we will calculate the simplest ratio of the elements by dividing the number of moles of each element wise smallest value 0.329. the empirical formula of the compound is AuCI3

Answer 2

Answer:

The empirical formula of the compound is gold chloride( $AuCl_{3}$)

Explanation:

Given:

A metal M having an atomic weight 197

Chloride contains 35.1 % of chlorine

To find: The empirical formula of the compound.​

Solution:

Mole:

A mole is the mass of a substance made up of the same number of fundamental components.

Given that

A compound contains chlorine = 35.1 %

And rest of them will be gold

Therefore,

100 gm of a compound will contain = 100 - 35.1

                                                           = 64.9 %

Gold = 64.9 %

Molar mass of chlorine = 35.5 g

We know that,

Number of Moles  = $\frac{Mass}{Molar mass}$

$N = \frac{m}{M}$

Number of moles in 35.1gm of chlorine = given mass/ molar mass

                                                            =  $\frac{35.1}{35.5}$

                                                            = 0.988 moles.

Molar mass of gold = 197 gm

Number of moles present in 64.9 gm of gold = given mass/ molar mass

                                                                        = $\frac{64.9}{197}$

                                                                        = 0.329

Now we can calculate the simplest ratio of the element by dividing the number of moles of each element by smallest value 0.329

Simplest ratio = Au $(\frac{0.329}{0.3289})$ Cl $(\frac{0.988}{0.329} )$

Hence the empirical formula of the compound is $AuCl_{3}$

Final answer:

The empirical formula of the compound is gold chloride( $AuCl_{3}$)

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