Lithium has a relative atomic mass of 7. Bromine has a relative atomic mass of 80. A compound was found to contain 16g of lithium and 183g of bromine. What is the empirical formula of this compound?

Respuesta :

6.015 multiplied by 0.0742 + 7.016 multiplied by 0.9258 = 6.94

To get empirical formula of the compound, we will follow certain steps:

Step 1: Find the number of moles of lithium that is Li and bromine that is Br.

To calculate number of moles, we will follow the equation,

[tex] Number of moles of a substance= \frac{Given mass of the substance}{Molar mass of the substance} [/tex]

Here the given mass of Li=16 g

Given mass of Br=183 g

Molar mass of Li= 7 g mol⁻¹

Molar mass of Br= 80 g mol⁻¹

So,

[tex] Number of moles of Li= \frac{Given mass of the Li}{Molar mass of the Li} [/tex]

[tex] Number of moles of Li= \frac{16}{7} [/tex]

Number of moles of Li= 2

Similarly,

[tex] Number of moles of Br= \frac{Given mass of the Br}{Molar mass of the Br} [/tex]

[tex] Number of moles of Br= \frac{183}{80} [/tex]

Number of moles of Br= 2

Step 2: Find the smallest number of moles:

Smallest number of moles is 2 as both moles are same

Step 3: Find the ratio of the atom by diving their moles with smallest number of moles:

Ratio of Li= \frac{2}{2}

=1

Ratio of Br= \frac{2}{2}

=1

So the empirical formula is Li₁Br₁ that is LiBr.