Find the percent composition of chromium(|||) fluoride (CrF3), which is commonly used as a dye in textiles

%Cr = 47.71%

%F = 52.29%

Explanation:

To do this, we need the atomic weights of all the elements in the chromiun fluoride, which are the following:

Cr: 51.996 g/mol

F: 18.998 g/mol

With these atomic weights, let's calculate the molecular mass of the chromium fluoride:

MM CrF₃ = (51.996) + (18.998 * 3) = 108.99 g/mol

Now, to get the percent composition, all we have to do is divide the atomic weight of each element by the molecular mass, and then, multiply by 100%:

% = (AW / MM) * 100

Replacing with each element:

%Cr = (51.996 / 108.99) * 100

%F = 100 - 47.71

Hope this helps

answer

the ions in solute supperate from eachouthere