Answer:
%Cr = 47.71%
%F = 52.29%
Step-by-step 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
%Cr = 47.71%
%F = 100 - 47.71
%F = 52.29%
Hope this helps