A compound containing chromium and sulfur contains 76.4 mass percent chromium.
Determine its empirical formula.
Divisiblity factor for Cr "= \\dfrac{weight percent}{molar mass } =\\dfrac{76.4}{52}=1.47"
Divisiblity factor for S "=" "\\dfrac{weight percent}{molar mass } =\\dfrac{23.6}{32}=0.73"
Now divide the above both terms with the smallest number between the two so and that will be the new divisiblity factor of the both.
So new factor for Cr = 2
So new factor for S = 1
Hence the emperical formula will became = "Cr_2 S"
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