Find the empirical formula of a compound that contains 68.4% chromium and 31.6% oxygen.

Question

Please help. I cried over this twice and I just don't understand. Thank you in advance.

Answer 1

Pretend like percentages are the masses in grams

68.4 g Cr

31.6 g O

Convert both to moles by dividing by periodic table mass

68.4gCr/51.996=1.316 mol Cr

31.6gO/16.00=1.975 mol O

Divide both mole values by the smaller number (1.316 is smaller than 1.975)

Cr

1.316/1.316=1

O

1.975/1.316=1.5

Since 1.5 is not a whole number, and multiplying 0.5 by 2 gets rid of the decimal, multiply both values by 2

1Cr (2)=2

1.5 O(2)=3

These are the subscripts

Cr₂O₃

If you divide the moles and you get a decimal of ... multiply by

0.2, 0.4, 0.6, 0.8 ... 5

0.25, 0.75 ... 4

0.33, 0.67 ... 3

0.5 ... 2