1.19 g of chromium(Cr) formed 1.74 g of chromium oxide. The molar mass of Cr is 52.00 g/mol.?

Question

What is the empirical formula of chromium oxide?

Answer 1

mass of oxygen=1.74-1.19=0.55
% mass(or mass in 100g) of Cr=1.19/1.74*100=68.4%
% mass of oxygen=0.55/1.74*100=31.6%

Cr and O
mass in 100g(%) 68.4 and 31.6
molar mass 52 and 16
mole in 100g 68.4/52=1.325 and 31.6/16=1.975
ratio of atoms 1.325/1.325 and 1.975/1.325
1 and 1.5
multiply thru by 2 to get whole number
2 and 3
empirical form.=Cr2O3

Answer 2

a million.seventy 4 g of the chromium oxide minus a million.19 g chromium leaves .fifty 5 g oxygen. Divide those numbers by technique of the atomic mass to alter them to mole-form numbers. a million.19 g Cr / fifty two = .02288 .fifty 5 g O / sixteen = ..034375 Then divide the two by technique of the smallest and you get a million.5 O to a million Cr. you desire total numbers, so the least confusing ration is 3 to 2. Cr2O3 is the empirical formula.

Answer 3

Divide: 1.19/52 (moles of Cr)

and (1.74 - 1.19)/16 (moles of O).

Now compare these two numbers, and divide the larger (O) by the smaller(Cr), which will give you the empirical formula.