how do i find the formula for this compound?

Question

an oxide of chromium is 68.4% chromium by mass. the molar mass of the oxide is 152g/mol. what is the formula of the compound?

Answer 1

lets assume the formula is Cr_xO_y
now molecular wt of Cr : 52
and that of O: 16

thus 52x + 16y = 152
x/(x+y) = 0.684

solving the two equations simultaneously, we get
x = 2, y = 3

so the oxide is Cr2O3

Answer 2

Let's say you had 100 grams of the mystery compound. 68.4% of the mass is chromium, or 68.4 grams. The remaining 31.6 grams must therefore be oxygen. Now we convert these masses into moles (dividing by the molar mass):

(68.4 g chromium) / (52.0 g/mol) = 1.315 mol chromium

(31.6 g oxygen) / (16.0 g/mol) = 1.975 mol oxygen

Now we try to find two small whole numbers that have the same ratio as these:

1.975 / 1.315 = 1.501

This is almost exactly 1.5, or 3/2. We can therefore guess that there are 3 oxygen atoms for every 2 chromium atoms.

However, this only tells us the ratio. We don't know if the formula is Cr2O3 or Cr4O6 or Cr6O9 or another multiple. That's where the molar mass of 152 g/mol comes in. It turns out that Cr2O3 has a molar mass of 152 g/mol, so that's the right answer.

Answer 3

Cr-51 atomic mass
O- 16 atomic mass

100%-68.4=31.6% for O

Cr- 68.4/51=1.34

O-31.6/16=1.9

divide by lowest amount
Cr= 1.34/1.34=1
O=1.9/1.34=1.4 (round to 2)

the molecular formula is CrO2

to find the emperical:
add both atomic mass=67
divide by g/mol- 152/67=(2)

therefore the empirical must be Cr2O4

Answer 4

you have to use stoichiometry. look it up on google.