Question

a compound found to contain 26.56% potassium, 35.41% chromium and the remainder oxygen. Find the empirical formula of the compound.

Answer 1

The remainder of oxygen is the remaining percentage:

`100-26.56-35.41=38.03.`

Now, let's calculate the number of moles based on the percentage as a mass:

`26.56g\text{ K}\cdot\frac{1\text{ mol K}}{39\text{ g K}}=0.68\text{ mol K.}`
`35.41\text{ g Cr}\cdot\frac{1\text{ mol Cr}}{52g\text{ Cr}}=0.68\text{ mol Cr.}`
`38.03\text{ g O}\cdot\frac{1\text{ mol O}}{1\text{6 g }O}=2.38\text{ mol O.}`

The next step is to divide every number of moles by the lower number of moles that we have. In this case, is 0.68 moles. So, it will be:

`\frac{0.68\text{ mol K}}{0.68}\text{ =1 mol K,}`
`\frac{0.6\text{8 mol Cr}}{0.68}=\text{1 mol Cr,}`
`\frac{\text{2.37 mol O}}{0.68}=3.5\text{ mol O.}`

But, oxygen cannot be written as a decimal, so by multiplying all the obtained moles by two, we're going to obtain:

`1\text{molK}\cdot2=2\text{ mol K, 1molCr}\cdot2=\text{2 mol Cr, 3.5mol O}\cdot2=7m\text{ol O.}`

And finally, we can build the empirical formula using the number of moles, like this:

`K_2Cr_2O_7.`