# What is the empirical formula for a compound containing 26.57 g potassium, 35.36 g chromium, and 38.07 g oxygen?

Apr 4, 2016

$K C r {O}_{3}$

#### Explanation:

Find the number of moles for each element by dividing the mass present with the relative mass of the atom.

$K \to \frac{26.57 g}{39.098 g m o {l}^{-} 1} = 0.68 m o l K$
$C r \to \frac{35.36 g}{51.996 g m o {l}^{-} 1} = 0.68 m o l C r$
$O \to \frac{38.07 g}{15.999 g m o {l}^{-} 1} = 2.34 m o l O$

Now divide each number of moles by the lowest one, in this case $0.68$.

$\frac{0.68 m o l K}{0.68} = 1 m o l K$
$\frac{0.68 m o l C r}{0.68} = 1 m o l C r$
$\frac{2.34 m o l O}{0.68} = 3.44 m o l O$

Round to the nearest whole number

${K}_{1} C {r}_{1} {O}_{3} = K C r {O}_{3}$

Apr 4, 2016

$\text{Empirical formula}$ $\equiv$ ${K}_{2} C {r}_{2} {O}_{7}$

#### Explanation:

We got $100 \cdot g$ of stuff, and we calculate the number of atoms element by element:

$K$ $=$ $\frac{26.57 \cdot g}{39.1 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.680 \cdot m o l$ $K$

$C r$ $=$ $\frac{35.36 \cdot g}{52.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.680 \cdot m o l$ $C r$

$O$ $=$ $\frac{38.07 \cdot g}{16.0 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.38 \cdot m o l$ $O$

Note that NORMALLY you would NEVER be given the percentage oxygen. Why not? Because there are very few ways to measure the proportion of this gas. At a 1st year undergrad level this question would have proposed an oxide of chromium that contained 26.57% potassium, and 35.36% chromium, and expected the student to twig that the missing percentage was due to the oxygen.

When we divide thru by the lowest number of moles ($0.68 \cdot m o l$) we get an $\text{empirical formula}$ of....

${K}_{\frac{0.680 \cdot m o l}{0.680 \cdot m o l}} C {r}_{\frac{0.680 \cdot m o l}{0.680 \cdot m o l}} {O}_{\frac{2.38 \cdot m o l}{0.680 \cdot m o l}}$, i.e. $K C r {O}_{3.5}$.

But by definition, the empirical formula is the simplest WHOLE number ratio defining constituent atoms in a species, so we must double this provisional formula to give:

${K}_{2} C {r}_{2} {O}_{7}$, $\text{i.e. potassium dichromate}$.

And this is a crystalline, orange powder, that is widely used in oxidations of organic materials....