# A 170.00 g sample of an unidentified compound contains 29.84 g sodium, 67.49 chromium, and 72.67 g oxygen. What is the compound's empirical formula?

May 24, 2017

$N {a}_{2} C {r}_{2} {O}_{7}$

#### Explanation:

So you have 29.84g of Sodium, 67.49g of chromium and 72.67g of oxygen

You would first find the number of mole of each substance in each given mass using:

$m o l s = \frac{M a s s}{M o l a r M a s s}$

$n \left(N a\right) = \frac{29.84 g}{22.9897 g} = 1.298 m o l s$

$n \left(C r\right) = \frac{67.49 g}{51.996 g} = 1.298 m o l s$

$n \left(O\right) = \frac{72.67 g}{16 g} = 4.54 m o l s$

Then you find the molar ratios between each element by dividing the moles by the lowest moles calculated

Molar ratio (Na) = $\frac{1.298}{1.298} = 1$

Molar ratio (Cr) $\frac{1.298}{1.298} = 1$

Molar ratio (O) $\frac{4.54}{1.298} = 3.5$

However 3.5 is not a whole number, to make a whole number we multiply it by 2 however if we do that we have to multiply each mol by 2 as well.

Molar ratio (Na) = $1 \cdot 2 = 2$

Molar ratio (Cr) = $1 \cdot 2 = 2$

Molar ratio (O) = $3.5 \cdot 2 = 7$

Then we can write the empirical formula

$N {a}_{2} C {r}_{2} {O}_{7}$