Question

checkpoint 1
find the empirical formula.
there is an oxide of chromium that is 68.4% chromium and 31.6% oxygen. what is the substances empirical formula? hint: remember, you can assume you have 100 grams of the unknown substance.
additional information:
chromium (cr) = 52.00 g/mol
oxygen (o) = 16.00 g/mol

Explanation:

Step1: Calculate moles of Cr

Given mass of Cr = 68.4g (since 68.4% of 100g). Molar mass of Cr = 52.00g/mol.
Moles of Cr = $\frac{68.4g}{52.00g/mol}=1.315mol$

Step2: Calculate moles of O

Given mass of O = 31.6g (since 31.6% of 100g). Molar mass of O = 16.00g/mol.
Moles of O = $\frac{31.6g}{16.00g/mol}=1.975mol$

Step3: Find mole - ratio

Divide each mole value by the smaller mole value (1.315).
For Cr: $\frac{1.315mol}{1.315mol}=1$
For O: $\frac{1.975mol}{1.315mol}\approx1.5$
Multiply by 2 to get whole - numbers. So, Cr: $1\times2 = 2$, O: $1.5\times2=3$

Answer:

$Cr_2O_3$