Question

a compound is found to contain 26.56% potassium, 35.41% chromium, and the remainder oxygen. find its empirical formula. a k2cro4 b kcro3 c k2cr2o7 question 4 of 5

Explanation:

Step1: Assume 100g of the compound

So we have 26.56g of potassium (K), 35.41g of chromium (Cr), and the mass of oxygen (O) is $100 - 26.56 - 35.41=38.03$g.

Step2: Calculate the moles of each element

The molar - mass of K is approximately 39.1 g/mol, so the moles of K: $n_{K}=\frac{26.56g}{39.1g/mol}\approx0.68$mol. The molar - mass of Cr is approximately 52 g/mol, so the moles of Cr: $n_{Cr}=\frac{35.41g}{52g/mol}\approx0.68$mol. The molar - mass of O is approximately 16 g/mol, so the moles of O: $n_{O}=\frac{38.03g}{16g/mol}\approx2.38$mol.

Step3: Find the mole - ratio of the elements

Divide each number of moles by the smallest number of moles (0.68 mol in this case). For K: $\frac{0.68mol}{0.68mol}=1$, for Cr: $\frac{0.68mol}{0.68mol}=1$, for O: $\frac{2.38mol}{0.68mol}\approx3.5$. Multiply each ratio by 2 to get whole - numbers. So the ratio of K : Cr : O is 2:2:7.

Answer:

A. $K_2Cr_2O_7$