Question

analysis of a compound indicates that it contains 1.04 grams K 0.70 g Cr and 0.86 g O. Find its empirical formula

Answer 1

1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = 0.0135 mol Cr   
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr  /0.0135= 1 mol Cr
 0.0538 mol O/0.035= 4 mol Cr
K2CrO4

Answer 2

Answer: The empirical formula for the given compound is $K_2CrO_4$

Explanation:

We are given:

Mass of Cr = 0.70 g

Mass of K = 1.04 g

Mass of O = 0.86 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Chromium =$\frac{\text{Given mass of Chromium}}{\text{Molar mass of Chromium}}=\frac{0.70g}{52g/mole}=0.013moles$

Moles of Potassium = $\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{1.04g}{39g/mole}=0.027moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.86g}{16g/mole}=0.054moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.013 moles.

For Chromium = $\frac{0.013}{0.013}=1$

For Potassium = $\frac{0.027}{0.013}=2.07\approx 2$

For Oxygen = $\frac{0.054}{0.013}=4.15\approx 4$

Step 3: Taking the mole ratio as their subscripts.

The ratio of K : Cr : O = 2 : 1 : 4

Hence, the empirical formula for the given compound is $K_2CrO_4$