A sample of iron was burned in a crucible at high heat after cooling iron oxide

Question

A sample of iron was burned in a crucible at high heat after cooling iron oxide formed was weighed. Calculated results of the mass of iron and mass of oxygen (O) are given below. mass of FP = 0.9694g and mass of O = 0.4166g Calculate empirical formula of iron oxide formed in the experiment. Show all steps of calculations with conversion factor A sample of 1350 g of an impure sample of KHP was titrated with 0.120 m NaoH standardized solution. This volume of NaOH to neutralize was 35.50 mL of NaOH. Calculate this percentage of KHP in this sample Show all steps with units and Cv and FV.

Explanation / Answer

Mass of Iron = 0.9694 g = 0.9694/55.85 = 0.017357 Moles

Mass of oxygen = 0.4166 g = 0.4166 /16 = 0.0260375 moles

Fe : Oxygen = 0.017357 : 0.0260375 = 17.3 : 26 = 1 : 1.5 = 2 :3

Hence Fe2O3 is empirical formula

Volume of NaOH = 35.5 ml

COncentration of NaOH = 0.12 M

= 0.12 x 35.5 / 1000 = 0.00426 M

Molar mass of KHP = 204.22

Actual mass of KHP = 0.00426 x 204.22 = 0.87 gm

Weight of KHP with Impurity = 1.35

Hence the percentage of purity = 0.87 x 100 / 1.35 = 64.44%

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