Answer to Question #181280 in Chemistry for Albert

Question

What will 628 mg of Zr (HPO4)2 weighed after ignition to ZrP2O7?

Answer 1

Solution:

Zr(HPO₄)₂ → ZrP₂O₇ + H₂O

Analyte - Zr(HPO₄)₂; Precipitate - ZrP₂O₇

The molar mass of Zr(HPO₄)₂ is 283.1826 g/mol.

The molar mass of ZrP₂O₇ is 265.1673 g/mol.

Zr(HPO₄)₂ → ZrP₂O₇

Hence,

GF = (283.1826 g Zr(HPO₄)₂ mol^-1) / 265.1673 g ZrP₂O₇ mol^-1) × (1/1) = 1.06794 g Zr(HPO₄)₂ / g ZrP₂O₇

Gravimetric factor (GF) = 1.06794 g Zr(HPO₄)₂ / g ZrP₂O₇

grams analyte = GF × grams precipitate

Hence,

628 mg of Zr(HPO₄)₂ = (1.06794 g Zr(HPO₄)₂ / g ZrP₂O₇) × grams of ZrP₂O₇

grams of ZrP₂O₇ = 628 mg / 1.06794 = 588.048 mg = 588 mg ZrP₂O₇

Answer: 588 mg