Answer to Question #180698 in Chemistry for Kyle

Question

Calculate the Gravimetric factor of the following. Show your Solution.

A. Al in Al2O3

B. Na in Na2C2O4

C. Cr2O3 in PbCrO4

D. K in K2PtCl6

E. Cl in AgCl

Answer 1

Solution:

The gravimetric factor (GF), represents the weight of analyte per unit weight of precipitate. It is obtained from the ratio of the formula weight of the analyte to that of the precipitate, multiplied by the moles of analyte per mole of precipitate obtained from each mole of analyte, that is:

Thus:

A) Al in Al₂O₃

Analyte - Al; Precipitate - Al₂O₃

2Al → Al₂O₃

The molar mass of Al is 26.98154 g/mol.

The molar mass of Al₂O₃ is 101.96 g/mol.

Hence,

GF = (26.98154 g Al mol^-1) / 101.96 g Al₂O₃ mol^-1) × (2/1) = 0.52926 g Al / g Al₂O₃

GF = 0.52926 (Al in Al₂O₃)

B) Na in Na₂C₂O₄

Analyte - Na; Precipitate - Na₂C₂O₄

2Na → Na₂C₂O₄

The molar mass of Na is 22.9897 g/mol.

The molar mass of Na₂C₂O₄ is 133.9985 g/mol.

Hence,

GF = (22.9897 g Na mol^-1) / 133.9985 g Na₂C₂O₄ mol^-1) × (2/1) = 0.34313 g Na / g Na₂C₂O₄

GF = 0.34313 (Na in Na₂C₂O₄)

C) Cr₂O₃ in PbCrO₄

Analyte - Cr₂O₃; Precipitate - PbCrO₄

Cr₂O₃ → 2PbCrO₄

The molar mass of Cr₂O₃ is 151.99 g/mol.

The molar mass of PbCrO₄ is 323.1937 g/mol.

Hence,

GF = (151.99 g Cr₂O₃ mol^-1) / 323.1937 g PbCrO₄ mol^-1) × (1/2) = 0.23514 g Cr₂O₃ / g PbCrO₄

GF = 0.23514 (Cr₂O₃ in PbCrO₄)

D) K in K₂PtCl₆

Analyte - K; Precipitate - K₂PtCl₆

2K → K₂PtCl₆

The molar mass of K is 39.0983 g/mol.

The molar mass of K₂PtCl₆ is 485.99 g/mol.

Hence,

GF = (39.0983 g K mol^-1) / 485.99 g K₂PtCl₆ mol^-1) × (2/1) = 0.1609 g K / g K₂PtCl₆

GF = 0.16090 (K in K₂PtCl₆)

E) Cl in AgCl

Analyte - Cl; Precipitate - AgCl

Cl → AgCl

The molar mass of Cl is 35.453 g/mol.

The molar mass of AgCl is 143.32 g/mol.

Hence,

GF = (35.453 g Cl mol^-1) / 143.32 g AgCl mol^-1) × (1/1) = 0.24737 g Cl / g AgCl

GF = 0.24737 (Cl in AgCl)

Answers:

A) GF = GF = 0.52926 (Al in Al₂O₃);

B) GF = 0.34313 (Na in Na₂C₂O₄);

C) GF = 0.23514 (Cr₂O₃ in PbCrO₄);

D) GF = 0.16090 (K in K₂PtCl₆);

E) GF = 0.24737 (Cl in AgCl).