Chapter 4.4, Problem 4.3CYU

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# One method for determining the purity of a sample of titanium(IV) oxide, TiO2, an important industrial chemical, is to react the sample with bromine trifluoride.3 TiO2(s) + 4 BrF3(ℓ) → 3 TiF4(s) + 2 Br2(ℓ) + 3 O2 (g)This reaction is known to occur completely and quantitatively. That is. all of the oxygen in TiO2, is evolved as O2 Suppose 2.367 g of a TiO2-containng sample evolves 0.143 g O2. What is the mass percent of TiO2 in the sample?

Interpretation Introduction

Interpretation:

The mass percentage of TiO2 in the given sample of compound should be determined.

Concept introduction:

• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• Mass percentage of reacted mass in the original sample is the ratio of mass of substance reacted to mass of whole sample taken.
• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
Explanation

Balanced chemical equation for the given reaction is,

Â Â Â Â 3TiO2(s)â€‰+â€‰4â€‰BrF3â€‰(l)â€‰â†’â€‰3TiF4(s)â€‰+â€‰2Br2(l)â€‰+â€‰3O2(g)

The amount of O2 was calculated from its mass. Â Because 1 mol of TiO2 was present in the sample for each mole of O2 isolated. Â Therefore amount of TiO2 should be known to calculate its mass and mass percentage in the sample.

The molar mass of O2 is 32.0â€‰gâ€‰/Â mol. Then the amount of oxygen can be calculated as follows,

â€‚Â 0.143â€‰gâ€‰O2â€‰Ã—1â€‰molâ€‰O232.0â€‰gâ€‰O2=â€‰0.00446â€‰â€‰molâ€‰O2

1 mol of O2 is produced from 1 mol of TiO2, the amount of TiO2 in the sample must also have been 0

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