Use the following data to answer Mass of crucible=24.77g Mass crucible and tungsten=28.45g Mass crucible and tungsten oxide product=29.29g

Question

A) Determine the empirical formula of the tungsten oxide product

B)Determine the % tungsten in the tungsten oxide product

C)Determine the % error given a literature value of 79.3% tungsten in the tungsten oxide product

Answer 1

Mass of crucible = 24.77 g

Mass of crucible + tungsten = 28.45 g

Mass of crucible + tungsten oxide = 29.29 g

Mass of tungsten = (Mass of crucible + tungsten) - (Mass of crucible) = 28.45 - 24.77 = 3.68 g

Mass of tungsten oxide = (Mass of crucible + tungsten oxide) - (Mass of crucible) = 29.29 - 24.77 = 4.52 g

Mass of oxygen in tungsten oxide = (Mass of tungsten oxide) - (Mass of tungsten) = 4.52 - 3.68 = 0.84 g

(a) Empirical Formula

To determine the empirical formula, FIRST convert grams of tungsten (W) and oxygen (O) to their respective # of moles.

3.68 g W x (1 mol W / 183.84 g W) = 0.0200 mol W

0.84 g O x (1 mol O / 16.00 g O) = 0.053 mol O

Next divide the moles of each element by the lowest # of moles to calculate the mole ratio of each element in tungsten oxide.

W = 0.0200 / 0.0200 = 1.00 mol W ≈ 1 mol W

O = 0.053 / 0.0200 = 2.65 mol O ≈ 3 mol O

The empirical formula is WO₃.

(b) Percent tungsten

Molar mass of W in WO₃ = 183.84 g/mol

Molar mass of O in WO₃ = 3 x 16.00 g/mol = 48.00 g/mol

Molar mass of WO₃ = 183.84 + 48.00 = 231.84 g/mol

%-W in WO₃ = (183.84 / 231.84) x 100% = 79.3%

(c) Percent error

%-error = (|experimental value - theoretical value| / theoretical value) x 100%

= (|79.3 - 79.3| / 79.3) x 100% = 0%