Question

14. A chemistry student determined the empirical formula for tungsten oxide (W Oy). To do so, he heated tungsten with oxygen in a crucible. The data that he recorded are shown below: Weight of crucible Weight of tungsten Weight of crucible and product 11.120 g 8.820 g 22.998 g a What is the percent composition by mass of tungsten oxide? D What is the empirical formula of tungsten oxide? (10 points)

Answer 1

answer:

the weight of crucible = 11.120 g

the weight of tungsten taken = 8.820 g

molar mass of tungsten = 183.84 g/mol

so mole fo tungsten in 8.820 g = 8.820/183.84 = 0.048 mol

mass of crucible and product = 22.998 g

so the mass of product = 22.998 - 11.120 = 11.878 g

mass of oxygen in the product = 11.878 - 8.820 =3.058 g

molar mass of oxygen = 16.0 g/mol

so moles of oxygen in 3.058 g = 3.058/16.0 = 0.1911 mol

a) % compostion by mass

tungsten = (mass of tungsten/ total mass) *100

= (8.820 /11.878)*100 = 74.25%

oxygen = (3.058/11.878)*100 = 25.75%

b) mole ration of tungsten to oxygen is = 0.048 / 0.1911

ration is 1:4

so empirical formula is WO₄ which is usually WO₄^2- which is tungstate ions