Empirical Formulas In your experiment, you reacted copper metal with excess sulf

Question

Empirical Formulas In your experiment, you reacted copper metal with excess sulfur to determine the coefficients x and y in Cu_x S_y. A similar experiment can be done using iron and oxygen in the air. When iron powder is heated in a crucible in air, an oxide of iron is formed. Identify it, using the experimental data below to justify your answer. Known iron oxides are: FeO, Fe_2O_3, and Fe_3O_4. Atomic masses: Fe = 55.845; O = 15.999 Mass of crucible 47.3946 g Mass of crucible and iron powder 48.8436 g After first heating in air 49.3988 g After second heating in air 49.4487 g After third heating in air 49.4493 g a) What is the mass of iron in the experiment? b) What is the mass of oxygen incorporated into the product? c) What is the formula of the iron oxide product produced? d) How do you know that the experiment is complete?

Explanation / Answer

a)

Mass of iron = (Mass of Crusible + iron)- Mass of Crucible

= 48.8436g - 47.3976g

= 1.446g

b)

Mass of oxygen incorporated in to product

= mass after third heating in air - Mass of crusible iron = 49.4493g - 48.8436g

= 0.6057g

c)

Mass of iron = 1.446g

Molar mass of Iron = 55.845g

No of mole of Iron = Mass/Molar mass = 1.446g/55.845g = 0.025893

Mass of Oxygen incorporated = 0.6057g

Molar mass of Oxygen = 15.999g

No of mole of Oxygen = 0.6057g/15.999g =0.037858

No of mole of Oxygen is 1.5 more than the No of mole of Iron as in Fe2O3

So, the formula is Fe2O3

d)

If further heatings give constant mass , then we could understand that the reaction is complated

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