Question

As a result of this procedure, anything that was in the crucible at the end of the experiment, along with the magnesium oxide product, would cause an error in the figure that is recorded as "mass of oxygen". Would extra mass it the crucible cause the "mass of oxygen" to come out too high or too low? Explain. Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical formula according to these data. mass of crucible, cover, and tin sample 21.76 g mass of empty crucible with cover 19.66 g mass of crucible and cover and sample, after prolonged heating gives constant weight 22.29 g

Answer 1

Answer 9)

The Mass (impurity) in crucible cause "Mass of Oxygen" to come out too low.

Explanation :

a) In this experiment of determination of Oxygen mass, a fixed mass of Mg stoichiometrically react wih O2 and gives corresponding Magnesium oxide product.

b) As a part of experimental procedure first a mass of Mg taken in crucible is measured.If the crucible already contain some impurity it will also get added to mass of Mg mass. This is the mass which we will use in calculation.

c) Then this fixed known mass (considering wholely Mg only which is not actually due to impurity) react with O2 and gives product.

Say Actual mass of Mg = 'A' g Mass of Impurity = 'I' g.

Hence Wrong mass of Mg = (A+I) g.

And obviously (A+I) > A.

Let mass of product formed = 'P' g.

From this we calculate mass of Oxygen reacted as,

Apperent Mass of Oxygen reacted = Mass of product - Mass of Mg (taken wrongly) = [P - (A+I) ] g.

But Actual mass of Oxygen reacted = Mass of product - Actual mass of Mg. = [P - A] g.

As (A+I) > A

We have,

[P - (A+I)] < [P - A]

i.e. Apperent Mass of Oxygen < Actual Mass of Oxygen.

I.e. Apperent mass is low than Actual mass of Oxygen calculated.

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