A 2.540 g sample of an oxide of tin was heated in the air until the mass of the

Question

A 2.540 g sample of an oxide of tin was heated in the air until the mass of the sample no longer changed, and 2.842 g of tin (IV) oxide (SnO2) was obtained as a result of the heating.

a) What is the mass percent of tin in SnO2?

b) What is the mass of tin in the final sample?

c) What was the mass of tin in the original sample?

d) What was the mass of oxygen in the original sample?

e) What was the number of moles of tin in the original sample?

f) What was the number of moles of oxygen in the original sample?

g) What is the formula (SnxOy) of the original oxide of tin?

Explanation / Answer

work out the moles of tin (IV) oxide.

Its formula is SnO2

moles = mass / molar mass
= 2.842 g / 150.7 g/mol
= 0.01886 mol

Now, all the Sn that was in the original ends up in the final oxide. The mass of Sn stays the same, we can work out mass of Sn present.

every SnO2 has 1 Sn
So moles Sn = moles SnO2 = 0.01886 mol
mass Sn = molar mass x moles
= 118.69 g/mol x 0.01886 mol
= 2.2383 g

Therefore mass O in the original sample = mass sample - mass tin
= 2.540 g - 2.2383 g
= 0.3017 g O

moles O = mass / molar mass = 0.3017 g / 16.00 g/mol = 0.01886 mol

Thus ratio moles Sn : O in original sample
= 0.01886 : 0.01886
= 1 : 1

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