Homework question L.35 part 1 [ENDORSED]

[Julia Lee]

Hi, can someone help me figure out how to solve L.35 part one of the homework question?

A 1.50-gram sample of metallic tin was placed in a 26.45 g crucible and heated until all the tin had reacted with the oxygen in air to form an oxide. The crucible and product together were found to weigh 28.35 g.

What is the empirical formula of the oxide?

[RachaelKoh3A]

Hi!

The mass of the crucible remains constant. We can find the mass of the product first.

Mass of product = 28.35 - 26.45 = 1.90 g (2dp)

Since a type of tin oxide was formed and the mass of tin in this tin oxide should not change, we can find the mass of oxygen in the product by subtract the mass of product from the mass of the tin sample.

Mass of oxygen = 1.90 - 1.50 = 0.40 g (2dp)

You can then use these masses of tin and oxygen to find the empirical formula. Mr of Sn is 118.71 g/mol and Mr of O is 16.00 g/mol.
Take the masses divide by the Mr of the elements to find the number of moles.
Number of moles of Sn = 1.50 / 118.71 = 0.0126 mol (3sf)
Number of moles of O = 0.40 / 16.00 = 0.025 mol (2sf)

Divide these two numbers by the smallest value, 0.025. You should obtain a ratio of Sn:O = 1:2. So the empirical formula will be SnO2.

Hope this helps! :)