Outline 1 Question L39

[Inderpal Singh 2L]

A 1.50-g 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. (a) What is the empirical formula of the oxide? (b) Write the name of the oxide.

Can someone help me understand how to solve this problem?

[shevanti_kumar_1E]

Ok, so you first calculate the g of Sn and O2 by subtracting the gram of the crucible from the total mass.
Total: 28.35 g
28.35g - 26.45 g = 1.9 g of oxide
You know the amount of tin so you subtract that from 1.9g to get the g of O2.
1.9g - 1.5g = 0.4 g O

Divide by molar mass to find mol of O and Sn in oxide.

0.4g/16g/mol = 0.025 mol O
1.5g/118.71 g/mol = 0.0126 mol Sn

Now that you have mol of O and Sn see their ratio to get the empirical formula.

0.025/0.0126 = 1.98 which is around 2

So the empirical formula is SnO2.

The name of the oxide is Tin (4) oxide (O has charge of -2 and there are 2 molecules so total charge of -4 so tin would have to have charge of +4).

[Chelsea_Guzman_3C]

Crucible + Sn = 28.35g
28.35g - 26.45g = 1.9g
1.90g - 1.50g = 0.40g Oxygen

0.40g O x 1mol/16g = 0.025 mol Oxygen
1.50g tin x 1mol/118.71g = 0.0126 mol Tin

a)
Sn0.0126/0.0126 _ Oxygen 0.025/0.0126 =
SnO2 (empirical formula)

b) name: tin dioxide

(Hope this helps!)

[AlbertGu_2C]

Given:
1.5g Sn
26.45g Crucible Mass
28.35g Mass of Crucible + Product

From here, you can calculate the mass of Oxygen,
28.35g (total) - 26.45g (crucible) = 1.9g product
1.9g product - 1.5g Sn = .4g O
Use this .4g O in order to find out the moles of O
.4g O/16.11g O/mol = .025
Use given 1.5g Sn in order to find out the moles of Sn
1.5g Sn/118.71g Sn/mol = .0126

Divide moles of Sn and moles of O
.025 O/.0126 Sn = 2:1 O:Sn

Sn+2O-> SnO2
a. SnO2
b. Tin (IV) Dioxide
Hope this helps!