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.

Work through the following questions. You only will be asked to submit your answers to some of them, but you may need to calculate all of the answers a) - f) to determine the final answer g).

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 (Sn_xO_y) of the original oxide of tin?

Round to the nearest 0.001 g please

Answer 1

a) mass percent of tin in SnO2 = (2.842/150.71)*(118.71)*(1/2.842)*100

   = 78.77%

b) mass of tin in the final sample = 2.842*78.77/100 = 2.24 grams

c) mass of tin in the original sample = 2.24 grams

d) mass of oxygen in the original sample = 2.54-2.24 = 0.3 grams

e) number of moles of tin in the original sample

   = 2.24/118.71 = 0.02 mol

f) number of moles of oxygen in the original sample

= 0.3/16 = 0.01875 mol

g) formula

sn = 0.02/0.01875 = 1

o = 0.01875/0.01875 = 1

formula = SnO