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

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

h let me think

2b2t

explanation:

2b2t