Empirical formula of the unknown compound?

2012-11-07 6:23 am
I know this is a lot but I would really appreciate the help because I am so lost with empirical formulas.
Thanks

A student heated a crucible containing titanium metal and allowed it to react with oxygen from the air:
mass of empty crucible and lid: 21.588g
mass of crucible, lid, &titanium metal: 26.153g
mass of crucible, lid, & contents after heating: 28.089g
mass of crucible, lid, & contents after re heating : 28.267g
mass of crucible, lid & contents after second re heating :28.433g
mass of crucible, lid & contents after third re heating: 28.434g

a. What was the mass of titanium metal used in this experiment? (show calculation)
b. What was the mass of oxygen that reacted with the metal? (show calculation)
c. How many moles of titanium reacted? (show calculation)
d. How many moles of oxygen reacted? (show calculation)
e. What was the empirical formula of the compound that formed? (show calculation)

Thanks again!

回答 (1)

2012-11-08 1:35 am
✔ 最佳答案
a.
26.153g - 21.588g = 4.565g Ti

b.
28.434g - 26.153g = 2.281g O

c.
(4.565g Ti) / (47.8671 g Ti/mol) = 0.09537 mol Ti

d.
(2.281g O) / (15.99943 g O/mol) = 0.1426 mol O

e.
0.1426 mol O / 0.09537 mol Ti = 1.495
So the mole ratio of O to Ti is 3:2 and the empirical formula is Ti2O3


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