I just worked the problem and got 79% for W and 21% for O.
For the empirical, I got WO4.
Is this correct?
A student made a determination of the empirical formula ot tungsten oxide (WO) produced by the total ignition of air of a smaple of metallic Tungsten. From the lab:
Weight of crucible 11.120g
weight of tungsten 8.820
weight of crucible and product (WO) 22.836
a. Calculate the percent comp by mass of tengsten oxide
b. Determine the empirical formula of tungsten oxide.
Really having trouble with this. Product weighs 11.719 is all I can get started.
2 answers
If you used 79% and 21% you should have obtained WO3, I think, however, I have
(8.82/11.716)*100 = 75.3%W and
(2.899/11.716)*100 = 24.7% O.
Using these numbers I get a ratio of 1W to 3.77O Oxygen and that makes no sense for the whole number ratio would be W4O15.
There is a W2O3, a WO2, and a WO3.
(8.82/11.716)*100 = 75.3%W and
(2.899/11.716)*100 = 24.7% O.
Using these numbers I get a ratio of 1W to 3.77O Oxygen and that makes no sense for the whole number ratio would be W4O15.
There is a W2O3, a WO2, and a WO3.
1 answer