Calculating Net Dipole Moment for Water - What Am I Doing Wrong?

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SUMMARY

The discussion centers on the calculation of the net dipole moment for water, specifically addressing the formula provided on ChemWiki, which states that the net dipole moment equals 2(1.5) x cos(104.5/2) = 1.84 D. A participant initially calculated a net dipole moment of -1.20606 D due to confusion between degrees and radians in their calculator settings. After correcting the angle from radians to degrees, the participant obtained a corrected value of approximately 1.836 D, confirming the accuracy of the original formula.

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HelloCthulhu
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Greeting PF!

http://chemwiki.ucdavis.edu/Physical_Chemistry/Atomic_Theory/Dipole_Moments

On the website above, I found an equation for the dipole moment of water:

"...the net dipole moment =2(1.5)xcos(104.5/2)=1.84 D."

I understand how to calculate the bond dipole. But after calculating for the net dipole, I keep getting the same wrong answer: -1.20606

Please tell me what I'm doing wrong! Thank you so much for your assistance! :]
 
Chemistry news on Phys.org
Have you tried to switch your calculator to DEG?
 
I'm using an iPhone app. It has a tab for trig where I found cos(, so I don't think I have to switch modes. I've also used Google's scientific calculator and received the same answer. Was your answer 1.84?
 
But you know the difference between degrees and radians, do you?
 
Oh, wow. Totally screwed that up. Changed 104.5 radians and now I've got 1.836. Thanks Borek! :D
 

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