# Center of mass of molecules

1. Oct 23, 2012

### evenmu

1. The problem statement, all variables and given/known data

NH3 (ammonia) that is shown in the picture. These three hydrogen-atoms is formed as triangle.
Center of the triangle has a distance d = 0.940 from each hydrogen atom.
The nitrogen atom located in the vertex of a pyramid in which the three hydrogens defines "the base". Vertex is vertically above the center of the triangle. The ratio of the atomic masses of nitrogen and hydrogen is MN/MH = 13.9 and nitrogen-hydrogen length is L = 1.014.

Data:

d = 0.940
MN/MH = 13.9
L = 1.014

picture of the situation: https://www.dropbox.com/s/ruakers7j0zgg1w/Photo%2023.10.12%2020%2054%2019.jpg

2. Relevant equations

Determine the x-and y-coordinate of the molecule center of mass ?

Last edited: Oct 23, 2012
2. Oct 23, 2012

### tiny-tim

welcome to pf!

hi evenmu! welcome to pf!
the centre of mass is obviously on that "vertical" line marked y

what proportion of the way along the line is it?

and how long is that line?

3. Oct 24, 2012

### evenmu

So, you mean that I can treat this as a right-angled triangle and find the center of mass from that ?

4. Oct 24, 2012

### tiny-tim

no, i mean that you can treat this as a straight line,

and find the center of mass as if the three hydrogen atoms were all together at one end

5. Oct 24, 2012

### evenmu

Okay i se it now, but i still need to find the length of the y-axis,
and to find it, i need to treat it like a triangle. Pluss the mass of these hydrogen will alsow be
multiplied by 3 if they goes together.

Am I on the right path here?

6. Oct 24, 2012

### tiny-tim

yup!

(and you can get the length from pythagoras )

7. Oct 24, 2012

### evenmu

One more thing :)

The ratio of mass, is it the same mass for the Nitrogen and Hydrogen atoms ?
Bothe uses the same mass (MN/MH = 13.9) ?

8. Oct 24, 2012

### tiny-tim

sorry, not following you

9. Oct 24, 2012

### evenmu

The Equation for Center of mass is:

XCM = 1/M Ʃ mi*xi

YCM = 1/M Ʃ mi*yi

What is the mass for hydrogen and nitrogen ?

In the problem it says: The ratio of the atomic masses of nitrogen and hydrogen is MN/MH = 13.9 ?

How do I interpret this ?

10. Oct 24, 2012

### tiny-tim

to find the centre of mass, you only need the ratios of the masses,

so put MN = 13.9 MH,

you'll get a fraction with MH on top and bottom, so it'll cancel, leaving you just a number!

11. Oct 24, 2012

### evenmu

Here is what I have done until now:

L = 1.014m
d= 0.940m

(pytagoras)
a =√(L2)-(d2) = 0.38m

Then i set up the Ceter of mass for X- and Y-axis. Like this:

XCM = 1/M (3*(MN/13.9)*(0.940m) + (13.9MH)*(0m)

And the same for Y-axis.

Is this right, or am I doing something wrong here?

12. Oct 24, 2012

### tiny-tim

fine so far

(are you sure about the figures? that makes a very short fat molecule that looks nothing like the diagram)

you needn't bother with the x and z coordinates …

from symmetry, aren't they obviously 0 ?

as to the y coordinate …

yCM = 1/(MN + 3MH) [(MN)*(a) + (3MH)*(0m)]

13. Oct 24, 2012

### evenmu

Yes i know, but its the exact same number that is on my homework task.

I calculate YCM and got YCM = 0.31. Do you think that this answer is realistic, in relation to the drawing ?

14. Oct 24, 2012

### tiny-tim

let's see …

a = .38, so it's roughly 14/(14+3) times .38, = .31 …

yup, looks ok

(but that's not metres, is it? )

15. Oct 24, 2012

### evenmu

16. Oct 24, 2012

### evenmu

Can you explain way the X- and Z-axis is zero(0) ?

17. Oct 24, 2012

### tiny-tim

isn't it obvious from the diagram?

the y axis is an axis of symmetry, so the centre of mass must lie on it

(and the y axis is the line x = z = 0)

18. Oct 26, 2012

### evenmu

I forgot to say, thank you very much for your help!