What is the Center of Mass of an NH3 Molecule?

[evenmu]

Homework Statement

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 M_N/M_H = 13.9 and nitrogen-hydrogen length is L = 1.014.

Data:

d = 0.940
M_N/M_H = 13.9
L = 1.014

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

Homework Equations

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

 

[tiny-tim]

welcome to pf!

hi evenmu! welcome to pf!

Center of the triangle has a distance d = 0.940 from each hydrogen atom.
…
The ratio of the atomic masses of nitrogen and hydrogen is M_N/M_H = 13.9 and nitrogen-hydrogen length is L = 1.014.

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?

 

[evenmu]

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

 

[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

 

[evenmu]

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

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?

 

[tiny-tim]

yup!

(and you can get the length from pythagoras )

 

[evenmu]

One more thing :)

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

 

[tiny-tim]

sorry, not following you

 

[evenmu]

The Equation for Center of mass is:


X_CM = 1/M Ʃ m_i*x_i

Y_CM = 1/M Ʃ m_i*y_i

What is the mass for hydrogen and nitrogen ?

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

How do I interpret this ?

 

[tiny-tim]

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

so put M_N = 13.9 M_H,

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

 

[evenmu]

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

so put M_N = 13.9 M_H,

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

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:

X_CM = 1/M (3*(M_N/13.9)*(0.940m) + (13.9M_H)*(0m)

And the same for Y-axis.

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

 

[tiny-tim]

Here is what I have done until now:


L = 1.014m
d= 0.940m

(pytagoras)
a =√(L²)-(d²) = 0.38m

fine so far

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

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

X_CM = 1/M (3*(M_N/13.9)*(0.940m) + (13.9M_H)*(0m)

And the same for Y-axis.

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

from symmetry, aren't they obviously 0 ?

as to the y coordinate …

y_CM = 1/(M_N + 3M_H) [(M_N)*(a) + (3M_H)*(0m)]

 

[evenmu]

fine so far




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

from symmetry, aren't they obviously 0 ?

as to the y coordinate …

y_CM = 1/(M_N + 3M_H) [(M_N)*(a) + (3M_H)*(0m)]

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

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

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

 

[tiny-tim]

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

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? )

 

[evenmu]

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

No no, it should be a konstant Å(Ångstrøm) behind the values.

Å = 1*10^-10

Like this:

L = 1.014Å
d = 0.940Å

Wikipedia: http://en.wikipedia.org/wiki/%C3%85ngstr%C3%B8m

 

[evenmu]

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

 

[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)

 

[evenmu]

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)

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