In ammonia (NH3) molecule, the three hydrogen atoms form an equilateral triangle, the distance between the centers of the atoms being 16.28x10^-11 m, so that the center of the triangle is 9.40x10^-11 m from each hydrogen atom. The nitrogen atom is at the apex of the pyramis, the three hydrogens consistuting the base. The nitrogen/hydrogen distance is 10.14x10^-11 m and the nitrogen/hydrogen atomic mass is 13.9. Locate the center of mass relative to the nitrogen atom

How can the nitrogen/hydrogen mass be 13.9amu? That is less than nitrogen.

the atomic mass ratio is 13.9 as nitrogen is a 14.008 and hydrogen is 1.008 approximately

To locate the center of mass relative to the nitrogen atom in the ammonia (NH3) molecule, we need to consider the positions of the hydrogen atoms and their respective masses.

The center of mass can be calculated using the formula:

Center of Mass = (m1 * r1 + m2 * r2 + m3 * r3) / (m1 + m2 + m3)

Where:
m1, m2, m3 are the masses of the three hydrogen atoms
r1, r2, r3 are the respective positions (vectors) of the hydrogen atoms relative to the nitrogen atom

First, let's assign some variables for simplicity:

r = 16.28x10^-11 m (distance between the centers of the hydrogen atoms)
d = 9.40x10^-11 m (distance between the center of the triangle and each hydrogen atom)
da = 10.14x10^-11 m (distance between the nitrogen atom and each hydrogen atom)
m = 13.9 (mass of the nitrogen atom)

Now, let's calculate the positions of the hydrogen atoms relative to the nitrogen atom:

r1 = (-r/2, 0, d) // This means the first hydrogen atom is to the left, in front, and above the nitrogen atom.
r2 = (r/2, 0, d) // This means the second hydrogen atom is to the right, in front, and above the nitrogen atom.
r3 = (0, r*cos(60°), -d) // This means the third hydrogen atom is directly below the nitrogen atom.

Next, plug these values into the formula to calculate the center of mass:

Center of Mass = (m1 * r1 + m2 * r2 + m3 * r3) / (m1 + m2 + m3)

Substituting the given values:

Center of Mass = [(m * r1) + (m * r2) + (m * r3)] / (m + m + m)
= [(13.9 * (-r/2, 0, d)) + (13.9 * (r/2, 0, d)) + (13.9 * (0, r*cos(60°), -d))] / (13.9 + 13.9 + 13.9)

Simplifying further, taking into account the values of r (16.28x10^-11 m), d (9.40x10^-11 m), and cos(60°) (0.5):

Center of Mass = [(-6.95x10^-11 m, 0, 13.9x10^-11 m) + (6.95x10^-11 m, 0, 13.9x10^-11 m) + (0, 14.14x10^-11 m, -9.4x10^-11 m)] / 41.7

= [0, 14.14x10^-11 m, 17.6x10^-11 m] / 41.7

Finally, divide each component by 41.7 to find the normalized center of mass relative to the nitrogen atom:

Center of Mass = [0/41.7, (14.14x10^-11 m)/41.7, (17.6x10^-11 m)/41.7]

= [0, 0.339x10^-11 m, 0.423x10^-11 m]

Therefore, the center of mass relative to the nitrogen atom in the ammonia molecule is located at coordinates (0, 0.339x10^-11 m, 0.423x10^-11 m).