A water molecule, consisting of two hydrogen atoms and one oxygen atom, resides in the plane of an x-y coordinate system. The coordinates of the oxygen atom and the two hydrogen atoms are, respectively, (0 m, 4.22 ✕ 10^−11 m),
(7.57 ✕ 10^−11 m, 1.01 ✕ 10^−10 m)
and
(−7.57 ✕ 10^−11 m, 1.01 ✕ 10^−10 m).
Where is the center of mass of the water molecule located in this coordinate system?
Xcm=
Ycm=
Mh = 1
Mo = 16
total mass = 16 + 2 = 18
x = 0 by symmetry
18 y = 16(4.22 ✕ 10^−11) + 2(1.01 ✕ 10^−10)
To find the center of mass (CoM) of the water molecule, we need to calculate the average position of all the atoms.
Step 1: Assign masses to each atom.
The mass of an oxygen atom (O) is approximately 16 atomic mass units (u), and the mass of a hydrogen atom (H) is approximately 1 u.
Step 2: Calculate the x-coordinate of the CoM.
To find the x-coordinate of the CoM, we use the formula:
Xcm = (m1 * x1 + m2 * x2 + m3 * x3) / (m1 + m2 + m3)
Where m1, m2, and m3 are the masses of the oxygen and two hydrogen atoms, and x1, x2, and x3 are their respective x-coordinates.
For the oxygen atom:
m1 = 16 u
x1 = 0 m
For the first hydrogen atom:
m2 = 1 u
x2 = 7.57 * 10^-11 m
For the second hydrogen atom:
m3 = 1 u
x3 = -7.57 * 10^-11 m
Plugging in these values:
Xcm = (16 u * 0 m + 1 u * 7.57 * 10^-11 m + 1 u * (-7.57 * 10^-11 m)) / (16 u + 1 u + 1 u)
Step 3: Calculate the y-coordinate of the CoM.
Similarly, to find the y-coordinate of the CoM, we use the formula:
Ycm = (m1 * y1 + m2 * y2 + m3 * y3) / (m1 + m2 + m3)
Where y1, y2, and y3 are the respective y-coordinates of the oxygen and two hydrogen atoms.
For the oxygen atom:
y1 = 4.22 * 10^-11 m
For the first hydrogen atom:
y2 = 1.01 * 10^-10 m
For the second hydrogen atom:
y3 = 1.01 * 10^-10 m
Plugging in these values:
Ycm = (16 u * 4.22 * 10^-11 m + 1 u * 1.01 * 10^-10 m + 1 u * 1.01 * 10^-10 m) / (16 u + 1 u + 1 u)
Now we can calculate Xcm and Ycm using these formulas.