Courses for Kids
Free study material
Offline Centres
Store Icon

The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is \[1.1\mathop A\limits^ \circ\] Given mass of carbon atom is \[12\] a.m.u. and mass of oxygen atom is \[16\] a.m.u., calculate the position of the centre of mass of the carbon monoxide molecule.
A.\[6.3 \mathop A\limits^ \circ\] from the carbon atom.
B.\[1 \mathop A\limits^ \circ\] from the carbon atom.
C.\[0.63 \mathop A\limits^ \circ\] from the carbon atom.
D.\[0.12 \mathop A\limits^ \circ\] from the carbon atom.

Last updated date: 16th Jul 2024
Total views: 346.8k
Views today: 6.46k
346.8k+ views
Hint: We know that the atomic mass of an element is the average mass of the atoms of an element measured in atomic mass units (amu, also known as Daltons, D). The atomic mass is a weighted average of all of the isotopes of that element, in which the mass of each isotope is multiplied by the abundance of that particular isotope.

Complete answer:
An atomic unit of mass is defined as accurately \[1/12\] the mass of a carbon\[-12\] atom. The carbon\[-12\] atom has six neutrons and six protons in its nucleus. The atomic unit mass is symbolized as AMU or amu. The mass of one mole of atoms of a pure element in grams is equivalent to the atomic mass of that element in atomic mass units (amu) or in grams per mole (g/mol). Although mass can be expressed as both amu and g/mol, g/mol is the most useful system of units for laboratory chemistry. Using this theory, we can solve this question.
The center of mass of a system is a hypothetical point, where the entire mass of the body is supposed to be concentrated. The velocity of the system's center of mass does not change, as long as the system is closed. The system moves as if all the mass is concentrated at a single point. The final l location will be at the weighted distance between the masses. In most mechanics’ problems the gravitational field is assumed to be uniform.
The distance of centre of mass from carbon atom is:
${{x}_{cm}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\dfrac{\left( 12\times 0 \right)\left( 16\times 1.1 \right)}{12+16}=0.63\mathop A\limits^ \circ$
Thus, the centre of mass is \[0.63\mathop A\limits^ \circ\] from the carbon atom.

Therefore, the correct answer is option C.

Remember that we can conclude that the centre of mass of an object is the point at which the object can be balanced. The centre of mass is useful because problems can often be simplified by treating a collection of masses as one mass at their common centre of mass. The weight of the object then acts through this point.