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# 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. Verified
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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.

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.
${{x}_{cm}}=\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\dfrac{\left( 12\times 0 \right)\left( 16\times 1.1 \right)}{12+16}=0.63\mathop A\limits^ \circ$
Thus, the centre of mass is $0.63\mathop A\limits^ \circ$ from the carbon atom.