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**Hint:**Since the ratio of atoms are given therefore the simplest form can be obtained of the compound i.e. the empirical formula. Then after determining the number of empirical formulas through molecular mass we can find the actual molecular formula of the compound.

**Complete step by step solution:**

The formula weight is the sum of the atomic weights of all atoms involved in a given chemical formula. It is generally applied to a substance that does not consist of individual molecules, such as the ionic compound like sodium chloride. Such a substance is generally represented by a chemical formula that describes the simplest ratio of the number of atoms of the constituent elements, i.e., an empirical formula.

-The molecular mass (molecular weight) of a molecule is its average mass as calculated by adding all together with the atomic weights of the atoms in the molecular formula.

-So, since the definitions differ, let us understand the difference between them.

-The molecular formula shows the type and number of atoms in a molecule. The molecular formula of glucose is\[{{C}_{6}}{{H}_{12}}{{O}_{6}}\], which indicates that one molecule of glucose contains 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen.

-The empirical formula is also known as the simplest formula of a compound. It is used to determine the mole ratio of elements present in a compound. The empirical formula of glucose would be\[~C{{H}_{2}}O\]

-The formula mass and molecular mass of water (\[{{H}_{2}}O\] ) are the same, while the formula and molecular mass of glucose is completely different. The formula mass (also known as formula weight) of glucose is 30 (say in gm per mole or no unit), while the molecular mass (also known as molecular weight) is 180.156 g/mol. So whenever there is a molecular formula then we have to divide the subscripts by a whole number (usually 2 or 3), you know to expect the formula mass will be different.

For this question, since the simple whole-number ratio of atoms is given thus empirical formula can be calculated easily from this.

Therefore empirical formula=${{C}_{3}}{{H}_{3}}O$

again number of moles, of the compound is given by n=$\dfrac{W}{M}$

Where, W= weight of the compound

M= Molecular weight of the compound

thus, \[M=\dfrac{W}{n}=\dfrac{1}{6.06\times {{10}^{-3}}}=165.0\]

Thus, Empirical mass=\[=12\times 3+3\times 1+16\times 1=55\]

Therefore, No of molecular mass present in Molecular weight is \[\dfrac{165}{55}=3\]

Therefore, molecular formula=\[{{\left( {{C}_{3}}{{H}_{3}}O \right)}_{3}}={{C}_{9}}{{H}_{9}}{{O}_{3}}\]

**Therefore option (C) is the correct option.**

**Note:**Only in the case the simple whole-number ratio of atoms in a molecule the empirical mass can be determined. Otherwise, the ratio has to be converted into a whole-number ratio first and then empirical weight can be determined.

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