Answer
Verified
95.7k+ views
Hint: Multiply the empirical formula with the ratio of the molecular formula weight to the Empirical formula weight to obtain the molecular formula.
Complete step by step answer:
The molecular formula gives the actual number of atoms of each element present in one molecule of a compound. The empirical formula gives the smallest whole number ratio of atoms of various elements present in a molecule of the given compound. In some cases, the empirical formula is same as the molecular formula. However in other compounds, empirical formula and the molecular formula are different. The relationship between the empirical formula and the molecular formula is as given below:
\[\text{Molecular formula = n }\times \text{ empirical formula }\]
Here, n is the ratio of the molecular formula weight to the empirical formula weight:
\[\text{n=}\dfrac{\text{Molecular formula weight}}{\text{Empirical formula weight}}\]
The atomic masses of carbon, hydrogen and oxygen are \[12\text{ }g/mol,\text{ }1\text{ }g/mol\text{ }and\text{ }16\text{ }g/mol\] respectively.
Empirical formula of a compound is \[\text{C}{{\text{H}}_{\text{2}}}\text{O}\].
Calculate the empirical formula mass.
\[12\text{ }+2\left( 1 \right)+16=30\text{ }g/mol\]
The molecular mass is \[90\text{ }g/mol\].
Divide molecular formula weight with empirical formula weight to calculate n.
\[\begin{align}
& \text{n=}\dfrac{90\text{ g/mol}}{30\text{ g/mol}} \\
& \text{n=3} \\
\end{align}\]
Multiply empirical formula with 3 to obtain the molecular formula.
\[\begin{align}
& \text{Molecular formula = n }\times \text{ empirical formula } \\
& \text{Molecular formula = 3 }\times \text{ C}{{\text{H}}_{2}}\text{O } \\
& \text{Molecular formula = }{{\text{C}}_{3}}{{\text{H}}_{6}}{{\text{O}}_{3}} \\
\end{align}\]
Hence, the molecular formula of the compound is \[{{\text{C}}_{3}}{{\text{H}}_{6}}{{\text{O}}_{3}}\]:
Hence, the option A) \[{{\text{C}}_{3}}{{\text{H}}_{6}}{{\text{O}}_{3}}\] is the correct option.
Note:
We can also calculate the mass of molecular given in the four options and match it with the options
For \[{{\text{C}}_{3}}{{\text{H}}_{\text{6}}}{{\text{O}}_{3}}\], \[3\left( 12 \right)+6\left( 1 \right)+3\left( 16 \right)=36+6+48=90\]
For \[{{\text{C}}_{2}}{{\text{H}}_{\text{4}}}{{\text{O}}_{2}}\], \[2\left( 12 \right)+4\left( 1 \right)+2\left( 16 \right)=24+4+32=60\]
For \[{{\text{C}}_{6}}{{\text{H}}_{\text{12}}}{{\text{O}}_{6}}\], \[6\left( 12 \right)+12\left( 1 \right)+6\left( 16 \right)=72+12+96=180\]
For \[\text{C}{{\text{H}}_{\text{2}}}\text{O}\] \[12+2\left( 1 \right)+16=30\]
The molecular mass of \[{{\text{C}}_{3}}{{\text{H}}_{\text{6}}}{{\text{O}}_{3}}\] matches with that of given compound hence it is correct.
Complete step by step answer:
The molecular formula gives the actual number of atoms of each element present in one molecule of a compound. The empirical formula gives the smallest whole number ratio of atoms of various elements present in a molecule of the given compound. In some cases, the empirical formula is same as the molecular formula. However in other compounds, empirical formula and the molecular formula are different. The relationship between the empirical formula and the molecular formula is as given below:
\[\text{Molecular formula = n }\times \text{ empirical formula }\]
Here, n is the ratio of the molecular formula weight to the empirical formula weight:
\[\text{n=}\dfrac{\text{Molecular formula weight}}{\text{Empirical formula weight}}\]
The atomic masses of carbon, hydrogen and oxygen are \[12\text{ }g/mol,\text{ }1\text{ }g/mol\text{ }and\text{ }16\text{ }g/mol\] respectively.
Empirical formula of a compound is \[\text{C}{{\text{H}}_{\text{2}}}\text{O}\].
Calculate the empirical formula mass.
\[12\text{ }+2\left( 1 \right)+16=30\text{ }g/mol\]
The molecular mass is \[90\text{ }g/mol\].
Divide molecular formula weight with empirical formula weight to calculate n.
\[\begin{align}
& \text{n=}\dfrac{90\text{ g/mol}}{30\text{ g/mol}} \\
& \text{n=3} \\
\end{align}\]
Multiply empirical formula with 3 to obtain the molecular formula.
\[\begin{align}
& \text{Molecular formula = n }\times \text{ empirical formula } \\
& \text{Molecular formula = 3 }\times \text{ C}{{\text{H}}_{2}}\text{O } \\
& \text{Molecular formula = }{{\text{C}}_{3}}{{\text{H}}_{6}}{{\text{O}}_{3}} \\
\end{align}\]
Hence, the molecular formula of the compound is \[{{\text{C}}_{3}}{{\text{H}}_{6}}{{\text{O}}_{3}}\]:
Hence, the option A) \[{{\text{C}}_{3}}{{\text{H}}_{6}}{{\text{O}}_{3}}\] is the correct option.
Note:
We can also calculate the mass of molecular given in the four options and match it with the options
For \[{{\text{C}}_{3}}{{\text{H}}_{\text{6}}}{{\text{O}}_{3}}\], \[3\left( 12 \right)+6\left( 1 \right)+3\left( 16 \right)=36+6+48=90\]
For \[{{\text{C}}_{2}}{{\text{H}}_{\text{4}}}{{\text{O}}_{2}}\], \[2\left( 12 \right)+4\left( 1 \right)+2\left( 16 \right)=24+4+32=60\]
For \[{{\text{C}}_{6}}{{\text{H}}_{\text{12}}}{{\text{O}}_{6}}\], \[6\left( 12 \right)+12\left( 1 \right)+6\left( 16 \right)=72+12+96=180\]
For \[\text{C}{{\text{H}}_{\text{2}}}\text{O}\] \[12+2\left( 1 \right)+16=30\]
The molecular mass of \[{{\text{C}}_{3}}{{\text{H}}_{\text{6}}}{{\text{O}}_{3}}\] matches with that of given compound hence it is correct.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
A flask contains a mixture of compound A and B Both class 12 chemistry JEE_Main
When three capacitors of equal capacities are connected class 12 physics JEE_Main
If the number of integral terms in the expansion of class 11 maths JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main