Answer
Verified
39.3k+ 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
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
The length of a potentiometer wire is 10m The distance class 12 physics JEE_MAIN
Other Pages
A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main
A convex lens is dipped in a liquid whose refractive class 12 physics JEE_Main
Iodoform can be prepared from all except A Acetaldehyde class 12 chemistry JEE_Main
The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main
448 litres of methane at NTP corresponds to A 12times class 11 chemistry JEE_Main
Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main