
The empirical formula of a compound is $C{H_2}O$.its vapor density is 45. Its molecular formula is
A.${C_2}{H_4}{O_2}$
B.${C_3}{H_6}{O_3}$
C.${C_4}{H_6}{O_4}$
D.$C{H_2}O$
Answer
512.7k+ views
Hint: An empirical formula does not always indicate the number of atoms in a compound's molecule; it merely indicates the ratio between those numbers. Despite the fact that they share the same empirical formula, they are two distinct compounds with separate molecular formulae.
Complete answer:
Molecular formula - The chemical symbols for the constituent elements are followed by numeric subscripts denoting the number of atoms of each element contained in the molecule in a molecular formula. The empirical formula represents the compound's simplest whole-integer atom ratio.
Vapor density - The weight of a volume of pure vapor or gas compared to an equal amount of dry air at the same temperature and pressure is referred to as vapor density. It is unitless because it is calculated by dividing the molecular weight of the vapor by the average molecular weight of air.
Molar mass – A substance's molar mass is the mass in grammes of one mole of the material. We may calculate a substance's molar mass by adding the molar masses of its constituent atoms, as shown in this video. The computed molar mass can then be used to convert between mass and number of moles of the material.
\[\
Molecular{\text{ }}weight\; = 2 \times \;Vapour{\text{ }}density\; \\
= 2 \times 45 = 90 \\
Let{\text{ }}the{\text{ }}molecular{\text{ }}formula{\text{ }}be\;{C_x}{H_{2x}}{O_x}. \\
Therefore,\;{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 90 = 12x + 2x + 16x = 30x \\
we{\text{ }}get,\;{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} x = 3. \\
\ \]
But putting the value of x as 3.
We get the compound to be ${C_3}{H_6}{O_3}$ .
So, the final answer is option (B) i.e. ${C_3}{H_6}{O_3}$
Note:
The average mass of an element's atoms, measured in atomic mass units, is its atomic mass (amu, also known as Dalton’s, D). The atomic mass of an element is a weighted average of all its isotopes, where the mass of each isotope is multiplied by the abundance of that isotope.
Complete answer:
Molecular formula - The chemical symbols for the constituent elements are followed by numeric subscripts denoting the number of atoms of each element contained in the molecule in a molecular formula. The empirical formula represents the compound's simplest whole-integer atom ratio.
Vapor density - The weight of a volume of pure vapor or gas compared to an equal amount of dry air at the same temperature and pressure is referred to as vapor density. It is unitless because it is calculated by dividing the molecular weight of the vapor by the average molecular weight of air.
Molar mass – A substance's molar mass is the mass in grammes of one mole of the material. We may calculate a substance's molar mass by adding the molar masses of its constituent atoms, as shown in this video. The computed molar mass can then be used to convert between mass and number of moles of the material.
\[\
Molecular{\text{ }}weight\; = 2 \times \;Vapour{\text{ }}density\; \\
= 2 \times 45 = 90 \\
Let{\text{ }}the{\text{ }}molecular{\text{ }}formula{\text{ }}be\;{C_x}{H_{2x}}{O_x}. \\
Therefore,\;{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 90 = 12x + 2x + 16x = 30x \\
we{\text{ }}get,\;{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} x = 3. \\
\ \]
But putting the value of x as 3.
We get the compound to be ${C_3}{H_6}{O_3}$ .
So, the final answer is option (B) i.e. ${C_3}{H_6}{O_3}$
Note:
The average mass of an element's atoms, measured in atomic mass units, is its atomic mass (amu, also known as Dalton’s, D). The atomic mass of an element is a weighted average of all its isotopes, where the mass of each isotope is multiplied by the abundance of that isotope.
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