
The empirical formula of a compound of molecular mass $120$ is $C{H_2}O$. The molecular formula of the compound is:
A. ${C_2}{H_2}{O_2}$
B. ${C_4}{H_8}{O_4}$
C. ${C_3}{H_6}{O_9}$
D. none of the above
Answer
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Hint:The number of atoms of each type in the empirical formula is directly proportional to the number of atoms of each type in the molecular formula, with the proportionality constant being a simple integer. Therefore, dividing the molecular mass of the compound with the empirical formula mass will give us the integer with which each element has to be multiplied in the empirical formula to get the molecular formula.
Formulas used: ${(E.F)_n} = (M.F)$
Where $E.F,M.F$ denote the empirical and molecular formulas respectively and $n$ is the integer.
Complete step by step answer:
As the empirical formula is the simplest integer ratio of the elements present in a compound, the actual number of each element in the compound will be an integral multiple of the number of atoms mentioned in the empirical formula. In other words,
${(E.F)_n} = (M.F)$
Where $E.F,M.F$ denote the empirical and molecular formulas respectively and $n$ is the integer.
The molecular masses will also follow this rule. Therefore, the actual molecular mass of a compound will be $n$ times the mass of the empirical formula.
Here, as the empirical formula is $C{H_2}O$, its mass:
Empirical formula mass $ = 12 + (1 \times 2) + 16 = 30g$
Dividing the molecular mass of the compound ($120$) with this will give us $n$. Therefore,
$n = \dfrac{{120}}{{30}} = 4$
Hence, we have to multiply each number of atoms of each element in the empirical formula with $4$ to get the number of atoms in the molecular formula. Hence:
$M.F = {(C{H_2}O)_4} = {C_4}{H_8}{O_4}$
Therefore, the molecular formula of the compound is ${C_4}{H_8}{O_4}$ and thus, we have to mark option B.
Note: In practical purposes, the empirical formula helps us to determine the molecular mass and molecular formula of unknown compounds, since the only measurable quantity is the mass composition. Note that two compounds can have the same empirical formula but different molecular formulas. A common example is glucose (${C_6}{H_{12}}{O_6}$) and formaldehyde ($HCHO$), which have the same empirical formula ($C{H_2}O$) .
Formulas used: ${(E.F)_n} = (M.F)$
Where $E.F,M.F$ denote the empirical and molecular formulas respectively and $n$ is the integer.
Complete step by step answer:
As the empirical formula is the simplest integer ratio of the elements present in a compound, the actual number of each element in the compound will be an integral multiple of the number of atoms mentioned in the empirical formula. In other words,
${(E.F)_n} = (M.F)$
Where $E.F,M.F$ denote the empirical and molecular formulas respectively and $n$ is the integer.
The molecular masses will also follow this rule. Therefore, the actual molecular mass of a compound will be $n$ times the mass of the empirical formula.
Here, as the empirical formula is $C{H_2}O$, its mass:
Empirical formula mass $ = 12 + (1 \times 2) + 16 = 30g$
Dividing the molecular mass of the compound ($120$) with this will give us $n$. Therefore,
$n = \dfrac{{120}}{{30}} = 4$
Hence, we have to multiply each number of atoms of each element in the empirical formula with $4$ to get the number of atoms in the molecular formula. Hence:
$M.F = {(C{H_2}O)_4} = {C_4}{H_8}{O_4}$
Therefore, the molecular formula of the compound is ${C_4}{H_8}{O_4}$ and thus, we have to mark option B.
Note: In practical purposes, the empirical formula helps us to determine the molecular mass and molecular formula of unknown compounds, since the only measurable quantity is the mass composition. Note that two compounds can have the same empirical formula but different molecular formulas. A common example is glucose (${C_6}{H_{12}}{O_6}$) and formaldehyde ($HCHO$), which have the same empirical formula ($C{H_2}O$) .
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