Answer
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Hint: To answer this question, we should know about determining empirical formulas. We should know that empirical formula is the simplest whole-number ratio of atoms in a compound.
Step by step solution:
We should know that the empirical formula does not necessarily represent the actual numbers of atoms present in a molecule of a compound; it represents only the ratio between those numbers. It is important to note that the molecular formula of a compound may be exactly like an empirical formula, or it may be a multiple of the empirical formula.
For example, the molecular formula of butene, \[{{C}_{4}}{{H}_{8}}\], shows that butene contains four atoms of carbon and eight atoms of hydrogen. Its empirical formula is \[C{{H}_{2}}\]. One molecule of ethylene (molecular formula \[{{C}_{2}}{{H}_{4}}\]) contains two atoms of carbon and four atoms of hydrogen. Its empirical formula is \[C{{H}_{2}}\]. Both have the same empirical formula, yet they are different compounds with different molecular formulas. Butene is \[{{C}_{4}}{{H}_{8}}\], or four times the empirical formula; ethylene is \[{{C}_{2}}{{H}_{4}}\], or twice the empirical formula.
Now let’s go back to our question, we have to find an empirical formula of ethanoic acid. We know that the molecular formula of ethanoic acid is \[{{C}_{2}}{{H}_{4}}{{O}_{2}}\].
So, now we should take common factors out of the molecular formula. The simplest formula that remains will be the empirical formula.
So, let’s take our formula of ethanoic acid.
\[\begin{align}
& {{C}_{2}}{{H}_{4}}{{O}_{2}} \\
& =2(C{{H}_{2}}O) \\
\end{align}\]
We will find out that the common factor that will come out is 2. And the simplest formula that will remain is \[C{{H}_{2}}O\]. It is our empirical formula. So, from the above option we will find out that option B is correct.
Note: We should know that we use chemical formulas to present information about the chemical proportions of atoms that form a particular chemical compound or molecule, using chemical element symbols, numbers, and sometimes also other symbols.
Step by step solution:
We should know that the empirical formula does not necessarily represent the actual numbers of atoms present in a molecule of a compound; it represents only the ratio between those numbers. It is important to note that the molecular formula of a compound may be exactly like an empirical formula, or it may be a multiple of the empirical formula.
For example, the molecular formula of butene, \[{{C}_{4}}{{H}_{8}}\], shows that butene contains four atoms of carbon and eight atoms of hydrogen. Its empirical formula is \[C{{H}_{2}}\]. One molecule of ethylene (molecular formula \[{{C}_{2}}{{H}_{4}}\]) contains two atoms of carbon and four atoms of hydrogen. Its empirical formula is \[C{{H}_{2}}\]. Both have the same empirical formula, yet they are different compounds with different molecular formulas. Butene is \[{{C}_{4}}{{H}_{8}}\], or four times the empirical formula; ethylene is \[{{C}_{2}}{{H}_{4}}\], or twice the empirical formula.
Now let’s go back to our question, we have to find an empirical formula of ethanoic acid. We know that the molecular formula of ethanoic acid is \[{{C}_{2}}{{H}_{4}}{{O}_{2}}\].
So, now we should take common factors out of the molecular formula. The simplest formula that remains will be the empirical formula.
So, let’s take our formula of ethanoic acid.
\[\begin{align}
& {{C}_{2}}{{H}_{4}}{{O}_{2}} \\
& =2(C{{H}_{2}}O) \\
\end{align}\]
We will find out that the common factor that will come out is 2. And the simplest formula that will remain is \[C{{H}_{2}}O\]. It is our empirical formula. So, from the above option we will find out that option B is correct.
Note: We should know that we use chemical formulas to present information about the chemical proportions of atoms that form a particular chemical compound or molecule, using chemical element symbols, numbers, and sometimes also other symbols.
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