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Hint:Combustion is a process of rapid heating of a chemical substance with oxygen, involving the production of heat and light.
Empirical Formula - A formula that gives the simplest whole-number ratio of atoms in a given compound is called Empirical Formula.
Complete step by step solution:
In the question it is given that 0.72 g of water and 3.08 gm of carbon dioxide ejected by the combustion of hydrocarbons.
We have to calculate the empirical formula of the hydrocarbon from the given data.
The reaction of hydrocarbons with the oxygen is as follows.
\[{{C}_{x}}{{H}_{y}}+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}+{{H}_{2}}O\]
Hydrocarbons react with oxygen under combustion and form carbon dioxide and water as products.
\[{{C}_{x}}{{H}_{y}}+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}+{{H}_{2}}O\]
3.08 gm 0.72 g
Number moles of carbon dioxide
\[\begin{align}
& {{n}_{C{{O}_{2}}}}=\frac{\text{given weight of carbon dioxide }}{\text{molecular weight of carbon dioxide}} \\
& \text{ }=\frac{3.08}{44} \\
& \text{ = 0}\text{.07} \\
\end{align}\]
Number moles of water
\[\begin{align}
& {{n}_{{{H}_{2}}O}}=\frac{\text{given weight of water }}{\text{molecular weight of water}} \\
& \text{ }=\frac{0.72}{18} \\
& \text{ = 0}\text{.04} \\
\end{align}\]
From the above calculations number moles of carbon \[{{n}_{C}}\] = 0.07moles
Number of moles of hydrogen \[{{n}_{H}}\] = 2×0.04 = 0.08 moles
The ratio of number moles of carbon and number moles of hydrogen gives the empirical formula of the hydrocarbon.
\[\begin{align}
& \text{empirical formula=}\frac{{{n}_{C}}}{{{n}_{H}}} \\
& \text{ = }\frac{0.07}{0.08} \\
& \text{ = }\frac{7}{8} \\
\end{align}\]
From the above calculation we can say that the number of carbons in the hydrocarbon is 7 and the number of hydrogen atoms are 8.
Therefore, the empirical formula of hydrocarbon is \[{{C}_{7}}{{H}_{8}}\].
So, the correction option is B.
Note: Don’t confuse empirical formula with molecular formula.
“Empirical formulas show the simplest whole-number ratio of atoms in a compound and molecular formula shows the number of each type of atom in a molecule”.
Empirical Formula - A formula that gives the simplest whole-number ratio of atoms in a given compound is called Empirical Formula.
Complete step by step solution:
In the question it is given that 0.72 g of water and 3.08 gm of carbon dioxide ejected by the combustion of hydrocarbons.
We have to calculate the empirical formula of the hydrocarbon from the given data.
The reaction of hydrocarbons with the oxygen is as follows.
\[{{C}_{x}}{{H}_{y}}+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}+{{H}_{2}}O\]
Hydrocarbons react with oxygen under combustion and form carbon dioxide and water as products.
\[{{C}_{x}}{{H}_{y}}+{{O}_{2}}\xrightarrow{{}}C{{O}_{2}}+{{H}_{2}}O\]
3.08 gm 0.72 g
Number moles of carbon dioxide
\[\begin{align}
& {{n}_{C{{O}_{2}}}}=\frac{\text{given weight of carbon dioxide }}{\text{molecular weight of carbon dioxide}} \\
& \text{ }=\frac{3.08}{44} \\
& \text{ = 0}\text{.07} \\
\end{align}\]
Number moles of water
\[\begin{align}
& {{n}_{{{H}_{2}}O}}=\frac{\text{given weight of water }}{\text{molecular weight of water}} \\
& \text{ }=\frac{0.72}{18} \\
& \text{ = 0}\text{.04} \\
\end{align}\]
From the above calculations number moles of carbon \[{{n}_{C}}\] = 0.07moles
Number of moles of hydrogen \[{{n}_{H}}\] = 2×0.04 = 0.08 moles
The ratio of number moles of carbon and number moles of hydrogen gives the empirical formula of the hydrocarbon.
\[\begin{align}
& \text{empirical formula=}\frac{{{n}_{C}}}{{{n}_{H}}} \\
& \text{ = }\frac{0.07}{0.08} \\
& \text{ = }\frac{7}{8} \\
\end{align}\]
From the above calculation we can say that the number of carbons in the hydrocarbon is 7 and the number of hydrogen atoms are 8.
Therefore, the empirical formula of hydrocarbon is \[{{C}_{7}}{{H}_{8}}\].
So, the correction option is B.
Note: Don’t confuse empirical formula with molecular formula.
“Empirical formulas show the simplest whole-number ratio of atoms in a compound and molecular formula shows the number of each type of atom in a molecule”.
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