Question:
The mass % of carbon in \(C_{57}H_{110}O_6\) is:
The mass % of carbon in \(C_{57}H_{110}O_6\) is:
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Percentage composition is calculated using:
\[
\%\text{Element}
=
\frac{\text{Mass of element in one mole}}
{\text{Molar mass of compound}}
\times100
\]
Updated On: Jun 26, 2026
- \(57.95\)
- \(62.35\)
- \(73.45\)
- \(76.85\)
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The Correct Option is D
Solution and Explanation
Step 1: Calculate the molar mass of \(C_{57}H_{110}O_6\).
Using atomic masses: \[ C=12,\quad H=1,\quad O=16 \] Molar mass: \[ =(57\times12)+(110\times1)+(6\times16) \] \[ =684+110+96 \] \[ =890 \] \[ \text{Molar mass}=890\ \text{g mol}^{-1} \]
Step 2: Calculate the mass of carbon present.
Mass contributed by carbon: \[ 57\times12 \] \[ =684 \] Thus, \[ \text{Mass of carbon}=684\ \text{g} \]
Step 3: Calculate percentage of carbon.
\[ \%\text{Carbon} = \frac{684}{890}\times100 \] \[ =76.85\% \]
Step 4: Final conclusion.
Therefore, \[ \boxed{76.85\%} \] Hence, the correct option is \[ \boxed{(4)} \]
Using atomic masses: \[ C=12,\quad H=1,\quad O=16 \] Molar mass: \[ =(57\times12)+(110\times1)+(6\times16) \] \[ =684+110+96 \] \[ =890 \] \[ \text{Molar mass}=890\ \text{g mol}^{-1} \]
Step 2: Calculate the mass of carbon present.
Mass contributed by carbon: \[ 57\times12 \] \[ =684 \] Thus, \[ \text{Mass of carbon}=684\ \text{g} \]
Step 3: Calculate percentage of carbon.
\[ \%\text{Carbon} = \frac{684}{890}\times100 \] \[ =76.85\% \]
Step 4: Final conclusion.
Therefore, \[ \boxed{76.85\%} \] Hence, the correct option is \[ \boxed{(4)} \]
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