How many molecules of sucrose, ${{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}$ are there in ${\text{454grams}}$ of sucrose$\left( {{{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}} \right)$?
Answer
575.1k+ views
Hint:The mole concept is used to determine the amount of the substance which is obtained by taking the ratio of the mass of the substance to the molecular mass of the substance.
One mole of the substance contains the Avogadro’s numbers of the particles that are atoms, ions, or molecules of the substance.
Avogadro’s number is represented as \[{{\text{N}}_{\text{A}}}\]. The value of Avogadro’s number is \[{\text{6}}{.023 \times 1}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}\].
Complete solution:
Here, the molecular weight of the sucrose given is \[342.30\,{\text{g/mol}}\] and the weight of the sucrose is ${\text{454grams}}$.
We have to determine the first moles of the sucrose in ${\text{454grams}}$ as follows:
\[{\text{moles = }}\dfrac{{{\text{weight}}}}{{{\text{Molecular}}\,{\text{weight}}}}\]
Substitute, \[342.30\,{\text{g/mol}}\] for molecular weight and ${\text{454grams}}$ for weight.
\[{\text{moles = }}\dfrac{{454\,{\text{g}}}}{{342.30\,{\text{g/mol}}}}\]
\[{\text{moles = 1}}{\text{.3263}}\,{\text{mol}}\]
Thus, the moles of the sucrose in ${\text{454grams}}$are \[{\text{1}}{\text{.3263}}\,{\text{mol}}\].
Now, determine the molecules of the sucrose in ${\text{454grams}}$as follows:
As we know that one mole of the substance contains Avogadro’s numbers of molecules.
\[{\text{1mol}}\,{\text{ = }}\,{{\text{N}}_{\text{A}}}{\text{ = 6}}{.023 \times 1}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}\]
\[{\text{1}}{\text{.3263}}\,{\text{mol = }}\dfrac{{{\text{1}}{\text{.3263}}\,{mol \times 6}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}}}{{{\text{1mol}}\,}}\]
\[{\text{1}}{\text{.3263}}\,{\text{mol = 7}}{.988 \times 1}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}\]
Thus, the molecules of sucrose in ${\text{454grams}}$are \[{\text{7}}{.988 \times 1}{{\text{0}}^{{\text{23}}}}\].
Note:Molecular weight is the average mass of the molecules and it differs from molecule to molecule. In the case of the atom, it is called atomic mass while in the case of the molecular mass it is calculated by taking the sum of the atomic masses of all the atoms in the molecule.
The SI unit of the molecular weight is gram per mole. Its unit itself indicates the ratio of the mass of the substance to the molar molecular mass of the substance.
One mole of the substance contains the Avogadro’s numbers of the particles that are atoms, ions, or molecules of the substance.
Avogadro’s number is represented as \[{{\text{N}}_{\text{A}}}\]. The value of Avogadro’s number is \[{\text{6}}{.023 \times 1}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}\].
Complete solution:
Here, the molecular weight of the sucrose given is \[342.30\,{\text{g/mol}}\] and the weight of the sucrose is ${\text{454grams}}$.
We have to determine the first moles of the sucrose in ${\text{454grams}}$ as follows:
\[{\text{moles = }}\dfrac{{{\text{weight}}}}{{{\text{Molecular}}\,{\text{weight}}}}\]
Substitute, \[342.30\,{\text{g/mol}}\] for molecular weight and ${\text{454grams}}$ for weight.
\[{\text{moles = }}\dfrac{{454\,{\text{g}}}}{{342.30\,{\text{g/mol}}}}\]
\[{\text{moles = 1}}{\text{.3263}}\,{\text{mol}}\]
Thus, the moles of the sucrose in ${\text{454grams}}$are \[{\text{1}}{\text{.3263}}\,{\text{mol}}\].
Now, determine the molecules of the sucrose in ${\text{454grams}}$as follows:
As we know that one mole of the substance contains Avogadro’s numbers of molecules.
\[{\text{1mol}}\,{\text{ = }}\,{{\text{N}}_{\text{A}}}{\text{ = 6}}{.023 \times 1}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}\]
\[{\text{1}}{\text{.3263}}\,{\text{mol = }}\dfrac{{{\text{1}}{\text{.3263}}\,{mol \times 6}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}}}{{{\text{1mol}}\,}}\]
\[{\text{1}}{\text{.3263}}\,{\text{mol = 7}}{.988 \times 1}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}\]
Thus, the molecules of sucrose in ${\text{454grams}}$are \[{\text{7}}{.988 \times 1}{{\text{0}}^{{\text{23}}}}\].
Note:Molecular weight is the average mass of the molecules and it differs from molecule to molecule. In the case of the atom, it is called atomic mass while in the case of the molecular mass it is calculated by taking the sum of the atomic masses of all the atoms in the molecule.
The SI unit of the molecular weight is gram per mole. Its unit itself indicates the ratio of the mass of the substance to the molar molecular mass of the substance.
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