\(\boxed{6.216 \times {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ atoms}}}\) of carbon is present in 1.50 g of butane.
Further Explanation:
Avogadro’s number indicates how many atoms or molecules a mole can have in it. In other words, it provides information about the number of units that are present in one mole of the substance. It is numerically equal to \({\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}\). These units can be atoms or molecules.
The formula to calculate the moles of \({{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}\) is as follows:
\({\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} = \dfrac{{{\text{Given mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}{{{\text{Molar mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}\) …… (1)
Substitute 1.50 g for the given mass and 58.1 g/mol for the molar mass of \({{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}\) in equation (1).
\(\begin{aligned}{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} &= \left( {{\text{1}}{\text{.50 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{58}}{\text{.1 g}}}}} \right)\\&= {\text{0}}{\text{.0258 mol}}\\\end{aligned}\)
Since one mole of \({{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}\) has \({\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}\) of \({{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}\). Therefore the formula to calculate the molecules of butane is as follows:
\({\text{Molecules of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} = \left( {{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}} \right)\left( {{\text{Avogadro's Number}}} \right)\) …… (2)
Substitute 0.0258 mol for the moles of \({{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}\) and \({\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}\) for Avogadro’s number in equation (2).
\(\begin{aligned}{\text{Molecules of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}{\mathbf{}}&=\left( {0.0258{\text{ mol}}} \right)\left( {\frac{{{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}{{{\text{1 mol}}}}} \right)\\&= 1.554 \times {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}} \\\end{aligned}\)
The chemical formula of butane is \({{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}\). This indicates one molecule of butane has four atoms of carbon. Therefore the number of carbon atoms can be calculated as follows:
\(\begin{aligned}{\text{Atoms of carbon}} &= \left( {1.554 \times {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}} \right)\left( {\frac{{{\text{4 C atoms}}}}{{{\text{1 molecule of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}}} \right)\\&= 6.216 \times {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ C atoms}} \\\end{aligned}\)
Learn more:
- Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603
- Calculate the moles of ions in the solution: https://brainly.com/question/5950133
Answer details:
Grade: Senior School
Chapter: Mole concept
Subject: Chemistry
Keywords: 1.50 g, 58.1 g/mol, butane, C4H10, Avogadro’s number, \(6.216*10^22\) C atoms, \(1.554*10^22\)molecules, moles, one mole, chemical formula, carbon atoms, molar mass of C4H10, given mass of C4H10.
