Question

The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How many carbon atoms are in 1.50 g of butane?

Answer 1

Answer:

              6.21 × 10²² Carbon Atoms

Solution:

Data Given:

                 Mass of Butane (C₄H₁₀)  =  1.50 g

                 M.Mass of Butane  =  58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  1.50 g ÷ 58.1 g.mol⁻¹

                 Moles  =  0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

            Moles  =  Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

             Number of C₄H₁₀ Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

     Number of C₄H₁₀ Molecules  =  0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

                 Number of C₄H₁₀ Molecules  =  1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

                            1 Molecule of C₄H₁₀ contains  =  4 Atoms of Carbon

So,

          1.55 × 10²² C₄H₁₀ Molecules will contain  =  X Atoms of Carbon

Solving for X,

 X =  (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X  =  6.21 × 10²² Atoms of Carbon

Answer 2

$\boxed{6.216 \times {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ atoms}}}$ 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:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603
2. Calculate the moles of ions in the solution: 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.