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

Question

Sveta_85

4 months ago

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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 3.50 g of butane?

Chemistry

2 answers:

castortr0y [3K] · Answer 1 · 2026-02-17T18:41:30+03:00

$\boxed{{\text{1}}{\text{.45}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ atoms}}}$ of carbon is present in 3.50 g of butane.

Explanation in detail:

The mole is a unit for quantifying the amount of a substance. It is defined as the mass of material that contains the same number of elementary entities as there are atoms in 12 g of carbon-12. Those elementary entities may be atoms, molecules, or formula units.

Avogadro’s number gives the count of entities contained in one mole of any substance. Its value is written as ${\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}$ per mole and can refer to electrons, atoms, molecules, or ions.

Use this relation to determine the moles of butane:

${\text{Moles of butane}}=\dfrac{{{\text{Given mass of butane}}}}{{{\text{Molar mass of butane}}}}$ ......(1)

The sample mass of butane provided is 3.50 g.

The molar mass of butane equals 58.12 g/mol.

Plug these numbers into equation (1).

$\begin{aligned}{\text{Moles of butane}}&=\left({{\text{3}}{\text{.50 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{58}}{\text{.12 g}}}}}\right)\\&={\text{0}}{\text{.060220234 mol}}\\&\approx {\text{0}}{\text{.0602 mol}}\\\end{aligned}$

Now find the number of butane molecules using:

${\text{Molecules of butane}}=\left({{\text{Moles of butane}}}\right)\left( {{\text{Avogadro's Number}}}\right)$ ......(2)

The amount in moles of butane is 0.060220234 mol.

The value for Avogadro’s number is ${\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ .

Insert those values into equation (2).

$\begin{aligned}{\text{Molecules of butane}}{\mathbf&{ = }}\left({0.060220234{\text{ mol}}} \right)\left( {\frac{{{\text{6}}{\text{.022}} \times{\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}{{{\text{1 mol}}}}}\right)\\&={\text{3}}{\text{.62646}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}\\&\approx {\text{3}}{\text{.626}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{molecules}}\\\end{aligned}$

Each butane molecule contains four carbon atoms. Therefore, the count of carbon atoms in ${\text{3}}{\text{.62646}}\times {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}$ of butane is computed by:

${\text{Atoms of carbon}}=\left( {{\text{Molecules of butane}}}\right)\left( {\dfrac{{{\text{4 carbon atoms}}}}{{{\text{1 butane molecule}}}}}\right)$

.......(3)

Replace the molecule quantity with ${\text{3}}{\text{.62646}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}$ in equation (3).

$\begin{aligned}{\text{Atoms of carbon}}&=\left( {{\text{3}}{\text{.62646}}\times {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}}\right)\left({\frac{{{\text{4 carbonatoms}}}}{{{\text{1 molecule}}}}} \right)\\&= 1.45 \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ atoms}}\\\end{aligned}$

Hence, the total number of carbon atoms is ${\mathbf{1}}{\mathbf{.45 \times 1}}{{\mathbf{0}}^{{\mathbf{23}}}}{\mathbf{ atoms}}$ .

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

Subject: Chemistry

Chapter: Mole concept

Keywords: mole, atoms of carbon, molecules of butane, carbon-12, 12 g, Avogadro’s number, butane, moles of butane, electrons, molecules, atoms.

KiRa [2.9K] · Answer 2 · 2026-02-17T18:38:23+03:00

First convert grams of C4H10 to moles using its molar mass of 58.1 g/mol: 3.50 g C4H10 × (1 mol C4H10 / 58.1 g C4H10) = 0.06024 mol C4H10 Next convert moles to molecules using Avogadro’s number: 0.06024 mol C4H10 × (6.022×10^23 molecules C4H10 / 1 mol C4H10) = 3.627×10^22 molecules C4H10 Each butane molecule contains 4 carbon atoms, so: 3.627×10^22 molecules C4H10 × (4 atoms C / 1 molecule C4H10) = 1.45×10^23 carbon atoms present.