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 3.50 g of butane?

Answer 1

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

Further Explanation:

Mole is a measure of the amount of substance. It is defined as the mass of a substance that has the same number of fundamental units as there are atoms in 12 g of carbon-12. Such fundamental units can be atoms, molecules or formula units.

Avogadro’s number is the number of units that are present in one mole of the substance. Its value is equal to ${\text{6}}{\text{.022}}\times{\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}$  per mole of substance. These units can be electrons, atoms, molecules or ions.

The formula to calculate the moles of butane is as follows:

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

The given mass of butane is 3.50 g.

The molar mass of butane is 58.12 g/mol.

Substitute these values in equation (1).

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

The molecules of butane are calculated as follows:

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

The moles of butane is 0.060220234 mol.

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

Substitute these values in equation (2).

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

One molecule of butane has four carbon atoms. So the number of carbon atoms in ${\text{3}}{\text{.62646}}\times {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}$ of butane is calculated as follows:

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

                                                                           .......(3)

Substitute ${\text{3}}{\text{.62646}}\times{\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}$ for the molecules of butane in equation (3).

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

Therefore the number of atoms of carbon 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.

Answer 2

First convert grams to moles using molar mass of butane that is 58.1 g

3.50g C4H10 x (1 mol C4H10)/(58.1g C4H10) = 0.06024 mol C4H10 

Now convert moles to molecules by using Avogadro’s number

0.06024 mol C4H10 x (6.022x10^23 molecules C4H10)/(1 mol C4H10) = 3.627x10^22 molecules C4H10 

And there are 4 carbon atoms in 1 molecule of butane, so use the following ratio: 

3.627 x 10^22 molecules C4H10 x (4 atoms C)/(1 molecule C4H10) 
= 1.45 x 10^23 atoms of carbon are present