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 4.00 g of butane?

Answer 1

Matter is made of very tiny particles. One mole of a substance contains a given amount of particles. This number is constant for all particles- 6.022×10^23 , and is called Avogadro's constant. To get the number of particles in 4g of butane, we need to calculate the number of moles represented by 4 g, then multiply by the Avogadro's constant (Since each mole has particles equivalent to the Avogadro's constant) 
Therefore, if 1 mole = 58.1g
                how many moles will be equal to 4g?
                we cross multiply  (4x1) divide by 58.1
                equals 0.068847moles
multiply by Avogadro constant = 4.1466 x 10^molecules
But remember the question is not about the number of molecules in butane, its about the number of carbon atoms. 
Since there are four carbons in butane, and each has  4.1466 x 10^22 atoms
to get the number of atoms, we multiply by four
equals 1.6586 x 10^atoms

Answer 2

$\boxed{1.657 \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ atoms}}}$ of carbon is present in 4.00 g of butane.

Further Explanation:

The number of atoms or molecules that can be present in one mole of any substance is determined by a number, known as Avogadro’s number. The numerical value of Avogadro’s number is ${\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}$. Such units can either 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)

The given mass of butane is 4.00 g.

The molar mass of butane is $58.1 \text{g/mol}.$

Incorporate these values in equation (1).

 $\begin{aligned}{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}} &= \left( {{\text{4}}{\text{.00 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{58}}{\text{.1 g}}}}} \right)\\&= {\text{0}}{\text{.0688 mol}}\\\end{aligned}$

The number of molecules present in one mole of butane is ${\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$. So the expression to calculate the number of molecules of butane is as follows:

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

The number of moles of butane is 0.0688 mol.

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

Incorporate these values in equation (2).

$\begin{aligned}{\text{Molecules of butane}}{\mathbf{ }}&=\left( {0.0688{\text{ mol}}} \right)\left( {\frac{{{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ molecules}}}}{{{\text{1 mol}}}}} \right)\\&= 4.143 \times {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}\\\end{aligned}$  

A molecule of butane consists of four carbon atoms in it. So the number of carbon atoms can be calculated as follows:

 $\begin{aligned}{\text{Atoms of carbon}}&= \left( {4.143 \times {\text{1}}{{\text{0}}^{{\text{22}}}}{\text{ molecules}}} \right)\left( {\frac{{{\text{4 C atoms}}}}{{{\text{1 molecule of butane}}}}} \right)\\&= 1.657 \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{ C atoms}}\\\end{aligned}$

Learn more:

1. How many grams of potassium were in the fertilizer? brainly.com/question/5105904
2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Senior School

Chapter: Mole concept

Subject: Chemistry

Keywords: $4.00 \text{g}, 58.1 \text{g/mol},$ butane, C4H10, Avogadro’s number, $1.657*10^22$C atoms, moles, one mole, chemical formula, carbon atoms, molar mass of C4H10, given mass of C4H10.