Answer:
1.07 × 10²⁴ carbon atoms.
Explanation:
We are given that a beaker contains 44.7 mL of butane and we want to determine how many carbon atoms the liquid contains.
To convert from mL to atoms of C, we can first convert mL to grams, grams to moles, moles to molecules, and molecules to C atoms.
Using its density, determine the amount of grams of butane is in the beaker:
[tex]\displaystyle 44.7 \text{ mL C$_4$H$_{10}$} \cdot \frac{0.579 \text{ g C$_4$H$_{10}$}}{1 \text{ mL C$_4$H$_{10}$}} = 25.9 \text{ g C$_4$H$_{10}$}[/tex]
Find the molar mass of butane:
[tex]\displaystyle \begin{aligned} \text{MM}_\text{C$_4$H$_{10}$} &= (4(12.01) + 10(1.01)) \text{ g/mol} \\ \\ &= 58.14 \text{ g/mol}\end{aligned}[/tex]
Convert from grams to moles:
[tex]\displaystyle 25.9 \text{ g C$_4$H$_{10}$} \cdot \frac{1 \text{ mol C$_4$H$_{10}$}}{58.14 \text{ g C$_4$H$_{10}$}} = 0.445 \text{ mol C$_4$H$_{10}$}[/tex]
Convert from moles to molecules:
[tex]\displaystyle \begin{aligned} 0.445 \text{ mol C$_4$H$_{10}$} & \cdot \frac{ 6.02\times 10^{23} \text{ molecules C$_4$H$_{10}$}}{1 \text{ mol C$_4$H$_{10}$}} \\ \\ &= 2.68 \times 10^{23} \text{ molecules C$_4$H$_{10}$}\end{aligned}[/tex]
And because there are (exactly) four carbon atoms per butane molecule:
[tex]\displaystyle \begin{aligned}2.68 \times 10^{23} \text{ molecules C$_4$H$_{10}$} & \cdot \frac{4 \text{ C}}{1 \text{ molecule C$_4$H$_{10}$}} \\ \\ &= 10.7 \times 10^{23}\text{ C} \\ \\ &= 1.07 \times 10^{24} \text{ C}\end{aligned}[/tex]
In conclusion, there are 1.07 × 10²⁴ carbon atoms in a beaker containing 44.7 mL of butane.
