Question

Question: Consider the following gas samples: 4.0 g of hydrogen gas, 4.0 g of helium gas, 1.0 mole of fluorine gas, 44.0 g of carbon dioxide gas, and 146 g of sulfur hexafluoride gas. Arrange the gas samples in order of increasing number of total atoms present.

Short answer

The gas samples can be arranged in order of increasingnumber of total atoms present as:

$4.0\mathrm{gHe}<1.0{\mathrm{molF}}_2<44.0{\mathrm{gCO}}_2<4.0{\mathrm{gH}}_2<146{\mathrm{gSF}}_6$

Step-by-step solution

Definition of mole

A mole is the amount of substance that contains many elementary entitiesof atoms in 0.012 kilogram of carbon 12.

Number of atoms in 4.0 g of hydrogen

$$\begin{gathered} =4.0{\mathrm{gH}}_2\times \frac{1{\mathrm{molH}}_2}{2.016{\mathrm{gH}}_2}\times \frac{2\mathrm{molH}}{1{\mathrm{molH}}_2}\times \frac{6.022\times {10}^{23}\mathrm{atomsH}}{1\mathrm{molH}} \\ =2.4\times {10}^{24}\mathrm{atoms} \end{gathered}$$

Number of atoms in 4.0 g of helium

$$\begin{gathered} =4.0\mathrm{gHe}\times \frac{1\mathrm{molHe}}{4.003\mathrm{gHe}}\times \frac{6.022\times {10}^{23}\mathrm{atomsHe}}{1\mathrm{molHe}} \\ =6.0\times {10}^{23}\mathrm{atoms} \end{gathered}$$

Number of atoms in 44.0 g of carbon dioxide

$$\begin{gathered} =1.0{\mathrm{molF}}_2\times \frac{2\mathrm{molF}}{1.0{\mathrm{molF}}_2}\times \frac{6.022\times {10}^{23}\mathrm{atomsHe}}{1\mathrm{molF}} \\ =1.2\times {10}^{24}\mathrm{atoms} \end{gathered}$$

Then, in 44.0 g as,

$=1.81\times {10}^{24}\mathrm{atoms}$

Number of atoms in 146 g of sulphur hexafluoride

$$\begin{gathered} =146{\mathrm{gSF}}_6\times \frac{1{\mathrm{molSF}}_6}{146.1{\mathrm{gSF}}_6}\times \frac{7\mathrm{molatoms}}{1{\mathrm{molSF}}_6}\times \frac{6.022\times {10}^{23}\mathrm{atomsHe}}{1\mathrm{molatoms}} \\ =4.21\times {10}^{24}\mathrm{atoms} \end{gathered}$$