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To determine the number of methane ([tex]$CH_4$[/tex]) molecules in 125 grams of [tex]$CH_4$[/tex] given that the molar mass of [tex]$CH_4$[/tex] is 16.05 g/mol, follow these steps:
1. Calculate the number of moles of methane:
First, use the relationship between mass, molar mass, and number of moles.
[tex]\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar Mass}} \][/tex]
Given the mass of [tex]$CH_4$[/tex] is 125 grams and its molar mass is 16.05 g/mol:
[tex]\[ \text{Number of moles of } CH_4 = \frac{125 \text{ g}}{16.05 \text{ g/mol}} \approx 7.788 \text{ mol} \][/tex]
2. Calculate the number of molecules of methane:
Use Avogadro's number, which is [tex]$6.022 \times 10^{23}$[/tex] molecules per mole, to convert moles to molecules.
[tex]\[ \text{Number of molecules} = (\text{Number of moles}) \times (\text{Avogadro's number}) \][/tex]
[tex]\[ \text{Number of molecules of } CH_4 = 7.788 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 46.900311526479754 \times 10^{23} \text{ molecules} \][/tex]
To express this in the form [tex]$[?] \times 10^{[?]}$[/tex]:
[tex]\[ = 4.6900311526479754 \times 10^{24} \text{ molecules} \][/tex]
Thus, there are approximately [tex]\( 4.69 \times 10^{24} \)[/tex] molecules of methane in 125 grams of [tex]$CH_4$[/tex].
1. Calculate the number of moles of methane:
First, use the relationship between mass, molar mass, and number of moles.
[tex]\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar Mass}} \][/tex]
Given the mass of [tex]$CH_4$[/tex] is 125 grams and its molar mass is 16.05 g/mol:
[tex]\[ \text{Number of moles of } CH_4 = \frac{125 \text{ g}}{16.05 \text{ g/mol}} \approx 7.788 \text{ mol} \][/tex]
2. Calculate the number of molecules of methane:
Use Avogadro's number, which is [tex]$6.022 \times 10^{23}$[/tex] molecules per mole, to convert moles to molecules.
[tex]\[ \text{Number of molecules} = (\text{Number of moles}) \times (\text{Avogadro's number}) \][/tex]
[tex]\[ \text{Number of molecules of } CH_4 = 7.788 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 46.900311526479754 \times 10^{23} \text{ molecules} \][/tex]
To express this in the form [tex]$[?] \times 10^{[?]}$[/tex]:
[tex]\[ = 4.6900311526479754 \times 10^{24} \text{ molecules} \][/tex]
Thus, there are approximately [tex]\( 4.69 \times 10^{24} \)[/tex] molecules of methane in 125 grams of [tex]$CH_4$[/tex].
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