Answer:
6.6253*10²¹molecules of ethane (C₂H₆) are present in 0.334 g of C₂H₆
Explanation:
Avogadro's Number is the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of that substance. Its value is 6.023 * 10²³ particles per mole. Avogadro's number applies to any substance.
In this case, being:
the molar mass of ethane C₂H₆ is:
C₂H₆: 2*12 g/mole + 6* 1 g/mole= 30 g/mole
Then you can apply the following rule of three: if 30 grams of C₂H₆ are present in 1 mole, 0.334 grams of C₂H₆ in how many moles are present?
[tex]moles of C_{2} H_{6} =\frac{0.334 grams*1 mole}{30 grams}[/tex]
moles of C₂H₆=0.011
Finally, taking into account the definition of Avogadro's number, you can apply the following rule of three: if there are 6.023 * 10²³ molecules of C₂H₆ in 1 mole, how many molecules are there in 0.011 moles?
[tex]molecules of C_{2} H_{6} =\frac{0.011 moles*6.023*10^{23}molecules }{1 mole}[/tex]
molecules of C₂H₆= 6.6253*10²¹
6.6253*10²¹molecules of ethane (C₂H₆) are present in 0.334 g of C₂H₆
