Question

How many molecules of ethane (C2H6) are present in 0.334 g of C2H6?

Answer 1

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:

• C: 12 g/mole
• H: 1 g/mole

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?

$moles of C_{2} H_{6} =\frac{0.334 grams*1 mole}{30 grams}$

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?

$molecules of C_{2} H_{6} =\frac{0.011 moles*6.023*10^{23}molecules }{1 mole}$

molecules of C₂H₆= 6.6253*10²¹

6.6253*10²¹molecules of ethane (C₂H₆) are present in 0.334 g of C₂H₆