Question

How many molecules are in 70.7 grams of C4H10?

Answer 1

Answer: 7.34 x 10∧23 molecules are in 70.7grams of C4H10

Explanation:

First,we convert to moles using the molar mass of Butane( C4H10).

The molar mass of Butane ( C4H10) = 58.12g/mol

It means;

58.12g of C4H10 --------> 1 mole of C4H10

Therefore; 70.7g of C4H10 ---------> 1/ 58.12 x 70.7 = 1.22moles

Using the Avogadro's number;

1 mole of C4H10 contains 6.02 x 10∧23 molecules.

1.22 moles would contain; 1.22 x 6.02 x 10∧23 = 7.34 x 6.02 x 10∧23 molecules

Answer 2

1.216mol

Step-by-step explanation:

The molar mass of C₄H₁₀ is (12 x4)+ (1x 10) = 48 + 10 = 58g

1 grams C4H10 is equal to 0.017205129881525 mole.

70.7 grams = 70.7 x 0.017205129881525 = 1.216mol