Answer to Mole Exercise 2
Answer to Mole Exercise 2
Butane, C4H10 is one of the common hydrocarbons used as fuel. How many atoms of C are there in 87 grams of butane?
MW = 4 (12.0 amu) + 10 (1.0 amu) = 58 amu
So, (87 g C4H10) ( 1 mol C4H10)/ (58 g C4H10) (4 mols C)/(1 mol C4H10)
(6.02x 1023 atoms C)/(1 mol C) = 3.61 x 1024 atoms C
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