Question

The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How many carbon atoms are in 3.00 g of butane?

Answer 1

1.2433 *10^23 carbon atoms

Step-by-step explanation:

First you need to calculate the grams of the empirical formula.

C4H10

Carbon = 12.01 g x 4

Hydrogen = 1.008 x 10

Sum those together to get = 58.12 g

Divide the 3.00g by the total g in the compound.

3.00/58.12 = 0.0516

Then times 0.0516 by the number of carbon atoms

0.0516 x 4 = .20646

Then multiple by Avogadro's number

.20646 x 6.022x10^23 = 1.2433x10^23

Rounded to 1.24x10^23 carbon atoms

Answer 2

First, calculate the number of mols of C4H10 in 3.00 g of C4H10.

3.00g C4H10 / 58.1 g C4H10 = .0516 mol C4H10
There are 4 atoms of carbon per mol of C4H10
Multiply .0516 mol C4H10 by 4 (number of atoms of carbon)=.20654 mol Carbon.
To convert to atoms, multiply that by avogadro's number, 6.022 x 10^23
which equals 1.2438 x 10^23 atoms of Carbon