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 1.00 g of butane?

Answer 1

Answer : The number of carbon atoms present in butane are
`4.14* 10^(22)`

Explanation : Given,

Molar mass of
`C_4H_(10)` = 58.1 g/mole

Mass of
`C_4H_(10)` = 1.00 g

First we have to calculate the number of moles of butane.

`\text{Moles of }C_4H_(10)=\frac{\text{Mass of }C_4H_(10)}{\text{Molar mass of }C_4H_(10)}`

`\text{Moles of }C_4H_(10)=(1.00g)/(58.1g/mole)=0.0172mole`

Now we have to calculate the number of carbon atoms present in butane.

In butane, there are 4 atoms of carbon and 10 atoms of hydrogen.

As we know that, 1 mole of substance contains
`6.022* 10^(23)` number of atoms.

As, 1 mole of butane contains
`4* 6.022* 10^(23)` number of carbon atoms.

So, 0.0172 mole of butane contains
`0.0172* 4* 6.022* 10^(23)=4.14* 10^(22)` number of carbon atoms.

Therefore, the number of carbon atoms present in butane are
`4.14* 10^(22)`