Question

A mixture of methane (CH₄) and butane (C₄H₁₀) at one atmosphere pressure and 25°C has a density of 1.713 g/L. Assuming ideal behavior, what is the mass in grams of carbon present in 1 liter of the mixture?

Answer 1

To find the mass of carbon present in 1 liter of the mixture, we need to calculate the moles of methane and butane in the mixture. The mass of carbon in 1 liter of the mixture is 2.094 g.

Step-by-step explanation:

To find the mass of carbon present in 1 liter of the mixture, we need to calculate the moles of methane and butane in the mixture. First, let's calculate the volume of each component in the mixture using their respective densities.

The volume of methane (CH4) in 1 liter of the mixture can be calculated using the density given for methane (0.668 g/L):

1. Volume of methane = (1 L) x (0.668 g/L) = 0.668 g

Similarly, the volume of butane (C4H10) in 1 liter of the mixture can be calculated using the density given for butane (2.493 g/L):

1. Volume of butane = (1 L) x (2.493 g/L) = 2.493 g

Now, let's calculate the moles of carbon in each component:

1. Moles of carbon in methane = (0.668 g) x (1 mol/16.04 g) x (1 mol/1 mol of carbon) = 0.0416 mol
2. Moles of carbon in butane = (2.493 g) x (1 mol/58.12 g) x (4 mol/1 mol of carbon) = 0.171 mol

Finally, we can calculate the mass of carbon in 1 liter of the mixture:

1. Mass of carbon = (0.0416 mol + 0.171 mol) x (12.01 g/mol) = 0.0416 g + 2.053 g = 2.094 g