# Calculate the density in grams per liter of an equi-molar mixture of CH4 and C2H6 at 100 degrees celicius and 700mmHg pressure.

*print*Print*list*Cite

### 1 Answer

Gasses obey the Ideal Gas Law. The Ideal Gas Law is

PV = nRT

In general the volume that gas will occupy is proportional to the number of moles present. In the stated problem, there is not a specific number of moles given. However, this is not critical as we are asked to find the density of the gas at a specific set of conditions. The density at a given temperature and pressure will be constant regardless of the number of moles present because the volume occupied is directly proportional to the number of moles. Thus doubling the moles (doubling the mass of the gas) will double the volume. Density being the ratio of mass to volume will remain the same.

We are free to choose any number of moles we wish as long as the methane (CH4) and ethane (C2H6) are equi-molar. Equi-molar simply means there must be the same number of moles of methane as there are moles of ethane. Being free to choose allows us to select one mole of each.

Applying this and the other conditions of the problem allows us to calculate the volume of the gas

P = 700 mmHg = (700/760)atm = 0.921 atm

n = 2 moles (1 mole CH4 + 1 mole C2H6 = 2 mole of gas)

T = 100 deg C = 373.15 K

R = 0.0821 (L atm)/(mol K)

Solving PV = nRT for V results in V = (nRT)/P

V = (2mol)X(0.0821 L atm/mol K)X(373.15K)/0.921 atm

V = 66.5 L

To get the density we must know the mass of the gas. This can be obtained by calculating the molar mass of each of the gasses.

The 1 mole of CH4 has a mass of 12.01g + 4(1.008g) =16.042 g

The 1 mole of C2H6 has a mass of 2(12.01g) + 6(1.008g) = 30.068 g

The total mass of the gas is thus 16.042g + 30.068g = 46.11 g

The density is determined from D = m/V = 46.11g/66.5L

**D = 0.693 g/L.**

Some attention should be given to the number of significant digits in the problem. There is some ambiguity in the problem concerning the number of signficant digit in each of the stated quantities, and thus the answer may need to be rounded to the nearest tenth of a gram/L.