Question

A gas is found to have a density of `"1.80 g/L"` at `"76 cm Hg"` and `27^@ "C"`. The gas is?

a)Oxygen b)Carbon Dioxide c)Ammonia d)Sulphur dioxide

Answer 1

I'd go for (b) `"CO"_2`

Explanation:

The idea here is that you need to figure out the molar mass of the gas by using its density under the given conditions for pressure and temperature.

Your tool of choice will be the ideal gas law equation

`color(blue)(ul(color(black)(PV = nRT)))`

Here

• `P` is the pressure of the gas
• `V` is the volume it occupies
• `n` is the number of moles of gas present in the sample
• `R` is the universal gas constant, equal to `0.0821("atm L")/("mol K")`
• `T` is the absolute temperature of the gas

Start by converting the pressure of the gas to atmospheres and the temperature to Kelvin. You will have

`76 color(red)(cancel(color(black)("cmHg"))) * "1.0 atm"/(76 color(red)(cancel(color(black)("cmHg")))) = "1.0 atm"`

and

`T["K"] = 27^@"C" + 273.15 = "300.15 K"`

Now, you know that the density of the gas, `rho`, can be expressed by using the mass of the sample, let's say `m`, and the volume it occupies, `V`

`color(blue)(rho = m / V)`

The number of moles of gas can be expressed by using the mass of the sample and the molar mass of the gas, let's say `M_M`

`n = m/M_M`

Plug this into the ideal gas law equation to get

`PV = m/M_M * RT`

Rearrange to isolate the molar mass of the gas

`M_M = color(blue)(m/V) * (RT)/P`

This is equivalent to

`M_M = color(blue)(rho) * (RT)/P`

Plug in your values to find

`M_M = "1.80 g" color(red)(cancel(color(black)("L"^(-1)))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 300.15color(red)(cancel(color(black)("K"))) )/(1.0color(red)(cancel(color(black)("atm"))))`

`M_M = "44 g mol"^(-1) ->` rounded to two sig figs

The closest match is carbon dioxide, which has a molar mass of

`M_ ("M CO"_ 2) = "44.01 g mol"^(-1)`