Calculate the molar mass in #"g"/"mol"#of diacetyl (butanedione) given that in the gas phase #100# degrees Celsius and #747# torr, a #0.3060# g sample of diacetyl occupies a volume of #0.111L#?

2 Answers
May 26, 2017

Answer:

The molar mass of diacetyl, #"C"_4"H"_6"O"_2"#, as calculated based on the question parameters is #"85.83 g/mol"#. Its actual molar mass is #"86.09 g/mol"#.

Explanation:

You can use the ideal gas law to answer this question. The equation is:

#PV=nRT#,

where #P# is pressure, #V# is volume, #n# is moles, #R# is the gas constant, and #T# is the temperature in Kelvins. Add #273.15# to the Celsius temperature to get the temperature in Kelvins: #100^@"C" + 273.15="373 K"#.

We will use the ideal gas law to determine moles of diacetyl gas, then divide the given mass by the moles of diacetyl gas to determine its molar mass.

Known

#m="0.3060 g"#

#P="747 torr"#

#V="0.111 L"#

#R=62.364color(white)(.)"L torr K"^(-1) "mol"^(-1)#
https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Laws/The_Ideal_Gas_Law

#T="373 K"#

Unknown

#n=?#

#"molar mass = ? g/mol"#

Solve for #n#.
Rearrange the equation to isolate #n#. Insert the known data and solve.

#n=(PV)/(RT)#

#n=(747color(red)cancel(color(black)("torr"))xx0.111color(red)cancel(color(black)("L")))/((62.364color(white)(.)color(red)cancel(color(black)("L"))color(red)cancel(color(black)("torr")) color(red)cancel(color(black)("K")))^(-1) "mol"^(-1)xx373color(red)cancel(color(black)("K")))="0.003565 mol"#

Now that you have moles, divide the mass of diacetyl given in the question by the moles.

#"Molar mass diacetyl" = (0.3060"g diacetyl")/(0.003565"mol diacetyl")="85.83 g/mol diacetyl"# rounded to four significant figures

The molecular formula for diacetyl is #("CH"_3"CO)"_2# or #"C"_4"H"_6"O"_2"# and its known molar mass to three sig figs is #"86.09 g/mol"#. https://www.ncbi.nlm.nih.gov/pccompound?term=%22diacetyl%22

#"Percent error" = abs(("known value"-"experimental value")/("accepted value"))xx100#

#"Percent error"=abs((86.09-85.83)/(86.09))xx100="0.3020%"#

May 26, 2017

Answer:

#85.9 "g"/"mol"#

Explanation:

What we can do here is calculate the density of the diacetyl, and use that to directly calculate the molar mass. We will use the equation

#M = (dRT)/P#

where #M# is the molar mass of the substance,
#d# is its density, in #"g"/"L"#,
#R# is the universal gas constant, equal to #0.08206 ("L"-"atm")/("mol"-"K")#,
#T# is the absolute temperature (in #"K"#), and
#P# is the pressure of the gas (in #"atm"#)

Where is this equation derived from? Read the steps below if you would like to know, otherwise, skip to the next step.

Well, let's recall our ideal-gas equation, and rearrange it to solve for units similar to that of density, #"mol"/"L"#, which is #n/V#:

#PV = nRT#

#P = (nRT)/V#

#P/(RT) = n/V#

Now, let's multiply both sides of the equation by #M#, the molar mass with units #"g"/"mol"#:

#(PM)/(RT) = (nM)/V#

If we list the right side of the equation in terms of units, we have

#cancel("mol")/"L" xx "g"/cancel("mol") = "g"/"L"#

Which is the units for density. Thus, the value #(nM)/V# is the density of the gas, and if we plug this back into our equation:

#(PM)/(RT) = (nM)/V = d#

Thus, #d = (MP)/(RT)#, and rearranging to solve for the molar mass yields our original equation, #M = (dRT)/P#.

The density of the diacetyl is

#d = (0.3060"g")/(0.111 "L") = 2.76 "g"/"L"#

The temperature, in Kelvin, is

#100^oC + 273 = 373 "K"#

and the pressure, in atmospheres, is

#747 cancel("torr")((1 "atm")/(760 cancel("torr"))) = 0.983 "atm"#

Finally, plugging in all our known variables, we have

#M = ((2.76 "g"/cancel("L"))(0.08206 (cancel("L")-cancel("atm"))/("mol"-cancel("K")))(373 cancel("K")))/(0.983 cancel("atm")) = color(red)(85.9 "g"/"mol"#

To check this, the molecular formula of diacetyl is #("CH"_3"CO")_2#, and from this the molar mass is found to be #86.1 "g"/"mol"#, which measures up well with our result.