Your strategy here will be to
So, you know that you're dealing with a hydrocarbon that contains
You know from the aforementioned that this sample will contain
Use the molar masses of the two to figure out how many moles of each you'd get in this sample
$92.3 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "7.6846 moles C"$
$7.7 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "7.6393 moles H"$
Divide both values by the smallest one to get the that exists between the two elements in the compound
$"For C: " (7.6846 color(red)(cancel(color(black)("moles"))))/(7.6393color(red)(cancel(color(black)("moles")))) = 1.006 ~~1$
$"For H: " (0.76393color(red)(cancel(color(black)("moles"))))/(0.76393color(red)(cancel(color(black)("moles")))) = 1$
The empirical formula for this hydrocarbon is
$"C"_1"H"_1 implies "CH"$
Now, STP conditions are characterized by a pressure of
The equation
$color(blue)(PV = nRT)$
can be rewritten using the definition of
$n = m/M_M$
Plug this into the ideal gas law equation to get
$PV = m/M_M * RT$
Rearrange to get
$M_M = overbrace(m/V)^(color(blue)(=rho)) * (RT)/P$
Since density is defined as mass per unit of volume, you can say that
$M_M = rho * (RT)/P$
Plug in the STP pressure and temperature and solve for
Also, convert the density of the gas from grams per milliliter to grams per liter
$0.00454 "g"/color(red)(cancel(color(black)("mL"))) * (1000color(red)(cancel(color(black)("mL"))))/"1 L" = "4.54 g/L"$
$M_M = 4.54"g"/color(red)(cancel(color(black)("L"))) * (0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm"))))$
$M_M = "103.2 g/mol"$
Now, the molecular formula will always be a multiple of the empirical formula. The molar mass of the empirical formula is
$1 xx "12./011 g/mol" + 1 xx "1.00794 g/mol" = "13.019 g/mol"$
This means that you have
$13.019 color(red)(cancel(color(black)("g/mol"))) xx color(blue)(n) = 103.2 color(red)(cancel(color(black)("g/mol")))$
This will get you
$color(blue)(n) = 103.2/13.019 = 7.93 ~~ 8$
Therefore, the molecular formula of the hydrocarbon will be
$("CH")_color(blue)(8) implies color(green)("C"_8"H"_8)$