# How would you determine the molecular formula of a substance that contains 92.3%C and 7.7%H by mass. This same substance when in the gas phase has a density of 0.00454g/ml at stp conditions?

##### 1 Answer

#### Explanation:

Your strategy here will be to

*use a sample of this compound to help you determine its***empirical formula***use the ideal gas law equation, the gas' density, and the known STP conditions for pressure and temperature to find th gas'***molar mass***use the empirical formula and the gas' molar mass to find its***molecular formula**

So, you know that you're dealing with a **hydrocarbon** that contains

You know from the aforementioned percent composition that this sample will contain

Use the molar masses of the two elements 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 mole ratio 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 ideal gas law equation

#color(blue)(PV = nRT)#

can be rewritten using the definition of **mass** and the **molar mass** of a compound

#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 **do not** forget to convert the pressure from *kPa* to *atm* and the temperature from *degrees Celsius* to *Kelvin*!

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)#