# A solution of 2.50 g of a compound having the empirical formula #C_6H_5P# in 25.0 g of benzene is observed to freeze at 4.3°C . What is the molar mass of the solute and its molecular formula?

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The idea here is that you need to use the *freezing-point depression* equation to determine what the molality of the solution.

Once you have the solution's molality, use it to find the number of moles of solute it contains.

As you know, the equation for *freezing-point depression* looks like this

#color(blue)(DeltaT_f = i * K_f * b)" "# , where

*van't Hoff factor*

*cryoscopic constant* of the solvent;

The cryoscopic constant of benzene is equal to

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

You're dealing with a **non-electrolyte**, which means that the van't Hoff factor will be equal to

The freezing-point depression is defined as

#color(blue)(DeltaT_f = T_f^@ - T_f)" "# , where

*pure solvent*

*Pure benzene* freezes at

#DeltaT_f = 5.5^@"C" - 4.3^@"C" = 1.2^@"C"#

Plug in your values and solve for

#DeltaT_f = i * K_f * b implies b = (DeltaT_f)/(i * K_f)#

#b = (1.2 color(red)(cancel(color(black)(""^@"C"))))/(1 * 5.12 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "0.2344 mol kg"^(-1)#

As you know, molality is defined as moles of solute per **kilograms of solvent**.

#color(blue)(b = n_"solute"/m_"solvent")#

In your case, you have a mass of

#b = n_"solute"/m_"solvent" implies n_"solute" = b xx m_"solvent"#

#n_"solute" = "0.2344 mol" color(red)(cancel(color(black)("kg"^(-1)))) * 25.0 * 10^(-3)color(red)(cancel(color(black)("kg")))#

#n_"solute" = "0.00586 moles"#

Now, **molar mass** is defined as the mass of **one mole** of a substance. In your case, the

#M_M = "2.50 g"/"0.00586 moles" = "426.6 g/mol"#

As you know, a compound's *empirical formula* tells you what the **smallest whole number ratio** that exists between the elements that make up said compound is.

The **molecular formula**, which tells you exactly how many atoms of each element are needed to form the compound, will always be a *multiple* of the empirical formula.

In this case, calculate the molar mass of the empirical formula by adding the molar masses of all of its constituent elements

#6 xx "12.011 g/mol" + 5 xx "1.00794 g/mol" + 1 xx "30.974 g/mol" = "108.08 g/mol"#

You can thus say that

#108.08 color(red)(cancel(color(black)("g/mol"))) * color(blue)(n) = 426.6color(red)(cancel(color(black)("g/mol")))#

This will get you

#color(blue)(n) = 426.6/108.08 = 3.95 ~~ 4#

The compound's molecular formula will be

#("C"_6"H"_5"P")_color(blue)(4) implies "C"_24"H"_20"P"_4 -># 1, 2, 3, 4 - tetraphenyltetraphosphetane