# Question #ee73c

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

**!! LONG ANSWER !!**

Ok, here's my take on this. The way I see it, you are actually supposed to determine the **molecular mass** of the compound, **not** its *molecular formula*.

In order to calculate its molecular formula you need to have some information about its percent composition. This will allow you to find its *empirical formula* first, then use its **molecular mass** to find its molecular formula.

Also, I am not really sure what *aqueous tension* means, although I do suspect that it's actually the **vapor pressure of water vapor** at that temperature, i.e.

http://www.endmemo.com/chem/vaporpressurewater.php

So, the idea here is that this problem is probably based on a **Victor Meyer experiment**, a laboratory technique used to find the *molecular mass* of a **volatile** compound.

The working principle revolves around the idea that the *vapor* of the unknown compound will **displace an equal volume** of dry air, which is then *collected over water* forming *wet air*.

A Victor Meyer apparatus

Your ultimate goal here will be to find the **number of moles** of your unknown substance present in the volume of vapor that displaced the **dry air**, **not** the wet air.

The easiest way to do that is to use the molar volume of a gas at STP. More specifically, you know that **one mole** of any ideal gas occupies exactly **STP** conditions, which imply

a pressure of#"100 kPa"# a temperature of#0^@"C"#

The first important thing to keep in mind here is that air **collected over water** is actually a mixture of dry air, which is what was displaced by the vapor of your substance, and **water vapor**.

This means that the pressure of wet air will be

#P_"wet air" = P_"dry air" + P_"water vapor"#

The partial pressure of **dry air** will be, using the **vapor pressure of water** at

#P_"dry air" = "725 mmHg" - overbrace("12 mmHg")^(color(purple)("vapor pressure of water")) = "713 mmHg"#

Now, you know that the substance displaced a volume of **STP**, use the combined gas law equation

#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)#

**Do not** forget to use matching units for this! Also, the temperature **must be** expressed in *Kelvin*! Rearrange the equation to solve for **STP**

#V_2 = P_1/P_2 * T_2/T_1 * V_1#

You can go from *mmHg* to *kPa* by going through *atm* first.

#713 color(red)(cancel(color(black)("mmHg"))) * (1color(red)(cancel(color(black)("atm"))))/(760color(red)(cancel(color(black)("mmHg")))) * "101.325 kPa"/(1color(red)(cancel(color(black)("atm")))) = 711/760 * "101.325 kPa"#

This means that you'll have

#V_2 = (713/760 * 101.325 color(red)(cancel(color(black)("kPa"))))/(100color(red)(cancel(color(black)("kPa")))) * ((0 + 273.15)color(red)(cancel(color(black)("K"))))/((273.15 + 14)color(red)(cancel(color(black)("K")))) * "24.2 cm"^3#

#V_2 = "21.883 cm"^3#

So, your unknown substance displaced **dry air** at STP. This of course implies that the volume of its vapor will be

#V_"substance" = V_"dry air" = "21.883 cm"^3#

The **number of moles** of your substance present in

#"21.883 cm"^3 = "21.883 mL"#

of vapor at STP will thus be

#21.883 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * overbrace("1 mole"/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas")) = "0.00096401moles"#

The **molecular mass** of a substance, which tells you the mass of **one mole** of that substance, can be calculated by dividing the **mass** of a sample by the **number of moles** it contains

#color(blue)("molar mass" = M_M = "mass"/"no. of moles" = m/n = ["grams"/"mol"])#

In your case, the sample had a mass of

#M_M = "0.188 g"/"0.00096401 moles" = color(green)("195 g/mol")#

I'll leave the answer rounded to three sig figs.

**SIDE NOTE** *Many textbooks and online resources still list STP conditions as a pressure of* *and a temperature of*

*Under these conditions, one mole of any ideal gas occupies* *If that's the value of the molar volume of a gas that was given to use, simply redo the final part of the calculations using* *instead of*