# Question 61c8a

May 17, 2016

The values given to you are incorrect.

#### Explanation:

You are right, the mass of nitrogen in the mixture comes out negative, which can only mean that some of the values given to you are incorrect.

The idea here is that you need to use the ideal gas law equation to find the total number of moles of as present in the mixture.

Once you know that, you can use the number of moles of helium present in the sample to find eventually the mass of nitrogen.

The ideal gas law equation looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

You know that your mixture is being kept in a $\text{9.79-L}$ flask at a pressure of $\text{2.58 atm}$ and a temperature of ${90}^{\circ} \text{C}$.

Use these values to find the number of moles of gas present in the flask -- do not forget to convert the temperature from degrees Celsius to Kelvin

$P V = n R T \implies n = \frac{P V}{R T}$

You will have

n = (2.58 color(red)(cancel(color(black)("atm"))) * 9.79color(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")))) * (90 + 273.15)color(red)(cancel(color(black)("K")))) = "0.06954 moles gas"

Now, use the molar mass of helium to find how many moles you have in that $\text{1.09-g}$ sample

1.09color(red)(cancel(color(black)("g"))) * "1 mole He"/(4.003color(red)(cancel(color(black)("g")))) = "0.2723 moles He"

At this point, it becomes obvious that the values are wrong, since you have more moles of helium than you have total moles of gas in the mixture.

This will lead you to a negative number of moles of nitrogen present in the mixture

${n}_{{N}_{2}} = \text{0.06954 moles" - "0.2723 moles}$

${n}_{{N}_{2}} = \textcolor{red}{\cancel{\textcolor{b l a c k}{- {\text{0.2028 moles N}}_{2}}}}$

which will of course lead to a negative mass of nitrogen gas. As you know, that's is not possible.

A quick fix would be to use $\text{0.109 g}$ as the mass of helium present in the mixture. This would get you

0.109color(red)(cancel(color(black)("g"))) * "1 mole He"/(4.003color(red)(cancel(color(black)("g")))) = "0.02723 moles He"

You'd now have

${n}_{{N}_{2}} = \text{0.06954 moles" - "0.02723 moles}$

${n}_{{N}_{2}} = {\text{0.04231 moles N}}_{2} \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}$

The mass that would contain this many moles can be determined by using nitrogen gas' molar mass

0.04231 color(red)(cancel(color(black)("moles N"_2))) * "28.0134 g"/(1color(red)(cancel(color(black)("mole N"_2)))) = "1.2 g"#