# Question eac5b

Jan 26, 2016

$\text{0.40 g}$

#### Explanation:

Your tool of choice here will be the ideal gas law equation

$\textcolor{b l u e}{P V = n R T} \text{ }$, 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, i.e. the temperature expressed in Kelvin

As you can see, this form of ideal gas law equation does not allow you to calculate the mass of the gas directly. However, you can use the molar mass of the gas to find a relationship between the mass of a sample and the number of moles it contains.

As you know, the molar mass of a substance tells you the mass of one mole of that substance.

$\textcolor{b l u e}{{M}_{M} = \frac{m}{n}} \text{ }$, where

$m$ - the mass of the sample
$n$ - the number of moles it contains

Notice that you can use this equation to find the number of moles

${M}_{M} = \frac{m}{n} \implies n = \frac{m}{M} _ M$

Plug this into the ideal gas law equation to get

$P V = \frac{m}{M} _ M \cdot R T$

Rearrange to solve for $m$

$P V \cdot {M}_{M} = m \cdot R T \implies m = {M}_{M} \cdot \frac{P V}{R T}$

Now, before you plug in your values, make sure that the units you're working with match those used by the universal gas constant.

 {:(color(red)("Need"), color(white)(aaaaa), color(blue)("Have")), (color(white)(aaaa), color(white)(aaaa), color(white)(aaaa)), (color(white)(aa)"L", color(white)(aaaa), color(white)(a)"mL"color(white)(aaaaaa)), ("atm", color(white)(aaaa), color(white)(a)"torr"color(white)(aaaaa)), (color(white)(aa)"K", color(white)(aaaa), color(white)(a)""^@"C"color(white)(aaaaaa)) :}

You will thus have to use the following conversion factors

$\text{1 atm " = " 760 torr}$

$\text{1 L" = 10^3"mL}$

$T = {t}^{\circ} \text{C} + 273.15$

Carbon dioxide, ${\text{CO}}_{2}$, has a molar mass of $\text{44.01 g/mol}$. Plug in your values to get

m = 44.01"g"/color(red)(cancel(color(black)("mol"))) * ( 625/760color(red)(cancel(color(black)("atm"))) * 27 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))#

$m = \text{0.03992 g}$

Rounded to two sig figs, the answer will be

$m = \textcolor{g r e e n}{{\text{0.40 g CO}}_{2}}$