# Question #eac5b

##### 1 Answer

#### Explanation:

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

#color(blue)(PV = nRT)" "# , where

*number of moles* of gas

**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.

#color(blue)(M_M = m/n)" "# , where

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

#M_M = m/n implies n = m/M_M#

Plug this into the ideal gas law equation to get

#PV = m/M_M * RT#

Rearrange to solve for

#PV * M_M = m * RT implies m = M_M * (PV)/(RT)#

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

#"1 atm " = " 760 torr"#

#"1 L" = 10^3"mL"#

#T = t^@"C" + 273.15#

Carbon dioxide,

#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 = "0.03992 g"#

Rounded to two sig figs, the answer will be

#m = color(green)("0.40 g CO"_2)#