If 2.30 g occupies 0.870 L at 690 torr and 37 degrees C. How would you calculate the molar mass?

Oct 25, 2015

$\text{74 g/mol}$

Explanation:

To get the molar mass of the gas, which tells you what the mass of one mole of the gas is, you need to determine how many moles you have in that sample.

SInce you are given volume, pressure, and temperature, you can use the ideal gas law

$\textcolor{b l u e}{P V = n R T}$

to solve for $n$, the number of moles of gas.

The universal gas constant, $R$, is usually given as

$R = 0.082 \left(\text{atm" * "L")/("mol" * "K}\right)$

which means that you will have to convert the pressure of the gas from torr to atm by using the conversion factor

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

So, plug in your values and solve for $n$

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

n = (690/760color(red)(cancel(color(black)("atm"))) * 0.870color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 37)color(red)(cancel(color(black)("K")))) = "0.03106 moles"

This means that the molar mass of the gas will be equal to

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

${M}_{\text{m" = "2.30 g"/"0.03106 moles" = "74.05 g/mol}}$

Rounded to two sig figs, the answer will be

M_"m" = color(green)("74 g/mol")

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