General Chemistry

Density and molar mass are calculated by modifying the ideal gas law equation. What we do is first replace the moles (n) in PV = nRT with mass over the molar mass:                                                 

\[{\rm{n}}\;{\rm{(moles)}}\,{\rm{ = }}\;\frac{{{\rm{m}}\;{\rm{(mass)}}}}{{{\rm{M}}\;{\rm{(molar}}\,{\rm{mass)}}}}\]

 

 

Next, we rearrange this formula to bring the V under the m which is the density of the gas (mass/volume = density):

 

 

From this equation, we can obtain an expression for the density and molar mass of an ideal gas:

 

 

An example of calculating the density of an ideal gas:

What is the density of CO2 gas at 386 K and 17.0 atm.

 

\[{\rm{d}}\;{\rm{ = }}\;\frac{{{\rm{PM}}}}{{{\rm{RT}}}}\]

 

\[{\rm{d}}\;{\rm{ = }}\;\frac{{{\rm{17}}{\rm{.0}}\;\cancel{{{\rm{atm}}}}\;{\rm{44}}{\rm{.0}}\;{\rm{g}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}}}{{{\rm{0}}{\rm{.08206}}\;{\rm{L}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}{\rm{ 386}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{23}}{\rm{.6}}\;{\rm{g/L}}\]

 

An example of calculating the molar mass of an ideal gas:

A scientist carries out an experiment to determine the molar mass of a 2.84-g sample of a colorless liquid which exerts 756 mmHg pressure when vaporized in a 260-mL flask at 142 oC. What is the molecular mass of this compound?

                                                                    

\[{\rm{PV}}\;{\rm{ = }}\;{\rm{nRT}}\]

\[{\rm{PV}}\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{M}}}{\rm{RT}}\]

\[{\rm{M}}\;{\rm{ = }}\;\frac{{{\rm{mRT}}}}{{{\rm{PV}}}}\]

\[{\rm{M}}\;{\rm{ = }}\;\frac{{{\rm{2}}{\rm{.84}}\;{\rm{g }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{415}}\;\cancel{{\rm{K}}}}}{{{\rm{0}}{\rm{.995}}\;\cancel{{{\rm{atm}}}}\;{\rm{0}}{\rm{.260}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{374}}\;{\rm{g/mol}}\]

 

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Practice

1.

Identify the unknown gas that weighs 17.75 grams in a 17.0 L cylinder held at 0.700 atm  pressure and 250°C.
a) NO2 b) CO2 c) H2 d) SO2 e) He

answer

d) SO2

Solution

To identify the unknown gas, we need to determine its molar mass.

\[{\rm{PV}}\;{\rm{ = }}\;{\rm{nRT}}\]

\[{\rm{PV}}\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{M}}}{\rm{RT}}\]

\[{\rm{M}}\;{\rm{ = }}\;\frac{{{\rm{mRT}}}}{{{\rm{PV}}}}\]

Before plugging the numbers, convert the units to L, K, and atm to match the units of R.

T =250 + 273 = 523 K

\[{\rm{M}}\;{\rm{ = }}\;\frac{{{\rm{17}}{\rm{.75}}\;{\rm{g }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{523}}\;\cancel{{\rm{K}}}}}{{{\rm{0}}{\rm{.700}}\;\cancel{{{\rm{atm}}}}\;{\rm{17}}{\rm{.0}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{64}}{\rm{.0}}\;{\rm{g/mol}}\]

2.

What is the density of CO2 gas at 386 K and 17.0 atm.

answer

23.6 g/L

Solution

The correlation between density and molar mass of a gas is derived from the ideal gas law equation:

\[{\rm{d}}\;{\rm{ = }}\;\frac{{{\rm{PM}}}}{{{\rm{RT}}}}\]

\[{\rm{d}}\;{\rm{ = }}\;\frac{{{\rm{17}}{\rm{.0}}\;\cancel{{{\rm{atm}}}}\;{\rm{44}}{\rm{.0}}\;{\rm{g}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}}}{{{\rm{0}}{\rm{.08206}}\;{\rm{L}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}{\rm{ 386}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{23}}{\rm{.6}}\;{\rm{g/L}}\]

3.

Determine the density of ammonia gas, NH3, at 36.0 oC and 695 mmHg. Report the density in grams per liter.

answer

0.613 g/L

Solution

The correlation between density and molar mass of a gas is derived from the ideal gas law equation:

\[{\rm{d}}\;{\rm{ = }}\;\frac{{{\rm{PM}}}}{{{\rm{RT}}}}\]

Before plugging the numbers, convert the units to L, K, and atm to match the units of R.

T = 36.0 + 273 = 309 K

P = 695 mmHg x 1 atm/760 mmHg = 0.914 atm

\[{\rm{d}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.914}}\;\cancel{{{\rm{atm}}}}\;{\rm{17}}{\rm{.0}}\;{\rm{g}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}}}{{{\rm{0}}{\rm{.08206}}\;{\rm{L}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;\cancel{{{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}}{\rm{ 309}}\;\cancel{{\rm{K}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.613}}\;{\rm{g/L}}\]

4.

A scientist carries out an experiment to determine the molar mass of a 2.84-g sample of a colorless liquid which exerts 756 mmHg pressure when vaporized in a 260-mL flask at 142 oC. What is the molecular mass of this compound?

answer

374 g/mol

Solution

\[{\rm{PV}}\;{\rm{ = }}\;{\rm{nRT}}\]

\[{\rm{PV}}\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{M}}}{\rm{RT}}\]

\[{\rm{M}}\;{\rm{ = }}\;\frac{{{\rm{mRT}}}}{{{\rm{PV}}}}\]

Before plugging the numbers, convert the units to L, K, and atm to match the units of R.

T =142 + 273 = 415 K

P = 756 mmHg x 1 atm/760 mmHg = 0.995 atm

V = 260 mL x 1L/1000 mL = 0.260 L

\[{\rm{M}}\;{\rm{ = }}\;\frac{{{\rm{2}}{\rm{.84}}\;{\rm{g }}\;{\rm{0}}{\rm{.08206}}\;\cancel{{\rm{L}}}\;\cancel{{{\rm{atm}}}}\;\cancel{{{{\rm{K}}^{{\rm{ – 1}}}}}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\;{\rm{415}}\;\cancel{{\rm{K}}}}}{{{\rm{0}}{\rm{.995}}\;\cancel{{{\rm{atm}}}}\;{\rm{0}}{\rm{.260}}\;\cancel{{\rm{L}}}}}\;{\rm{ = }}\;{\rm{374}}\;{\rm{g/mol}}\]

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