density at standard conditions -...

Density at Standard Conditions

Density is the ratio of mass to volume.

Density of a gas is generally given in g/L.

The mass of 1 mole = molar mass.

The volume of 1 mole at STP = 22.4 L

Calculate the density of N2(g) at STP.

MM d

Gas Density

Density is directly proportional to molar mass

“n”

gg/mol

( )

( )gL

gmol

L atmmol K

( )( ) atm

( )K=

Density is directly proportional to molar mass

Calculate the density of N2 at 125°C and 755 mmHg .

P = 755 mmHg, t = 125 °C, dN2 = ?

P, MM, T, R d

Calculate the density of a gas at 775 torr and 27 °C if 0.250 moles weighs 9.988 g.

m = 9.988g, n = 0.250 mol, P = 775 torr, T = 27ºC, density = ?

P, n, T, R V V, m d

K = 273+ºC=273+(27)=300

Calculate the density of a gas at 775 torr and 27 °C if 0.250 moles weighs 9.988 g.

m = 9.988g, n = 0.250 mol, P = 775 torr, T = 27ºC, density = ?

V, m d P, n, T, R V

V= 6.0355 L

Molar Mass of a Gas

One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law.

Determining the Molar Mass of an Unknown Volatile Liquid.

JBA Dumas (1800-1884)

Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg.

m=0.311g, V=0.225 L, P=886 mmHg, T = 55 ºC, molar mass = ?

P, V, T, R n n, m MM

K = 273+ºC=273+(55)=328

Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg.

m=0.311g, V=0.225 L, P=886 mmHg, T = 55 ºC, molar mass = ?

n, m MM P, V, T, R n

n = 9.7454 x 10-3 mol

What is the molar mass of a gas if 12.0 g occupies 197 L at 3.80 x 102 torr and 127 °C?

m=12.0 g, V= 197 L, P=380 torr, t=127°C, molar mass = ?

P, V, T, R n n, m MM

K = 273+ºC=273+(127)=400

Mixtures of GasesWhen gases are mixed together, their molecules

behave independent of each other.

All the gases in the mixture have the same volume.

All gases in the mixture are at the same temperature.

In certain applications, the mixture can be thought of as one gas.

We can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules.

Partial PressureThe pressure of a single gas in a mixture of gases

is called its partial pressure.

Dalton’s Law of Partial Pressures - The sum of the partial pressures of all the gases in the mixture equals the total pressure

We can calculate the partial pressure of a gas if

we know what fraction of the mixture it composes and the total pressure,or

we know the number of moles of the gas in a container of known volume and temperature

The partial pressure of each gas in a mixture can be calculated using the ideal gas law.

Dalton's Law of Partial Pressures

Find the partial pressure of neon in a mixture of neon and xenon with total pressure 3.9 atm, volume 8.7 L,

temperature 598 K, and 0.17 moles Xe.

Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol , PNe = ?

nXe, V, T, R PXe Ptot, PXe PNe

Mole FractionThe fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes.

The ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction, χ.

The partial pressure of a gas is equal to the mole fraction of that gas times the total pressure.

Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K.

mHe = 24.2 g, mO2 = 4.32 g, V = 12.5 L, T = 298 K

χHe, χO2, PHe, PO2, Ptotal = ?????

mgas ngas χgas

χgas, Ptotal Pgas

ntot, V, T, R Ptot

Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K .

6.185 mol

6.185 mol

ntotal = 6.185 mol

Collecting Gas By Displacement of Water

To collect a gas by displacement of water, the bottle is filled completely with water and inverted in a large vessel of water (a). As the gas to be collected is released into the bottle, gas bubbles rise to fill the bottle as water in the bottle is displaced.

When the water levels inside and outside the bottle are equal, pressures inside and outside the bottle are also equal (b). At this point Patm = Pgas collected + Pwater vapor.

Collecting GasesGases are often collected by having them

displace water from a container.

Because water evaporates, there is also water vapor in the collected gas.

The partial pressure of the water vapor, called the vapor pressure, depends only on the temperature, so you can use a table to find out the partial pressure of the water vapor in the gas you collect.

Vapor Pressure of Water

For example, if you collect a gas sample with a total pressure of 758.2 mmHg* at 25 °C, the partial pressure of the water vapor will be 23.78 mmHg – so the partial pressure of the dry gas will be 734.4 mmHg.

1.02 L of O2 collected over water at 293 K with a total pressure of 755.2 mmHg. Find the mass of O2.

V=1.02 L, P=755.2 mmHg, T=293 K, mass O2 = ?

Ptot, PH2O PO2 PO2,V,T nO2 gO2

0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure in torr.

V=10.0 L, nH2 = 0.12 mol, T = 323 K, Ptotal = ?

nH2,V,T PH2 PH2, PH2O Ptotal

Reactions Involving Gases

Reactions Involving Gases

The principles stoichiometry can be combined with the gas laws for reactions involving gases.

In reactions of gases, the amount of a gas is often given as a volume instead of moles.

The ideal gas law allows us to convert from the volume of the gas to moles; then we can use the coefficients in the equation as a mole ratio.

P, V, T of Gas A mole A mole B P, V, T of Gas B

What volume of H2 is needed to make 37.5 g of CH3OH at 738 mmHg and 355 K?

CO(g) + 2 H2(g) → CH3OH(g)

mCH3OH = 37.5g, P=738 mmHg, T=355 K , VH2 = ?

g CH3OH mol CH3OH mol H2 P, n, T, R V

How many grams of H2O form when 1.24 L H2 reacts completely with O2 at STP?O2(g) + 2 H2(g) → 2 H2O(g)

VH2 = 1.24 L, P = 1.00 atm, T = 273 K massH2O = ?

g H2O L H2 mol H2 mol H2O

What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO?

2 HgO(s) → 2 Hg(l) + O2(g)(MMHgO = 216.59 g/mol)

mHgO = 10.0g, P=0.750 atm, T=313 K VO2 = ?

g HgO mol HgO mol O2 P, n, T, R V