Question

Exercise 11.76

A mixture of helium, nitrogen, and oxygen has a total pressure of 755 mmHg . The partial pressures of helium and nitrogen are 241 mmHg and 194 mmHg , respectively.

Part A

What is the partial pressure of oxygen in the mixture?

​Part B

The hydrogen gas formed in a chemical reaction is collected over water at 30 ∘C at a total pressure of 730 mmHg .

What is the partial pressure of the hydrogen gas collected in this way?

Exercise 11.78 - Part C

The oxygen gas emitted from an aquatic plant during photosynthesis is collected over water at a temperature of 25 ∘C and a total pressure of 748 torr

Part C

What is the partial pressure of the oxygen gas?

Exercise 11.104---- Part D

A gaseous compound containing hydrogen and carbon is decomposed and found to contain 85.63 % C and 14.37 % H by mass. The mass of 258 mL of the gas, measured at STP, is found to be 0.646 g.

Part D

What is the molecular formula of the compound?

Express your answer as a chemical formula.

.

Answer 1

A mixture of helium, nitrogen, and oxygen has a total pressure of 755 mmHg . The partial pressures of helium and nitrogen are 241 mmHg and 194 mmHg , respectively.

A)

P-O2 --> from

Ptotal = Pgases

755 = 241+194 + P-O2

P-O2 = 755 -( 241+194)

P-O2 = 320 mm Hg

b)

Ptotal = PGas + Pvapor

730 = PGas + PVapor

Pvapr (30°C) = 31.8

Pgas = 730-31.8 = 698.2 torr

C)

Pgas = Ptotal - Pvapor

Pgas = 748 - 23.8 = 724.2 torr

D)

find MW of molecule:

1 mol at STP = 22.4 L

x mol = 0.258

x = 0.258/22.4

x = 0.01151 mol of species

MW empirical:

mol of C = 85.63/12 = 7.14

mol of H = 14.37/1 = 14.37

C::H ratio --> 1:2 --> Empricial -->C1H2

MW of empirical = 12+2 = 14

MW formula = mass/mol = 0.646 / 0.01151 = 56.125

ratio = 56.125/14 = 4x

CH2 x 4 = C4H8