Question

Complete combustion of a 0.0150 mol sample of a hydrocarbon, CxHy, gives 3.024 L of CO2 at STP and 1.891 g of H2O.

(a) What is the molecular formula of the hydrocarbon?

(b) What is the empirical formula of the hydrocarbon?

A 0.500-L bulb containing Ne at 685 torr is connected by a valve to a 2.50-L bulb containing CO2 at 375 torr. The valve between the two bulbs is opened and the two gases mix. The initial gas pressures as known to three significant figures.

(a) What is the partial pressure (torr) of Ne?

(b) What is the partial pressure (torr) of CO2?

(c) What is the total pressure?

(d) What is the mole fraction of Ne?

Answer 1

QUES

moles of CO2 produced can be calculated from the given data as,

22.4 L of CO2 at STP will be1 mole of CO₂
3.024 L of CO2 at STP will be => (3.024/22.4) = 0.135 moles of CO2
moles of C in burned sample = 0.135 moles
Now calculating the moles of H;
The fraction by mass of H in water = 2(1.0)/(2(1.0) +16)
                                                     = 2/18
                                                     = 0.111
mass of H in compound = (0.111)1.891 = 0.210g
Hence, moles of H in sample = 0.210/1 = 0.210 moles
Thus, (moles of C : moles of H) = 0.135:0.210 = 1:1.55 (approx 2:3)
Hence the empirical formula will be C2H3
Since 0.0150 moles of CxHy contains 0.210 moles of H
And1.0 mole of CxHy contains y moles of H
thus, y/1.0 = 0.210/0.015

that is y = 14
now since x:y = 2:3, hence x = 9.33 (approx 9)
Molecular formula of the compound is C9H14

QUES

(a) The partial pressure (torr) of Ne= P1*V1 / (V1+V2)
(685 torr Ne) x (0.50 L) / (0.50 L + 2.50 L) = 117.16 torr Ne

(b) The partial pressure (torr) of CO2 = P2*V2 / (V1+V2)
(375 torr CO2) x (2.50 L) / (2.50 L + 0.50 L) = 312.5 torr CO2

(c) The total pressure will be the summation of partial pressure of both the gases, that is;
117.16 torr + 312.5 torr Cl2 = 429.66 torr total

(d) The mole fraction of Ne will be = (P1*V1) / (P1*V1 + P2*V2)
(0.50 L x 685 torr Ne) / ((0.50 L x 685 torr) + (2.50 L x 375 torr)) = 0.267