50 QUANTITATIVE ASPECTS OF CHEMISTRY
By measurement it is found that 1 liter of the vapor at 136.5°C.
and 760 mm. weighs 2.203 grams. Reduce the volume to stand-
ard conditions:
273
V.t = 1.000 X = 0.6667 liter
273 + 13o.5
Thus 0.6667 liter under standard conditions weighs 2.203 grams.
22.4 liters will weighX 2.203 = 74.02 gramsThus the molecular weight is 74.02.
Chemical analysis shows that the composition of ethyl ether is:
C, 64.87 per cent; H, 13.53 per cent; O, 21.60 per cent
The weight.of each of these elements in one mole is:
C: 0.6487 X 74.02 = 48.02 = 4 X 12.01 grams
H: 0.1353 X 74.02 = 10.01 = 10 X 1.008 grams
O: 0.2160 X 74.02 = 15.99 = 1 X 16.00 grams
Therefore the mole of ethyl ether contains 4 gram atomic weights
of caxbon, 10 gram atomic weights of hydrogen, and 1 gram atomic
weight of oxygen and the formula is C 4 Hi 0 O.PROBLEMS- One liter of a certain gas under standard conditions
weighs 2.25 grams. Calculate the molecular weight of this
gas. - The molecular weight of a certain volatile substance
is to be determined: 0.435 gram of the substance is placed
in an evacuated vessel and the whole is heated to 136.5°, at
which temperature the substance is entirely converted to gas.
The pressure and volume of the gas are now found to be
380 mm. and 405.6 cc, respectively. Calculate the molecu-
lar weight of the gas. - When a certain liquid substance is vaporized, its vapor
is found to weigh 5.413 times as much as an equal volume of
air under the same conditions. Assuming the average mo-
lecular weight of air to be 28.955, find the molecular weight
of the substance.