Three volatile compounds X,Y, and Z each contain elements Q. The precent by weight of element Q in each compound was determined. Some of the data obtained are given below
Percent by weight Molecular
Compound of Element Q Weight
X 64.8% ?
Y 73.0% 104.
Z 59.3% 64.0
(a) the vapor density of compound X at 27 degress C and 750.mm Hg was determined to be 3.53 grams per litre. Calculate the molecular weight of compound X.
(b) Determine the mass of the element Q contained in 1.00 mole of each of the three compounds.
(c) Calculate the most probable value of the atomic weight of element Q
(d) Compound Z contains carbon, hydrogen, and element Q. When 1.00 gram of compound Z is oxidized and all of the carbon and hydrogen are converted to oxidized, 1.37 grams of CO2 and 0.281 gram of water are produced. determine the most prbable molecular formula of compound Z
Hey, this is what I found on another website, actually. Part A seems to be fine, but I can't fathom how they got 64.2 as 73 percent of 104 in Part B. Part C is pretty confusing to me as well; hope you can make something of it:
a) For x; MM = dRT/P = 88
b) m = for x = 88. x 0.648 = 57 g
m = for y = 104 x 0.730 = 64.2g
m = for z = 64 x 0.593 = 37.95
c) % = M x 100/aw
M = (aw x%)/100
x = 37.95
y = 75.9
z = 37.95
Atomic weight is probably 37.95 ~= 38
this is stupid
a.) THE MOLAR MASS IS 88
c.) The mass of the element is 19
d.) C2H2F2
i got 75.92 as 73% of 104. i have the same values for x and z though. i found Q to have an atomic mass of 19.
I know right.
how the heck did this person get .730*104 to equal that? it should definitely be 75.92. also his/her logic for c makes no sense...
This has a few incorrect calculations. Also incorrect copying. Please do not use these answers.
what about part d
In order to understand how they obtained the values for Part B and Part C, let's break down each step:
a) For calculating the molecular weight (MM) of compound X, they used the formula MM = dRT/P, where d is the vapor density, R is the ideal gas constant, T is the temperature in Kelvin, and P is the pressure. They provided the values for d (3.53 grams per liter), T (27 degrees Celsius, which is 300 Kelvin), and assumed the pressure to be 750 mm Hg. They plugged these values into the formula to find the molecular weight of compound X as 88 grams per mole.
b) To determine the mass of element Q contained in 1.00 mole of each compound, they multiplied the molecular weight of each compound by the percentage of element Q in that compound.
For compound X: MM_X = 88 g/mol * 0.648 = 57 g/mol
For compound Y: MM_Y = 104 g/mol * 0.730 = 75.92 g/mol (rounded to 64.2 g/mol in the given information, which seems to be a typo)
For compound Z: MM_Z = 64.0 g/mol * 0.593 = 37.95 g/mol (rounded to 37.95 in the given information)
So, the mass of element Q in 1.00 mole of each compound is approximately:
For compound X: 57 grams
For compound Y: 64.2 grams (assuming the typo was corrected)
For compound Z: 37.95 grams
c) To calculate the most probable value of the atomic weight of element Q, they used the formula % = M x 100/aw, where % is the percentage by weight of element Q in the compound, M is the mass of element Q in grams, and aw is the atomic weight of element Q. They rearranged the formula to solve for aw, which gives aw = (M x 100)/%. They substituted the values to calculate the most probable atomic weight of element Q for each compound.
For compound X: aw_X = (57 g x 100)/64.8% = 88 approx
For compound Y: aw_Y = (64.2 g x 100)/73.0% = 88.11 approx
For compound Z: aw_Z = (37.95 g x 100)/59.3% = 64 approx
Based on these calculations, they concluded that the most probable atomic weight of element Q is approximately 88 for compounds X and Y, and 64 for compound Z.
d) Unfortunately, the explanation for Part D is not provided in the given information. To determine the most probable molecular formula of compound Z, more information about the elements present in the compound is needed.