A 0.2857 mL sample of pure, unknown compound in the liquid state which as a dens
ID: 977735 • Letter: A
Question
A 0.2857 mL sample of pure, unknown compound in the liquid state which as a density of 1.049 g/ml is analyzed. The following results are obtained..
1. Liquid sample is heated to 525.0 K, it vaporizes into an ideal gas with a pressure of 1.001 atm occupying a volume of 214.8 mL.
2. When the liquid sample is combusted in the presence of an excess of pure oxygen gas, 0.4393 grams of co2 and 0.17 grams of h20 are obtained. No other products form in the combustion reaction.
3. When the liquid sample is combusted in a calorimeter (constant 425.0 j/C) containing 126.1 g water (specific heat is 4.184 J/gC), the temperature increases from 22.65 to 26.77
Questions:
1. How many moles of compound are obtained in the sample?
2. What is the molar mass of the sample?
3. What is the empirical formula of the compound?
4. What is the molecular formula of the compound?
5. Write a balanced equation showing phases of products and reactants
6. What is the heat of combustion of this compound (kj/mol)?
Explanation / Answer
1. Volume = 0.2857 ml Denisty = 1.049 g/ml Hence mass = .2857 X 1.049 = 0.3g
Molar volume = RT/P = 0.082 X 525 / 1 = 43.05 L/mol
Molar mass = Molar vol X Density = 43.05 X 1.049 = 45.16 g/mol
Number of moles = 0.3/45.16 = 6.6 X 10-3
2.Molar mass = 45.16 g/mol
3. From the combustion data, we know that the compound contains only C, H and O.
Given : CxHyOz + O2 ------------> CO2 + H2O
(0.01mol) (0.01mol)
Solving we get percentage of C, H and O as 20, 33.2 and 46.8 respectively.
Hence the emperical formula is CH2O2
4. Mass of emperical formula =46 which is same as the molecular mass. Hence molecular formula is also CH2O2
5. 2CH2O2 (l) + O2 (g) -------> 2CO2 (g)+ 2H2O(l)
6. Heat evolved = q= C dT =425 X 4.12 = 1751J
Heat evolved for 1 mol = (1751 X 1) / 6.6 X 10-3 = 263.6 kJ/mol
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