Analysis of a compound gave 39.50% C, 2.21% H, and 58.30% Cl. When 0.855 g of this solid was dissolved in 7.50 g of naphthalene, the solution had a freezing point of 78.0°C. The pure solvent freezes at 80.0°C; its molal freezing point constant is 6.8°C/ m . a What is the molecular formula of the compound? b What is its molecular weight to the nearest 0.1 g?
Analysis of a compound gave 39.50% C, 2.21% H, and 58.30% Cl. When 0.855 g of this solid was dissolved in 7.50 g of naphthalene, the solution had a freezing point of 78.0°C. The pure solvent freezes at 80.0°C; its molal freezing point constant is 6.8°C/ m . a What is the molecular formula of the compound? b What is its molecular weight to the nearest 0.1 g?
Analysis of a compound gave 39.50% C, 2.21% H, and 58.30% Cl. When 0.855 g of this solid was dissolved in 7.50 g of naphthalene, the solution had a freezing point of 78.0°C. The pure solvent freezes at 80.0°C; its molal freezing point constant is 6.8°C/m.
a What is the molecular formula of the compound?
b What is its molecular weight to the nearest 0.1 g?
(a)
Expert Solution
Interpretation Introduction
Interpretation:
A compound contains 39.50% Carbon , 2.21% Hydrogen and 58.30% Chlorine. 0.855 g of this substance is dissolved in 7.50 g of naphthalene and the solution has freezing point of 78°C. Molecular formula of the compound has to be determined.
Concept Introduction:
Depression in freezing point is the phenomenon of lowering of freezing point of a substance, which is considered as solvent, by adding a non-volatile solute. It is expresses as –
ΔTf=Kf×m
Where,
ΔTf = depression in freezing pointKf = cryoscopic constantm = molality of the solute
Number of moles of a substance is related to molar mass of the substance as,
no.of moles= massmolar mass
Answer to Problem 12.108QP
Molecular formula of the compound is determined as C12H8Cl6.
Explanation of Solution
Given that freezing point of the solution is 78°C. freezing point of pure naphthalene is 80°C. Kf for naphthalene is 6.8°C/m. rewriting the depression in freezing point equation, determine the molality of the solution.
m=ΔTfKf=[80°C−78°C)]6.8°C/m= 0.294 m
Number of moles of the compound considering 1 kg of solvent,
moles=0.294 mol1 kg of solvent×0.00750 kg= 2.21×10−3 mol
Molar mass of the compound is calculated as,
molar mass= 0.855 g2.21×10−3 mol= 387 g/mol
Number of moles of elements in 100 g of the compound are calculated as follows -
moles of C= 39.50 g C ×1 mol12.01 g C= 3.2889 molmoles of H= 2.21 g H ×1 mol1.008 g H= 2.192 molmoles of Cl= 58.30 g Cl ×1 mol16.00 g Cl= 1.6466 mol
From the above calculation we could deduce that mole ratio of the elements Carbon, Hydrogen and Chlorine is 2:1.33:1. Multiply this by three to ease the calculation. Thus the mole ratio of the elements is 6:4:3. Hence the empirical formula of the compound could be C6H4Cl3. Unit formula mass of this compound is approximately equal to 182.5 amu. Previously we have calculated the molar mass of the compound as 387 g/mol. this is approximately equal to twice the value of 74 amu. hence the molecular formula of the compound could be (C6H4Cl3)2 which is C12H8Cl6.
(b)
Expert Solution
Interpretation Introduction
Interpretation:
A compound contains 39.50% Carbon , 2.21% Hydrogen and 58.30% Chlorine. 0.855 g of this substance is dissolved in 7.50 g of naphthalene and the solution has freezing point of 78°C. Molar mass of the compound nearest to the tenth of a gram has to be determined.
Concept Introduction:
Depression in freezing point is the phenomenon of lowering of freezing point of a substance, which is considered as solvent, by adding a non-volatile solute. It is expresses as –
ΔTf=Kf×m
Where,
ΔTf = depression in freezing pointKf = cryoscopic constantm = molality of the solute
Number of moles of a substance is related to molar mass of the substance as,
no.of moles= massmolar mass
Answer to Problem 12.108QP
Molar mass of the compound nearest to tenth of a gram is calculated to be 364.9g/mol.
Explanation of Solution
Molar mass of the compound nearest to tenth of a gram is calculated as,
12C(12.01)+8 H (1.008)+6 Cl (16.00)= 364.884 g/mol= 364.9 g/mol
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.