Concept explainers
Interpretation: From the given data, empirical formula and molecular formula of the compound should be determined.
Concept introduction:
- According to Thomas Graham the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.
The relative rate of effusion of two gases at the same temperature and pressure are inverse ratio of the square root of the masses of the gases particles. That is,
Rate of effusion for gas 1Rate of effusion for gas 2 =√M2√M1orRate1Rate2 = (M2M1)1/2
M1 and M2 are the molar masses of gas1and gas2
This equation is known as Graham’s law of effusion.
- Empirical formula can be determined from the mass per cent as given below
-
- 1. Convert the mass per cent to gram
- 2. Determine the number of moles of each elements
- 3. Divide the mole value obtained by smallest mole value
- 4. Write empirical formula by mentioning the numbers after writing the
symbols of respective elements
- Molecular formula can be write by the following steps
-
- 1. Determine the empirical formula mass
- 2. Divide the molar mass by empirical formula mass
Molar massEmpirical formula=n
- 3. Multiply the empirical formula by n
Answer to Problem 145AE
Answer
Empirical formula and molecular formula of the compound=C2H3N
Explanation of Solution
Explanation
- To find: the empirical formula of the compound
Convert the mass percent to gram
58.51% Carbon= 58.51 g C7.37 % Hydrogen=7.37 g H34.12% Nitrogen =34.12g N
First step to determine the empirical formula is to convert the mass percent to gram.
Here the given gasses are carbon, hydrogen, nitrogen
Percentage composition of carbon and hydrogen are, 58.51% and 7.37% respectively.
Out of 100 g of compound, percent composition of Nitrogen is 34.12%
- To determine: the number of moles of each element
Number of moles of carbon=4.872 molNumber of moles of hydrogen= 7.31 molNumber of moles of nitrogen=2.435 mol
Number of moles from its given mass is,Number of moles = Given mass in gramMolecular massMasses of carbon,nitrgen, and hydrgrogen are given above.Molecular masses of Carbon = 12.01gNitogen = 14.01gHydrogen =1.008gNumber of moles of carbon =58.51 g12.01 g = 4.872 molNumber of moles of hydrogen =7.37g1.008g = 7.31 molNumber of moles of nitrogen= 34.12g14.01g = 2.435 mol
- To divide: the mole value obtained by smallest mole value
In the case of carbon,Ratio of mole value to smallest mole value= 2.001In the case of hydrogen,Ratio of mole value to smallest mole value= 3.00In the case of nitrogen,Ratio of mole value to smallest mole value = 1.00
From the mole values, smallest mole value is 2.435
Number of moles of carbon=4.872 molNumber of moles of hydrogen= 7.31 molNumber of moles of nitrogen=2.435 molIn the case of carbon,Ratio of mole value to smallest mole value= 4.8722.435= 2.001In the case of hydrogen,Ratio of mole value to smallest mole value = 7.312.435=3.00In the case of nitrogen,Ratio of mole value to smallest mole value = 2.4352.435=1.00
- To write: the empirical formula by mentioning the numbers after writing the symbols of respective elements
The empirical formula is C2H3N
Empirical formula can be determined by dividing the mole value of each gas to smallest mole value gas.
Ratio of mole value to smallest mole value of each gases isC : H : N= 2 : 3 :1
So, the empirical formula is C2H3N
- To find: molecular formula of the given compound
Determine the molar mass of the given compound.
Molar mass of the given compound is = 41.0g/mol
According to Thomas Graham,Rate of effusion for gas 1Rate of effusion for gas 2 =√M2√M1orRate1Rate2 = (M2M1)1/2M1 and M2 are the molar masses of gas1and gas2Let gas1= HeliumM1 = 4.003Helium effuses through a porus 3.20 times as fast as the compound doesSo,3.20 = (M24.003)1/2M2=(3.20)2×4.003 = 10.24 × 4.003 = 41.0 g/mol
- To determine: the empirical formula mass
Empirical formula mass = 41.0g/mol
Empirical formula is C2H3N.The empirical formula mass of C2H3N ≈ 2×12 + 3×1 + 1×14 = 41.0 g/mol
- To divide: the molar mass by empirical formula
n = Molar massEmpirical formula = 1
n =Molar massEmpirical formulaMolar mass of the given compound is= 41.0g/molEmpirical formula mass= 41.0g/moln = 41.0g/mol41.0g/mol =1
- To multiply: the empirical formula by n
By multiplying the empirical formula by n,
molecular formula of the compound=C2H3N
By multiplying the empirical formula by n molecular formula of the compound should obtain.
n = 1 and empirical of the compound=C2H3NC2H3N × 1 = C2H3NSo,the molecular formula of the compound is C2H3N
Conclusion
Empirical formula of the compound is determined from the mass per cent as given below
1. Converted the mass per cent to gram
2. Determined the number of moles of each element
3. Divided the mole value obtained by smallest mole value
4. Wrote empirical formula by mentioning the numbers after writing the symbols of respective elements
Molecular formula of the given compound is determined by the following steps
1. Determined the empirical formula mass
2. Determined the molar mass from rate effusion
3. Divided the molar mass by empirical formula mass
Molar massEmpirical formula=n
- 4. Multiplied the empirical formula by n
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Chapter 5 Solutions
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