Classify the following compounds as binary ionic, ternary ionic, or molecular.
1 answer:
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Answer:
The molecular formula =
Explanation:
% of C = 11.79
Molar mass of C = 12.0107 g/mol
<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>
% of Cl = 69.57
Molar mass of Cl = 35.453 g/mol
<u>% moles of Cl = 69.57 / 35.453 = 1.9623</u>
Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,
% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64
Molar mass of F = 18.998 g/mol
<u>% moles of F = 18.64 / 18.998 = 0.9812</u>
Taking the simplest ratio for C, Cl and F as:
0.9816 : 1.9623 : 0.9812
= 1 : 2 : 1
The empirical formula is =
Also, Given that:
Pressure = 21.3 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 21.3 / 760 = 0.02803 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
Volume = 458 mL = 0.458 L (1 mL = 0.001 L)
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 0.00052445 moles
Given that :
Amount = 0.107 g
Molar mass = ?
The formula for the calculation of moles is shown below:
Thus,
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol
Molar mass = 204.0233 g/mol
So,
Molecular mass = n × Empirical mass
204.0233 = n × 101.9147
⇒ n = 2
<u>The molecular formula = </u>
The mass of ammonium chloride that must be added is : ( A ) 4.7 g
<u>Given data :</u>
Volume of water ( V ) = 250 mL = 0.25 L
pH of solution = 4.85
Kb = 1.8 * 10⁻⁵
Kw = 10⁻¹⁴
Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :
NH₄CI + H₂O ⇄ NH₃ + H₃O⁺
where conc of H₃O⁺
[ H₃O⁺ ] = and Ka = Kw / Kb
∴ Ka = 5.56 * 10⁻¹⁰
Next step : Determine the concentration of H₃O⁺ in the solution
pH = - log [ H₃O⁺ ] = 4.85
∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵
Next step : Determine the concentration of NH₄CI in the solution
C = [ H₃O⁺ ]² / Ka
= ( 1.14125 * 10⁻⁵ )² / 5.56 * 10⁻¹⁰
= 0.359 mol / L
Determine the number of moles of NH₄CI in the solution
n = C . V
= 0.359 mol / L * 0.25 L = 0.08979 mole
Final step : determine the mass of ammonium chloride that must be added to 250 mL
mass = n * molar mass
= 0.08979 * 53.5 g/mol
= 4.80 g ≈ 4.7 grams
Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g
Learn more about ammonium chloride : brainly.com/question/13050932