If the CaCO3 weighed 983 g and the CaO weighed 551 g, how many grams of CO2 were formed in the reaction?
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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>