Question

40.532 g of an unknown compound X is dissolved in 351.11 mL. What is the molarity of X? Here is some information to determine the identity of X. When 1.00g of X is dissolved in water and allowed to react wtih AgNO3, all the chlorine in X precipitates and 1.95g of AgCl is collected. When 1.00g of X undergoes complete combustion, 0.900 g of CO2 and 0.735 g of H2O were collected. The empirical and molecular formula are identical. Compound X only contains carbon, hydrogen, nitrogen, and chlorine.

Answer 1

Firsttttt, let's use the data from the reaction with AgNO3 to determine the amount of chlorine in X:

1.95 g AgCl = (1.95 g AgCl) / (143.32 g/mol AgCl) = 0.0136 mol AgCl
0.0136 mol AgCl = 0.0136 mol Cl

Since we know the empirical formula and the molar mass of X is 40.532 g/mol, we can use the empirical formula to calculate the molecular formula. Let's assume that the empirical formula is CₐHᵦNᵧClᵹ:

From the combustion reaction, we can calculate the moles of carbon and hydrogen in X:

0.900 g CO2 = (0.900 g CO2) / (44.01 g/mol CO2) = 0.0205 mol CO2
0.735 g H2O = (0.735 g H2O) / (18.02 g/mol H2O) = 0.0408 mol H2O
0.0205 mol CO2 = 0.0205 mol C
0.0408 mol H2O = 0.0408 mol H
Using the law of conservation of mass, we can find the moles of nitrogen and chlorine in X:

0.0136 mol Cl = 0.0136 mol Clᵹ
0.0205 mol C + 0.0408 mol H + 0.0136 mol Clᵹ = 0.0754 mol total
0.0754 mol total = 0.0754 mol X

Now we can calculate the mole ratio of each element in X:

C: 0.0205 mol / 0.0754 mol = 0.272
H: 0.0408 mol / 0.0754 mol = 0.542
N: 0 / 0.0754 mol = 0
Cl: 0.0136 mol / 0.0754 mol = 0.181

The empirical formula is therefore C₀.₂₇H₀.₅₄N₀.₀Cl₀.₁₈.

The molar mass of the empirical formula is:
0.272 g/mol C + 0.542 g/mol H + 0.018 g/mol Cl = 0.832 g/mol
The molecular formula mass is 40