Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When 1.00 g of X is dissolved in water and allowed to react with excess silver nitrate, AgNO3, all the chlorine in X reacts and 1.95 g of solid AgCl is formed. When 1.00 g of X undergoes complete combustion, 0.900 g of CO2 and 0.735 g of H2O are formed. What is the empirical formula X?
I'm not really sure how to approach this question.
To find the empirical formula of compound X, we need to determine the ratio of the atoms present in the compound. The given information about the reaction with silver nitrate (AgNO3) and combustion will help us determine this ratio.
First, let's analyze the reaction with silver nitrate:
1.00 g of compound X reacts completely with silver nitrate to form 1.95 g of silver chloride (AgCl). From this, we can determine the mass of chlorine (Cl) present in 1.00 g of X.
The molar mass of AgCl is the sum of the atomic masses of Ag (silver) and Cl (chlorine), which is 107.87 g/mol.
Using this information, we can calculate the moles of AgCl produced from the given mass:
moles of AgCl = mass of AgCl / molar mass of AgCl
moles of AgCl = 1.95 g / 107.87 g/mol
Next, we need to convert the moles of AgCl to moles of chlorine (Cl) since 1 mole of AgCl contains 1 mole of Cl atoms.
moles of Cl = moles of AgCl
Now, we can calculate the mass of chlorine (Cl) in X:
mass of Cl = moles of Cl x molar mass of Cl
Next, let's analyze the combustion reaction:
1.00 g of X undergoes complete combustion to produce 0.900 g of carbon dioxide (CO2) and 0.735 g of water (H2O). We can use this information to calculate the mass of carbon (C) and hydrogen (H) present in 1.00 g of X.
The molar mass of CO2 is the sum of the atomic masses of C (carbon) and 2 atoms of O (oxygen), which is 44.01 g/mol.
Using this information, we can calculate the moles of CO2 produced from the given mass:
moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 0.900 g / 44.01 g/mol
Similarly, from the given mass of water (H2O), we can calculate the moles of water and, consequently, the moles of hydrogen (H):
moles of H2O = mass of H2O / molar mass of H2O
moles of H2O = 0.735 g / 18.02 g/mol
Next, we need to convert the moles of CO2 to moles of carbon (C) since 1 mole of CO2 contains 1 mole of carbon atoms. We also need to convert the moles of H2O to moles of hydrogen (H) since 1 mole of H2O contains 2 moles of hydrogen atoms.
moles of C = moles of CO2
moles of H = 2 x moles of H2O
Now, we can calculate the mass of carbon (C) and hydrogen (H) in X:
mass of C = moles of C x molar mass of C
mass of H = moles of H x molar mass of H
Finally, calculate the mass of nitrogen (N) in X:
mass of N = mass of X - mass of C - mass of H - mass of Cl
Now that we have determined the masses of carbon (C), hydrogen (H), nitrogen (N), and chlorine (Cl) in 1.00 g of X, we can calculate the empirical formula by finding the ratio of these elements.
To do this, divide each mass by the corresponding molar mass of each element, and then divide each result by the smallest value obtained to get the simplest whole-number ratio. This will give us the subscripts for the empirical formula.
The empirical formula will be expressed as CxHyNzClw, where x, y, z, and w represent the subscripts for carbon, hydrogen, nitrogen, and chlorine, respectively.