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 of X?

To determine the empirical formula of compound X, we need to find the ratio of atoms present in the compound based on the given information.

1. Finding the moles of chlorine:
From the reaction with silver nitrate, we know that 1.00 g of X reacts to form 1.95 g of AgCl. We can convert the grams to moles using the molar mass of AgCl.
Molar mass of AgCl = 107.87 g/mol (atomic masses of Ag = 107.87 g/mol and Cl = 35.45 g/mol)
Moles of AgCl = 1.95 g / 107.87 g/mol = 0.018 mol AgCl

Since AgCl is formed from the chlorine in compound X, the moles of chlorine can be equated to the moles of AgCl:
Moles of Cl = 0.018 mol Cl

2. Finding the moles of carbon dioxide:
From the combustion reaction, we know that 1.00 g of X produces 0.900 g of CO2. We can convert the grams to moles using the molar mass of CO2.
Molar mass of CO2 = 44.01 g/mol (atomic mass of C = 12.01 g/mol and O = 16.00 g/mol)
Moles of CO2 = 0.900 g / 44.01 g/mol = 0.020 mol CO2

3. Finding the moles of water:
From the combustion reaction, we know that 1.00 g of X produces 0.735 g of H2O. We can convert the grams to moles using the molar mass of H2O.
Molar mass of H2O = 18.02 g/mol (atomic mass of H = 1.01 g/mol and O = 16.00 g/mol)
Moles of H2O = 0.735 g / 18.02 g/mol = 0.041 mol H2O

4. Finding the empirical formula:
Now, we can determine the ratio of atoms based on the moles obtained.
Moles of C: 1.00 g / molar mass of C
Moles of H: 0.735 g / molar mass of H
Moles of N: 0 (since the compound does not contain nitrogen)
Moles of Cl: 0.018 mol
To find the simplest whole number ratio, divide each mole value by the smallest number of moles, which in this case is 0.018 mol.
The ratio becomes:
Moles of C: (1.00 g / molar mass of C) / 0.018 mol
Moles of H: (0.735 g / molar mass of H) / 0.018 mol
Moles of N: 0 (since the compound does not contain nitrogen)
Moles of Cl: 1.00 mol / 0.018 mol

After dividing and simplifying, you will have the mole ratio. Multiply by a whole number to get the simplest ratio:

Let's assume, for example, the mole ratio for carbon is 1:
Moles of C: (1.00 g / molar mass of C) / 0.018 mol = 1

Then, find the ratios for hydrogen and chlorine using the same approach. Once you have the simplified mole ratio for each element, you can write the empirical formula for compound X.

Keep in mind that this is just a hypothetical example for explaining the steps. You will need to perform the actual calculations to find the empirical formula of compound X based on the given data.

you are given two reactions, in which you can determine the grams of C, H, Cl, and then the N is 1gram minus those.

figure the mole ratio from that.