when a 328 mg sample of gold and chlorine was treated with silver nitrate, Cl was converted to AgCl producing 464 mg of silver chloride. what is the empirical formula?
AuCl + Ag(NO3) -> AgCl + Au(NO3)
First, you can NOT write an equation because you don't know the formula of gold chloride not the formula of gold nitrate..
The Cl in gold chloride was converted to AgCl. So how much Cl must have been there with the gold? That's 464 mg x (atomic mass Cl/molar mass AgCl) = approx 115 mg Cl.
That leaves 328-115 = approx 213 mg for Au.
mols Au = grams/atomic mass Au = ?
mols Cl = grams/atomic mass Cl = ?
Now find the ratio of Au to Cl with the lowest number being 1.00. The easy way to do that is to divide the smaller number by itself which makes sure that is 1.00. Then divide the other number by the same small number. Round to a whole number. That gives you the x and y to make the empirical formula AuxCly. I get AuCl3.
To determine the empirical formula, we need to find the mole ratio between gold (Au) and chlorine (Cl) in the compound.
First, let's calculate the number of moles of silver chloride (AgCl) formed using the given mass of 464 mg.
The molar mass of AgCl is equal to the sum of the atomic masses of silver (Ag) and chlorine (Cl), which is 107.87 g/mol + 35.45 g/mol = 143.32 g/mol.
To convert the mass of AgCl from mg to grams, divide by 1000:
464 mg = 464/1000 = 0.464 g
To find the number of moles of AgCl, divide the mass by the molar mass:
0.464 g / 143.32 g/mol = 0.003238 mol
Now, we know that for every mole of AgCl, one mole of chlorine is present. Therefore, the number of moles of chlorine (Cl) in the sample is also 0.003238 mol.
Next, let's find the number of moles of gold (Au) in the sample. We can subtract the moles of chlorine from the total moles of the compound.
The total moles of the compound can be calculated using the molar mass of AuCl, which is the sum of the atomic mass of gold (Au) and chlorine (Cl), which is 196.97 g/mol + 35.45 g/mol = 232.42 g/mol.
To find the moles of Au:
Total moles - moles of Cl = 0.003238 mol - 0.003238 mol = 0 mol
Based on the calculation, there are 0 moles of gold (Au) present. This suggests that gold (Au) did not react with silver nitrate and chlorine (Cl) in this compound. Therefore, the empirical formula would only be AgCl.