A compound was found to contain copper and chlorine. The mass of the compound was 351 mg. The amount of copper in the sample was found to be 225 mg. What is the empirical formula for the compound?
So if the mass of the compound is 351 mg the mass of chlorine is (351-225) mg = 126 mg
Thus number of moles of Cu =mass/RAM
Thus number of moles of Cl =mass/RAM
divided by smallest number of moles to get the ratio.
I'll let you work it out.
To determine the empirical formula of a compound, we need to know the mass or percentage composition of each element present. Given that the mass of the copper-chlorine compound is 351 mg and the mass of copper is 225 mg, we can calculate the mass of the chlorine in the compound.
Mass of chlorine = Total mass of compound - Mass of copper
Mass of chlorine = 351 mg - 225 mg
Mass of chlorine = 126 mg
Now that we have the masses of both copper (225 mg) and chlorine (126 mg), we can calculate the mole ratio between them. This will give us the empirical formula of the compound.
Step 1: Convert the mass of each element to moles using their respective molar masses.
The molar mass of copper (Cu) is approximately 63.55 g/mol, so:
Moles of copper = Mass of copper / Molar mass of copper
Moles of copper = 225 mg / (63.55 g/mol x 1000 mg/g)
Moles of copper = 0.0035398 mol
The molar mass of chlorine (Cl) is approximately 35.45 g/mol, so:
Moles of chlorine = Mass of chlorine / Molar mass of chlorine
Moles of chlorine = 126 mg / (35.45 g/mol x 1000 mg/g)
Moles of chlorine = 0.0035462 mol
Step 2: Divide the mole ratios of each element by the smallest value to obtain simplified whole-number ratios.
Dividing both moles of copper and chlorine by 0.0035398 mol (smallest value), we get:
Moles of copper = 0.0035398 mol / 0.0035398 mol = 1
Moles of chlorine = 0.0035462 mol / 0.0035398 mol ≈ 1.0018
Since the values are very close to each other, we can approximate them as a 1:1 ratio.
Therefore, the empirical formula for the compound containing copper and chlorine is CuCl.