In determing the simplest formula of lead sulfide, 2.46 grams of of lead are placed in a crucible with 2.00 grams of sulfur. when the reaction is complete, the product has a mass of 3.22 grams. What mass of sulfur should be used in the simplest formula calculation? Find the simplest formula of lead sulfide
To find the simplest formula of lead sulfide, we need to determine the ratio between the moles of lead and sulfur in the compound.
First, let's calculate the moles of lead and sulfur in the reaction using their masses.
The molar mass of lead (Pb) is 207.2 g/mol, and the molar mass of sulfur (S) is 32.1 g/mol.
The moles of lead (Pb) can be calculated as:
moles of Pb = mass of Pb / molar mass of Pb = 2.46 g / 207.2 g/mol
Similarly, the moles of sulfur (S) can be calculated as:
moles of S = mass of S / molar mass of S = 2.00 g / 32.1 g/mol
Now, we need to determine the ratio between the moles of lead and sulfur in the compound. Let's call this ratio x.
x = moles of Pb / moles of S
Now, we'll use the given information that when the reaction is complete, the product has a mass of 3.22 grams to find the mass of sulfur that should be used in the simplest formula calculation.
To do that, we'll assume that the mass of the lead sulfide compound is the sum of the masses of lead and sulfur used in the reaction:
mass of lead sulfide = mass of Pb + mass of S
Since the mass of lead sulfide is 3.22 grams, we can rearrange the equation as follows:
mass of S = mass of lead sulfide - mass of Pb
Substituting the given values, we get:
mass of S = 3.22 g - 2.46 g
Now, we can calculate the mass of sulfur that should be used in the simplest formula calculation.
Finally, to find the simplest formula of lead sulfide, we need to divide the moles of lead (Pb) and sulfur (S) by their greatest common factor to simplify the ratio.
Let's say the ratio x is simplified as x:y, where x and y represent the moles of lead (Pb) and sulfur (S) respectively. The simplest formula of lead sulfide would then be PbₓSᵧ, where ₓ and ᵧ represent the respective subscripts.