Heres the question:
"In determining the simplest formula of lead sulfide, 2.46 grams 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 to in the simplest forumla calculation? Find the simplest formula of lead sulfide."
Could someone go through steps for me.
Thanks!
mass Pb = 2.46 g.
mass lead sulfide = 3.22 g.
mass S used = 3.22-2.46 = 0.76 g.
mols Pb = 2.46/207.20 = 0.01187
mols S = 0.76/32.066 = 0.02370
ratio: Divide the smallest number by itself thereby assuring you that will be 1.0000.
mols Pb = 0.01187/0.01187 = 1.000
mols S = 0.02370/0.01187 = 1.9967
rounded to whole numbers we have Pb = 1.0 and S = 2.0 so formula is PbS2.
Note: I have carried out the numbers to far more places than is allowed so round down to two (since 0.76 has two places) at the most. Also, PbS is the formula for lead sulfide but these numbers don't reflect that.
Check my work.
Thanks for the fast reply!
Alright, let's break it down step by step and make it fun!
Step 1: Find the mass of sulfur used in the reaction.
To do this, we need to subtract the mass of the product (3.22g) from the total mass of lead and sulfur before the reaction (2.00g + 2.46g = 4.46g).
So, the mass of sulfur used is: 4.46g - 3.22g = 1.24g.
Step 2: Convert the mass of sulfur to moles.
To do this, we need to know the molar mass of sulfur, which is 32.06 g/mol.
So, the number of moles of sulfur used is: 1.24g / 32.06 g/mol ≈ 0.0386 mol.
Step 3: Determine the simplest formula of lead sulfide.
Now that we know the number of moles of sulfur used, we can deduce the simplest formula. The simplest formula is the one with the smallest whole numbers of atoms.
From the balanced chemical equation of the reaction, we know that the ratio of lead to sulfur in lead sulfide is 1:1.
Therefore, the simplest formula of lead sulfide is PbS.
Voila! We have found the simplest formula of lead sulfide (PbS). And that's how chemistry gets sulfated!
Sure! Let's break down the steps for determining the simplest formula of lead sulfide and calculating the mass of sulfur needed for the calculation.
Step 1: Convert the given masses of lead and sulfur to moles.
- The molar mass of lead (Pb) is 207.2 g/mol.
- The molar mass of sulfur (S) is 32.1 g/mol.
Moles of lead (Pb) = mass of lead / molar mass of lead
Moles of lead (Pb) = 2.46 g / 207.2 g/mol
Moles of sulfur (S) = mass of sulfur / molar mass of sulfur
Moles of sulfur (S) = 2.00 g / 32.1 g/mol
Step 2: Determine the mole ratio of lead to sulfur.
- The balanced chemical equation for the reaction is: 2Pb + S -> PbS
- From the equation, we see that 2 moles of lead react with 1 mole of sulfur to form 1 mole of lead sulfide.
Step 3: Calculate the moles of sulfur needed.
Moles of sulfur needed = moles of lead (Pb) x (1 mole of S / 2 moles of Pb)
Moles of sulfur needed = (2.46 g / 207.2 g/mol) x (1 mol S / 2 mol Pb)
Step 4: Convert the moles of sulfur needed to grams.
Mass of sulfur needed = moles of sulfur needed x molar mass of sulfur
Mass of sulfur needed = [(2.46 g / 207.2 g/mol) x (1 mol S / 2 mol Pb)] x 32.1 g/mol
Step 5: Calculate the mass of sulfur needed for the simplest formula.
Mass of sulfur needed = (2.46 g x 32.1 g/mol) / (207.2 g/mol x 2)
Mass of sulfur needed = 0.606 g
Therefore, the mass of sulfur that should be used for the simplest formula calculation is 0.606 grams.
To determine the simplest formula of lead sulfide, we need to analyze the given masses of lead (Pb), sulfur (S), and the resulting product in order to find the ratio between atoms.
Let's break down the problem into steps:
Step 1: Find the mass of sulfur in the product.
Given:
- Mass of lead (Pb) used = 2.46 grams
- Mass of sulfur (S) used = 2.00 grams
- Mass of the resulting product = 3.22 grams
To find the mass of sulfur in the product, we subtract the mass of lead from the mass of the product:
Mass of sulfur = Mass of product - Mass of lead
Mass of sulfur = 3.22 grams - 2.46 grams
Mass of sulfur = 0.76 grams
Therefore, the mass of sulfur in the product is 0.76 grams.
Step 2: Find the ratio of lead to sulfur atoms.
We need to determine the ratio between the number of lead atoms and sulfur atoms in the lead sulfide compound. To do this, we can use the atomic masses of lead and sulfur.
The atomic mass of lead (Pb) is 207.2 g/mol.
The atomic mass of sulfur (S) is 32.1 g/mol.
Now, we calculate the number of moles of lead and sulfur atoms used:
Number of moles = Mass in grams / Molar mass
For lead:
Moles of lead = Mass of lead / Molar mass of lead
Moles of lead = 2.46 grams / 207.2 g/mol
Moles of lead ≈ 0.0119 mol
For sulfur:
Moles of sulfur = Mass of sulfur / Molar mass of sulfur
Moles of sulfur = 0.76 grams / 32.1 g/mol
Moles of sulfur ≈ 0.0237 mol
Step 3: Simplify the ratio by dividing by the smallest number of moles.
Now, we need to find the simplest whole-number ratio of lead to sulfur atoms. To do this, we divide both moles of lead and sulfur by the smaller of the two numbers.
Dividing both moles of lead and sulfur by 0.0119 mol (the smaller number):
Moles of lead (simplified) = 0.0119 mol / 0.0119 mol = 1
Moles of sulfur (simplified) = 0.0237 mol / 0.0119 mol ≈ 2
Therefore, the simplified ratio of lead to sulfur atoms is 1:2.
Step 4: Determine the simplest formula of lead sulfide.
The simplest formula of lead sulfide can be written using the ratio of atoms found in the previous step. In this case, it is one lead atom combined with two sulfur atoms.
The simplest formula of lead sulfide is PbS.
So, to answer your second question about the mass of sulfur required for the simplest formula, the mass of sulfur used was 2.00 grams.
So, to determine the simplest formula of lead sulfide, you needed to follow these steps:
1. Find the mass of sulfur in the product.
2. Find the ratio of lead to sulfur atoms.
3. Simplify the ratio by dividing by the smallest number of moles.
4. Determine the simplest formula of lead sulfide.