How many grams of Sulfur are there in 6.0 grams of Fe2(SO4)3
How many grams of sulfur are there in 6.0 g of Fe2(SO4)3?
To determine the number of grams of sulfur in Fe2(SO4)3, we need to find the molar mass of Fe2(SO4)3 and calculate the percentage of sulfur in the compound.
Here is a step-by-step solution:
Step 1: Find the molar mass of Fe2(SO4)3.
The molar mass of an element or compound is the sum of the atomic masses of each element present.
The atomic masses are:
Iron (Fe) = 55.845 g/mol
Sulfur (S) = 32.06 g/mol
Oxygen (O) = 16.00 g/mol
Fe2(SO4)3 consists of 2 iron atoms, 3 sulfur atoms, and 12 oxygen atoms.
The molar mass of Fe2(SO4)3 can be calculated as follows:
(2 x atomic mass of Fe) + (3 x atomic mass of S) + (12 x atomic mass of O)
= (2 x 55.845 g/mol) + (3 x 32.06 g/mol) + (12 x 16.00 g/mol)
= 111.69 g/mol + 96.18 g/mol + 192.00 g/mol
= 399.87 g/mol
So, the molar mass of Fe2(SO4)3 is 399.87 g/mol.
Step 2: Calculate the percentage of sulfur in Fe2(SO4)3.
The percentage of a particular element in a compound can be calculated using the following formula:
Percentage of element = (Number of atoms of the element x Atomic mass of the element) / Molar mass of the compound x 100
For sulfur in Fe2(SO4)3:
Number of sulfur atoms = 3
Atomic mass of sulfur = 32.06 g/mol
Molar mass of Fe2(SO4)3 = 399.87 g/mol
Percentage of sulfur = (3 x 32.06 g/mol) / 399.87 g/mol x 100
= 96.18 g/mol / 399.87 g/mol x 100
= 24.05%
Step 3: Calculate the grams of sulfur in 6.0 grams of Fe2(SO4)3.
To determine the grams of sulfur in 6.0 grams of Fe2(SO4)3, we can use the mole-to-mole ratio.
Since the percentage of sulfur in Fe2(SO4)3 is 24.05%, we multiply it by 6.0 grams to find the grams of sulfur:
Grams of sulfur = (24.05 / 100) x 6.0 g
= 1.443 g
Therefore, there are approximately 1.443 grams of sulfur in 6.0 grams of Fe2(SO4)3.
To determine the number of grams of sulfur in Fe2(SO4)3, we need to calculate the molar mass of Fe2(SO4)3 and then determine the proportion of sulfur in that compound.
Here's how to do it step by step:
1. Start by finding the molar mass of Fe2(SO4)3. To do this, we need to know the atomic masses of iron (Fe), sulfur (S), and oxygen (O). The atomic masses are as follows:
- Atomic mass of Fe = 55.845 g/mol
- Atomic mass of S = 32.06 g/mol
- Atomic mass of O = 16.00 g/mol
2. Since Fe2(SO4)3 has two iron atoms, three sulfur atoms, and 12 oxygen atoms, we can calculate the molar mass as follows:
Molar mass = (2 × atomic mass of Fe) + (3 × atomic mass of S) + (12 × atomic mass of O)
= (2 × 55.845) + (3 × 32.06) + (12 × 16.00)
= 111.69 + 96.18 + 192.00
= 399.87 g/mol
3. Now that we know the molar mass of Fe2(SO4)3 is 399.87 g/mol, we can calculate the proportion of sulfur in the compound. Since sulfur accounts for three out of the total 399.87 g/mol, we can set up the following proportion:
Sulfur mass / Fe2(SO4)3 mass = (3 × atomic mass of S) / molar mass of Fe2(SO4)3
Plugging in the known values:
Sulfur mass / 6.0 g = (3 × 32.06 g/mol) / 399.87 g/mol
4. Now, we can solve for the mass of sulfur:
Sulfur mass = (6.0 g) × [(3 × 32.06 g/mol) / 399.87 g/mol]
Calculating this:
Sulfur mass ≈ 0.484 g
Therefore, there are approximately 0.484 grams of sulfur in 6.0 grams of Fe2(SO4)3.
6.0/molar mass Fe2(SO4)3 = moles Fe2(sO4)3.
moles Fe2(SO4)3 x (3 atoms S/1 mole Fe2(SO4)3 = moles S.
Then moles S x atomic mass S = grams S.