Find the empirical formula of a compound that is 53.7% iron and 46.3% sulfur.What is the empirical mass of the said compound in g/mol?Use:Fe = 56 amuS = 32 amu
Question
Find the empirical formula of a compound that is 53.7% iron and 46.3% sulfur.
What is the empirical mass of the said compound in g/mol?
Use:
- Fe = 56 amu
- S = 32 amu
Solution
Step 1: Assume you have 100g of the compound. This means you have 53.7g of iron (Fe) and 46.3g of sulfur (S).
Step 2: Convert the mass of each element to moles.
For iron: 53.7g Fe * (1 mol Fe / 56g Fe) = 0.96 mol Fe
For sulfur: 46.3g S * (1 mol S / 32g S) = 1.45 mol S
Step 3: Write the ratio of the moles of each element to the smallest number of moles calculated.
For iron: 0.96 mol Fe / 0.96 = 1 mol Fe
For sulfur: 1.45 mol S / 0.96 = 1.51 mol S
Step 4: Since the empirical formula is the simplest whole number ratio of atoms in a compound, we round 1.51 to the nearest whole number, which is 2.
So, the empirical formula of the compound is FeS2.
Step 5: To find the empirical mass of the compound, add the atomic masses of all atoms in the empirical formula.
For FeS2: (1 * 56g/mol Fe) + (2 * 32g/mol S) = 56g/mol + 64g/mol = 120g/mol
So, the empirical mass of the compound is 120g/mol.
Similar Questions
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
A compound is found to contain 23.3% magnesium, 30.7% sulfur, and 46.0% oxygen. What is the empirical formula of this compound?
A compound is found to contain 24.0 g carbon, 4.00 mol hydrogen and 1.204 × 1024 atoms of oxygen. What is the empirical formula of this compound?
What is the empirical formula of a compound that has a molar mass of 42 grams and an empirical formula of CH2?Group of answer choicesCH2C2H4C4H12C3H6
A compound has 40% Carbon 6.6% Hydrogen & 53.3% Oxygen. Calculate the Empirical Formula of the compound.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.