What is the empirical formula of a compound that contains 46.3% lithium and 53.7% oxygen?
Question
What is the empirical formula of a compound that contains 46.3% lithium and 53.7% oxygen?
Solution
Sure, here's how you can find the empirical formula of a compound:
Step 1: Assume you have 100g of the compound. This means you have 46.3g of lithium (Li) and 53.7g of oxygen (O).
Step 2: Convert the mass of each element to moles. The atomic mass of Li is approximately 6.94g/mol and the atomic mass of O is approximately 16.00g/mol.
For Li: 46.3g / 6.94g/mol = 6.67 mol For O: 53.7g / 16.00g/mol = 3.36 mol
Step 3: Write the ratio of the moles. This gives us Li6.67O3.36.
Step 4: Divide each number in the ratio by the smallest number to get the simplest, whole number ratio. In this case, 3.36 is the smallest number.
For Li: 6.67 / 3.36 = 1.99 ≈ 2 For O: 3.36 / 3.36 = 1
Step 5: Write the empirical formula. The empirical formula of the compound is Li2O.
Similar Questions
determine the empirical formula of lithium oxide.
What is the empirical formula of a compound containing 40.0% sulfur and 60.0% oxygen 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?
What is the empirical formula for a compound containing 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen?
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.