Question

what is the empirical formula for a compound containing 37.5% carbon, 12.6% hydrogen, and 49.9% oxygen? c₄ho₅ c₂h₈o₂ c₂h₁₂o₃ ch₄o

Explanation:

Step1: Assume 100g of the compound

So, we have 37.5g of C, 12.6g of H, and 49.9g of O.

Step2: Calculate the moles of each element

• Moles of C: $n_{C}=\frac{37.5g}{12.01g/mol}\approx3.12mol$
• Moles of H: $n_{H}=\frac{12.6g}{1.008g/mol}\approx12.5mol$
• Moles of O: $n_{O}=\frac{49.9g}{16.00g/mol}\approx3.12mol$

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (3.12mol)

• For C: $\frac{3.12mol}{3.12mol} = 1$
• For H: $\frac{12.5mol}{3.12mol}\approx4$
• For O: $\frac{3.12mol}{3.12mol}=1$

Answer:

$CH_{4}O$