Question

Determine the empirical formula of a compound that contains 2.61 g of carbon, 0.65 g of hydrogen, and 1.74 g of oxygen.

Answer 1

\[2.61g C \times \frac{ 1 mol C }{ 12.01g C }=0.217 mol C\]\[.65g H \times \frac{ 1mol H }{ 1.008g H }=0.645mol H \]\[1,74g O \times \frac{ 1 mol O }{ 16 g O = }0.109mol O\]Divide all by the smalles amount\[\frac{ 0.217 }{ 0.109 }=2\] \[\frac{ 0.645 }{ 0.109 }=6\]\[\frac{ 0.109 }{ 0.109 }=1\]\[C _{6}H _{6}O\] should be the answer. Do you understand how I got this?

Answer 2

Empirical formula should be C2H6O2. You want to convert from: grams--> to moles--> to mole ratio. I often form a chart to make things easier. Refer to attachment- Sara