Question

Question #ed77e

Answer 1

`C_1H_2O_1`

Explanation:

Let's try to work this through.

`color(blue)("1) Assume 100 g sample and multiply by percentages to find mass of each element")`

`C = 39.99g`
`H = 6.79g`
`O = 53.28g`

`color(blue)("2)Take mass of each element and convert to moles using molar mass from periodic table")`

`"Carbon"->(39.99g)/(1)*(1 "mol")/(12g) = 3.33" mol C"`

`"Hydrogen"->(6.79g)/(1)*(1 "mol")/(1.00g) = 6.79" mol H"`

`"Oxygen"->(53.28g)/(1)*(1 "mol")/(16g) = 3.33" mol O"`

`color(blue)("3)Divide the moles by the smallest mole number of the three values")`

`"Carbon"-> (3.33)/(3.33) = 1`
`"Hydrogen"->(6.79)/(3.33) = 2.03 approx 2`
`"Oxygen"->(3.33)/(3.33) = 1`

`color(blue)("4)Write out the element with their respected whole number mole ratios")`

`C_1H_2O_1`

Answer:`C_1H_2O_1`