Question

What is the empirical formula of a compound that contains 50.0% carbon, 6.7% hydrogen, and 43.3% oxygen by mass?

CH3O
C3H5O2
C2H2O5
C3H205
C1H3O5

Answer 1

In the layman's terms, air?
but from the list C3H2O5 looks right to me

Answer 2

`C_3H_5O_2` the empirical formula of a compound.

Step-by-step explanation:

Suppose in 100 g of compound contains 50.0% carbon, 6.7% hydrogen, and 43.3% oxygen;

Then mass of carbon =
`100\%* (50.0)/(100)=50.0 g`

Moles of carbon =
`(50.0 g)/(12 g/mol)=4.16 mol`

Then mass of hydrogen=
`100\%* (6.7)/(100)=6.7 g g`

Moles of hydrogen=
`(6.7 g)/(1 g/mol)=6.70 mol`

Then mass of oxygen =
`100\%* (43.3)/(100)=43.3 g`

Moles of oxygen =
`(43.3 g)/(16 g/mol)=2.70 mol`

To determine the empirical formula compound dived the smallest value of moles from each moles of element.

Carbon =
`(4.1666 mol)/(2.7062 mol)=1.5`

Hydrogen=
`(6.70 mol)/(2.7062 mol)=2.5`

Oxygen=
`(2.7062 mol)/(2.7062 mol)=1`

The empirical formula ;
`C_(1.5)H_(2.5)O_(1)=C_3H_5O_2`

`C_3H_5O_2` the empirical formula of a compound.