What is the percent composition of a carbon-oxygen compound, given that a 95.2 g sample of the compound contains 40.8 g of carbon and 54.4 g of oxygen?

1 Answer
Jul 3, 2017

Answer:

#"%element"="mass of element"/"mass of compound"xx100%#

Explanation:

And thus.........

#"%C"=(40.8*g)/(95.2*g)xx100%=42.9%#.

#"%O"=(54.4*g)/(95.2*g)xx100%=57.1%#.

Why should the individual percentage sum to #100%#? Do they......

We can divide the elemental percentages by the atomic masses if we assume #100*g# of unknown compound, and get the so-called #"empirical formula"#, which is the simplest whole number ratio defining constituent elements in a species.

And thus for #100*g# of this compound..........there are....

#(42.9*g)/(12.011*g*mol^-1)=3.57*mol*C#. And........

#(57.1*g)/(15.999*g*mol^-1)=3.57*mol*O#.

We divide thru by the element with the smallest number of moles, and thus we get an empirical formula of #CO#.........

This could be #CO# or cyclopropanetrione, #C_3O_3#, or #C_4O_4#; these beasts are fiercely reactive, and it would be hard to put them in a bottle.