Question

Question #affe9

Answer 1

The formula is `"C"_ 6"H"_ {12}"O"_ 6`.

Explanation:

You are supposed to convert mass to moles using atomic and molecular weights.

Start with one mole of the carbohydrate. That is given as `180.2"g"`. Then the amount of carbon is

`({40%"C"}/{100%})×({180.2"g"}/"mol")=72.1"g C"`

Since to the nearest tenth, `100-53.3-6.7=40.0`, I assume that the `40%` carbon has one significant zero after the decimal point (`40.0%`). Hence the use of three significant digits.

And we then convert the carbon to moles by dividing by the atomic weight. Most periodic tables give this as `12.01` so:

`(72.1"g C")({1"mol"}/{12.01"g"})=6.00"mol C"`

So one mole of the carbohydrate has six moles of carbon. That means the molecular formula has `"C"_6`.

For oxygen we do the same thing, but with the oxygen numbers:

`({53.3%"O"}/{100%})×({180.2"g"}/"mol")=96.0"g O"`

`(96.0"g O")({1"mol"}/{15.999"g"})=6.00"mol O"`

So the formula will contain six oxygen atoms. Now we have `"C"_ 6"H" _{??}"O" _6`, as formulas for organic compounds are usually written with carbon and hydrogen first.

I will let you do the calculations for hydrogen. You should find the formula will contain `"H"_ {12}`. Thus #"C"_ 6"H"_ {12}"O"_ 6#.