Question

Question #3e1fd

Answer 1

`46.5 "g C"_6"H"_12"O"_6`

Explanation:

We can convert the given number of `"C"` atoms to moles using Avogadro's number, `N_"A"`:

`9.33 xx 10^23 cancel("atoms C")((1 "mol C")/(6.022 xx 10^23 cancel("atoms C"))) = 1.55 "mol C"`

There are `6` moles of carbon in one mole of glucose, so the number of moles of glucose in this sample is

`1.55 cancel("mol C")((1 "mol C"_6"H"_12"O"_6)/(6cancel( "mol C"))) = 0.258 "mol C"_6"H"_12"O"_6`

Lastly, we'll use the molar mass of glucose (`180.18 "g"/"mol"`) to convert this to grams:

`0.258 cancel("mol C"_6"H"_12"O"_6)((180.18 "g C"_6"H"_12"O"_6)/(1 cancel("mol C"_6"H"_12"O"_6))) = color(red)(46.5 "g C"_6"H"_12"O"_6`