Question #3e1fd

1 Answer
May 24, 2017

Answer:

#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#