How many atoms are present in 21.59 g of #C_6H_12O_6#?

1 Answer
anor277
Feb 10, 2016

We need to (i) work out the molar quantity of glucose; and (ii) convert this molar quantity into an actual number of the constituent elements.

Explanation:

#"Moles of glucose"# #=# #(21.59*cancelg)/(180.15*cancelg*mol^-1)# #=# #0.120# #mol#.

Now in one MOLECULE of glucose there are #6# #xx# #C#, and #6# #xx# #O#, and #12# #xx# #H# #=# #24# #"atoms"# in total.

We have a #0.120# #mol# quantity, i.e. #0.120# #mol# #xx# #N_A#, (where #N_A# #=# #"Avogadro's Number"# #=# #6.022# #xx# #10^23# #mol^-1#), is the number of MOLECULES.

So #"number of atoms"# #=# #24xx0.120*molxx10^23# #mol^-1# #=# #??#

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