Calculate the number of C, H, and O atoms in 4.50 g of glucose (C6H12O6), a sugar.
okay i did it and i got just the C right and not the H and O
help?
ok.
4.50 g x (1 mol/180.156 = 0.02498 mols C6H12O6.
atoms C = 6*0.02498*6.02 x 10^23 = 9.022 x 10^22 which rounds to 9.02 x 10^22.
atoms H = 12*0.02498*6.02 x 10^23 = 1.8044 x 10^23 which rounds to 1.80 x 10^23.
atoms O = same as atoms C.
You method is ok. I suspect, without knowing, that our answers are just a little apart because you probably rounded mols to 0.0250 from your first calculation, then used that for the remaining work. However, I left the answer in my calculator and multiplied by 6, then by 6.02 x 10^23, read the final answer and rounded that. If you are typing your answer in to a computer base to check your answers, try the ones I calculated and see if that is the problem. The computer may not accept 9.03 instead of 9.02 and 1.81 instead of 1.80.
Show what you did (include the C please) and I will try to find the error.
okay first i found C6H12O6=180.156g/mole
then i changed 4.50g to moles by mult. by the 180.156g/mole and got .0250 mole
then for the C i did .0250 x 6
for H .0250 x 12
and for O .0250 x 6
then i multiplied all of them by 6.022e23atoms/mole
i got 9.03e22 for C and its right
1.81e23 for H and its wrong
9.033e22 and its wrong...
molar mass C6H12O6 is ok.
The first error is to convert grams to mols you DIVIDE.
mols = grams/molar mass.
The remainder of the procedure looks ok at first glance. Let me know if you don't get the right answer.