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?
7 answers
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...
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.
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.
then how would i get one right and not the others??
i checked my work and i did divide i just typed that wrong
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.
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.
okay thanks that worked :)
2 answers