Question is: If you gain your iron from an iron citrate (Fe(C6H7O7)2) supplement, how many grams of iron citrate do you need to take to gain 8.88 mg of iron?

I tried setting this problem up, but i don't think i did it right. can someone please tell me if i did it right or not? Thanks!!

Here's what i got so far:

(8.88 mg Fe)(1g Fe/1000mg Fe)(1 mol Fe/55.85g Fe)(2 mol C6H7O7/1 mol Fe)(191g C6H7O7/1 mol C6H7O7)= .0607 g citrate

check you units...

mg Fe divides out
mole Fe divides out
mole C6H7O7 divdes out.
you are left with g C6H7O7.

Has to be wrong.

Try this...
(8.88 mg Fe)(1g Fe/1000mg Fe)(1 mol Fe/55.85g Fe)(2 mol C6H7O7/1 mol Fe)(55.85/(55.85 +2*191 )) =

To solve this problem, you can set up a dimensional analysis by multiplying conversion factors to cancel out the units you don't want and keep the units you do want.

To find out how many grams of iron citrate you need to take to gain 8.88 mg of iron, you can use the molar mass of iron citrate to convert from moles of iron to grams of iron citrate.

First, convert 8.88 mg of iron to grams:
8.88 mg Fe x (1 g Fe / 1000 mg Fe) = 0.00888 g Fe

Next, convert grams of iron to moles of iron using the molar mass of iron (55.85 g/mol):
0.00888 g Fe x (1 mol Fe / 55.85 g Fe) = 0.000159 mol Fe

Since there are 2 moles of iron citrate for every 1 mole of iron, we need to multiply by a conversion factor of (2 mol C6H7O7 / 1 mol Fe):

0.000159 mol Fe x (2 mol C6H7O7 / 1 mol Fe) = 0.000318 mol C6H7O7

Finally, convert moles of iron citrate to grams by multiplying by the molar mass of iron citrate (Fe(C6H7O7)2), which is the sum of the molar masses of all the elements in the compound:

0.000318 mol C6H7O7 x [(55.85 g Fe / (55.85 g Fe + 2 * (12 g C + 1 g H + 7 g O)) = 0.000318 mol C6H7O7 x (55.85 g Fe / 191 g C6H7O7) = 0.00925 g C6H7O7

Therefore, to gain 8.88 mg of iron, you would need to take approximately 0.00925 grams of iron citrate.