Verified answer

the other answerers are very close to the best method, but not quite

you want to use "factor label method" aka "dimensional analysis" ... "DA"... for all your chemistry problems. this, conversion to moles, stoichiometry.. etc

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some things to think about in DA.

.. (1) sort through the problem to identify the FROM, the TO and all the

... .. .unit factors

.. (2) start with the FROM on the left and convert to the right

.. (3) setup the problem first, double check everything, then run the calcs

.. (4) to double check.. these are the two types of common mistakes

... .. .(a) verify everything cancels properly

... .. .... ..example.. 88g x (1mol / 44g) vs 88g x (44g / mol)

... .. .(b) verify the unit factors are correct

.... .. .... .example... (2.54cm / 1in) vs (1cm / 2.54in)

... .... .. .... .. the first is correct.. 2.54cm = 1in

... ..... .... .....the second is backwards

.. (5) make sure the unit factors have as many sig figs as the FROM

... ... .and preferably.. 1 extra.

... .. . this ensures the precision in the answer is due to the FROM

... .. ..and not your choice of conversion factors

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got all that?

this problem

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FIRST... a correction to make

.. ..thyroxine is C15H11NO4I4.. H11.. not H17 !!!!!!!!

google it if you don't believe me

next.. we're converting

.. FROM 15.0g C15H11NO4I4

.. TO grams I

so let's use the unit factor g I per g C15H11NO4I4

.. that's called "mass fraction"

...and it's a very very very common unit factor.. REMEMBER THIS

and the setup

.. 15.0g C15H11NO4I4 x (507.6g I / 776.9g C15H11NO4I4) = 9.80g Iodine

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now look at the units canceling

... g C15H11NO4I4... . ... .... ...g I

.. .------ ------ ------ --- x ------ ------ ------ --- ====> g Iodine

.... ... ..... .... .... .... .... ..g C15H11NO4I4

that checks out

next verify the unit factor is correct

.. does 507.6g I = 776.9g thryoxine

.. or is it the other way around.

.. this is correct

so we run the calcs

.. 15 x 507.6 / 776.9 = ?

and we're done

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notice the sig figs in my unit factor

.. 15.0 has 3

.. so my mass I (507.6) has 4... 1 more than the "3"

.. and my mass C15H11NO4I4 has 4 also..

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questions?

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can you spot where the guy "REAL" made an error? That's a common mistake with these big formulas.. if you can, use google to verify your calculated molar masses. And if you did that, you would also realize the formula was wrong.

This is much easier. You don't even have to use moles. The goal is to first calculate what percentage does iodine take in C15 H17 NO4 I4. First, get the relative masses of each of atoms of those elements from the periodic table:

C 12.01

H 1.008

N 14.01

O 16

I 126.9

Now, how many of those atoms are in C15 H17 NO4 I4?

C 12.01 15 atoms

H 1.008 17 atoms

N 14.01 4 atoms

O 16 4 atoms

I 126.9 4 atoms

The goal is to calculate what percentage does iodine take in C15 H17 NO4 I4. Calculate the relative mass of those elements altogether by multiplying the number of atoms by their relative mass.

C 12.01 * 15 = 60.05

H 1.008 * 17 = 17.136

N 14.01 * 4 = 56.04

O 16 * 4 = 64

I 126.9 * 4 = 507.6

Add up all of those relative masses together:

60.05 + 17.136 + 56.04 + 64 + 507.6 = 704.826

Then calculate what is the percentage of mass of all the atoms of iodine in that sum:

507.6 / 704.826 * 100 = 72.0178%.

And finally, calculate mass of iodine in thyroxine using the mass of thyroxine and percentage of iodine:

0.720178 * 15 = 10.8026

So, from 15 grams of thyroxine, 10.8026 grams of iodine can be exterted. Or 10,802.6 milligrams.