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Topic: Pre-Lab Homework Help  (Read 3908 times)

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Offline JesseTCarter

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Pre-Lab Homework Help
« on: September 29, 2019, 10:30:22 AM »
Calculate salt (100 mg/kg salt and 39 mg/kg salt). Find the moles of iron for the high and low values of iron present in Himalayan pink salt. Use correct Sig Figs and include units.

I’m confused as to where I should start. I know salt is NaCl and I can get it’s MM but I’m not sure about Himalayan pink salt. Or where I would start to calculate the moles of iron, because wouldn’t there be other moles of Cl in the 100mg/kg and 39mg/kg of salt? Maybe I’m just overcomplicating it.
« Last Edit: September 29, 2019, 02:18:36 PM by JesseTCarter »

Offline chenbeier

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Re: Pre-Lab Homework Help
« Reply #1 on: September 29, 2019, 12:21:24 PM »
The 100 mg/kg is the amount of Iron in NaCl. You should calculate the moles of it. The unit mg/kg can be expressed differently.

Offline JesseTCarter

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Re: Pre-Lab Homework Help
« Reply #2 on: September 29, 2019, 01:00:30 PM »
So then would it be for 100mg/kg

584.4277 mol of iron mg/kg?

Offline JesseTCarter

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Re: Pre-Lab Homework Help
« Reply #3 on: September 29, 2019, 01:06:35 PM »
Wait sorry that was for NaCl not Fe.

I got 55845 mol of iron mg/kg

Offline AWK

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Re: Pre-Lab Homework Help
« Reply #4 on: September 29, 2019, 01:13:59 PM »
Stop fooling around - 100 mg is less than 0.002 mole.
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Offline JesseTCarter

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Re: Pre-Lab Homework Help
« Reply #5 on: September 29, 2019, 01:26:56 PM »
Sorry, I’m not trying to fool around. I’m just confused, my Lab and Lecture are on two different parts. I know that the molar mass of Fe is 55.845g/mol that’s 55845mg and what’s that divided by 100mg/kg of iron? That’s how I got my answer. I didn’t think it was right because I know moles would be an extremely smaller number. Avogadro’s number. 6.022e23. We just haven’t learned this yet, so I’m a little confused and lost. I’m used to converting a mol of something into grams or the other way around. When balancing equations and doing stoichiometry.

Offline AWK

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Offline JesseTCarter

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Re: Pre-Lab Homework Help
« Reply #7 on: September 29, 2019, 02:05:03 PM »
100mg/55845mg makes so much more sense. I have no idea how I could miss that. 0.00179067061mg/kg mol of iron. That looks so much better.


Offline AWK

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Re: Pre-Lab Homework Help
« Reply #8 on: September 29, 2019, 02:08:12 PM »
Now the second number and significant digits.
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Offline JesseTCarter

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Re: Pre-Lab Homework Help
« Reply #9 on: September 29, 2019, 02:10:09 PM »
0.000698361536mg/kg mol of iron

and

0.00179067061mg/kg mol of iron

With Sig Figs and rounding wouldn’t it just be

0.0007mg/kg mol of iron

And

0.002mg/kg mol of iron

Offline AWK

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Re: Pre-Lab Homework Help
« Reply #10 on: September 29, 2019, 02:18:34 PM »
You did calculation only in nominator.
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Offline JesseTCarter

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Re: Pre-Lab Homework Help
« Reply #11 on: September 29, 2019, 02:20:20 PM »
Do you mean by that, that it’s only mg? And not mg per kg

Offline JesseTCarter

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Re: Pre-Lab Homework Help
« Reply #12 on: September 29, 2019, 02:22:39 PM »
Or wouldn’t it just be moles?

Offline AWK

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Re: Pre-Lab Homework Help
« Reply #13 on: September 29, 2019, 02:24:35 PM »
Now mol/kg but you did not use kg in calculations.
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Offline JesseTCarter

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Re: Pre-Lab Homework Help
« Reply #14 on: September 29, 2019, 02:59:18 PM »
That’s true, yes I only did mg because it’s mg of “something” per kg. We got it down to moles coming from mg; so we now have mol of “something” per kg. Would we leave it at mol/kg or does it boil down even further? The answers would have to be

0.0018mol/kg and 0.0007mol/kg

Because we’re calculating moles of iron from 100mg of iron per 1kg of salt and 39mg of iron per 1kg of salt. (It’s 1kg of salt, stated from a previous question in the pre-lab. I’m still unsure, but it sounds correct?)

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