The Earth's crust value for Fe are 47,200 ppm (Shales), 9800 ppm (Sandstones), and 3800 ppm (Carbonates). Note that ppm is equal to mg/kg.

See attached files for the distribution of other elements in earth's lithic crust.  

Dear Saiful: formally ppm is defined as: mg/L, because for dilute solutions density of solution is equal to unity, the same of the water, then mg/kg(dilute solution) = mg/L. D(water) = 1 Kg/L. Mr. Benson say the true.

Regards

mg/Kg is equal to ppm, meaning parts per million. However, mg/L is a density unit. You should not consider 'mg/kg' and 'mg/L' as the same. 4.09 wt% is equal to 40, 900 ppm. There is 13% difference between 40,900 ppm and 47,200 ppm. If so, you need to search more reference to get a confirmed average value. 

You have written: 'mg/kg' and 'mg/L' as the same. I have written : for dilute solutions density of solution is equal to unity, "the same of the water", if density "D" is the same of the water, then mg/kg(dilute solution) = mg/L, because 1 L of dlilute solution weighs 1 Kg. You can read over and over and you will never find that afirmation.

But, if you have some doubt, we have treated this topic in Research Gate in the past, but too you can review the bibliography about this issue in books of Básic Chemistry. However, is not the same density(g/ml or Kg/L) to ppm(mg/L), as you would seem to have.

Greetings

Thanks to above all for honest collaboration and precious time.

Just explain 'ppm' based on what I understand one more time. At first, Nsikak has provided a perfect answer for Saiful's question. Both of us agree with that. ppm is unitless or its unit is 1, which means 0.0001% (with unit 1). 'ppm' can be used for solid, aqueous solution with density (>,=,

Dear Dr. Hongzhu Liao

Thank u very much for clear appearance. 

Dear Saiful,

1 kg is equal to 1000000 mg and Fe is 4.09% of it. So the calculation is  1000000mg*0.0409 per kg, which is 40900 mg/kg (or ppm).

I hope that was helpful.

Kind regards,

Manos

 Dear Dr. Manos

Thank you very much,really it was helpful.

Sincerely,

Saiful

@Manousos-Ioannis Manousakas Is this the shale value or otherwise for Fe? What is the reference? I do not think the value is correct as quoted by Saiful in his question. It is appropriate to understand the matrix (sandstone,shale, carboneous, igneous rock, or deep-sea sediments) to know the specific value to use for his calculation. The comprehensive table for all elements has been attached in my first response. All the same, your explanation is correct but I am not certain of the 40900 mg/kg value.

Dear Dr. Manos,

In dilute solutions, the unit mg/L can be matched to "ppm" as stated by La Rosa. In the case of solids the correct unit is mg/kg. Because the parallel with liquid samples, people use "ppm" as an equivalent unit but this is totally wrong.

On the Fe value you asked, Bowen (1979) gives the value of 4,1% for mean crust. Iron content increases for clay and sediments (Turekian and Whedepol, 1961) and is lower in sandstone and limestone (Bowen, 1979).

@Nsikak Benson I just used the information Saiful provided to help him with the conversion. I am not suggesting that the value quoted is the most appropriate to be used. Just indicating how the conversion is made.

Kind regards to all!

The concentration of Fe in earth crust is 5% to convert  % to ppm(mg/kg) we must multiply  it by 10000 (Kabata Pandeais 2011)

To convert to ppm, it depends on whether the value is given as "Fe"  or "FeO". If the former, ppm is simply 5% of a million or 50,000 ppm. If the latter, you must convert FeO to Fe. To do this simply divide the atomic weight of Fe (55.85) by the atomic weight of FeO (71.85) and mutliply your FeO value by the result. For 5% FeO you now have 3.88% Fe or 38,800 ppm. You can do this for any element given (as is typical in geological analysis) as wt.% oxide.

Dear James: A little correction, I quote....To do this simply divide the atomic weight of Fe (55.85) by the "atomic weight" of FeO ....wanted to say " molecular weight or formula weight" FeO.

Regards