Hi! If the measurement results are given in mg/L (i.e. m/V), you need to know the volume (V) of the sample solution that was measured to calculate the mass (m) of the component being determined. If you are not using volumetric flasks, you should know the mass of the solution and its density to calculate its volume.

By the way. If the measurement technique was ICP-OES spectrometry, your solutions probably had to be diluted many times first to be suitable for measurements.

ZJ

Dear Jong:

Your information is not clear enough for us to help you. Please don’t worry about the level of your English. We will try our best to help you if you can answer the following questions:

What do you try to measure Na, and Li in your Na2SO4?

You have 27g of Na2SO4 sample.

The mass of the sample used for dissolution = ??

(Is it 0.1xxx g of Na2SO4? Please give the decimal numbers as accurately as you can for xxx)

Mass of HNO3 used for dissolution = ??

(Is it 9.9xxx g? that, is total mass = 10.0xxxg?)

It appears that you have further diluted the above solution for ICPES measurement. Then,

What is the mass of solution used for dilution = ??

(0.1xxx g?)

What is the final mass of solution after dilution = ??

(9.9xxx + 0.1xxx g? )

What is the concentration of Li measured by ICP = ??

(10.72 mg/L?)

What is the concentration of Na measured by ICP = ??

(10.72 mg/L?)

I don’t understand what are the following numbers, or where you got them from:

Na - 255.9 mg/L to mg/kg???

Q2. I have 337.54g and 234ml of solution

If you can fill in the above information, we will be able to help you.

Thanks,

Nimal

Nimal De Silva

I am so grateful for your help. I'll try to give you enough information.

I have 27g Na2SO4 powder sample.

I used 0.100 g of Na2SO4 and 9.900 g HNO3 for dissolution

total mass = 10.000 g ( I call it D_S (diluted solution) )

And used 1.000 g of D_S and 9.000 g of HNO3 for dilute

total mass = 10.000 g

So dilute the Na2SO4 1000 times..

The Na concentration of the diluted solution as measured by ICP-OES is 255.9 mg/L.

Concentration of Li measured by ICP is 10.72 mg/L

I can convert solution sample from mg/L to mg/Kg by knowing its mass and volume

but I don't know how to do it for a solid sample( Na2SO4 powder sample ).

I hope I give you enough information

Thank you.

Jong

Zbigniew Jońca

Hi ZJ. Thank you so much for your answer.

I use flasks for solution samples but I don't know how to do it for solid samples..

Dear Jong:

Now, your information is very clear. Thanks. Below is the calculation.

Step 1:

You used 0.100 g of Na2SO4 and 9.900 g HNO3 for dissolution

total mass = 10.000 g ( I call it D_S (diluted solution) )

Dilution factor at Step 1 (DF1):

= Final mass of solution / initial mass (solid)

= 10.000 g / 0.100g = 100

Step 2:

You used 1.000 g of D_S and 9.000 g of HNO3 for dilute

total mass = 10.000 g

DF2. = 10

Final dilution factor (for the measured solution):

DF final = DF1 x DF2 = 1000. (g/g; i.e. no units)

So far, we have done all the above calculations of dilution factors on mass/mass basis.

Step 3 - ICP-ES measurement:

It appears that you have prepared your calibration standards and done the measurements on mass/volume basis (mg/L or ug/mL).

Measured Na concentration = 255.9 mg/L = 0.2559 mg/mL

Measured Li concentration = 10.72 mg/L = 0.01072 mg/mL

However, our dilution factors are based on mass/mass basis. So, strictly speaking, we have to express (convert) the above concentrations also in mass/mass basis (mg/mL -> mg/g). For that, we need the density (d g/mL) of the measured solution to use the following relationship:

C in mg/g = C mg/mL / d g/mL = C/d mg/g

Suppose, the density of your measured solution was 1.025 g/mL

Then, in the measured solution:

The concentration of Na = 0.2559 mg/mL / 1.025 g/mL = 0.2496 mg/g

The concentration of Li = 0.01072 mg/mL / 1.025 g/mL = 0.01045 mg/g

Now, to convert the above concentrations in solution to the “concentration in the original solid” you simply multiply them by the final DF.

The Na concentration in solid Na2SO4 = 0.2496 mg/g x 1000

= 2496 mg/g = 0.2496 g/g = 24.96%.

The Li concentration in solid Na2SO4 = 0.01045 mg/g x 1000

= 10.45 mg/g = 0.01045 g/g = 1.045%

If you don’t know the density (d) of the solution, you may reasonably assume it to be 1.00 g/mL. (Please replace my assumed number 1.025 g/mL with the actual, for your calculation).

I agree with Dr. Zbigniew Jońca's comment that you may dilute the solutions further for ICP measurement, which will produce more accurate data (with lower matrix effects).

Hope, the above clarifies your question.

PS: Your English seems to be perfect!!

Best wishes,

Nimal

Hi Lee,

You may use the web tool to calculate

https://www.alfa-chemistry.com/solution-dilution-calculator.htm

https://www.alfa-chemistry.com/mass-molarity-calculator.htm