[Kelly_Kapowski]

I'm not in the chemistry field, I work a front office, but my boss told me that I need to figure out how much Silicon is in Silica. We are getting some results in tomorrow that are similar to this:

SiO₂
Mean corrected Intesity: 12345.4
Std. Dev.: 0.306
Sample conc.: 21.40 mg/L
RSD: 1.43%

Is it possible for someone to explain to me how to do this because I might have to do similar Silicon/Silica problems in the next few weeks. Thank you for your help.

[opsomath]

Hi! I would be glad to help you. So I guess this is some kind of a suspension or slurry of silica in water?

Silica is almost exactly 50% by mass silicon. (49.983%) So, if your sample is 21.4 mg/L of silica, it has 10.7 mg/L of silicon in it. (IE you multiply by 50%, which is dividing by 2).

Does that help?

[Kelly_Kapowski]

"So I guess this is some kind of a suspension or slurry of silica in water?"

It's not really a slurry.  Usually a customer will send us a water sample with nitric acid in it to "preserve it" and we'll ship it out to a lab and get the report out to our customer (we're the middle man).

"Silica is almost exactly 50% by mass silicon."

If I tell my boss this he's going to ask how I know this, can you explain in more detail?

[opsomath]

Your boss sounds like a charmer....so this is a water sample. Okay, cool.

I'm glad you asked, because I realized I made a mistake.

The molar mass of silicon is 28.0855 g/mol. The molar mass of silicon dioxide is 60.08 g/mol. For every unit (mole, or molecule, whatever...it's a ratio) of silicon dioxide, you have one silicon, since the formula of silicon dioxide is SiO2. (So, it's made of one silicon and two oxygens)

So, silicon dioxide contains one unit of silicon. That means we can figure out what the mass of silicon is in silicon dioxide by dividing the mass of ONE silicon by the mass of ONE silicon dioxide. That number is 28.0855/60.08 = 46.75%. In other words, silica is 48.75% silicon. (This was the mistake I made earlier, I accidentally copied the wrong number in.)

So, you can multiply 21.4 mg/ml silica * (0.4675) = 10.00 mg/ml silicon.

How's that?

[Borek]

That number is 28.0855/60.08 = 46.75%. In other words, silica is 48.75% silicon.

You may want to correct it again 

[opsomath]

Dad gummit! Don't drink and derive, people...

The 46.75% figure should be correct, IIRC. Apologies.

[Kelly_Kapowski]

Thank you very much for your help.  The way that you have explained it makes it easy for me to wrap my head around it but I have one last question.  You gave me a result in mg/ml in my final report I have to use mg/L how do I adjust this?  I figure it's a matter of moving a decimal but in what direction?

[almfranc]

Thank you very much for your help.  The way that you have explained it makes it easy for me to wrap my head around it but I have one last question.  You gave me a result in mg/ml in my final report I have to use mg/L how do I adjust this?  I figure it's a matter of moving a decimal but in what direction?

So you have an answer in mL, but would like to be in L - think of the difference between these two quantities. A L is 1000x larger in volume than a mL (i.e. 1000mL in every 1L). If you scale the denominator (the volume) by 1000, you must also scale the numerator (the mass in mg) by 1000 as well. If you had an answer like 5mg/mL, this would scale by a factor of 1000 on top and bottom to 5000mg/L. It's basically multiplying your answer by 1 (which is totally allowed), but your 1 is 1000/1000. Hope that helps!

[Borek]

The 46.75% figure should be correct, IIRC. Apologies.

Correct enough.

[Kelly_Kapowski]

Ok so I got our first report in today and I wanted to run this past you guys.  The report shows SiO2 at 11.29 mg/L.  This means that I would multiply (11.29 * 0.4675) which gives me 5.278 mg/L of Silicon in this sample. 

The explanation being that Silicon is a percentage of Silica.  We can figure out the percentage by taking the molar mass of the two and dividing them.  Then we just multiply our report result by the percentage to get our value.  Also I had asked a question about the mg/L and mg/mL earlier and although the explanation made sense to me it I don't think it applies, because I'm already working with mg/L, am I right?

[Borek]

Looks OK to me.