December 05, 2021, 09:02:02 AM
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### Topic: Determining The Gold Content In A Quartz and Calcite Specimen  (Read 315 times)

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#### SteveE

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##### Determining The Gold Content In A Quartz and Calcite Specimen
« on: November 05, 2021, 09:06:17 AM »
Hello Everyone.

Hope everybody is doing well these days.

I'm not sure this topic belongs here but didn't know where else to post it. Feel free to move it if needed

I recently found a nice gold in quartz and calcite specimen. I'm trying to determine the weight of the gold in it without destroying the specimen or buying too much equipment. Does anyone have an idea on how to go about this?

Thanks

#### Corribus

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##### Re: Determining The Gold Content In A Quartz and Calcite Specimen
« Reply #1 on: November 05, 2021, 09:47:17 AM »
You could get a rough estimate based on a density measurement.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### SteveE

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##### Re: Determining The Gold Content In A Quartz and Calcite Specimen
« Reply #2 on: November 05, 2021, 09:51:10 AM »
You could get a rough estimate based on a density measurement.

Can you provide the formula... preferably a simple one as I am math challenged.

#### Corribus

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##### Re: Determining The Gold Content In A Quartz and Calcite Specimen
« Reply #3 on: November 05, 2021, 10:39:16 AM »
For a binary mixture you can start with the fact that volumes are additive and the volume of any component is equal to the density divided by its mass.

For a mixture of A and B, the total volume

$$V_t = V_A + V_B$$

Therefore

$$V_t = \frac{m_A}{\rho_A} + \frac{m_B}{\rho_B}$$

Where mA and mB are the masses of components A and B, respectively, and ρ values are the respective densities. But the masses are also additive, so the mass of B is just difference between the total mass (mt) and the mass of A.

$$V_t = \frac{m_A}{\rho_A} + \frac{m_t-m_A}{\rho_B}$$

You are looking to find the mass of A in the mixture. Therefore the mass of A is equal to:

$$m_A=\frac{V_t \rho_A \rho_B}{\rho_B - \rho_A} - \frac{\rho_A m_t}{\rho_B - \rho_A}$$

The densities are known. You can measure the total mass with a scale and the total volume using submersion in a liquid.

Let's assume your solid measured 10 mL in volume and weighed 100 g. Using densities for gold of 19.3 g/mL and for the other component (quartz) of 2.5 g/mL*, I find that the mass of the gold component is 86.16 g. This seems about right because 100 grams of pure gold would be ~5 mL and of pure quartz would be 40 mL. The true volume is closer to that of pure gold, so we should have a majority of gold.

Note: this only works for binary mixture. If you have more than two components you would need more information, but you can approximate this as a binary mixture because quarts and calcite have similar densities and they are very different from gold. There are a few other areas where this would break down (if the two phases are really well mixed, such that the bulk densities aren't appropriate), so as I said: rough estimate. Also, be careful of your units.

(Someone check all that, I did it rather quickly.)

(I took a value around the densities of quartz and calcite.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### SteveE

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##### Re: Determining The Gold Content In A Quartz and Calcite Specimen
« Reply #4 on: November 06, 2021, 01:05:45 PM »
Is this formula another way to get the same answer? I found it on another forum.

Dry weight × 3.07

Wet weight × 1.16

Subtract the smaller number and the result will equal Gold weight

#### Corribus

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##### Re: Determining The Gold Content In A Quartz and Calcite Specimen
« Reply #5 on: November 07, 2021, 06:36:47 PM »
No idea what those numbers are or where they come from.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman