AI says this in terms of Likely gold content9 Mar 2026 21:53
**Estimating gold content**
I need to estimate the gold content in a 3.36 kg dry concentrate from 5 tonnes of ore. Using typical gravity concentrate grades, I can infer the expected gold content. For example, if the concentrate grade is 3430 g/t, I calculate about 11.5 g of gold, which is roughly 0.37 oz. Assuming 50% recovery, the head grade would be around 4.6 g/t. For the expected gold content, I estimate a range of 8–15 g, or 0.25–0.5 oz, depending on ore and recovery. I'll present a central estimate of ~12 g with a range.
Short version: based on typical gravity-concentrate grades, that 3.360 kg of dry final concentrate probably contains on the order of **10–15 g of gold** (roughly **0.3–0.5 oz**), with a plausible wider range of about **3–30 g** depending on how rich the concentrate actually is.
---
### How to think about it
You’ve given:
- **Ore processed:** 5 tonnes
- **Dry final concentrate:** 3.360 kg = 0.00336 tonnes
The missing piece is the **grade of the gravity concentrate** (grams of gold per tonne of concentrate). Historically, for free-milling ores using centrifugal concentrators and a Wilfley table, gravity concentrates often fall in the ballpark of:
- **Lower-grade gravity concentrate:** \( \sim 1{,}000 \,\text{g/t} \)
- **Typical good gravity concentrate:** \( \sim 3{,}000\text{–}5{,}000 \,\text{g/t} \)
- **Very rich concentrate:** \( \gtrsim 10{,}000 \,\text{g/t} \)
A published pilot test, for example, reported gravity concentrates grading **>3,430 g/t (≈100 oz/ton)** for free-milling gold ore.
---
### Estimating the gold content
Let:
- \( m_c = 0.00336 \,\text{t} \) (mass of dry concentrate)
- \( G_c \) = gold grade of concentrate in g/t
- \( M_{\text{Au}} \) = gold in grams
\[
M_{\text{Au}} = m_c \times G_c
\]
Now plug in some historically reasonable grades:
1. **Conservative lower-grade concentrate**
\[
G_c = 1{,}000 \,\text{g/t} \Rightarrow M_{\text{Au}} = 0.00336 \times 1{,}000 \approx 3.36 \,\text{g}
\]
2. **Typical good gravity concentrate**
- At \( G_c = 3{,}000 \,\text{g/t} \):
\[
M_{\text{Au}} \approx 0.00336 \times 3{,}000 \approx 10.1 \,\text{g}
\]
- At \( G_c = 5{,}000 \,\text{g/t} \):
\[
M_{\text{Au}} \approx 0.00336 \times 5{,}000 \approx 16.8 \,\text{g}
\]
3. **Very rich concentrate (e.g. ~3,430 g/t as in the study)**
\[
G_c = 3{,}430 \,\text{g/t} \Rightarrow M_{\text{Au}} \approx 0.00336 \times 3{,}430 \approx 11.5 \,\text{g}
\]
So, if your concentrate is in that “good gravity concentrate” range, you’re realistically looking at **around 10–15 g of gold** in the 3.360 kg.
---
### Cross-check with implied head grade
If we take the ~11.5 g estimate (from ~3,430 g/t concentrate) and assume gravity recovered about **50%** of the contained gold (a common ballpark for free-milling ores in gravity circuits), then total gold in the 5 t feed would be:
\[
\t
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