Lab 2

Calcuations

Bucket Fill Fraction

If you go into the field to collect a sample, you very well may not have a scale. If you have a 55 gallon drum, or a 1 gallon bucket, it is helpful to understand what fraction of that bucket to fill in order to obtain a large enough sample.

For this lab exercise, we will collecting a 4 kg sample in a 2 gallon bucket with an ore to then split into 4 1 kg lots. What fraction of the bucket should we fill up to obtain a 4 kg sample? We are given the following information:

• 2 gallon bucket
• 3.785 L = 1 gal
• Crushed ore has 30% void space
• Density of the solid ore is about 2.65 g/cm³

We can assume the mass of air is negligible.

Using the above information, we know that:

$$ \rho_\mathrm{ore} = \frac{m_\mathrm{ore}}{V_\mathrm{ore}} $$

We also know that the mass of air is negligible, so:

$$\begin{aligned} m_\mathrm{sample} &= m_\mathrm{ore} + \xcancel{m_\mathrm{air}} \\ m_\mathrm{sample} &= m_\mathrm{ore} \end{aligned}$$

Because $m_\mathrm{sample} = m_\mathrm{ore}$, we can simply use the formula for density and given information to find $V_\mathrm{ore}$:

$$ \begin{aligned} \rho_\mathrm{ore} &= \frac{m_\mathrm{ore}}{V_\mathrm{ore}} \\
2.65 \frac{g}{cm^3} &= \frac{4,000 g}{V_\mathrm{ore}}\\
V_\mathrm{ore} & \approx 1509 cm^3 \approx 1.51 L \end{aligned} $$

Next, we can use the information about void space to calculate the $V_\mathrm{air}$ and then $V_\mathrm{sample}$. First writing down the volume relationship for void space, converting the 30% void space into a fraction:

$$ 0.3 = \frac{V_\mathrm{air}}{V_\mathrm{ore} + V_\mathrm{air}} $$

Substituting in $V_\mathrm{ore} = 1.51L$ and solving for $V_\mathrm{air}$ gives:

$$ \begin{aligned} 0.3 &= \frac{V_\mathrm{air}}{1.51 + V_\mathrm{air}} \\
0.3 \times 1.51 + 0.3V_\mathrm{air} &= V_\mathrm{air} \\
0.3 \times 1.51 &= (1-0.3) V_\mathrm{air}\\
V_\mathrm{air} &= \frac{0.3 \times 1.51}{(1-0.3)}\\
V_\mathrm{air} &= 0.647 L \end{aligned} $$

So now we can calculate our total volume:

$$ \begin{aligned} V_\mathrm{sample} &= V_\mathrm{ore} + V_\mathrm{air}\\
V_\mathrm{sample} &= 1.51L + 0.647L \\
V_\mathrm{sample} &= 2.16 \end{aligned} $$

And converting 2 gal = 7.57 L we can see that:

\begin{align} pct_\mathrm{bucket} &= \frac{V_\mathrm{sample}}{V_\mathrm{bucket}} \times 100 \\
pct_\mathrm{bucket} &= \frac{2.16 L}{7.57 L} \times 100 \\
pct_\mathrm{bucket} &= 28.5 pct \\
\mathrm{frac}_\mathrm{bucket} &\approx \frac{1}{3} \end{align}

So for a two gal bucket, we need to fill up about $\frac{1}{3}$ to obtain approximately a 4 kg sample.