Chapter 17, Problem 14PEB

A sandstone building stone measuring 50.0 cm long by 20.5 cm wide by 15 cm thick is 70 percent quartz with the rest of its volume being empty pore space. Based on the specific gravity of quartz determined in exercise 3, what is the mass of this building stone in kg? What would be its mass if it were quartzite? (Recrystallization of metamorphism eliminates the pore space.)

To determine

The mass of building stone in kg and mass of the Quartzite, if sandstone building measuring is $50.0\text{ cm}$ long by $20.5\text{ cm}$ wide by $15\text{ cm}$ thick is 70 percent quartz with the rest of its volume being empty pore space.

Answer to Problem 14PEB

Solution:

$29\text{ kg}$ and $41.5\text{ kg}$

Explanation of Solution

Given data:

Length of the sandstone building is $50.0\text{ cm}$.

Width of the sandstone building is $20.5\text{ cm}$.

Height of the sandstone building is $15\text{ cm}$.

Quartz of the sandstone building percent is $70\%$.

Formula used:

Write the formula for density $\left(\rho \right)$:

$\rho =\frac{m}{V}$

Here, $V$ is the volume and $m$ is the mass.

Write the formula for volume and multiply by the percentage that is quartz:

$V=L\times W\times H\left(\%\text{ quartz}\right)$

Here, $L$ is the length, $W$ is the width and $H$ is the height.

Explanation:

For sandstone:

Convert length m to cm.

$\begin{array}{c}4.70\text{ m}=4.70\text{ m}\left(\frac{100\text{ cm}}{1\text{ m}}\right)\\ =470\text{ cm}\end{array}$

Convert width m to cm.

$\begin{array}{c}0.80\text{ m}=0.80\text{ m}\left(\frac{100\text{ cm}}{1\text{ m}}\right)\\ =80\text{ cm}\end{array}$

Recall the formula for density $\left(\rho \right)$:

$\rho =\frac{m}{V}$

Rearrange:

$V=\frac{m}{\rho }$

Recall the formula for volume and multiply by the percentage, i.e., quartz:

$V=L\times W\times H\left(\%\text{ quartz}\right)$

Substitute $\frac{m}{\rho }$ for $V$.

$\begin{array}{l}\frac{m}{\rho }=L\times W\times H\left(\%\text{ quartz}\right)\\ m=\rho \left(L\times W\times H\right)\left(\%\text{ quartz}\right)\end{array}$ …… (1)

Substitute $2.7\frac{\text{g}}{{\text{cm}}^3}$ for $\rho$, $50\text{ cm}$ for $L$, $\text{20}\text{.5 cm}$ for $W$, $15\text{ cm}$ for $H$ and $70\%$ for $\%\text{ quartz}$ in equation (1).

$\begin{array}{c}m=\left(2.7\frac{\text{g}}{{\text{cm}}^3}\right)\left(50\text{ cm}\times \text{20}\text{.5 cm}\times 15\text{ cm}\right)\left(70\%\right)\\ =\left(2.7\frac{\text{g}}{{\text{cm}}^3}\right)\left(50\text{ cm}\times \text{20}\text{.5 cm}\times 15\text{ cm}\right)\left(70\right)\left(\frac{1}{100}\right)\\ =2.9\times {10}^4\text{ g}\end{array}$

Convert mass g to kg.

$\begin{array}{c}2.9\times {10}^4\text{ g}=2.9\times {10}^4\text{ g}\left(\frac{1\text{ kg}}{{10}^3\text{ g}}\right)\\ =29\text{ kg}\end{array}$

For Quartzite:

Substitute $2.7\frac{\text{g}}{{\text{cm}}^3}$ for $\rho$, $50\text{ cm}$ for $L$, $\text{20}\text{.5 cm}$ for $W$, $15\text{ cm}$ for $H$ and $100\%$ for $\%\text{ quartz}$ in equation (1).

$\begin{array}{c}m=\left(2.7\frac{\text{g}}{{\text{cm}}^3}\right)\left(50\text{ cm}\times \text{20}\text{.5 cm}\times 15\text{ cm}\right)\left(100\%\right)\\ =\left(2.7\frac{\text{g}}{{\text{cm}}^3}\right)\left(50\text{ cm}\times \text{20}\text{.5 cm}\times 15\text{ cm}\right)\left(100\right)\left(\frac{1}{100}\right)\\ =4.15\times {10}^4\text{ g}\end{array}$

Convert mass g to kg.

$\begin{array}{c}4.15\times {10}^4\text{ g}=4.15\times {10}^4\text{ g}\left(\frac{1\text{ kg}}{{10}^3\text{ g}}\right)\\ =41.5\text{ kg}\end{array}$

Conclusion:

The mass of building stone in kg is $29\text{ kg}$ and the mass of Quartzite is $41.5\text{ kg}$.