Question

Red gold is a gold‑copper alloy used to make jewelry. A piece of jewelry made of red gold weighs 7.24 g and has a volume of 0.436 cm3 . Gold has a density of 19.3 g/cm^3 and copper has a density of 8.96 g/cm^3 . Calculate the percentage by mass of each metal in the jewelry. Assume the total volume of the jewelry is the sum of the volumes of the two metals it contains.

Answer 1

Answer:

%cu = 13.95%

%au=86.05%

Explanation:

Since the total volume is the sum of both volumes (gold, copper) we can say:

$V_{T} = V_{AU} + V_{CU}$

We know that $V_{AU} = \frac{m_{AU}}{\rho_{AU}}$  and $V_{CU} = \frac{m_{CU}}{\rho_{CU}}$. Replacing these values into our volume equation:

$V_{T} = \frac{m_{AU}}{\rho_{AU}} + \frac{m_{CU}}{\rho_{CU}}$

We also know that: $m_{T}=m_{AU}+m_{CU}$

Solving this equation system:

$\left \{ {{V_{T} = \frac{m_{AU}}{\rho_{AU}} + \frac{m_{CU}}{\rho_{CU}}} \atop {m_{T}=m_{AU}+m_{CU}}} \right.$

$m_{CU}=1.01g = 13.95%$

$m_{AU}=6.23g = 85.94%$

Answer 2

Percentage by mass of Gold is about 86%

Percentage by mass of Copper is about 14%

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Further explanation

Let's recall Density formula as follows:

$\boxed {\rho = m \div V}$

where:

ρ = density of object ( kg/m³ )

m = mass of object ( kg )

V = volume of object ( m³ )

Let us now tackle the problem!

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Given:

total mass of jewelry = m = 7.24 g

total volume of jewelry = V = 0.436 cm³

density of gold = ρ_g = 19.3 g/cm³

density of copper = ρ_c = 8.96 g/cm³

Asked:

percentage by mass of gold = m_g = ?

percentage by mass of copper = m_c = ?

Solution:

$\texttt{Total Volume = Volume of Gold + Volume of Copper}$

$V = V_g + V_c$

$\frac{m}{ \rho} = \frac{m_g}{ \rho_g} + \frac{m_c}{\rho_c}$

$\frac{m_g + m_c}{ \rho} = \frac{m_g}{ \rho_g} + \frac{m_c}{\rho_c}$

$\frac{m_g}{\rho} + \frac{m_c}{\rho} = \frac{m_g}{ \rho_g} + \frac{m_c}{\rho_c}$

$\frac{m_g}{\rho} - \frac{m_g}{ \rho_g} = \frac{m_c}{\rho_c} - \frac{m_c}{\rho}$

$m_g ( \frac{1}{\rho} - \frac{1}{\rho_g}) = m_c ( \frac{1}{\rho_c} - \frac{1}{\rho})$

$m_g ( \frac{ \rho_g - \rho}{ \rho\ \rho_g} ) = m_c ( \frac{ \rho - \rho_c} { \rho\ \rho_c} )$

$\frac{m_g}{m_c} = \frac{\rho_g}{\rho_c}\ \frac{\rho - \rho_c}{ \rho_g - \rho}$

$\frac{m_g}{m_c} = \frac{19.3}{8.96}\ \frac{ \frac{7.24}{0.436} - 8.96}{ 19.3 - \frac{ 7.24}{0.436} }$

$\frac{m_g}{m_c} = 6.112$

$\boxed {m_g = 6.112 m_c}$

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$\texttt{Percentage by mass of Gold} = \frac{m_g}{m_g + m_c} \times 100\%$

$\texttt{Percentage by mass of Gold} = \frac{6.112m_c}{6.112m_c + m_c} \times 100\%$

$\texttt{Percentage by mass of Gold} = \frac{6.112}{6.112 + 1} \times 100\%$

$\texttt{Percentage by mass of Gold} = \frac{6.112}{7.112} \times 100\%$

$\texttt{Percentage by mass of Gold} \approx 86 \%$

$\texttt{Percentage by mass of Copper} \approx 100\% - 86 \% \approx 14 \%$

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Answer details

Grade: High School

Subject: Physics

Chapter: Physical Quantities