# A 67.2 g sample of a gold and palladium alloy contains 2.49 times 10^23 atoms. What is the composition (by mass) of the alloy?

Sep 10, 2017

The alloy is 75% by mass with respect to goldo.....

#### Explanation:

Well, the molar mass of palladium is $106.4 \cdot g \cdot m o {l}^{-} 1$, and the molar mass of gold is $197.0 \cdot g \cdot m o {l}^{-} 1$.

We know that ${x}_{1} \times 106.4 \cdot g \cdot m o {l}^{-} 1 + {x}_{2} \times 197.0 \cdot g \cdot m o {l}^{-} 1$ $\equiv 67.2 \cdot g$. And here ${x}_{1} = \text{moles of palladium}$, and ${x}_{2} = \text{moles of golds}$.

We further know that $\left({x}_{1} + {x}_{2}\right) \times 6.02214 \times {10}^{23} = 2.49 \times {10}^{23}$, by the problem's specification. And thus we gots the REQUIRED two equation to solve for TWO UNKNOWNS......

And so $\left({x}_{1} + {x}_{2}\right) = \frac{2.49 \times {10}^{23}}{6.02214 \times {10}^{23}} = 0.414$...

And thus, ${x}_{1} = 0.414 - {x}_{2}$, and we substitute this back into ${x}_{1} \times 106.4 \cdot g \cdot m o {l}^{-} 1 + {x}_{2} \times 197.0 \cdot g \cdot m o {l}^{-} 1$ $\equiv 67.2 \cdot g$.

So....$\left(0.414 - {x}_{2}\right) \times 106.4 \cdot g \cdot m o {l}^{-} 1 + {x}_{2} \times 197.0 \cdot g \cdot m o {l}^{-} 1$ $\equiv 67.2 \cdot g$, and to simplify.....

....$\left(0.414 - {x}_{2}\right) \times 106.4 + {x}_{2} \times 197.0 = 67.2$.

$44.0 - 106.4 \cdot {x}_{2} + 197.0 \cdot {x}_{2} = 67.2$

$44.0 + 90.8 \cdot {x}_{2} = 67.2$

${x}_{2} = \frac{67.2 - 44.0}{90.8} = 0.256$

And back-substituting, ${x}_{1} = 0.414 - 0.256 = 0.158$.

I think that is enuff arifmetick for anyone, but we should check on the consistency of our results......

If we gots $0.256 \cdot m o l$ with respect to gold we got a mass of $0.256 \cdot m o l \times 197 \cdot g \cdot m o {l}^{-} 1 = 50.4 \cdot g$, and a mass of $0.158 \cdot m o l \times 106.42 \cdot g \cdot m o {l}^{-} 1 = 16.81 \cdot g$ with respect to palladium. The sum of the masses is $67.2 \cdot g$ as required by the boundary conditions of the problem. Are you happy with this treatment?